Hi everyone,
I have not been sharing for a while but continued my research in the background and some out of public forum.
I've decided to open all my unlisted video demos to public so anyone can see my research if it interests you.
Here is a link to my videos if you would like to watch what was not public: https://www.youtube.com/user/gotoluc/videos (https://www.youtube.com/user/gotoluc/videos)
I have title most of them with "Testing ideas of more efficient Coil Flyback use (test 1 to 9)
There is also a few others and you can tell as they would have a low view count.
Here is the latest video demo where I share what I find most interesting and suggest how to better use Coil Inductive kickback (flyback)
https://www.youtube.com/watch?v=XfLcBD3Fy7M&feature=youtu.be (https://www.youtube.com/watch?v=XfLcBD3Fy7M&feature=youtu.be)
If you wish to post in this topic please keep it on topic and constructive as I reserve the right to edit or delete any post that are not so.
Luc
That's awesome Luc thanks. Really interesting. I've only watched the last one that was published so far but I'll be checking out the others soon. It really got me thinking about the secondary in the mot with a cap. I'll be messing with it tomorrow.
I was particularly impressed with the magnetic action you demonstrated from the back EMF. Could you use that principle to pull or push the rotor "for free"? You had that transformer on your bench, and I was thinking: Move that over near the rotor and put it to work...
Quote from: wopwops on November 12, 2015, 11:04:52 AM
I was particularly impressed with the magnetic action you demonstrated from the back EMF. Could you use that principle to pull or push the rotor "for free"?
You had that transformer on your bench, and I was thinking: Move that over near the rotor and put it to work...
Yes, you've got it!... you can re-create a very powerful magnetic field using off pulse (flyback) of a motor coil, and
for free... if you use the correct geometry and timing. And why not put it back to further assist the motor. In a universal motor when the brushes move from one commutator segment to the other they've designed it to short the flyback so not to arc the commutator as motor designers have been trained that flyback spikes are useless and must be neutralized.
Making modifications to the existing design to allow this useful flyback to come out and maybe redirecting back in the stator coils may assist the motor... no?
There are other things that can be done. Take all these trick and you may end up with something we are looking for.
Keep thinking
Luc
I've been following this area of inquiry since the mid 1990s and this last group of videos from you is easily the most impressive thing I've seen so far. The instant I saw the physical result you were getting !!!outside of the duty cycle!!! I thought: Woh. I think Luc's got it.
The strength of that magnetic field seems so strong vs. what was going in. And for free...
It's probably a good sign that hardly anyone is paying attention to this thread. :)
very well done there!
seeing the block of metal want to lift that weight shows there is use to this.
ya now to find ways to recycle that into motion maybe a kind of hub motor or some storage.
you are our new leader.
Quote from: wopwops on November 12, 2015, 07:51:49 PM
The strength of that magnetic field seems so strong vs. what was going in. And for free...
It's probably a good sign that hardly anyone is paying attention to this thread. :)
If you think that was good wait till you see the next video 8)
Yes, I agree, it's probably a good thing I'm now mostly ignored. Make things much easier.
Luc
Hi gotoluc, hi to all.
As sometimes happen in life, I was studding for the last couple of weeks the same area you are exploring now.
My testing rig was somehow different but still the same. I was focusing to see how much energy can be recovered as BEMF from an inductor pulsed with DC. The testing circuit was the one from the attachment noted 1. The trace on scope was the one noted 2 from same attachment. I was testing different kinds of inductors (air core, iron core, different number of turns, also different voltages, different pulse width and the list can continue). All test results where noted down into an excel data sheet. Important notice: all tests where done in mW ranges.
By far the best results where obtain from an air core inductor (not saying that this is the way ... air core). Conditions for testing that particular coil where the following: 12V supply voltage, pulse width 7-8% at around 1.5 kHz. With this setup I was able to recover an astonishing 90% of the energy invested in the coil, from back EMF. Better yet, with or without the recovering circuit, the current draw from the source was the same.
For the rest of the tests (different coils, different working conditions) a media of 50% recovery can be taken into account.
My understanding of this: you put into an inductor some energy. As a result the inductor exhibit some magnetic effect and also BEMF. Or with another words I put in 1 (one) unit of energy, get some magnetic effect and recover 0.9 units of energy from the same inductor as BEMF.
Let's go with the idea one step further. Let's consider two identical circuits as the one noted 1 in the attachment. For the 1st circuit we need to make it run, 1 unit of energy. One can recovers from that input 50%, or 0.5 units of energy. 2nd circuit is the same. Invest 1 recover 0.5. Now let's combine the 2 of them using the following logic: the recovered energy from 1st one is half of the required energy for the 2nd one to run. The other half comes from the power supply. The second recovery circuit became the half of the energy requirements for the 1st circuit to run and the other half comes from the second power supply. To better understand what I'm trying to say, see the section 3 from my attachment.
What's the gain? One can get twice the magnetic effect for the same input: 1 (one) unit of energy. Things get better and better if the recovery circuit can recover more. This was just an example, to better explain the way I see it. I hope I'm not wrong.
Some other approach to confirm (at least for me) that BEMF is capable of doing some useful work was the following: take one usual relay that has normal closed contacts. Connect the normal closed contacts in series with the coil of the relay. Connect this setup to a power supply (see 4 from attachement). What u will get is some sort of buzzer that is ringing with some frequency. Nothing new here. Now connect a diode to kill BEMF. The ringing frequency is lower now. This is BEMF expanding the duration of the magnetic field inside the coil. So YES, BEMF can do useful work for us.
I'll end my post with a question. Please see this if u have the time: https://www.youtube.com/watch?v=LAtPHANEfQo It is a great animation that explain how a DC motor works, different types of DC motors and so on. At around 3:40 it is explained that BEMF is appearing on a DC motor also. But the explanation (and I saw it so many times over the years) that BEMF is a good thing in a DC motor cause otherwise the motor will overheat and burn is still shocking me every time I hear it. My question: Is this true? What if BEMF doesn't exist? Or because it is there (and we cannot remove it), then better yet, use it to do some useful work?
The way I see it is like this: u have a 6 volts lamp and a 12 volts battery. How can u make the lamp work without burning it?
Solution 1: use a series resistor connected with the lamp and the 12 volts battery;
Solution 2: use another 6 volts battery in series with the lamp and the 12 volts battery but connected with the opposite polarity (this is the case of BEMF don't you think?)
Solution 3: use a 12 volts lamp;
Solution 4: use a 6 volts battery;
Obviously one of the solution 3 or 4 are the logic (and most efficient) one to be used. Hell now, said the DC motor constructors. Solution 2 is the right way. Huh ... really?
mihai
Hi everyone,
here is a new test video with an accurate way of testing the magnetic flux amplification effect.
Link to video: https://www.youtube.com/watch?v=7dmKEOWOhQA
Luc
Hi gotoluc, thank you for sharing.
For an easy to build useful test device, the old joule thief with high voltage secondary.
Place bifilar oscillator on one side of toroid, then place high voltage secondary on other side to receive the flyback pulses.
With either a low or high voltage pickup wind over that, depending on desired final output volts needed.
I realize that's not an ideal setup, since it's all on the same core, though it's a start.
It might give us a more efficient jenna led light circuit.
peace love light
Hi Luc
Thank's very much for sharing your experiment, and it seems to work very well.
What puzzle me is the fact that the very sharp fly back spike normally has almost no energy, it has very high voltage peak, but very low current, and if you connect a diode and a capacitor, it takes a lot of pulsse to charge it to obtain some power. But in your design the cap get a real power at each pulse.
So i think that we have to think of what happen in the fly back spike itself.
How does the flybackspike "capture " this energy ?
https://youtu.be/tag5OlvPi54
Thank's
Laurent
Hi Laurent,
Very clever experiment, thanks for showing it!
To explain your question, the first thing to consider is that the actual load for the sharp flyback pulse is not only your 1 uF capacitor but the 1.8 Henry coil, i.e a parallel LC combination. So they have an AC impedance and although they do not seem to be in resonance with the frequency the rotor RPM establishes, they cannot represent as a low impedance for the spike as a solely (discharged) capacitor normally would.
So the "trick" is to load the flyback spike with a relatively high impedance component or circuit like a multiturn coil or a parallel LC circuit brought near or very near to the spike's frequency, these insure a high impedance load. An empty capacitor is just the opposite: it is a very low impedance load first (when fully discharged) and then it represents an increasing impedance as it charges up.
Addition: a HV coil has many number of turns so a small current a flyback can insure is still able to perfom a decent magnetic force. Amper times turns is involved, the relatively low current is backed up by the many fine turns of the HV coil.
To widen the pulse width of the captured energy in the HV coil, try to use a 2.2 uF capacitor instead of the 1 uF and see which direction it changes. Or you already tested this and arrived at the 1 uF value as the best choice?
The 2.2 uF with the 1.8 H coil could resonate pretty close to the rotor speed what I took from the scope shot as cca at 77 Hz.
This would be 4620 RPM, but we have to divide this by 3 (number of magnets) and we get 1540, pretty close to your tacho metered 1551. (With the 1 uF capacitor the 1.8 H coil resonates at 118.6 Hz.) Notice that I did not consider that the input coil "sees" the HV coil during the ON time of the flyback diode, this may mean a different value capacitor, so it needs to check some other values besides the 2.2 uF. Nevertheless, the tuning of the LC circuit need not be very sharp in this given case because the 210 Ohm DC resistance of the HV coil is a loss and reduces the loaded Q (figure of merit, Q=XL/R) of the LC circuit. So the voltage amplitude is limited by this transformed loss across the LC circuit, and the lower the DC resistance of the HV coil is, the higher the resonant impedance is hence the higher the voltage amplitude could be across it.
Perhaps the reed switch could be placed a little bit further away from the rotor magnets to shorten the ON time. This would probably involve less RPM but this may be compensated with placing the LV coil a bit closer to the rotor magnets, and this may be true for the HV coil too. I suggest these, if you feel like doing such refinements. 8)
Greetings,
Gyula
I only looked at Laurent's clip, and it's pretty clear that the strategy proposed by Luc is working. I am not going to disagree with Gyula, but instead give a different take on describing what is happening.
For starters, when the capacitor discharges into the high-voltage coil, you notice that there is no LC tank resonance taking place due to the 210 ohm resistance of the coil, creating an "over damped" condition. The nice linear decreasing ramp voltage is a bit of a surprise, but that is a surprise in favour of the experiment. It's telling you that a nice even pulse of current is flowing through the coil with a more or less flat top. So that means that the high-voltage coil is giving you a nice ON-OFF pulse of current and an ON-OFF pulse of attraction-repulsion to drive the rotor. This could easily be verified by putting a one-ohm current sensing resistor in series with the drive coil.
Here are the circuit dynamics: The reed switch switches ON and current ramps up in the low-voltage drive coil. When the reed switch switches OFF, the drive coil has to discharge its stored energy. You can never forget that this energy is discharged in the form of a pulse of current at a to-be-determined voltage, and NOT as a pulse of voltage.
The pulse of current first flows through the diode. We know that when you try to pulse the current through the high-voltage coil, it acts like an open circuit and refuses to let current flow though it. We know that when you try to pulse the current into the capacitor, it acts like a short circuit and readily accepts the current flow. Therefore, when the reed switch switches OFF, essentially all of the current pulse flows into the capacitor, and none of the current pulse flows into the low-voltage coil.
When the capacitor is at its peak voltage, that means that the low-voltage drive coil has completely discharged its current pulse energy. Then the capacitor starts to discharge through the high-voltage coil in a nice linear fashion as described above.
How much energy is in the pulse from the low-voltage coil? That's easy, you can calculate it based on the peak voltage of the capacitor and the value of the capacitor.
Where does that pulse energy go? It goes to two places in and around the high-voltage coil: 1) some of the energy pushes on the rotor, and 2) some of the energy is lost in the resistance of the high-voltage coil.
So, that leads up to the question, how much of the pulse energy pushes on the rotor and how much of the pulse energy is lost due to the resistive losses in the high-voltage coil?
That's an interesting question because it tells you how efficient your setup is in recovering the recycled pulse energy from the low-voltage drive coil. I will leave that as an exercise for the enthusiasts to work out.
Finally, Laurent did not try changing the angle of the high-voltage coil to look for a sweet spot. It would have been a worthwhile exercise to try that. In fact, I would have moved the reed switch angle first to find the sweet spot for the reed switch only (no high-voltage coil) and then I would have added the high-voltage coil and then looked for the sweet spot angle for that.
Quote from: woopy on November 16, 2015, 01:28:17 PM
Hi Luc
Thank's very much for sharing your experiment, and it seems to work very well.
What puzzle me is the fact that the very sharp fly back spike normally has almost no energy, it has very high voltage peak, but very low current, and if you connect a diode and a capacitor, it takes a lot of pulsse to charge it to obtain some power. But in your design the cap get a real power at each pulse.
So i think that we have to think of what happen in the fly back spike itself.
How does the flybackspike "capture " this energy ?
https://youtu.be/tag5OlvPi54 (https://youtu.be/tag5OlvPi54)
Thank's
Laurent
I'm glad people are starting to look at this using the many different methods to harvest and use the spike. This is what John Bedini and others have been talking about for years. Try it on the window motor with the full or even half bipolar Bedini/Cole switch or even the zero force... - you will all be amazed!
Cheers - Patrick
Quote from: woopy on November 16, 2015, 01:28:17 PM
Hi Luc
Thank's very much for sharing your experiment, and it seems to work very well.
What puzzle me is the fact that the very sharp fly back spike normally has almost no energy, it has very high voltage peak, but very low current, and if you connect a diode and a capacitor, it takes a lot of pulsse to charge it to obtain some power. But in your design the cap get a real power at each pulse.
So i think that we have to think of what happen in the fly back spike itself.
How does the flybackspike "capture " this energy ?
https://youtu.be/tag5OlvPi54
Thank's
Laurent
Very nice experiment Laurent.
Quote from: minoly on November 16, 2015, 06:07:53 PM
I'm glad people are starting to look at this using the many different methods to harvest and use the spike. This is what John Bedini and others have been talking about for years. Try it on the window motor with the full or even half bipolar Bedini/Cole switch or even the zero force... - you will all be amazed!
Cheers - Patrick
No it's not. There is no magical radiant energy entering the system here. This is showing real energy being captured and reused that would otherwise be wasted or dissipated as things like arc's-as could be seen clearly in Laurents video when he disconnected the cap,and an arc could be seen in the reed switch-->no unicorn power here.
Hi Laurent,
I wrote that to widen the pulse width for the HV coil the 1 uF capacitor needs to be increased. Now that I watched the scope waveforms on your scope when you removed the 1 uF capacitor, the pulse width actually increased to cca 2.2 ms (video time 9:17-9:18) from the cca 1.8-1.9 ms pulse width when the cap was included. The pulse amplitude got reduced to cca 36 Vpp no cap from the cca 80 Vpp amplitude with the 1 uF cap in place.
So very probably my suggestion of using 2.2 uF will not widen the pulse width, sorry for this, albeit it may influence the amplitude a little.
The bottom line is that the high DC resistance of the HV coil prevents getting higher gain in RPM, it would be worth to replace it with a less lossy HV coil. Unfortunately, this would involve the use of thicker wire than the present one.
Gyula
Quote from: woopy on November 16, 2015, 01:28:17 PM
Hi Luc
Thank's very much for sharing your experiment, and it seems to work very well.
What puzzle me is the fact that the very sharp fly back spike normally has almost no energy, it has very high voltage peak, but very low current, and if you connect a diode and a capacitor, it takes a lot of pulsse to charge it to obtain some power. But in your design the cap get a real power at each pulse.
So i think that we have to think of what happen in the fly back spike itself.
How does the flybackspike "capture " this energy ?
https://youtu.be/tag5OlvPi54 (https://youtu.be/tag5OlvPi54)
Thank's
Laurent
Hi Woppy, It's easy to capture the flyback spike.
But timing is everything.
Introduce lenz when you want it.
push instead of pull.
Gravity , needs to be considered.
I'm trying to find the balance between the two.
I think magnetic fields should be used as shearing forces.
artv
I think that it is important to reiterate that this is not a "discovery" and has never been a discovery. People in the 1920s fully understood the dynamics of a discharging inductor. It's just the mirror image of a discharging capacitor. There is no mystery, no secret, no discovery.
In the real world you would be very hard pressed to find a pulse motor running in a practical application. I can't think of one offhand.
There is a story that I am sure that you have read many times out of the JB universe and that of his associates or former associates. The story goes something like this: "We set up our pulse motor demo system at the show and were showing it powering LED lights and charging batteries. An engineer came over to our table and we showed him the battery charging and he just scratched his head and did not understand what he was looking at and wandered away with a lost look on his face."
I have read that story at least five times. The story is complete crap. Really, truly, complete nonsensical crap. And there is an ironic twist to the story: JB does not explain to his followers how the discharging coil that charges the charging battery works. He does not teach this. Therefore, all of the hardcore JB devotees that say it's "radiant energy" have no clue what is actually happening. They don't understand why the neons light up when the charging battery is disconnected and they don't understand why the "voltage spike" almost completely disappears when the charging battery is reconnected. You can pay $350 to go to a conference and spend 3/4 of a day building a pulse motor and they will not hold a seminar to teach you how it actually works. I find that very annoying. Just like I find it annoying that they don't tell you that when you transfer energy from the source battery to the charging battery via the famous spike that you lose about 40% of your energy. There is no "chemistry magic" when the misnamed "radiant spike" charges the charging battery. Likewise, measuring battery voltages is not a legitimate way to measure the state of charge of the charging battery. That's just a smokescreen to disguise the fact that in actual reality you are losing about 40% of your source battery energy when you transfer it to the charging battery using pulses of current.
I attached a 120 volt A.C. LED bulb directly to the charge circuit of a Bedini SSG and measured the Lumens. Only perhaps 17 percent of the input power was recovered this way. Woopy's capacitor is receiving a charge from the magnet rotor which is apparently bouncing back to the output coil in synchronized timing of the back spike. The power driving the rotor is mostly generated by the rotor and timed by the back spike. The actual power transferred to the output coil from the power coil is relatively insignificant. The back spike timing is the critical factor in the exceptional performance of this ingenious build.
Quote from: woopy on November 16, 2015, 01:28:17 PM
Hi Luc
Thank's very much for sharing your experiment, and it seems to work very well.
What puzzle me is the fact that the very sharp fly back spike normally has almost no energy, it has very high voltage peak, but very low current, and if you connect a diode and a capacitor, it takes a lot of pulsse to charge it to obtain some power. But in your design the cap get a real power at each pulse.
So i think that we have to think of what happen in the fly back spike itself.
How does the flybackspike "capture " this energy ?
https://youtu.be/tag5OlvPi54 (https://youtu.be/tag5OlvPi54)
Thank's
Laurent
Merci Laurent!... I could not of asked for a better replicator. You understood everything I have shared and you have made an excellent demo device and video documentary.
Thanks for sharing and continue thinking of all the other possibilities of using this most amazing and powerful magnetic field in our favor.
Here is a hint for your excellent demo ;) ... imagine if Art Porter would of used this to assist his GAP PM Flux Gate: https://www.youtube.com/watch?v=fnWuPzAKigs (https://www.youtube.com/watch?v=fnWuPzAKigs)
I don't think he did anything with his flyback and it looks like he used mid impedance coils to lock the flux gate which is costly in input power as much is wasted in coil resistance.
I'm suggesting a high Impedance coils can hold a magnetic field for much long then a mid or low impedance coils
but you want to load them with flyback of a super low impedance coils (below 1 ohm).
Initially you start the magnetic field with a short pulse in the low impedance coil which will make a good magnetic field and not waste power in resistance but in return it converts the current to a super fast high voltage spike which is in perfect condition for a high impedance coil and will create a powerful and
long lasting magnetic field (chose ideal cap for holding time).
So let's use it to our advantage in hopes to change the world.
See videos below on how a high impedance coil
alone can be used to hold a magnetic field. So after the peak pulse they can also be shorted to hold the flux even longer. So imagine how little input power a PM flux gate needs to allow magnets to do work!!!
Part1 https://www.youtube.com/watch?v=SHbQXnXK6Xc (https://www.youtube.com/watch?v=SHbQXnXK6Xc)
Part2 https://www.youtube.com/watch?v=BsN2sr3U0PY (https://www.youtube.com/watch?v=BsN2sr3U0PY)
Collect every MOT you can
Luc
Quote from: minoly on November 16, 2015, 06:07:53 PM
I'm glad people are starting to look at this using the many different methods to harvest and use the spike. This is what John Bedini and others have been talking about for years. Try it on the window motor with the full or even half bipolar Bedini/Cole switch or even the zero force... - you will all be amazed!
Cheers - Patrick
Hi Patrick,
I am quite familiar with JB work.
What is being shared here is very different from SSG and other things that have been built by him or Peter L. for that matter.
If you want flyback that can do work like I and woopy have demonstrated it must come out of a low Impedance coil (very low DC resistance). So this is not the same.
I will ask you and everyone else to refrain from posting any comments related to their work as it is not the same or taken from them.
I'm also not associated with the business of selling secrets for money.
Everything I do I give it free of charge in hope to make this world a better place.
You won't even find advertising on any of my videos.
Regards
Luc
EDIT: I have gone though all the posts and have deleted the posts that are off topic. Please keep it on topic and don't introduce other peoples work.
ADDED:
To MileHigh and others who wish to debate about other thing that are not directly related to this topic. I have split the topic and you can debate to your hearts content there: http://overunity.com/16186/ (http://overunity.com/16186/debates-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg465996/#msg465996)
debates-on-how-to-make-a-more-efficent-motor-using-flyback-moderated (http://overunity.com/16186/debates-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg465996/#msg465996)/msg465996/#msg465996 (http://overunity.com/16186/debates-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg465996/#msg465996)
From now on, those who wish to debate please use the debate topic above or I'll delete your post if I find it doesn't fit here.
Luc
Hi Luc,
Wow! you deleted my post... amazing. I better watch what I say - I thought it was very tame. Others have said worse.
In your first videos you show how the cap is necessary and without it the spike is "lost" and can not be converted.
Have you been able harness the spike to assist in rotation without a cap then? Also, in your last video you said this is all you will share with the public???
Thanks Luc,
Patrick
Quote from: gotoluc on November 17, 2015, 12:23:11 AM
Hi Patrick,
I am quite familiar with JB work.
What is being shared here is very different from SSG and other things that have been built by him or Peter L. for that matter.
If you want flyback that can do work like I and woopy have demonstrated it must come out of a low Impedance coil (very low DC resistance). So this is not the same.
I will ask you and everyone else to refrain from posting any comments related to their work as it is not the same or taken from them.
I'm also not associated with the business of selling secrets for money.
Everything I do I give it free of charge in hope to make this world a better place.
You won't even find advertising on any of my videos.
Regards
Luc
EDIT: I have gone though all the posts and have deleted the posts that are off topic. Please keep it on topic and don't introduce other peoples work.
ADDED:
To MileHigh and others who wish to debate about other thing that are not directly related to this topic. I have split the topic and you can debate to your hearts content there: http://overunity.com/16186/ (http://overunity.com/16186/debates-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg465996/#msg465996)debates-on-how-to-make-a-more-efficent-motor-using-flyback-moderated (http://overunity.com/16186/debates-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg465996/#msg465996)/msg465996/#msg465996 (http://overunity.com/16186/debates-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg465996/#msg465996)
From now on, those who wish to debate please use the debate topic above or I'll delete your post if I find it doesn't fit here.
Luc
Quote from: minoly on November 17, 2015, 10:33:17 AM
Hi Luc,
Wow! you deleted my post... amazing. I better watch what I say - I thought it was very tame. Others have said worse.
Hope you understand it's not a personal thing. It's about keeping the topic very focused so future replicators won't get confused.
Quote from: minoly on November 17, 2015, 10:33:17 AM
In your first videos you show how the cap is necessary and without it the spike is "lost" and can not be converted.
Have you been able harness the spike to assist in rotation without a cap then?
In this video I demonstrated it can also assist without a cap: https://www.youtube.com/watch?v=7dmKEOWOhQA (https://www.youtube.com/watch?v=7dmKEOWOhQA)
Depends on may factors. In time or if you experiment you will understand and chose what you want.
Quote from: minoly on November 17, 2015, 10:33:17 AM
Also, in your last video you said this is all you will share with the public???
Thanks Luc,
Patrick
Woopy made me happy so I gave one idea away.
So are you going to build something?
Luc
Luc, understand this, I build everything I can. I'm not one of those fly-by-night commenters who thinks they know everything. I have experimented with this for some time and have many videos. I still don't understand, you say you share for free - yet you still have many secrets.
regarding the video you show the high voltage coil being used without a CAP, that is a different type of experiment from my question. my question was "
Have you been able harness the spike to assist in rotation without a cap then?"[/size]
I'm not taking away from that experiment that was very nice!
no worries,
Patrick
Quote from: gotoluc on November 17, 2015, 10:53:06 AM
Hope you understand it's not a personal thing. It's about keeping the topic very focused so future replicators won't get confused.
In this video I demonstrated it can also assist without a cap: https://www.youtube.com/watch?v=7dmKEOWOhQA (https://www.youtube.com/watch?v=7dmKEOWOhQA)
Depends on may factors. In time or if you experiment you will understand and chose what you want.
Woopy made me happy so I gave one idea away.
So are you going to build something?
Luc
Quote from: minoly on November 17, 2015, 11:10:25 AM
Luc, understand this, I build everything I can. I'm not one of those fly-by-night commenters who thinks they know everything. I have experimented with this for some time and have many videos.
Please post the link to your video experiments so we can see your work.
Quote from: minoly on November 17, 2015, 11:10:25 AM
I still don't understand, you say you share for free - yet you still have many secrets.
Have I not shared the above to woopy and anyone else who reads this topic?... I don't appreciate your tone or unfounded accusation.
Let's see what you have shared over the years
Quote from: minoly on November 17, 2015, 11:10:25 AM
regarding the video you show the high voltage coil being used without a CAP, that is a different type of experiment from my question. my question was "Have you been able harness the spike to assist in rotation without a cap then?"
Maybe you need to do the experiment?
Regards
Luc
Hi Luc
Thank's for your encouragement. Very intersting and inspiring video.
I have made a test where it is clear that the flybackspike is really powerfull, as it can spin the rotor quite esaily without the magnetic field of the main motor coil. Very encouraging
.
https://youtu.be/y4S3XvloAnM
Hi Guyla
Thank's for your input. I have made a second video (just above) which will answer a part of your question. Concerning the capacitance, the best rpm i can get with my device as it is is 0.3 uF. Now concerning the impedance of the assistant coil i will see what i can do.
Hi all
Thank's for kind word, and please feel free to replicate
Laurent
Thanks again Laurent for this new video demo.
I would suggest that the cap is used to adjust the pulse width (magnetic field on time) of the HV coil. Too much on time can slow the rotor but may prove to have more torque for certain applications and too little on time would not hold the magnetic field long enough to pull or push the rotor magnet or core through.
Thanks for sharing
Luc
Thanks Woopy, that was an excellent tutorial & demo. I haven't watched your last one Luc but I have been testing a few of the things you've demonstrated on a couple of motors. I'll be doing a replication tonight and testing various coil configs. Great work guys.
Quote from: gotoluc on November 17, 2015, 02:16:18 PM
Thanks again Laurent for this new video demo.
I would suggest that the cap is used to adjust the pulse width (magnetic field on time) of the HV coil. Too much on time can slow the rotor but may prove to have more torque for certain applications and too little on time would not hold the magnetic field long enough to pull or push the rotor magnet or core through.
I watched Laurent's video and there were some interesting things revealed.
However, for starters, I think Laurent made some mistakes. He needed longer wires connected to the main drive coil so that he could move it much farther away from the rotor when he wanted to eliminate the influence of the main drive coil on the spinning rotor. Perhaps adding another 30 cm of wire would have allowed him to move the main drive coil a safe distance away from the rotor. The second issue is that I don't think he gave the rotor enough time spin down and stabilize at a new RPM after the main drive coil was partially removed from influencing the rotor. It's a hard drive bearing motor, so perhaps waiting at least 45 seconds to a minute would be required to let the rotor stabilize at a new speed when only being driven by the high-voltage coil.
You can clearly see a trend in Laurent's clip with respect to the changing of the capacitor: If you increase the value of the capacitor the maximum voltage decreases and the discharge time increases. Likewise if you decrease the value of the capacitor the maximum voltage increases and the discharge time decreases.
Now, if your goal is to get the highest RPM for your rotor, that suggests you want to have the high-voltage coil switched ON so the rotor magnet gets a push from one pole and a pull from the other pole of the high-voltage coil. You want the high-voltage coil to switch OFF at the end of the pull phase. This is a function of the ON time, the angle subtended by the two poles of the high-voltage coil, and the RPM of the rotor when it stabilizes at maximum speed.
One more time, this all is an investigation into the timing analysis of the motor. It's hard to find that "Goldilocks" timing with fixed components, but nothing is stopping you from playing with the value of the capacitor live while the motor is running. What you do need though is an accurate timing reference. The easiest way to do that would be to use a small independent pick-up coil to sense the rotor magnet fly-bys and trigger on that. You "sacrifice" a scope channel for that but then you are not "flying blind." You can use your second scope channel to look at other events and know exactly what the rotor position is relative to those other events. Timing is everything when it comes to a pulse motor.
Another option if you are hard core is to look at TK's MHOP clips and make an "LED strobe-scope" or "LED timing gun" and look at the rotor spinning in dim light with the LED illumination revealing the timing to you.
QuoteToo much on time can slow the rotor but may prove to have more torque for certain applications and too little on time would not hold the magnetic field long enough to pull or push the rotor magnet or core through.
I think that you are overlooking one fundamental thing. If the ON time is long or short, you are still dealing with exactly the same amount of energy in the pulse. So the question is how do you get the maximum amount of rotor bang for your fixed pulse energy buck? In my previous posting I am suggesting a precise "Goldilocks" pulse timing that may get the most (push + pull) from the high-voltage coil if the pulse timing is exactly the right width and exactly the right position relative to the high-voltage coil in time but who knows, perhaps a shorter more powerful pulse (same energy) would be better.
There are more suggested questions and investigations leading out of Laurent's clip:
Why does the capacitor voltage get higher when he moves the main drive coil away from the rotor?
Laurent exclaims that the high-voltage coil gives a comparable push to the main low-voltage drive coil. That suggests an important question: How much electrical energy per pulse can you estimate goes into the main low-voltage drive coil vs. how much energy goes into the high-voltage drive coil? How do you make measurements to compare the two? This will give you hard numbers about the relative propulsive force of the high-voltage coil as compared to the low-voltage coil. This is the very essence of your experiment.
Finally, there is an outstanding question that becomes even more important considering what was just mentioned above: The high-voltage coil is quite a high resistance and therefore there are certainly a lot of resistive losses in the high-voltage coil. How high are the resistive losses in the high-voltage coil and how do you measure them? How does that compare to the pulse energy that you put into the high-voltage coil.
These are serious real-world questions about what you guys are doing. If you choose to ignore them because you don't know what to do that's your choice. What I think you guys should really do is really discuss these issues and get your juices flowing and try to figure these things out among yourselves. It's the difference between doing an "experiment" that is little more than observing your motor in action vs. truly trying to understand the energy dynamics and trying to optimize your motor. This is all up to you guys for yourselves.
Thanks again Woopy! Very nice info!
And what about the back-spike coming off of the back-spike coil... Could a second reed switch be positioned so as to use it's back-spike ALSO on a third coil? (etc. etc. etc.)
Main coil - Reed switch - back-spike to (coil & Cap) - Reed switch - back-spike to (coil & Cap) - Reed switch - back-spike to (coil & Cap)
Just thinking of ideas here.
Hi Laurent,
Thanks for your new video with the interesting tests.
So my first estimation on increasing the 1 uF to make the pulse width wider was good but then this reduced the rotor RPM. :(
With your second 1 uF added to the first 1 uF, the pulse width widened to 3 ms from the 2 ms but the peak amplitude went down to cca 62 Vpp from the cca 90 Vpp. This makes sense because the flyback pulse via the diode has a more or less given energy content from the collapsing field and the 2 uF capacitor with the same 1.8 H coil clearly represents a heavier load to the pulse than the 1 uF cap with the same 1.8 H does as MileHigh described this.
This means that if the goal is to get the highest pushing or pulling force at the ends of the C core it is the peak voltage across the HV coil which is to be maximized as you mentioned in the video because max peak voltages can give max peak currents which participate in the Amper-turns.
You mentioned you found the 0.3 uF cap value where the rotor RPM was the highest. This surely means that the peak voltage across the HV coil must have gone well above the earlier 90 Vpp value. The 0.3 uF gives resonance at as high as 216.5 Hz with the 1.8 H coil, this means the parallel LC circuit remains a capacitive load for the flyback pulse even at the increased rotor RPM too.
Greetings,
Gyula
Another Idea,
If we are looking to make a magnified magnetic motive force from the back-spike, then maybe try running the back-spike through a step-up transformer first to increase the voltage spike to even higher voltages, then send that spiked-up energy to multiple High voltage electromagnet cores (stators).
As Newman said, he runs his motors off of voltage.
The higher voltage overrides the wire resistance to create a magnetic field through many many many turns of wire. Once the magnetic field exist though, evidently the choice of capacitor will determine the rate of magnetic field collapse. In the mean time, the magnetic field can be used to produce great torque.
I don't have the ability to test the above, but sure would like to see from Luc or Woopy if stepping up the voltage will in fact allow multiple High voltage electromagnet cores vs. one electromagnet core without a voltage step-up.
MagnaMoRo
Quote from: woopy on November 17, 2015, 01:01:55 PM
Hi Luc
Thank's for your encouragement. Very intersting and inspiring video.
I have made a test where it is clear that the flybackspike is really powerfull, as it can spin the rotor quite esaily without the magnetic field of the main motor coil. Very encouraging
.
https://youtu.be/y4S3XvloAnM (https://youtu.be/y4S3XvloAnM)
Hi Guyla
Thank's for your input. I have made a second video (just above) which will answer a part of your question. Concerning the capacitance, the best rpm i can get with my device as it is is 0.3 uF. Now concerning the impedance of the assistant coil i will see what i can do.
Hi all
Thank's for kind word, and please feel free to replicate
Laurent
Hey Woopy.
Probably seen it before and may not be what you are doing with the caps, but it does show a single drive coil powered by a battery with a reed switch and the bemf is taken from the first coil and sent to other a cap bank to power a second drive coil. And then the bemf from the second drive coil to charge a very bad battery.
The second drive coil here also helps the rotor to pick up speed.
https://www.youtube.com/watch?v=nXxvAQ_mdUk
Mags
Quote from: Magluvin on November 17, 2015, 11:05:44 PM
Hey Woopy.
Probably seen it before and may not be what you are doing with the caps, but it does show a single drive coil powered by a battery with a reed switch and the bemf is taken from the first coil and sent to other a cap bank to power a second drive coil. And then the bemf from the second drive coil to charge a very bad battery.
The second drive coil here also helps the rotor to pick up speed.
https://www.youtube.com/watch?v=nXxvAQ_mdUk (https://www.youtube.com/watch?v=nXxvAQ_mdUk)
Mags
Sorry Mags, but this is not what we are doing here and overcomplicated.
I have always respected you and your research. However, I need to keep this topic simple and not a consortium of different experiments. So I hope you understand if I delete your post and this one tomorrow.
Kind regards and please feel free to replicate what is being shared here.
Luc
Quote from: gotoluc on November 17, 2015, 11:56:57 PM
Sorry Mags, but this is not what we are doing here and overcomplicated.
I have always respected you and your research. However, I need to keep this topic simple and not a consortium of different experiments. So I hope you understand if I delete your post and this one tomorrow.
Kind regards and please feel free to replicate what is being shared here.
Luc
Luc
Mags system is much the same-just a little more advanced. He collects the backEMF,and drives a second coil with it-this is what Woopy is doing. The difference is that Mags can trigger the second coil with that captured energy with the second reed switch. He then also captures the backEMF from the second coil,and uses that to charge a battery,where as Woopy looses that within the cap/inductor set up on the second coil.
Mag's is well within the thread title-more so than others ATM as far as i can see
Quote title: Sharing ideas on how to make a more efficent motor using Flyback
Brad
I concur with tinman mags is one step ahead of the average BEAR BOBO.
I looked at Luc's first clip: Sharing ideas, how to build a more efficient motor (p.1)
https://www.youtube.com/watch?v=XfLcBD3Fy7M
When he has the high-voltage coil in the MOT core and no capacitor there is no perceptible attraction with the transformer core top plate. Then when he ads the 11 uF capacitor to the circuit then there is fairly strong attraction with the transformer core top plate.
It begs the question, why is there no perceptible attraction when there is no capacitor? Note that in both cases the same BEMF pulse energy is coming from the motor drive coil. I just find it somewhat unusual that there is no perceptible attraction with no capacitor in the circuit. Also, what happens to the attractive force as the capacitor gets larger and larger?
This is for Luc and all others interested in this subject to ponder and try to explain. You are trying to use a second coil to do work on the rotor to make it spin faster so it behooves you to want to understand the why and how behind these issues.
One more time, this is about the difference between just observing the effects and taking note of them (a "demonstration"), and observing the effects and then understanding the how and why that explain the observed effects (an "experiment.")
This is what I was trying to point out, let's see if your words stick better than mine...
Magluvin is putting the spike to a cap and rather than using 100% of it on one pulse, he is switching it with another reed to another "same voltage coil" and conserving even more to a second battery.
This is a different nuance than what Luc is doing however, as MH states one has to do all the experiments to observe each effect. You don't however have to post a vid each time to prove anything to anyone.
It's all about the spike.
I feel this post is on topic however you might delete it anyway either way I'm still getting a lot out of reading every post in this thread.
Thanks!
Quote from: MileHigh on November 18, 2015, 07:42:31 AM
I looked at Luc's first clip: Sharing ideas, how to build a more efficient motor (p.1)
https://www.youtube.com/watch?v=XfLcBD3Fy7M (https://www.youtube.com/watch?v=XfLcBD3Fy7M)
When he has the high-voltage coil in the MOT core and no capacitor there is no perceptible attraction with the transformer core top plate. Then when he ads the 11 uF capacitor to the circuit then there is fairly strong attraction with the transformer core top plate.
It begs the question, why is there no perceptible attraction when there is no capacitor? Note that in both cases the same BEMF pulse energy is coming from the motor drive coil. I just find it somewhat unusual that there is no perceptible attraction with no capacitor in the circuit. Also, what happens to the attractive force as the capacitor gets larger and larger?
This is for Luc and all others interested in this subject to ponder and try to explain. You are trying to use a second coil to do work on the rotor to make it spin faster so it behooves you to want to understand the why and how behind these issues.
One more time, this is about the difference between just observing the effects and taking note of them (a "demonstration"), and observing the effects and then understanding the how and why that explain the observed effects (an "experiment.")
Quote from: tinman on November 18, 2015, 05:23:30 AM
Luc
Mags system is much the same-just a little more advanced. He collects the backEMF,and drives a second coil with it-this is what Woopy is doing. The difference is that Mags can trigger the second coil with that captured energy with the second reed switch. He then also captures the backEMF from the second coil,and uses that to charge a battery,where as Woopy looses that within the cap/inductor set up on the second coil.
Mag's is well within the thread title-more so than others ATM as far as i can see
Quote title: Sharing ideas on how to make a more efficent motor using Flyback
Brad
Thanks for your input Brad.
I understand quite well what Mags has done and I have made similar tests and circuits many years back but have not shared it as the Low impedance to Low impedance to once again Low impedance process did not produce favorable results. Much is lost in the conversions.
The idea I'm sharing is simple, it's a Low impedance to High impedance motor assistance. It does not involve any circuitry, just 2 passive components.
You're suggesting that both can do the same, so I challenge TinMan Power ;) to build both and report which one you find to best assist a motor.
Think of it as a PMBO challenge ;D
Thanks for your input
Luc
Quote from: MileHigh on November 18, 2015, 07:42:31 AM
I looked at Luc's first clip: Sharing ideas, how to build a more efficient motor (p.1)
https://www.youtube.com/watch?v=XfLcBD3Fy7M
When he has the high-voltage coil in the MOT core and no capacitor there is no perceptible attraction with the transformer core top plate. Then when he ads the 11 uF capacitor to the circuit then there is fairly strong attraction with the transformer core top plate.
It begs the question, why is there no perceptible attraction when there is no capacitor? Note that in both cases the same BEMF pulse energy is coming from the motor drive coil. I just find it somewhat unusual that there is no perceptible attraction with no capacitor in the circuit. Also, what happens to the attractive force as the capacitor gets larger and larger?
This is for Luc and all others interested in this subject to ponder and try to explain. You are trying to use a second coil to do work on the rotor to make it spin faster so it behooves you to want to understand the why and how behind these issues.
One more time, this is about the difference between just observing the effects and taking note of them (a "demonstration"), and observing the effects and then understanding the how and why that explain the observed effects (an "experiment.")
With and without the capacitor. It's much the same as to why a vehicle going a set speed can be stopped quicker if you dont lock up the brakes and skid the wheels. Voltage leads current in an inductor,and if the applied voltage pulse is to quick and short,then there will be very little current that follows. However,if that spike is sent to a capacitor where the current leads the voltage,then you can be assured that the inductor will receive all of the !now stored! energy from that inductive kickback that was stored within the cap.
Quote from: gotoluc on November 18, 2015, 09:26:23 AM
Thanks for your input Brad.
I understand quite well what Mags has done and I have made similar tests and circuits many years back but have not shared it as the Low impedance to Low impedance to once again Low impedance process did not produce favorable results. Much is lost in the conversions.
The idea I'm sharing is simple, it's a Low impedance to High impedance motor assistance. It does not involve any circuitry, just 2 passive components.
You're suggesting that both can do the same, so I challenge TinMan Power ;) to build both and report which one you find to best assist a motor.
Think of it as a PMBO challenge ;D
Thanks for your input
Luc
I accept that challenge.
But i can tell you now with some certainty that a low impedance secondary coil will perform better due to lower resistive losses ;)
Hi all
Thank's for info and proposals.
This setup in my videos is made to test Gotoluc's proposition concerning the flybackspike recovery.
Sofar i can confirm that the rotor spins easily and steadily when motorised by the "assistant" coil only.
Now for sure, those first and crude results are encouraging and motivating to go to a much better documented experiment.
Perhaps there will be people there to go in this direction.
Now one thing continues to puzzle me.
My beginner's knowledge indicates that, during the pulse, the main (low voltage ) motor coil, get a certain amount of electric energy (voltage and current) from the power station. This energy is dissipated for a small part in heat due to the impedance of the coil and for the main part in building a magnetic field . During the building of the magnetic field its magnetic strength increases with the increasing current in the inductor and propels the rotor magnet which get kinetic energy.
So it seems to me, that at the end of the pulse, all the electrical input energy should have been transformed in some heat and the kinetic energy of the rotor, yes or not? And the magnetic field seems to stay there at its max strength until the end of the pulse. yes or not?
Than it seems commonly admitted that at the end of the pulse, the magnetic field brutally collapses and creates the flybackspike.
But if the input energy is totally dissipated at the end of the pulse, how does the collapsing magnetic field create this powerfull flybackspike ? What is the process ?
Thank' to pardon my ignorance, but i am missing something here.
Laurent
Quote from: woopy on November 18, 2015, 10:27:30 AM
Hi all
Thank's for info and proposals.
This setup in my videos is made to test Gotoluc's proposition concerning the flybackspike recovery.
Sofar i can confirm that the rotor spins easily and steadily when motorised by the "assistant" coil only.
Now for sure, those first and crude results are encouraging and motivating to go to a much better documented experiment.
Perhaps there will be people there to go in this direction.
Now one thing continues to puzzle me.
My beginner's knowledge indicates that, during the pulse, the main (low voltage ) motor coil, get a certain amount of electric energy (voltage and current) from the power station. This energy is dissipated for a small part in heat due to the impedance of the coil and for the main part in building a magnetic field . During the building of the magnetic field its magnetic strength increases with the increasing current in the inductor and propels the rotor magnet which get kinetic energy.
So it seems to me, that at the end of the pulse, all the electrical input energy should have been transformed in some heat and the kinetic energy of the rotor, yes or not? And the magnetic field seems to stay there at its max strength until the end of the pulse. yes or not?
Than it seems commonly admitted that at the end of the pulse, the magnetic field brutally collapses and creates the flybackspike.
But if the input energy is totally dissipated at the end of the pulse, how does the collapsing magnetic field create this powerfull flybackspike ? What is the process ?
Thank' to pardon my ignorance, but i am missing something here.
Laurent
@Laurent,
This goes back to Nicola's very first invention: The "Spark Gap Generator". Simply a capacitor with two electrodes over head. The capacitor charge begins to increase naturally, one plate grounded and the other attached to an antenna and collector. The resistance between the air gap spark electrodes begins to decrease as the intensity of the electro magnetic field from the capacitor plates increases resulting in a sudden spark discharge and magnetic field collapse. The "Arc of the Covenent" was just such a generator.
Tesla theorized that there were two ways to generate power: One; Magnetic field collapse and the other; "Faraday Induction". The field collapse generates a "Longitudinal Wave" that reaches the limits of the Universe in all dimensions down to the Quanta and to the ends of the Cosmos instantaneously.
This invokes the "Broadcast Power Equation".
The power of a violent field collapse is infinite! Think about this!
Radio had it's beginnings when European experimenters, long before Tesla, discovered that a field collapse could excite iron filings at a remote distance. The B.C. "Arc" could play this trick! This combination could signal a coordinated attack command to separate concealed regiments!
Hi Synchro
Very nice explication
Have you some links on the subject for my info
Thank's
Laurent
Quote from: tinman on November 18, 2015, 10:20:52 AM
I accept that challenge.
But i can tell you now with some certainty that a low impedance secondary coil will perform better due to lower resistive losses ;)
Glad you accept!
And I would not argue with you that low impedance coils are more efficient since they have next to no resistance but the tradeoff is they have next to no inductance.
I'm suggesting we use the ideal qualities of each (tuned) at the appropriate time to achieve a longer flux holding time to switch a PM flux gate with less input power then just a low or mid impedance coil.
So no need to build a motor, just a flux gate will do. As we know a flux gate will make a motor work and what we need is one that can work with less power input so we get the magnets to do most of the work.
This is what I'm ideally proposing and thought I would make it clear before you start your build.
Thanks for your participation and looking forward to your results
Luc
Quote from: woopy on November 18, 2015, 10:27:30 AM
....
My beginner's knowledge indicates that, during the pulse, the main (low voltage ) motor coil, get a certain amount of electric energy (voltage and current) from the power station. This energy is dissipated for a small part in heat due to the impedance of the coil and for the main part in building a magnetic field . During the building of the magnetic field its magnetic strength increases with the increasing current in the inductor and propels the rotor magnet which get kinetic energy.
So it seems to me, that at the end of the pulse, all the electrical input energy should have been transformed in some heat and the kinetic energy of the rotor, yes or not? And the magnetic field seems to stay there at its max strength until the end of the pulse. yes or not?
Than it seems commonly admitted that at the end of the pulse, the magnetic field brutally collapses and creates the flybackspike.
But if the input energy is totally dissipated at the end of the pulse, how does the collapsing magnetic field create this powerfull flybackspike ? What is the process ?
....
Hi Laurent,
During the time a magnetic field is building up in your coil (input current is ON) and starts to propel your rotor, the field is not fully used up at all in the process (some small part of the field strays away, this depends also on the mechanical construction), just like the field of a permanent magnet is not used up in a normal magnetic interaction.
Of course, normally the action-reaction force rules between the rotor and the magnetic field created from the input energy: if you try to brake the movement of the rotor, input energy consumption increases during the ON time of the input current (Lenz law).
It is known that an inductance tries to resist to any current change. (A capacitor tries to resist to any voltage change across it).
In this link I think the "why" i.e. why a coil resists to any current change) is correctly explained:
http://www.allaboutcircuits.com/textbook/direct-current/chpt-15/magnetic-fields-and-inductance/
Quote from the text:
"Because inductors store the kinetic energy of moving electrons in the form of a magnetic field, they behave quite differently than resistors (which simply dissipate energy in the form of heat) in a circuit. Energy storage in an inductor is a function of the amount of current through it. An inductor's ability to store energy as a function of current results in a tendency to try to maintain current at a constant level. In other words, inductors tend to resist changes in current. When current through an inductor is increased or decreased, the inductor "resists" the change by producing a voltage between its leads in opposing polarity to the change."
Perhaps these can explain better what may happen and helps you.
Gyula
Quote from: tinman on November 18, 2015, 10:15:35 AM
With and without the capacitor. It's much the same as to why a vehicle going a set speed can be stopped quicker if you dont lock up the brakes and skid the wheels. Voltage leads current in an inductor,and if the applied voltage pulse is to quick and short,then there will be very little current that follows. However,if that spike is sent to a capacitor where the current leads the voltage,then you can be assured that the inductor will receive all of the !now stored! energy from that inductive kickback that was stored within the cap.
Yes, but you still have the mystery of where the pulse of energy from the drive coil goes and why at least when Luc holds the metal slab in his hands the magnetic attraction is imperceptible. It doesn't mean it's not there, but at least to his hands it's imperceptible. When you do a preliminary analysis in your head you would have an expectation that he might feel something.
QuoteVoltage leads current in an inductor,and if the applied voltage pulse is to quick and short,then there will be very little current that follows.
Let's try to be a bit more precise with your explanation. You have to remember, it's
not a voltage pulse, it's a
current pulse.
For starters, what happens when coil A discharges into coil B, where the initial conditions are there is some current is flowing in coil A, let's say it's one amp, and zero current flowing in coil B. Let's say both coils are one Henry of inductance. That analogous to what the situation is like in Luc's setup when there is no capacitor. Anybody want to try to answer that?
Quote from: woopy on November 18, 2015, 10:27:30 AM
My beginner's knowledge indicates that, during the pulse, the main (low voltage ) motor coil, get a certain amount of electric energy (voltage and current) from the power station. This energy is dissipated for a small part in heat due to the impedance of the coil and for the main part in building a magnetic field . During the building of the magnetic field its magnetic strength increases with the increasing current in the inductor and propels the rotor magnet which get kinetic energy.
So it seems to me, that at the end of the pulse, all the electrical input energy should have been transformed in some heat and the kinetic energy of the rotor, yes or not? And the magnetic field seems to stay there at its max strength until the end of the pulse. yes or not?
Than it seems commonly admitted that at the end of the pulse, the magnetic field brutally collapses and creates the flybackspike.
But if the input energy is totally dissipated at the end of the pulse, how does the collapsing magnetic field create this powerfull flybackspike ? What is the process ?
Laurent,
Your analysis is basically correct and I will "fill in some of the blanks" for you. For starters, this statement, "This energy is dissipated for a small part in heat due to the impedance of the coil and for the main part in building a magnetic field." should read as, "This energy is dissipated for a small part in heat due to the
resistance of the coil and for the main part in building a magnetic field."
The impedance of the coil is associated with its inductance and stores energy, it does not dissipate energy and turn it into heat like a resistance. So "impedance" and "resistance," although somewhat related because they are both associated with the blocking of the flow of current, in fact they are separate and distinct terms that should not be interchanged with each other.
QuoteSo it seems to me, that at the end of the pulse, all the electrical input energy should have been transformed in some heat and the kinetic energy of the rotor, yes or not? And the magnetic field seems to stay there at its max strength until the end of the pulse. yes or not?
You are essentially correct. The electrical input energy goes to three places: 1) heat energy, 2) kinetic energy of the rotor, and 3) the BEMF pulse energy.
The magnetic field is
not at it's max strength at the end of the pulse. If the drive coil gives a big kinetic energy push to the rotor, then at the end of the constant-voltage energizing pulse from the reed switch there will be
less current flowing through the coil. In other words, without the rotor in place, then there will be more current flowing through the drive coil at the end of the constant-voltage energizing pulse from the reed switch.
You actually see this effect in your own clip, and I made reference to it in the form of a question but nobody responded. Please review your clip and see if you can find any evidence for this effect.
QuoteBut if the input energy is totally dissipated at the end of the pulse, how does the collapsing magnetic field create this powerfull flybackspike ? What is the process ?
As indicated above, the input energy is not totally dissipated at the end of the constant-voltage energizing pulse from the reed switch. Some of the input energy remains at the end of the pulse on the rotor and it becomes the flyback pulse.
It goes full circle to what I said before: How much energy is in the main drive pulse on the rotor from the drive coil? How much energy is in the flyback pulse? How do I measure and compare the two?
Quote from: MileHigh on November 18, 2015, 02:07:40 PM
Laurent,
Your analysis is basically correct and I will "fill in some of the blanks" for you. For starters, this statement, "This energy is dissipated for a small part in heat due to the impedance of the coil and for the main part in building a magnetic field." should read as, "This energy is dissipated for a small part in heat due to the resistance of the coil and for the main part in building a magnetic field."
The impedance of the coil is associated with its inductance and stores energy, it does not dissipate energy and turn it into heat like a resistance. So "impedance" and "resistance," although somewhat related because they are both associated with the blocking of the flow of current, in fact they are separate and distinct terms that should not be interchanged with each other.
You are essentially correct. The electrical input energy goes to three places: 1) heat energy, 2) kinetic energy of the rotor, and 3) the BEMF pulse energy.
The magnetic field is not at it's max strength at the end of the pulse. If the drive coil gives a big kinetic energy push to the rotor, then at the end of the constant-voltage energizing pulse from the reed switch there will be less current flowing through the coil. In other words, without the rotor in place, then there will be more current flowing through the drive coil at the end of the constant-voltage energizing pulse from the reed switch.
You actually see this effect in your own clip, and I made reference to it in the form of a question but nobody responded. Please review your clip and see if you can find any evidence for this effect.
As indicated above, the input energy is not totally dissipated at the end of the constant-voltage energizing pulse from the reed switch. Some of the input energy remains at the end of the pulse on the rotor and it becomes the flyback pulse.
It goes full circle to what I said before: How much energy is in the main drive pulse on the rotor from the drive coil? How much energy is in the flyback pulse? How do I measure and compare the two?
@Milehigh,
This is a quote from you from above:
"Some of the input energy remains at the end of the pulse on the rotor and it becomes the flyback pulse".
This is nonsense! Here's what happens:
The coil is energized by the electrical current. A magnetic field appears in the coil. The current is stopped and the magnetic field collapses.
There is no left over current in the coil at this point!The magnetic field collapse generates a new hi-voltage current. As the field collapses the field is moving in the opposite direction from the expanding field and generates a new current in the opposite direction as it passes inward across the windings.
This is the flyback pulse! The faster the current is cut off to the coil the higher the flyback voltage. The magnet rotor can't have any effect on the power coil when the Reed switch is open.
Additionally, the magnetic field collapse generates a longitudinal scaler wave that has an "Impulse Magnetizing" effect. This power is infinite!
Quote from: gotoluc on November 18, 2015, 11:38:21 AM
Glad you accept!
And I would not argue with you that low impedance coils are more efficient since they have next to no resistance but the tradeoff is they have next to no inductance.
I'm suggesting we use the ideal qualities of each (tuned) at the appropriate time to achieve a longer flux holding time to switch a PM flux gate with less input power then just a low or mid impedance coil.
So no need to build a motor, just a flux gate will do. As we know a flux gate will make a motor work and what we need is one that can work with less power input so we get the magnets to do most of the work.
This is what I'm ideally proposing and thought I would make it clear before you start your build.
Thanks for your participation and looking forward to your results
Luc
So what you are proposing is different than what Laurent is doing then? I'm new to this Fluxgate making a motor work can anyone explain this or point me in the right direction?
Quote from: synchro1 on November 18, 2015, 05:01:21 PM
@Milehigh,
This is a quote from you from above:
"Some of the input energy remains at the end of the pulse on the rotor and it becomes the flyback pulse".
This is nonsense! Here's what happens:
The coil is energized by the electrical current. A magnetic field appears in the coil. The current is stopped and the magnetic field collapses. There's no left over current in the coil at this point!
This is the flyback pulse! The faster the current is cut off to the coil the higher the flyback voltage. The magnet rotor can't have any effect on the power coil when the Reed switch is open.
Additionally, the magnetic field collapse generates a longitudinal scaler wave that has an "Impulse Magnetizing" effect. This power is infinite!
QuoteThe magnetic field collapse generates a new hi-voltage current. As the field collapses the field is moving in the opposite direction from the expanding field and generates a new current in the opposite directions as it passes inward across the windings.
No Synchro,the current dose not reverse direction-->only the voltage across the coil/inductor inverts. You apply a current in one direction to build the magnetic field around the inductor= current to form magnetic field. Then when the supply current is cut off,the magnetic field collapses,and creates a current flow=magnetic field changing in time to create current flow through the inductor. As the magnetic field around that inductor has not inverted (swapped polarities),then the current flow remains in the same direction.
Brad.
Quote from: minoly on November 18, 2015, 05:45:59 PM
So what you are proposing is different than what Laurent is doing then? I'm new to this Fluxgate making a motor work can anyone explain this or point me in the right direction?
As the thread is entitled making a more efficient motor using the flyback,then that is what i will do.
The difference between the fluxgate and motor is that you have magnetic fields changing with time with regards to the inductor-this comes from the permanent magnets on the rotor.
I will simply use a generating coil with a variable resistive load to calculate motor torque and RPM. This way we can see how the motor performs under different loads.
Quote from: synchro1 on November 18, 2015, 05:01:21 PM
@Milehigh,
This is a quote from you from above:
"Some of the input energy remains at the end of the pulse on the rotor and it becomes the flyback pulse".
This is nonsense! Here's what happens:
The coil is energized by the electrical current. A magnetic field appears in the coil. The current is stopped and the magnetic field collapses. There's no left over current in the coil at this point! The magnetic filed collapse generates new hi-voltage current. As the field collapses the field is moving in the opposite direction from the expanding field and generates a new current in the opposite directions as it passes inward across the windings. The faster the current is cut off to the coil the higher the fly back voltage.
Absolutely correct.
Additionally, a soft iron core will amplify any field polarity produced by the windings. This is because the 'magnetic domains' align themselves to the external field and add to the sum total of the field.
Voltage is equivalent to pressure, it forces current through the wire. So the higher the voltage, the longer the wire can be through which one can 'move' current.
However,
current is NOT movement of anything trough the conductor (wire) end-to-end. It should be thought of as an out-of-phase alignment of conductor atoms by way of force (voltage pressure). The direction of the phase changing pressure relative to the conductor will determine the polarity of the resultant magnetic field around the conductor.
During the time that the phase is changing, the magnetic field will appear to be 'moving'. During the collapse, the out-of-phase atomic condition is springing back into place. So, the field is moving in the opposite direction when collapsing, however the field itself will still be in the same polarity (between 0 and 9 as apposed to -9 and 0).
The amount of out-of-phase atomic alignment is determined by the voltage and length of wire giving rise to sum total resistance to the out-of-phase condition. Once the available voltage causes a maximized out-of-phase condition, the magnetic field no longer appears to be moving. The voltage must be maintained to maintain a certain out-of-phase degree.
We are looking to get high voltage (pressure) to generate an out-of-phase atomic condition throughout a lengthy piece of wire with many turns rapped around the magnetic domains of a soft iron core, to create a strong usable magnetic field.
I say: increase the back-spike voltage (by way of a step-up transformer) so it works across a longer conductor (multiple electromagnets) (stators). ::)
MagnaMoRo
Quote from: MoRo on November 18, 2015, 06:13:34 PM
Absolutely correct.
Additionally, a soft iron core will amplify any field polarity produced by the windings. This is because the 'magnetic domains' align themselves to the external field and add to the sum total of the field.
Voltage is equivalent to pressure, it forces current through the wire. So the higher the voltage, the longer the wire can be through which one can 'move' current.
However, current is NOT movement of anything trough the conductor (wire) end-to-end. It should be thought of as an out-of-phase alignment of conductor atoms by way of force (voltage pressure). The direction of the phase changing pressure relative to the conductor will determine the polarity of the resultant magnetic field around the conductor.
During the time that the phase is changing, the magnetic field will appear to be 'moving'. During the collapse, the out-of-phase atomic condition is springing back into place. So, the field is moving in the opposite direction when collapsing, however the field itself will still be in the same polarity (between 0 and 9 as apposed to -9 and 0).
The amount of out-of-phase atomic alignment is determined by the voltage and length of wire giving rise to sum total resistance to the out-of-phase condition. Once the available voltage causes a maximized out-of-phase condition, the magnetic field no longer appears to be moving. The voltage must be maintained to maintain a certain out-of-phase degree.
We are looking to get high voltage (pressure) to generate an out-of-phase atomic condition throughout a lengthy piece of wire with many turns rapped around the magnetic domains of a soft iron core, to create a strong usable magnetic field.
I say: increase the back-spike voltage (by way of a step-up transformer) so it works across a longer conductor (multiple electromagnets) (stators). ::)
MagnaMoRo
QuoteAbsolutely correct.
Not correct at all-->the current dose not reverse direction when the field collapses around the inductor.
@Tinman,
Please watch the first portion of this video. Watch where the arrows change direction:
https://www.youtube.com/watch?v=hNHXtIWaaig
Quote from: synchro1 on November 18, 2015, 06:27:22 PM
@Tinman,
Please watch the first portion of this video. Watch where the arrows change direction:
https://www.youtube.com/watch?v=hNHXtIWaaig
You have shown a tank circuit where the current is being rocked back and forth between an inductor and a capacitor--different situation.
When a current is sent to an inductor,the inductor builds a magnetic field around it. When that current supply to the inductor is cut off,the field around that inductor collapses. The voltage across that inductor will invert,but the current flow through that inductor will continue to flow in the same direction. Think about what happens in a water pipe that causes water hammer when the tap that supplies that water flow is abruptly switched off. The pressure (voltage) in that pipe after the tap invert's,but the water(current) wants to keep flowing in the same direction through that pipe.
If we are to help Luc in making a more efficient motor using the flyback,then it is important that this is understood.
Voltage invert's across the inductor,but current keeps flowing in the same direction.
The reason a coil inductance takes time for the field to build and current to eventually fully flow is the fact that the initial expanding field 'generates a reverse current of the input. Due to resistance and losses, the input overcomes the reverse emf. So as Tinman said, when the field collapses, opposite of expanding, it generates a current in the same direction as the input before it is disconnected. For me, thats where the term BEMF is not correct or mistaken, and should be called forward emf. As long as the reference is the direction of the input.
Mags
Quote from: tinman on November 18, 2015, 06:25:34 PM
Not correct at all-->the current dose not reverse direction when the field collapses around the inductor.
True. -->> "the current dose NOT reverse direction"
Especially in light of what I have explained 'current' to be: -->> The out-of-phase degree.
However, during field collapse (which is actually: -->> spring-back to normal phase alignment) the detected voltage in the conductor end-to-end will be opposite polarity until the normal phase angle is reached.
MagnaMoRo
Quote from: tinman on November 18, 2015, 06:42:28 PM
You have shown a tank circuit where the current is being rocked back and forth between an inductor and a capacitor--different situation.
When a current is sent to an inductor,the inductor builds a magnetic field around it. When that current supply to the inductor is cut off,the field around that inductor collapses. The voltage across that inductor will invert,but the current flow through that inductor will continue to flow in the same direction. Think about what happens in a water pipe that causes water hammer when the tap that supplies that water flow is abruptly switched off. The pressure (voltage) in that pipe after the tap invert's,but the water(current) wants to keep flowing in the same direction through that pipe.
If we are to help Luc in making a more efficient motor using the flyback,then it is important that this is understood.
Voltage invert's across the inductor,but current keeps flowing in the same direction.
@Tinman,
Please look at the first minute of this video where Igor explains his schematic:
https://www.youtube.com/watch?v=vWvI7T7h3tk
Hi Luc
I haven't been able to watch any vids since my data is used up for the month.
But I have found that you can collect 5 times the amount of any coil ,whether it is drive or generating.
The lower the UF of the cap , along with a paralleled cap bank can catch the spike.
The spike can feed as many as you can switch into collection.
Woppy add some diodes to the out put of your coil ,feed them into seperate banks, using the 1UF as a buffer in front of the bank.
If it's off topic feel free to delete.
Great thread.
Thanks artv
Yes Shylo, great thread!
Please feel free to share more if you can as 5x more sounds very interesting.
I'm very happy to see everyone's participation and some of us coming to new understandings.
Even if my ideas don't improve a motors efficiency we will all come out of it with something which should be the spirit of this forum.
Thanks to all for sharing
Luc
@Tinman,
https://www.youtube.com/watch?v=y4S3XvloAnM
Here's Woopyjump's latest video above: His schematic below shows the power coming into the top of the coil through the reed switch from the positive electrode of the source. The schematic then shows where the flyback spike exits out through the bottom of the coil and travels to the auxiliary coil and capacitor through the diode: That's to the right!
It may look like the current's traveling in the same direction but instead of going from positive to negative the flyback's going from negative to positive! The flyback current is traveling in the reverse direction, up the auxiliary coil from bottom to top! There's merely the illusion that the flyback current's traveling in the same direction. The power pulse current originated from the positive electrode, then after the collapse and polarity reversal, the flyback runs backwards to return to the positive plate of the capacitor, not the negative ground! In order for the flyback current to be considered as traveling in the same direction, it would have to continue on the same path towards the ground, where the original power pulse current was formerly headed.
That's to the left!
The flyback current "REVERSES DIRECTION" at that junction! The flyback current in no way continues to travel in the same direction as the original power pulse current.
Luc Take a coil , spin a magnetic field over the coil, the coil produces an emf.
Hook a bridge to the coil, now you have DC output.
Connect a cap , now you have stored DC.
Where the bridge is hooked to the coil add half way bridges, but they have to be hooked to the same cap as before but a different one , But through a 1UF cap ,or it won't collect.
It will with a higher UF rating but not as good.
I'm still doing testes and need to build more switches , I think this thing will self run.
The half bridges with the low UF and high voltage caps , buffer the spike so it can be collected by a normal cap.
Probably bad wording.
artv
In woopjump's circuit, the flyback current needs the positively charged plate of the auxiliary capacitor to run toward while the Reed switch is open! This positively charged capacitor plate acts as the opposite current's new "Reverse Biased Ground"!
The Hi-voltage flyback spike travels through the auxiliary coil first to reach the positive cap plate and lowers the coil inductance triggering a capacitor discharge pulse backwards into the auxiliary coil. This discharge pulse powers the magnet rotor. The stored capacitor power is generated by the magnet rotor. This creates a timing event that has to be adjusted for by carefully positioning the auxiliary coil! The auxiliary output coil's polarity determines which of the capacitor plates is positive. This capacitor with no diode is a "Swinging Door".
Woopyjump removes the power coil and motors the magnet rotor solely with the auxiliary coil. He could easily run a second magnet rotor with the freed up power coil!
The question is: How did Luc get this all figured out to begin with?
Quote from: synchro1 on November 18, 2015, 09:02:24 PM
In woopjump's circuit, the flyback current needs the positively charged plate of the auxiliary capacitor to run toward while the Reed switch is open! This positively charged capacitor plate acts as the opposite current's new "Reverse Biased Ground"!
The Hi-voltage flyback spike travels through the auxiliary coil first to reach the positive cap plate and lowers the coil inductance triggering a capacitor discharge pulse backwards into the auxiliary coil. This discharge pulse powers the magnet rotor. The stored capacitor power is generated by the magnet rotor. This creates a timing event that has to be adjusted for by carefully positioning the auxiliary coil! The auxiliary output coil's polarity determines which of the capacitor plates is positive. This capacitor with no diode is a "Swinging Door".
Woopyjump removes the power coil and motors the magnet rotor solely with the auxiliary coil. He could easily run a second magnet rotor with the freed up power coil!
The question is: How did Luc get this all figured out to begin with?
Like I said, we are talking about two different things.
I am talking about the flyback from the primary coil, not the secondary coil tank circuit.
The current in the primary coil continues to flow in the same direction once the reed switch becomes open. BackEMF is an incorrect term for what is known as the inductive kickback. BackEMF is the voltage in an inductor that apposes that which created it.
Like Mags said, it should be called an inductive kick foward, but as the voltage inverts across the inductor, it got the name inductive kickback of flyback.
As the title of the thread says using flyback to increase motor efficiency, then I thought it important that the flybacks current flow direction is correctly known.
Quote from: tinman on November 18, 2015, 11:56:27 PM
Like I said, we are talking about two different things.
I am talking about the flyback from the primary coil, not the secondary coil tank circuit.
The current in the primary coil continues to flow in the same direction once the reed switch becomes open. BackEMF is an incorrect term for what is known as the inductive kickback. BackEMF is the voltage in an inductor that apposes that which created it.
Like Mags said, it should be called an inductive kick foward, but as the voltage inverts across the inductor, it got the name inductive kickback of flyback.
As the title of the thread says using flyback to increase motor efficiency, then I thought it important that the flybacks current flow direction is correctly known.
@Tinman,
Pay attention Bub! The current from the primary coil generates a magnetic field that collapses and reverses current polarity and travels to the right at the junction when the Reed switch opens. When the Reed switch is closed the current from the primary coil travels to the left toward the negative ground! The yellow marker's pointing at the general area in the schematic below!
Reading this thread and looking at the different videos gave me an idea.. for a project I have been working on.
Using a classical bedini circuit and running a second (identical one) on the back-emf alone.
Result
http://youtu.be/IZPmdtTTmz0
It also works with identical coils, although they dont pulse at the same time as this is going to be a 3 phase motor generator
But it will be more efficient than running it on one pulse motor alone..
Sorry the video was very quick (and dirty) but i had to do the experiment before i leave for holliday until next tuesday
Luciano
This ties directly into my 4th paper at https://sites.google.com/site/nilrehob/home/elementary-physics which is really cool :-)
The same thing can be done with charge and momentum etc, also in my papers.
I have a video showing it with charge at https://www.youtube.com/watch?v=xZcvOWSXcbU
/Hob
Recycling unused energy that would normally be wasted is a good thing. But how many times do you want to recycle that energy is a valid question. For example, when you look at Laurent's clips you notice when he dumps the drive coil pulse into the capacitor and then the capacitor discharges through the second drive coil it completely discharges and you are done. That's arguably an "elegant" solution - do a single recycle, burn off all of the unused recycled energy and you are done. There is arguably little to gain by adding a second energy recycling stage. It makes the circuit more complex and at each stage there are losses. More stages equals more losses.
There is a somewhat ironic situation happening. This thread is about a "more efficient motor" and only myself and Tinman have mentioned that the high-voltage coil has a relatively high resistance and that means more losses. I have challenged you guys about measuring the losses in the high-voltage coil and nobody wants to touch it. Also, nobody really knows what a true "Goldilocks" pulse would be for the high-voltage coil. Perhaps a longer push-pull pulse that subtends the angle between the limits of the "U" core of the high-voltage coil? (Relative to the passing rotor magnet.) What about a shorter stronger pushing pulse that is focused on one end of the "U" core only where the magnetic field is stronger? It's an unknown.
On the side of the good news, you are free to move the high-voltage coil around and hunt for a sweet spot. That is your very easy variable timing system. You don't actually need to scope the timing or make an LED timing gun. What I can say through is that if I was a builder, I would add this to my bag of tricks: In the TK MHOP clips he over-drives a "regular" LED and has a circuit to flash it at the start and end of the drive coil ON time. To make it simpler, I would get a big mean very high-power LED, and use the op-amp to switch on and off a power transistor. Then I would paint little white dots on top of the rotor magnets, and voila, you have your LED strobe timing gun to see the conduction angle for the drive coil. You could make a little jig and then use your LED strobe timing gun on any pulse motor that you build.
So here is a variation on what you see in Laurent's clip to consider: You need to have a high-voltage coil with multiple taps. Then you can experiment with different capacitors, and different tap points on the high-voltage coil. Can you shape the pulse to get more rotor bang for your pulse that comes from the drive coil? Here is the big question: When I change taps on the high-voltage coil, I lower the resistance of the coil. Will that improve my overall setup because there is less resistance in the secondary coil? The presumption is that you will get a shorter length pulse. But don't forget that you are dealing with less turns, and therefore a weaker generation of a magnetic field for the same amount of current. It's hard to know, but it could be an interesting investigation.
Now I am going to play devil's advocate. Sometimes in life you arrive at a zero sum game, and that may apply to the multiple-tap high-voltage coil. In one case you have higher turns, higher resistance, and a higher strength of magnetic field generated. In the other case you have lower turns, lower resistance, and a lower strength of magnetic field generated. What happens to the resistive losses in the coil when we compare the two and we want the same strength of magnetic field generated? Well, we know that low current, high turns, high resistance will be equal to high current, low turns, low resistance for the same strength of magnetic field. So what if we look at the resistive losses in these two cases? You are comparing (low current x high resistance) losses to (high current x low resistance) losses for the same strength of magnetic field generation. That sounds like a zero sum game to me. You can crunch the numbers on paper without even building anything to check this.
If you assume that it is indeed a zero sum game with respect to the resistive losses, (and we can't be 100% sure of that in the real world actual build) then the key to extracting the maximum rotor bang for your pulse buck may me more directly related to pulse strength, pulse timing and pulse duration. Let's not even discuss the pulse strength because you have limited control over it. So, to adjust the timing is trivial, because you have the luxury of physically being able to move the high-voltage coil around to effectively change the pulse timing. Pulse duration is controllable with the size of the capacitor with the caveat that the larger the capacitor, the longer the pulse duration, and the weaker the pulse strength.
As you can see, it's an interesting study in trade-offs when you play with the various parameters. This is all presuming that you have a fixed amount of pulse energy to work with. And I will leave you with another question to ponder: You are trying to make the most efficient pulse motor by recycling the pulse energy that comes from the drive coil when the reed switch (or transistor/MOSFET) switches off. But perhaps a big pulse of energy from the drive coil is not as good as optimizing the drive coil pulse first and foremost, and then working with a smaller pulse for the high-voltage coil. What is the best energy mix between the low-voltage drive coil and the high-voltage secondary coil? 70-30? 50-50? 30-70? It's hard to know that one but at least being conscious of it is worthwhile.
I had looked through my vids to see if i had one, but dont.... I remember trying to get a bemf spike into a higher henry coil and the higher H coil seemed to block most of the spike rather than take advantage of the full potential. Like a subwoofer crossover coil, it blocks out the high frequencies. So the capacitor across your higher inductance coil probably loads up first then delivers it charge to the parallel coil?
Mags
One other comment that just occurred to me. We notice in Laurent's clips that the capacitor simply discharges through the high-voltage coil and we observe the voltage decreasing in an almost linear fashion. That means there is a fairly even pulse of current going through the coil because i = C dv/dt. This is somewhat strange, because the circuit is an LC tank, and normally it's supposed to resonate. My assumption is that between the EMF induced into the high-voltage coil from the passing rotor magnet, and the relatively high resistance of the coil, you "stumble" upon this favourable capacitor discharge curve for the pulse motor operation.
However, if the secondary coil resistance starts to get lower because you change to a tap on the coil with a lower number of turns and a lower resistance, then you might start to see the LC circuit start to resonate. It's pretty safe to assume that a resonating LC tank circuit will NOT help in the operation of the pulse motor.
Quote from: synchro1 on November 19, 2015, 12:03:58 AM
@Tinman,
Pay attention Bub! The current from the primary coil generates a magnetic field that collapses and reverses current polarity and travels to the right at the junction when the Reed switch opens. When the Reed switch is closed the current from the primary coil travels to the left toward the negative ground! The yellow marker's pointing at the general area in the schematic below!
Bub ???
No, the current dose not revers polarity in the primary inductor-only the voltage dose.
You have confused current flow direction with current flow path.
OK,i will try one more time.
Please see attached pic below.
Diagram at top shows conventional current flow when reed switch is closed.
Diagram at bottom shows conventional current flow when reed switch opens.
As you can see by the red arrows,the current continues to flow in the !same! direction through the primary coil when the reed switch opend. You only have to look at the diode direction in the picture you posted of Woopy's circuit to see that this is true. If the current flow reversed as you say,then the diode would have to be turned around in order for that flyback current to flow into the cap/inductor combo on the secondary.
The cap will charge first with the flyback current,as it has a lower series resistance than that of the high turn coil. The cap will then start to discharge into the secondary coil as it reaches peak volatge. This is the reason for the !almost! linear discharge through the secondary coil,and the absence of resonant oscillations.
Hope that clears things up.
People tend to neglect preciously listening to details. When input current is switched on at a moment, from that moment on till 5 times L/R time the current increases and magnetic field in and around the coil also increases in strength. After the 5*L/R time elapses, a steady current flows in the coil and its magnetic field settles at a constant value. (R means the total DC resistance of coil and its closed circuit via the battery and the energy content of the field is E= L*I*I/2 where I is the instantaneous value of the current, L is the coil inductance.)
When you switch off the current, the magnetic field starts to collapse. This means if it was just increasing in strength, then it suddenly starts decreasing in strength but it cannot change its earlier poles at the coil's end: what was North pole at one end it still remains North pole, just the intensity of the North pole reduces drastically at the moment of the current switch-off.
This flux change i.e. from an increase to a decrease direction (or from a steady state value to a much lower value down to zero) of same poles is which flips the induced voltage polarity across the coil with respect to the voltage polarity the input battery established. To compare this increase-decrease field change, imagine you approach a magnet towards the end of a ferromagnetic core (no coil is needed for this comparison) and then you stop the magnet say 2mm from the core then you start to move the magnet away from the core along the same straith line the approach happened, then the change of the magnetic filed strength in the core represents the event that happens when you switch the current off in a coil. The rest of the process is also included in Tinman's nice explanation in his above post, no need to repeat.
Dear MileHigh,
you wrote:
"There is a somewhat ironic situation happening. This thread is about a "more efficient motor" and only myself and Tinman have mentioned that the high-voltage coil has a relatively high resistance and that means more losses."
Well, please be more attentive because I also wrote about the 210 Ohm DC resistance as a loss in the HV coil in Reply 11, in the last but one paragraph, just after Laurent showed his first video replication:
http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg465969/#msg465969 (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg465969/#msg465969)
Of course I do not consider this forum as a pissing contest, just I ask you to be more attentive, that is all.
One more thing: Laurent just tested the concept as he said and obviously did not design for low loss, I think he just used coils, components at hand.
Thanks,
Gyula
Hi all
Yes very intersting development here, thank's to all for participating.
As i have the motor , i spent some time to place a 1 ohm current sensing resistor all arround the flyback ciscuit.
1- direct in the flybackspike just between the output of the main (low voltage) coil and the diode, there is a very very narrow and strong current spike (peak at arround 1.3 A ) directed from coil to diode
2- Almost the same between the diode and the cap entry, but just after the spike,there is a very small and wide "opposed direction" current trace followed by very little ringing (difficult to say because the trace is noisy)
3 almost the same between the output of the cap and the entry of the main (low voltage ) coil,
4- between the entry of the cap and the high voltage coil (assistant) there is a very small and wide current trace corresponding to (2) or the discharge of the cap in the assistant coil.
So this confirm what Timan showed in his graphic.
Now the fact that there is almost no ringing, confirm also that the high voltage coil seems to prevent almost totally the resonance of the tank circuit and maintain the slow and regular discharge of the cap which provide the usable secondary pulse .
So we will see if Tinman can get the same or better result with a lower voltage assistant coil.
Just for info i have tried to calculate the average power of the pulse of the main coil and the assistant coil. (calculation of the pulse duration only ,not the complete cycle) As said this is quite difficult because the noisy trace of the very low current input in the assistant coil.
I have edited this part because i have to redo my calculation, because the main pulse and his son ( the pulse of the assistant Coil) are not of the same duration.
Hope this helps
laurent
Quote from: tinman on November 19, 2015, 05:12:11 AM
Bub ???
No, the current dose not revers polarity in the primary inductor-only the voltage dose.
You have confused current flow direction with current flow path.
OK,i will try one more time.
Please see attached pic below.
Diagram at top shows conventional current flow when reed switch is closed.
Diagram at bottom shows conventional current flow when reed switch opens.
As you can see by the red arrows,the current continues to flow in the !same! direction through the primary coil when the reed switch opend. You only have to look at the diode direction in the picture you posted of Woopy's circuit to see that this is true. If the current flow reversed as you say,then the diode would have to be turned around in order for that flyback current to flow into the cap/inductor combo on the secondary.
The cap will charge first with the flyback current,as it has a lower series resistance than that of the high turn coil. The cap will then start to discharge into the secondary coil as it reaches peak volatge. This is the reason for the !almost! linear discharge through the secondary coil,and the absence of resonant oscillations.
Hope that clears things up.
@Tinman,
There's no path back to the positive electrode for the reversed current with the Reed switch open. Positioning the diode at the top of the coil and connecting it to the positive electrode would send the current in the other direction backwards through the Primary to source. The destination determines the current polarity. The negative ground is the destination of the power pulse; Then the reversed current polarity seeks a positive ground!
You caused this same kind of confusion in that Universal motor schematic you passed to Chris Sykes on the bucking coil thread.
Charged capacitors resist change in voltage just like inductors. The sitting charge on the capacitor would determine the destination of the flyback, either through the inductor or to the cap. The LC tank is not isolated but part of a magnet rotor charging system. The spinning magnet rotor is causing a "Lenz Related" issue with the inductor that's outside the other factor. One the other or both, the flyback causes the capacitor to discharge into the Hi-voltage coil to accelerate the magnet rotor with a power pulse.
You're trying to maintain that a current can exist with a reversed voltage polarity. This is impossible! The current changes direction headed towards an opposite ground albeit by a circuitous path. The path is illusory, the direction of the current is switched from negative to positive inside the primary after the magnetic field collapse.
Hi Laurent,
I indicated with red X where I think you meant the positionings of the 1 Ohm resistor around the circuit, please correct me if I made any mistake and I will redraw it.
Would it be possible to take photos (snapshots) of the 4 waveforms on the scope you described in points 1 to 4? That would be better for everyone I suppose.
Thanks
Gyula
I wish I was coming to new understandings however I'm still having problems grasping the terminology.
A lot can happen in 24 hours of missed reading... I notice many are off topic. Tinman - thanks for trying to explain how a fluxgate can make a motor work to me. I'm so new to that terminology though, I'm hoping someone can break it down for me in simple step by step terms.
This is what Luc is trying to do right? Luc is not taking the spike from a coil on a rotor and then putting it to another coil to make it spin faster. He is using a fluxgate to make a motor work. However, I've looked up fluxgate and still can not figure out how to make a motor work with one.
So, putting the spike to a "high voltage coil" makes the coil stay magnetized longer which can electromagnetically hold onto a load longer, which has it's own uses, but I can't think of how to make a fluxgate make a motor work. Perhaps stronger at slower speeds? More torque? Or am I completely off and is what Laurent is doing exactly what you are after?
Can someone point me in the right direction as I believe this is the topic?
Quote from: gotoluc on November 18, 2015, 07:19:20 PM
Yes Shylo, great thread!
Please feel free to share more if you can as 5x more sounds very interesting.
I'm very happy to see everyone's participation and some of us coming to new understandings.
Even if my ideas don't improve a motors efficiency we will all come out of it with something which should be the spirit of this forum.
Thanks to all for sharing
Luc
Hi All,
I have written about this CLL system in my 4th paper at:
https://sites.google.com/site/nilrehob/home/elementary-physics
named 4-increasing-magnetic-flux.pdf. Increasing magnetic flux in the inductance domain is the same as increasing charge in the capacitance domain which I show in my video https://www.youtube.com/watch?v=xZcvOWSXcbU which in turn is the same as increasing momentum in the mass domain which I wrote about in my 2nd paper called 2-increasing-momentum-or-charge.pdf.
It is really simple and I love the way you guys have proved it!
You can think of it as a small mass colliding elastically with a big mass. In your case the inductors are the masses and the capacitor is the spring that makes the colission elastic. If the big mass is standing still before the collision, as the coil with hi inductance is with no current, the sum of the absolute value of the momentums increase, in this case the sum of the absolute value of the magnetic fluxs increase.
In the video above I show it for two capacitors and a coil. One big cap with no charge and one small cap discharging through a coil to make the collision elastic. Its the same thing, the sum of the absolute value of the charges increase.
/Hob
Hi Guyla
You are correct
Here some pics
your X1 is pic 1 and 2
your X2 is pic 3 and 4
your X3 is pic 5 and 6
your X4 is pic 7 and 8
Hope this helps
Laurent
Quote from: minoly on November 19, 2015, 11:07:16 AM
I wish I was coming to new understandings however I'm still having problems grasping the terminology.
A lot can happen in 24 hours of missed reading... I notice many are off topic. Tinman - thanks for trying to explain how a fluxgate can make a motor work to me. I'm so new to that terminology though, I'm hoping someone can break it down for me in simple step by step terms.
This is what Luc is trying to do right? Luc is not taking the spike from a coil on a rotor and then putting it to another coil to make it spin faster. He is using a fluxgate to make a motor work. However, I've looked up fluxgate and still can not figure out how to make a motor work with one.
So, putting the spike to a "high voltage coil" makes the coil stay magnetized longer which can electromagnetically hold onto a load longer, which has it's own uses, but I can't think of how to make a fluxgate make a motor work. Perhaps stronger at slower speeds? More torque? Or am I completely off and is what Laurent is doing exactly what you are after?
Can someone point me in the right direction as I believe this is the topic?
SUBJECT:
Sharing ideas on how to make a more efficient motor using Fly-backBy 'more efficient' we mean: 'over-unity'; COP > 1
By 'using Fly-back' we mean: by using the energy received back from a primary coil after the initial source energizing said primary coil has been switched off.
Problems to overcome: how to best use the energy received back.
The terms:
Yes, understanding what is actually happening in a circuit can be difficult to understand. We can't see electricity. Our understanding comes from observable phenomena. This requires experimentation and deep thought. Many diagrams do not reflect what is actually happening. I myself try to visualize what is actually happening at any given moment on an atomic level with any given section of a circuit, including the connecting wires. So, for example, I don't really like the words 'current' or 'flow'. these words imply that something is flowing through the wire. I don't believe that assumption to be correct, because the wires are solid. To best explain what is observed I believe there is simply 'voltage pressure' which causes a phase change away from normal in every single atom of a conductor that is in-between a potential difference. Apply voltage pressure one way and the atoms push back (resist). The resistance is exposed as a polarized field around the conductor, Shut off the voltage quickly and the atoms snap back to neutral phase quickly, perhaps even 'ringing' a little like a stiff spring. The greater the voltage the grater the atom goes out-of-phase, until you apply enough voltage to blow the atom itself off.
My other observations:
Luc's microwave core is more efficient. Why? because the copper windings don't just have an iron core passing through the middle of it, but the windings actually have soft iron material all the way around it creating a magnetic loop when the cover plate is placed on top. So the core surface area (the area against the core windings) is very important in the design of the electromagnet to achieve greater power from the available area.
Quote from: MoRo on November 19, 2015, 02:27:33 PM
SUBJECT: Sharing ideas on how to make a more efficient motor using Fly-back
By 'more efficient' we mean: 'over-unity'; COP > 1
By 'using Fly-back' we mean: by using the energy received back from a primary coil after the initial source energizing said primary coil has been switched off.
Problems to overcome: how to best use the energy received back.
The terms:
Yes, understanding what is actually happening in a circuit can be difficult to understand. We can't see electricity. Our understanding comes from observable phenomena. This requires experimentation and deep thought. Many diagrams do not reflect what is actually happening. I myself try to visualize what is actually happening at any given moment on an atomic level with any given section of a circuit, including the connecting wires. So, for example, I don't really like the words 'current' or 'flow'. these words imply that something is flowing through the wire. I don't believe that assumption to be correct, because the wires are solid. To best explain what is observed I believe there is simply 'voltage pressure' which causes a phase change away from normal in every single atom of a conductor that is in-between a potential difference. Apply voltage pressure one way and the atoms push back (resist). The resistance is exposed as a polarized field around the conductor, Shut off the voltage quickly and the atoms snap back to neutral phase quickly, perhaps even 'ringing' a little like a stiff spring. The greater the voltage the grater the atom goes out-of-phase, until you apply enough voltage to blow the atom itself off.
Very good ;)
Quote from: MoRo on November 19, 2015, 02:27:33 PM
My other observations:
Luc's microwave core is more efficient. Why? because the copper windings don't just have an iron core passing through the middle of it, but the windings actually have soft iron material all the way around it creating a magnetic loop when the cover plate is placed on top. So the core surface area (the area against the core windings) is very important in the design of the electromagnet to achieve greater power from the available area.
Excellent!... glad someone has their thinking cap on.
To capitalize on this effect, surface area is King.
The next thing I would recommend is a device that needs a long flux holding time. This is why I recommend a PM flux switch.
I have already designed and built it. I'll call it the GTL (short for "go to luc") Gate. The GTL Gate uses the same design principle of my mostly magnet motor, where both inner and outer fields of the coil is used which gives you double the magnetic field and obviously double the core surface area.
I will be sharing it soon but I was holding off till TinMan shares his results.
If you don't understand you will after I demonstrate it.
Thanks for everyone's participation
Luc
Gyula:
Yes I apologize for not acknowledging that you made reference to the efficiency issue because of the resistance of the coil. I actually did read your comments and I did remember them. That was just an unconscious "filter" in my mind acting because I take it for granted that you really know your stuff - hence I filtered out your contribution to the discussion.
All:
Congratulations to Laurent for attempting some power calculations. If he shows his measurements and calculations then we can then discuss it more. You all have to realize that spoon feeding you information is not the way to learn. Nor am I an expert at this stuff, but I am not afraid of being corrected if I am wrong. You can't ignore major issues about a circuit because you are unsure about what to do. Trying something is better than doing nothing. Share your ideas and see what comes of it. All of you must be interested in knowing what the resistive losses in the two coils are compared to the energy that you put into the coils. You have to be interested in this - "ideas on how to make a more efficient motor." Yet the silence from many of you guys about this and other issues is sometimes deafening.
Going back to my "zero sum gain" discussion, I realize that the only way to get lower resistance and accompanying lower power dissipation in a coil for the same amount of magnetic field strength is to increase the gauge of the wire. So multiple taps on the same coil when you want the same strength of magnetic field will all burn off approximately the same amount of power. I will just repeat myself that lower resistance in the high-voltage coil "risks" the onset of LC resonance which presumably will not have a positive effect on the operation of the pulse motor.
Minoly correctly uses quotations when he says "high voltage coil." It's very poor terminology. There is no such thing as a "high voltage coil." Any coil can generate high voltage or be driven by high voltage, hence the use of that term as well as "low voltage coil" does not really make sense. From now on I will use the terms "drive coil" or "low inductance coil" and "secondary coil" or "high inductance coil" to refer to the two coils.
Nilrehob's comments about the mechanical analogies for circuit components are excellent and I am a big fan of that. For inductors, I like the heavy spinning flywheel on a ball bearing analogy, but his moving mass analogy is equally valid. I asked and got no answer about what happens when the drive coil outputs the current pulse into the secondary coil and there is no capacitor present. Many of you guys are too shy and you only feel safe doing the same old tests and discussing the same old things that you have done many times before. Time to break out of the box. How would you frame the question and then answer the question if you substitute the drive coil and secondary coil for a pair of flywheels? Do that correctly and the answer comes very easily.
Quote from: MoRo on November 19, 2015, 02:27:33 PM
SUBJECT: Sharing ideas on how to make a more efficient motor using Fly-back
By 'more efficient' we mean: 'over-unity'; COP > 1
By 'using Fly-back' we mean: by using the energy received back from a primary coil after the initial source energizing said primary coil has been switched off.
Nope, by "more efficient" we mean trying to put as much of the supply power as possible into making the rotor spin. What goes hand-in-hand with that is we want to reduce the resistive losses as much as possible within reason.
Your analogies are somewhat bizarre and don't really make sense to me. You talk about a "phase change" and a "polarized field" without defining them. This is an exercise in basic electronics - a basic pulse circuit. A much more applicable analogy to the circuit would be moving masses or spinning flywheels, mechanical springs, and calipers with brake pads. It's more mundane sounding but much more real.
Quote from: MileHigh on November 19, 2015, 03:11:46 PM
Minoly correctly uses quotations when he says "high voltage coil." It's very poor terminology. There is no such thing as a "high voltage coil." Any coil can generate high voltage or be driven by high voltage, hence the use of that term as well as "low voltage coil" does not really make sense. From now on I will use the terms "drive coil" or "low inductance coil" and "secondary coil" or "high inductance coil" to refer to the two coils.
Nilrehob's comments about the mechanical analogies for circuit components are excellent and I am a big fan of that. For inductors, I like the heavy spinning flywheel on a ball bearing analogy, but his moving mass analogy is equally valid. I asked and got no answer about what happens when the drive coil outputs the current pulse into the secondary coil and there is no capacitor present. Many of you guys are too shy and you only feel safe doing the same old tests and discussing the same old things that you have done many times before. Time to break out of the box. How would you frame the question and then answer the question if you substitute the drive coil and secondary coil for a pair of flywheels? Do that correctly and the answer comes very easily.
MH
I'm going to ask you to be more tolerant when it comes to using the correct schooled engineering terminology. Nilrehob's is obviously an engineer, so that's why what he says or writes works for you (same school)
However, the reality on free energy sites like this one is, most experimenters are not school trained, as they are looking for something that is not taught in schools.
So don't nitpick on details like you did in your post, as you know quite well what he means.
I'll share my thoughts on Nilrehob's work... I can't understand most of what he is sharing as I have no school training. I never even finished high school. I quite on my 16th birthday as it was hell for 10 years. But that doesn't make me less then others who have schooling, it actually allows me to be creative and not just stick to what I was taught.
So be more tolerant and don't criticize as you were never taught what we are looking for, have you?
Kind regards
Luc
Luc,
I don't understand why you keep deleting my posts. I'm trying to understand more clearly exactly what you are doing. You even go on to answer my question yet you still delete my post???
Quote from: gotoluc on November 19, 2015, 03:43:48 PM
MH
I'm going to ask you to be more tolerant when it comes to using the correct schooled engineering terminology. Nilrehob's is obviously an engineer, so that why what he says or writes works for you (same school)
However, the reality on free energy sites like this one is, most experimenters are not schooled trained as they are looking for something that is not taught in schools.
So don't nitpick on details like you did in your post, as you know quite well what he means.
I'll share my thoughts on Nilrehob's work... I can't understand most of what he is sharing as I have no school training. I never even finished high school. I quite on my 16th birthday as it was hell for 10 years. But that doesn't make me less then others who have schooling, it actually allows me to be creative and not just stick to what I was taught.
So be more tolerant and don't criticize as you were never taught what we are looking for, have you?
Kind regards
Luc
Hi Laurent,
Thanks for showing the waveforms across the 4 different placement of the 1 Ohm (current shunt) resistor around the flyback circuit. In the meantime I figured that we could arrive at energy calculations in a relatively easy way, see below, albeit this would involve one more measurement from you...
I think if we wish to estimate the energy in the secondary (assistant) coil, then you could use a 10 Ohm non-inductive resistor instead of the 1 Ohm and repeat step 4 to learn about the current in the secondary coil. I suggest this because then we could calculate the energy in the coil by the (1.8*I*I)/2 and we can also calculate the (heat) loss in the 210 Ohm coil resistance. I suggest the 10 Ohm because it is still negligible with respect to the 210 Ohm and the scoped waveform will be 10 times bigger, this helps evaluating the coil current better than with the 1 Ohm. Try to make sure the scope should display the average or at least the rms value of the coil current waveform, this would make it easier to get the loss.
I think you used the 0.3 uF capacitor for the scope waveforms, right? This involved a cca 130 V flyback pulse across the drive coil if I recall correctly, is this correct? If yes, then we know that the peak voltage across the 0.3 uF capacitor must have been also about 130 V (let's neglect the 0.7 V forward voltage drop of the flyback diode). If this is correct, then the energy stored in this capacitor in the moment of the full peak voltage across it is Ec=(C*V*V)/2.
So it is (0.0000003*130*130)/2= 0.00253 Joule or 2.53 mJ. This energy will be going into the secondary coil. This energy could be used as a cross-checking for the energy in the secondary coil received from the 10 Ohm current shunt measurement.
Now to estimate the input power if you wish to know about it, then probably the input DC current times the input DC voltage should give a good approximation, read from your power supply meters. Perhaps the use of an electrolytic filter capacitor right across the voltage input of the circuit, (say at least a 220 or 470 uF or higher) would help reduce the current meter fluctuations. The capacitor's positive leg would directly connect to the reed switch positive end and the negative leg of the capacitor would go to the negative end of the drive coil via the shortest connecting wire. You could also build the RC filter circuit for the input as Luc uses such with the two electrolytic caps with a 0.1 Ohm resistor in beween. Of course, a some mH choke coil with low DC resistance would also be good here instead of the R member of the filter and the current meter of the power supply would "calm down". 8) But possibly the single parallel capacitor will hopefully suffice.
Thanks, Gyula
PS MileHigh: okay, thanks.
Quote from: MileHigh on November 19, 2015, 03:11:46 PM
Nilrehob's comments about the mechanical analogies for circuit components are excellent and I am a big fan of that. For inductors, I like the heavy spinning flywheel on a ball bearing analogy, but his moving mass analogy is equally valid. I asked and got no answer about what happens when the drive coil outputs the current pulse into the secondary coil and there is no capacitor present. Many of you guys are too shy and you only feel safe doing the same old tests and discussing the same old things that you have done many times before. Time to break out of the box. How would you frame the question and then answer the question if you substitute the drive coil and secondary coil for a pair of flywheels? Do that correctly and the answer comes very easily.
I believe I posted what happens if the cap were not there one page back.
http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg466132/#msg466132
"I had looked through my vids to see if i had one, but dont.... I remember trying to get a bemf spike into a higher henry coil and the higher H coil seemed to block most of the spike rather than take advantage of the full potential. Like a subwoofer crossover coil, it blocks out the high frequencies. So the capacitor across your higher inductance coil probably loads up first then delivers it charge to the parallel coil?
Mags"
Mags
Quote from: minoly on November 19, 2015, 04:09:42 PM
Luc,
I don't understand why you keep deleting my posts. I'm trying to understand more clearly exactly what you are doing. You even go on to answer my question yet you still delete my post???
The only posts I deleted were your first posts relating to J Bedini. Please tell me which post you believe to be deleted and why I would answer your question to a post you believe to be deleted?
BTW, you never replied to my post to you on page 2: http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg466026/#msg466026 (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg466026/#msg466026)
Now on page 7 and still waiting
Luc
Quote from: synchro1 on November 19, 2015, 09:51:11 AM
@Tinman,
QuoteThere's no path back to the positive electrode for the reversed current with the Reed switch open.
1st,there is no reverse current.
2nd,i have not drawn a current path with the reed switch open back to the source.
QuoteThe destination determines the current polarity.
No it dose not. The current flow direction through the primary inductor is determind by the inflow current direction.
Once again,the current flow through the primary inductor will remain the same once the reed switch is opened.
QuoteThe negative ground is the destination of the power pulse; Then the reversed current polarity seeks a positive ground!
No.
The current flow will seek the path of lowest resistance once the reed switch is opened. In this case,it is the capacitor,and this forms a current loop-->first through the capacitor and primary inductor,and the secondary loop once the cap is charged is from the cap through the secondary inductor--as i have drawn.
QuoteYou caused this same kind of confusion in that Universal motor schematic you passed to Chris Sykes on the bucking coil thread.
Only those that do not understand that the current continues to flow in the same direction through the primary coil once the current supply to that coil is switch off,are the only one's that get confused.
QuoteCharged capacitors resist change in voltage just like inductors. The sitting charge on the capacitor would determine the destination of the flyback, either through the inductor or to the cap.
If you look at Woopy's scope shot's,you can see that the cap is discharged each cycle,and thus the current would flow into the cap first in the next cycle.
QuoteOne the other or both, the flyback causes the capacitor to discharge into the Hi-voltage coil to accelerate the magnet rotor with a power pulse.
The flyback current charges the capacitor,and the capacitor then discharges via the secondary inductor. This you can see in the linear discharge scope trace in Woopy's video's.
QuoteYou're trying to maintain that a current can exist with a reversed voltage polarity. This is impossible!
This is not impossible at all,and is exactly what happens across the primary inductor when the reed switch opens-->the voltage across that inductor inverts,but the current flow continues in the same direction.
QuoteThe current changes direction headed towards an opposite ground albeit by a circuitous path. The path is illusory, the direction of the current is switched from negative to positive inside the primary after the magnetic field collapse.
No again.
The current dose not change direction,but the voltage polarity dose invert across the primary coil.
So when the reed switch is closed,the positive voltage side of the inductor is the same as the positive side of the supply(battery/power supply). When the reed switch opens,the current will continue to flow through that primary coil in the same direction,but the voltage across that coil has now inverted-->so what was the positive voltage side of the primary coil when the reed switch was closed in now the negative(ground) side once the reed switch opens.
Quote from: gotoluc on November 17, 2015, 12:58:24 PM
Please post the link to your video experiments so we can see your work.
Have I not shared the above to woopy and anyone else who reads this topic?... I don't appreciate your tone or unfounded accusation.
Let's see what you have shared over the years
Maybe you need to do the experiment?
Regards
Luc
this post was gone, I checked from two computers and 2 different browsers, now it's back... http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg466152/#msg466152
My mistake.
you answered indirectly when you replied to MoRo's attempt to reply to my question:
http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg466166/#msg466166
"
The next thing I would recommend is a device that needs a long flux holding time. This is why I recommend a PM flux switch.[/size]
[/size]
I have already designed and built it. I'll call it the GTL (short for "go to luc") Gate. The GTL Gate uses the same design principle of my mostly magnet motor, where both inner and outer fields of the coil is used which gives you double the magnetic field and obviously double the core surface area."[/size]
I still do not understand it though.
You know me from past. I've often commented in your vids. I've always had good things to say about your work. https://www.youtube.com/channel/UCrIHB1lv7XElAD_ztL-1FPw
You might not appreciate my tone, but I did not appreciate the smack when all I did was share my opinion in the first post about JB.
I hope I'm understanding this correctly, it is not discharging into a cap that is in parallel to a coil that will aid in spinning a rotor (to put it simply) that you are interested in; however, your interest is in creating a fluxgate using only high voltage/impedance coils that will hold/pull with more force using the spike? This GTL sounds very interesting. I have lot's of ideas how to use the spike to provide more torque, I cannot think of any that would aid in a mostly magnet motor - very interesting indeed.
Quote from: tinman on November 19, 2015, 06:15:18 PM
1st,there is no reverse current.
2nd,i have not drawn a current path with the reed switch open back to the source.
No it dose not. The current flow direction through the primary inductor is determind by the inflow current direction.
Once again,the current flow through the primary inductor will remain the same once the reed switch is opened.
No.
The current flow will seek the path of lowest resistance once the reed switch is opened. In this case,it is the capacitor,and this forms a current loop-->first through the capacitor and primary inductor,and the secondary loop once the cap is charged is from the cap through the secondary inductor--as i have drawn.
Only those that do not understand that the current continues to flow in the same direction through the primary coil once the current supply to that coil is switch off,are the only one's that get confused.
If you look at Woopy's scope shot's,you can see that the cap is discharged each cycle,and thus the current would flow into the cap first in the next cycle.
The flyback current charges the capacitor,and the capacitor then discharges via the secondary inductor. This you can see in the linear discharge scope trace in Woopy's video's.
This is not impossible at all,and is exactly what happens across the primary inductor when the reed switch opens-->the voltage across that inductor inverts,but the current flow continues in the same direction.
No again.
The current dose not change direction,but the voltage polarity dose invert across the primary coil.
So when the reed switch is closed,the positive voltage side of the inductor is the same as the positive side of the supply(battery/power supply). When the reed switch opens,the current will continue to flow through that primary coil in the same direction,but the voltage across that coil has now inverted-->so what was the positive voltage side of the primary coil when the reed switch was closed in now the negative(ground) side once the reed switch opens.
Thinking about it more...
If I recall correctly, my coils in my vid are 2mh .45ohm. They wont do much with a spike from another coil because it wont let high freq through. So you are most likely correct to try a coil like mine in the circuit with the cap in parallel.
Mags
Gyula:
Excellent post about the power measurements and quite similar to what I was thinking. I don't have the real bench measurement skills that you have but permit me to give an outline of my line of thinking and analysis for this task.
Just for fun, I am going to keep a running tally of all of the measurements and assign each one a capital letter.
For the input power I was thinking of just putting a one-ohm CVR before the reed switch and measuring across that with a quality multimeter to get the average voltage to compute the average input current for the entire circuit [A]. A non-inductive CVR would be preferable but not necessarily critical in this application. This assumes that the voltage output from the power supply is constant and then we can measure the input power for the whole circuit.
Then, if you measured the RMS voltage across the same CVR it would be giving you the RMS current through the drive coil . Therefore you have the power dissipation in the 0.5 ohm resistance of the drive coil [C].
The average input power would be the average input current times the input voltage [D].
Therefore you are already in a position to estimate the power in the drive coil that is exported to the "outside world." The outside world includes two components, 1) the mechanical power that is imparted on the spinning rotor, and 2) the power that goes through the diode and drives the LC circuit. So the "exported power from the drive coil" [E], would be the average input power [D] minus the power dissipation in the resistance of the drive coil [C].
You can make an accurate measurement of the pulse frequency [F], therefore you can measure the input pulse energy based on the average input power [G].
You can also measure the "exported pulse energy from the drive coil" [H].
Here is where we are so far:
A - Average input current for entire circuit
B - RMS current through the drive coil
C - Power dissipation in the resistance of the drive coil
D - Average input power
E - Exported power from the drive coil, ([D] - [C])
F - Pulse frequency
G - Input pulse energy ([D]/[F])
H - Exported pulse energy from the drive coil ([E]/[F])
For Synchro
I think what is confusing you about the current not reversing after the reed opens is you are not realizing the coil becomes the source of current and not a load.
You can easily prove this to yourself if you have a 2 channel scope. Simply add a low value resistor like 1 ohm or so to the circuit Tinman has provided in series with only the coil. Now watch the scope channels as you turn the coil on and off. You will see the signal across the resistor only goes up as the coil is charged and discharged. But the channel across the coil will show a reversal of polarity. But obviously if the channel across the resistor doesn't change polarity then the current does NOT reverse.
If we use the old school idea of electron flow you can look at it this way. The negative terminal of the battery has a surplus of electrons since electrons are negatively charged. So when the reed switch is turned on electrons move from the negative of the battery into the bottom of the coil. And electrons flow out the top of the coil back to the battery. If we put a meter on the coil we see the top as positive and the bottom as negative because the excess of electrons is at the bottom of the coil.
Now when we turn off the reed switch the collapsing magnetic field wants to keep the electrons flowing in the same direction. Since the coil is now the source and the current is still flowing in the SAME direction as before we now see the top of the coil as being more negative than the bottom. The current has not changed direction but the polarity of the voltage has changed.
Respectfully,
Carroll
Power measurements continued...
For the LC circuit by looking at Laurent's waveforms we can see that essentially all of the drive coil's pulse energy first goes into charging the one-uF capacitor. This is what was expected because the secondary coil will be blocking the initial current flow from the pulse of current from the drive coil.
So, we have initial maximum voltage measurement across the capacitor . From that we can calculate the pulse energy that goes into the capacitor [J]. If Laurent has a capacitance meter it should be used to get a more accurate value of this capacitor.
Now, we know the "exported pulse energy from the drive coil" [H], and we know the maximum energy that is transferred into the one-uF capacitor [J]. Therefore we can make a reasonable inference for how much pulse energy gets imparted onto the rotor from the drive coil [K], which is ([H]-[J]).
Here is where we are:
A - Average input current for entire circuit
B - RMS current through the drive coil
C - Power dissipation in the resistance of the drive coil
D - Average input power
E - Exported power from the drive coil, ([D] - [C])
F - Pulse frequency
G - Input pulse energy ([D]/[F])
H - Exported pulse energy from the drive coil ([E]/[F])
I - Maximum voltage across the 1 uF capacitor
J - Maximum energy in the 1 uF capacitor
K - Main drive coil pulse energy that is imparted onto the rotor ([H]-[J])
Thinking a bit more, and on topic....
I reposted Woopys circuit for reference below.
If you see the red x labeled 4, if you put a diode there, same direction as the one to the left of the x, the cap will discharge all of its charge through the coil only once.
Now for possibly an even better effect. Havnt tried but I think it would be a good thing if it works....
When the cap charges, then discharges through the coil with the added diode, the cap will fully discharge, and then some due to the flywheel effect of the coil, meaning the cap will end up with a charge reverse of what it started with. The possible advantage is, when the next spike comes from the primary drive coil, it will be facing a cap with a charge that is well willing to accept the spike as the polarities will be in series. So the spike enters the cap with help of the charge already in the cap. The possibility is that the cap may charge higher than without that added diode. ;) Im pretty sure of it from my experience in this stuff. ;D
Mags
Thinking more, again. I need to try what I just wrote in my last post. There might be something really good there. possibly more than I imagined.
If the cap does get an increased voltage level, the next pass may get even bigger and so on. Just went through it in my head. It just might be so. :o
Mags
Quote from: minoly on November 19, 2015, 06:39:53 PM
this post was gone, I checked from two computers and 2 different browsers, now it's back... http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg466152/#msg466152 (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg466152/#msg466152)
My mistake.
Yes, your mistake as it's impossible for me to delete a post and then re-post it using your user name.
That's the second time you accuse me of something that is not true. So please chill out and stop jumping to conclusions.
Quote from: minoly on November 19, 2015, 06:39:53 PM
you answered indirectly when you replied to MoRo's attempt to reply to my question:
http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg466166/#msg466166 (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg466166/#msg466166)
"The next thing I would recommend is a device that needs a long flux holding time. This is why I recommend a PM flux switch.[/size] [/size]I have already designed and built it. I'll call it the GTL (short for "go to luc") Gate. The GTL Gate uses the same design principle of my mostly magnet motor, where both inner and outer fields of the coil is used which gives you double the magnetic field and obviously double the core surface area."[/size]
I still do not understand it though.
You didn't add the part I wrote saying if you don't understand you will once I demonstrate the GTL Gate.
Quote from: minoly on November 19, 2015, 06:39:53 PM
You know me from past. I've often commented in your vids. I've always had good things to say about your work. https://www.youtube.com/channel/UCrIHB1lv7XElAD_ztL-1FPw (https://www.youtube.com/channel/UCrIHB1lv7XElAD_ztL-1FPw)
So you're cool Joule
Quote from: minoly on November 19, 2015, 06:39:53 PM
You might not appreciate my tone, but I did not appreciate the smack when all I did was share my opinion in the first post about JB.
My first post has this warning:
"If you wish to post in this topic please keep it on topic and constructive as I reserve the right to edit or delete any post that are not so"So how can this be a shock when you post about J Bedini stuff and then more posts arguing with others that it's the same thing?... I think you're over-dramatic. And right now me taking the time to write this is taking away from experiments and also adding pages with useless posts. Do you not see this?
Quote from: minoly on November 19, 2015, 06:39:53 PM
I hope I'm understanding this correctly, it is not discharging into a cap that is in parallel to a coil that will aid in spinning a rotor (to put it simply) that you are interested in; however, your interest is in creating a fluxgate using only high voltage/impedance coils that will hold/pull with more force using the spike? This GTL sounds very interesting. I have lot's of ideas how to use the spike to provide more torque, I cannot think of any that would aid in a mostly magnet motor - very interesting indeed.
Yes correct, my experiment results lead me to believe a flux gate would be best use of this effect but that doesn't mean there is no other use for it.
The mostly magnet motor was a good learning experiment. It has taught me many things that will be incorporated in the GTL Gate.
I see no reasons you have not started experiments as we I can see you have plenty of stuff to use and make a video of your findings.
Looking forward to seeing ways you can think of using this and your experimenl results
Regards
Luc
Quote from: citfta on November 19, 2015, 07:15:49 PM
For Synchro
I think what is confusing you about the current not reversing after the reed opens is you are not realizing the coil becomes the source of current and not a load.
You can easily prove this to yourself if you have a 2 channel scope. Simply add a low value resistor like 1 ohm or so to the circuit Tinman has provided in series with only the coil. Now watch the scope channels as you turn the coil on and off. You will see the signal across the resistor only goes up as the coil is charged and discharged. But the channel across the coil will show a reversal of polarity. But obviously if the channel across the resistor doesn't change polarity then the current does NOT reverse.
If we use the old school idea of electron flow you can look at it this way. The negative terminal of the battery has a surplus of electrons since electrons are negatively charged. So when the reed switch is turned on electrons move from the negative of the battery into the bottom of the coil. And electrons flow out the top of the coil back to the battery. If we put a meter on the coil we see the top as positive and the bottom as negative because the excess of electrons is at the bottom of the coil.
Now when we turn off the reed switch the collapsing magnetic field wants to keep the electrons flowing in the same direction. Since the coil is now the source and the current is still flowing in the SAME direction as before we now see the top of the coil as being more negative than the bottom. The current has not changed direction but the polarity of the voltage has changed.
Respectfully,
Carroll
Excellent post and test suggestion Carroll
Thanks for taking the time to help
Luc
Quote from: Magluvin on November 19, 2015, 07:27:03 PM
Thinking a bit more, and on topic....
I reposted Woopys circuit for reference below.
If you see the red x labeled 4, if you put a diode there, same direction as the one to the left of the x, the cap will discharge all of its charge through the coil only once.
Now for possibly an even better effect. Havnt tried but I think it would be a good thing if it works....
When the cap charges, then discharges through the coil with the added diode, the cap will fully discharge, and then some due to the flywheel effect of the coil, meaning the cap will end up with a charge reverse of what it started with. The possible advantage is, when the next spike comes from the primary drive coil, it will be facing a cap with a charge that is well willing to accept the spike as the polarities will be in series. So the spike enters the cap with help of the charge already in the cap. The possibility is that the cap may charge higher than without that added diode. ;) Im pretty sure of it from my experience in this stuff. ;D
Mags
Sounds very interesting Mags 8)
Thanks for sharing
Luc
Quote from: citfta on November 19, 2015, 07:15:49 PM
For Synchro
I think what is confusing you about the current not reversing after the reed opens is you are not realizing the coil becomes the source of current and not a load.
You can easily prove this to yourself if you have a 2 channel scope. Simply add a low value resistor like 1 ohm or so to the circuit Tinman has provided in series with only the coil. Now watch the scope channels as you turn the coil on and off. You will see the signal across the resistor only goes up as the coil is charged and discharged. But the channel across the coil will show a reversal of polarity. But obviously if the channel across the resistor doesn't change polarity then the current does NOT reverse.
If we use the old school idea of electron flow you can look at it this way. The negative terminal of the battery has a surplus of electrons since electrons are negatively charged. So when the reed switch is turned on electrons move from the negative of the battery into the bottom of the coil. And electrons flow out the top of the coil back to the battery. If we put a meter on the coil we see the top as positive and the bottom as negative because the excess of electrons is at the bottom of the coil.
Now when we turn off the reed switch the collapsing magnetic field wants to keep the electrons flowing in the same direction. Since the coil is now the source and the current is still flowing in the SAME direction as before we now see the top of the coil as being more negative than the bottom. The current has not changed direction but the polarity of the voltage has changed.
Respectfully,
Carroll
@Citfta,
That sounds pretty scientific. What both you and Tinman are proposing is that it's possible for electrical power to share current and voltage of opposite polarities. This is physically impossible!
There is no "Left Over" old current in the primary coil after the violence of the magnetic field collapse; There is only new current and voltage of opposite polarity. There is no cause and effect between the old current and the new current. The old current is in the past behind the event horizon. The violence of the field collapse has utterly and with absolute finality obliterated any trace of the old current along with any trace of previous electron paths for eternity. The new current can go in either direction depending on the pathway. Given a pathway of less resistance it will travel that way to it's newly biased ground. That includes an opposite direction. There is no force tendency of any kind at work on the new current from the old current. That's just superstition.
just tried on circuit sim. The reverse charge on the cap feeds back to primary. So we end up with a alternating oscillation. Unless the timing of the proceeding spikes are at the point where the cap is in full reverse charge before it is fed back to the pri. Just solid state at the moment to get the hang of it Will play with it more to see what I can do to fix that. Love that sim for times like this. Real time saver on simple stuff.
Mags
I will try to finish off the measurement discussion....
The only thing left to do is look at the capacitor discharging through the secondary coil. I agree with Gyula that changing the resistor at position #4 from one ohm to 10 ohms would be a good idea. When you look at Laurent's existing scope shots at position #4 you see a linearly decreasing ramp of current flow. That looks just like a resistor. That would suggest that the 210-ohm resistance of the coil is dominating over the inductance, but I am not sure how much secondary coil pulse energy will be imparted onto the rotor. The measurements and the number crunching should give us a handle on that.
I am going to assume a quality multimeter or Laurent's scope will give an accurate measurement for the RMS voltage across a 10-ohm CVR at position #4. When in doubt, do the research yourself to check these things otherwise you might be a victim of GIGO.
This time I am not going to give a description because it would just be a repeat of what was stated for other components.
Here are the new measurements:
L - RMS current through 10-ohm resistor at position #4
M - Average power dissipated in the resistance of the secondary coil
N - Pulse energy dissipated in the resistance of the secondary coil ([M]/[F])
O - Secondary coil pulse energy that is imparted onto the rotor ([J]-[N])
Here is the whole measurement shebang with the more interesting stuff highlighted. Some of the measurements are derived from other measurements, and you can easily determine other efficiency metrics, etc.
A - Average input current for entire circuit
B - RMS current through the drive coil
C - Power dissipation in the resistance of the drive coil
D - Average input power
E - Exported power from the drive coil, ([D] - [C])
F - Pulse frequency
G - Input pulse energy ([D]/[F])H - Exported pulse energy from the drive coil ([E]/[F])
I - Maximum voltage across the 1 uF capacitor
J - Maximum energy in the 1 uF capacitor
K - Main drive coil pulse energy that is imparted onto the rotor ([H]-[J])L - RMS current throuugh 10-ohm resistor at position #4
M - Average power dissipated in the resistance of the secondary coil
N - Pulse energy dissipated in the resistance of the secondary coil ([M]/[F])
O - Secondary coil pulse energy that is imparted onto the rotor ([J]-[N])P - Percentage of input pulse energy that becomes resistive losses (([C]/[F] + [M}) /[G]) * 100Q - Percentage efficiency (total rotor pulse energy/input pulse energy) * 100 == (([K]+- )/[G]) * 100[/b]
R - Ratio of secondary coil pulse energy imparted to rotor to drive coil pulse energy imparted to rotor - /[K]
Quote from: Magluvin on November 19, 2015, 05:46:55 PM
I believe I posted what happens if the cap were not there one page back.
http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg466132/#msg466132 (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg466132/#msg466132)
"I had looked through my vids to see if i had one, but dont.... I remember trying to get a bemf spike into a higher henry coil and the higher H coil seemed to block most of the spike rather than take advantage of the full potential. Like a subwoofer crossover coil, it blocks out the high frequencies. So the capacitor across your higher inductance coil probably loads up first then delivers it charge to the parallel coil?
Mags"
Mags
Magluvin.
Sorry for not acknowledging your comments. You are right, the capacitor loads up first and the inertia of the coil prevents any significant current flow. Note from Laurent's scope shots that the capacitor loads up in about 75 microseconds, which is quite fast. The "flywheel of the coil" has so much inertia that it has barely even budged after 75 microseconds.
With respect to the situation where there is no capacitor, you can simply extrapolate from Laurent's scope shots and imagine the capacitor getting smaller and smaller. The capacitor voltage will get higher and higher and the pulse width will get narrower and narrower. At the idealized limit you have an infinitely high voltage spike and an an infinitely narrow pulse with - and yet there is still a finite amount of energy in this idealized pulse.
The flywheel version: If you have a spinning flywheel next to a stationary flywheel on the same shaft, and they are the same size and weight, what happens when the two of them come into contact?
There are two cases you can examine for this:
1) Imagine there are friction plates like a clutch setup when the two flywheels make contact. There will be friction losses as the clutch plates come together. If flywheel A was spinning at 100 RPM and flywheel B was stationary, then after contact the pair of flywheels will be spinning together at 50 RPM. You will have lost one-half of the rotational energy through mechanical friction when the clutch plates make contact. This is also identical to shorting two capacitors together and when the current rushes and voltages balance you lose one-half of the energy in the wire resistance. This is also equivalent to a perfectly inelastic collision between two masses.
2) Imagine the impossible, just like magic the two flywheels are instantly fused together as one. The spinning flywheel will impart an infinite amount of torque on the stationary flywheel for an infinitely short amount of time. There is your infinitely high voltage spike for an infinitely short amount of time. The final speed in this case will not be 50 RPM, it will be 70.71 RPM. No energy will be lost after the magical creation of the unified flywheel. This is equivalent to a perfectly elastic collision between two masses.
On the bench, when the drive coil outputs its current pulse into the secondary coil and there is no capacitor, you get an approximation to case 2) above. You can assume that you lose some energy, but much less than losing 50%. The parasitic capacitance in the interconnect wiring will absorb the voltage spike that comes as a result of the current pulse from the drive coil. The resulting high-voltage spike is the "extreme torque for a very short time" that will get the secondary coil "rolling." Naturally, there will be a risk of dielectric breakdown in the air when this happens.
MileHigh
@Milehigh,
What effect does the EMF induced into the high-voltage coil from the passing rotor magnet have on the capacitor discharge curve?
Quote from: MileHigh on November 19, 2015, 09:04:45 PM
Magluvin.
Sorry for not acknowledging your comments. You are right, the capacitor loads up first and the inertia of the coil prevents any significant current flow. Note from Laurent's scope shots that the capacitor loads up in about 75 microseconds, which is quite fast. The "flywheel of the coil" has so much inertia that it has barely even budged after 75 microseconds.
With respect to the situation where there is no capacitor, you can simply extrapolate from Laurent's scope shots and imagine the capacitor getting smaller and smaller. The capacitor voltage will get higher and higher and the pulse width will get narrower and narrower. At the idealized limit you have an infinitely high voltage spike and an an infinitely narrow pulse with - and yet there is still a finite amount of energy in this idealized pulse.
The flywheel version: If you have a spinning flywheel next to a stationary flywheel on the same shaft, and they are the same size and weight, what happens when the two of them come into contact?
There are two cases you can examine for this:
1) Imagine there are friction plates like a clutch setup when the two flywheels make contact. There will be friction losses as the clutch plates come together. If flywheel A was spinning at 100 RPM and flywheel B was stationary, then after contact the pair of flywheels will be spinning together at 50 RPM. You will have lost one-half of the rotational energy through mechanical friction when the clutch plates make contact. This is also identical to shorting two capacitors together and when the current rushes and voltages balance you lose one-half of the energy in the wire resistance. This is also equivalent to a perfectly inelastic collision between two masses.
2) Imagine the impossible, just like magic the two flywheels are instantly fused together as one. The spinning flywheel will impart an infinite amount of torque on the stationary flywheel for an infinitely short amount of time. There is your infinitely high voltage spike for an infinitely short amount of time. The final speed in this case will not be 50 RPM, it will be 70.71 RPM. No energy will be lost after the magical creation of the unified flywheel. This is equivalent to a perfectly elastic collision between two masses.
On the bench, when the drive coil outputs its current pulse into the secondary coil and there is no capacitor, you get an approximation to case 2) above. You can assume that you lose some energy, but much less than losing 50%. The parasitic capacitance in the interconnect wiring will absorb the voltage spike that comes as a result of the current pulse from the drive coil. The resulting high-voltage spike is the "extreme torque for a very short time" that will get the secondary coil "rolling." Naturally, there will be a risk of dielectric breakdown in the air when this happens.
MileHigh
I hadnt ever put it together that the flywheel and inductor are both not going to react heavily to impulse hits. Like it would be very hard to get a flywheel to oscillate one way then the other at high freq. So the cap with the coil would be the flywheel with a spring. The smaller the cap, similarly the tighter the spring.
;)
For some reason the word inductor is underlined in red as a misspell as I write it here. ::)
Mags
Quote from: synchro1 on November 19, 2015, 11:45:52 PM
@Milehigh,
What effect does the EMF induced into the high-voltage coil from the passing rotor magnet have on the capacitor discharge curve?
Was thinking about that earlier, in a way. Like how a larger inductance will block an 'electrical' pulse, but a magnetic pulse induction would have no problem getting current to flow in the large inductance.
Mags
Quote from: Magluvin on November 20, 2015, 12:48:54 AM
Was thinking about that earlier, in a way. Like how a larger inductance will block an 'electrical' pulse, but a magnetic pulse induction would have no problem getting current to flow in the large inductance.
Mags
@Mags,
Woopjump produces a scope shot with the auxiliary coil separated from the rotor for comparison in his first video. A vastly greater amount of voltage appears on the scope when he repositions it close to the rotor. Woopy takes wide notice of it but doesn't offer any explanation.
Quote from: synchro1 on November 19, 2015, 11:45:52 PM
@Milehigh,
What effect does the EMF induced into the high-voltage coil from the passing rotor magnet have on the capacitor discharge curve?
It should slow down the capacitor discharge curve. This is because the passing rotor magnet induces counter-EMF in the secondary coil as compared to the voltage that the capacitor is putting across the coil. So that counter-EMF is "stealing" or "eating" some of the voltage put across the coil by the charged capacitor. Therefore there is less net voltage available to push current through the coil so the capacitor discharge slows down. It seems counter-intuitive when you first think about it.
In the previous posting to this posting you make reference to something that Laurent did that is apparently opposite to what I am saying. All that I can say is earlier in the thread I made reference to something similar that Laurent did in his clip that supports what I am saying.
The following comments are about the
drive coil and you see it in the first(?) clip: He is running the motor normally and the capacitor gets charged to say 60 volts. Then he moves the drive coil away from the spinning rotor, and you notice that now the capacitor gets charged to say 80 volts. So to repeat the logic: With the drive coil in place, the passing rotor magnets induce a CEMF in the coil that acts against the applied voltage from the power supply. So that CEMF "steals" some of the power supply voltage and that reduces the final current flow in the coil before the reed switch opens -> cap charges to 60 volts. When the drive coil is not in place, there is no CEMF "inside" the drive coil, and therefore there is a "normal" level of current flow when the reed switch opens -> cap charges to 80 volts.
Quote from: Magluvin on November 20, 2015, 12:44:32 AM
I hadnt ever put it together that the flywheel and inductor are both not going to react heavily to impulse hits. Like it would be very hard to get a flywheel to oscillate one way then the other at high freq. So the cap with the coil would be the flywheel with a spring. The smaller the cap, similarly the tighter the spring.
You are exactly right about the spring. In my little treatise the two flywheels stick together, hence they have to have the same RPM after they make contact, which translates into the same current flow through the two coils in series.
Just for the heck of it, let's look at Laurent's clip and model it in the physical domain but let's put it on a frictionless linear track this time and use moving masses for the coils and springs for the capacitors. Note in Laurent's circuit you have the diode, and the current flow in the two coils is separate and distinct.
So you have a long frictionless linear track. At the beginning of the track you have a small mass with a firecracker attached with the end cut off to act like a rocket motor. So when you light the firecracker you get the small mass moving. That's the drive coil and the initial application of voltage on it when the reed switch is closed.
When the firecracker burns out, that's the equivalent of the reed switch opening.
The small mass is moving and it's on a track with a ratchet so it can't go backwards, only forwards. That's the diode.
The instant the firecracker burns out, the small moving mass bumps into a spring attached to a large mass. As you can imagine the spring is the capacitor and the large mass is the secondary coil.
So the small mass starts to compress the spring and the large mass slowly starts to budge, but not by very much.
Eventually the small mass comes to a complete stop, and since this part of the track is ratcheted, it stops dead against the compressed spring and can't move backwards.
Let's look more closely at the instant of time where the small mass comes to a dead stop: The spring is at maximum compression, and the large mass is moving at a very low velocity. The large mass got a small "nudge" as the spring was compressing, but it is barely moving. At this instant in time assume that 99% of the available energy is stored as potential energy in the compressed spring and 1% of the available energy is stored in the very slowly moving large mass.
So now the spring decompresses and starts to transfer all of it's stored energy into the larger mass. The larger mass is sent on its merry way, and naturally it starts out moving at a slower maximum speed as compared to the maximum speed of the small mass.
The spring keeps pushing the mass along the track at it releases its stored energy.
But, alas, the large mass is not on a frictionless track for this final stretch. There are added friction pads on the track so the large mass lumbers along and in short order it comes to a dead stop and the experiment is over.
In this case for Laurent's setup the moment the spring gives up all of its stored energy the mass also lumbers to a dead stop. So that's the capacitor pushing current through the resistance of the secondary coil where the resistance seems to predominate. Once the capacitor is discharged, the current also stops flowing.
So, it took some thinking to get that right, and I don't think that I made a major mistake. "Seeing" the circuit operate in the physical domain in the form of moving masses and springs and friction pads on a linear track might help some people understand how the electrical circuit works, and it's kind of fun.
Quote from: gotoluc on November 19, 2015, 07:48:47 PM
Yes, your mistake as it's impossible for me to delete a post and then re-post it using your user name.
That's the second time you accuse me of something that is not true. So please chill out and stop jumping to conclusions.
You didn't add the part I wrote saying if you don't understand you will once I demonstrate the GTL Gate.
My first post has this warning: "If you wish to post in this topic please keep it on topic and constructive as I reserve the right to edit or delete any post that are not so"
So how can this be a shock when you post about J Bedini stuff and then more posts arguing with others that it's the same thing?... I think you're over-dramatic. And right now me taking the time to write this is taking away from experiments and also adding pages with useless posts. Do you not see this?
Yes correct, my experiment results lead me to believe a flux gate would be best use of this effect but that doesn't mean there is no other use for it.
The mostly magnet motor was a good learning experiment. It has taught me many things that will be incorporated in the GTL Gate.
I see no reasons you have not started experiments as we I can see you have plenty of stuff to use and make a video of your findings.
Looking forward to seeing ways you can think of using this and your experimenl results
Regards
Luc
QuoteSo you're cool Joule
Lol-yea. Wonder where he got that name from ::)
Quote from: synchro1 on November 19, 2015, 08:08:04 PM
@Citfta,
That sounds pretty scientific. What both you and Tinman are proposing is that it's possible for electrical power to share current and voltage of opposite polarities. This is physically impossible!
There is no "Left Over" old current in the primary coil after the violence of the magnetic field collapse; There is only new current and voltage of opposite polarity. There is no cause and effect between the old current and the new current. The old current is in the past behind the event horizon. The violence of the field collapse has utterly and with absolute finality obliterated any trace of the old current along with any trace of previous electron paths for eternity. The new current can go in either direction depending on the pathway. Given a pathway of less resistance it will travel that way to it's newly biased ground. That includes an opposite direction. There is no force tendency of any kind at work on the new current from the old current. That's just superstition.
Synchro-your just not getting it,so i will try again one last time.
Picture 1 below shows the current flowing through the primary coil when the reed is closed,and the coil is getting it's current supplied via the battery. The red arrows show the current flow path--this is conventional current flow,as if we use true current flow,then that would lead only to more confusion.
Picture one shows the current flow from the positive of the battery,through the primary coil,and to the negative side of the battery. Current is flowing from the positive potential to the negative potential via the primary coil.
Picture two.
Picture two is when the reed switch opens,and current flow from the battery is interrupted.
So in picture one,we see the current flowing into the top of the primary coil,and exiting out through the bottom. Now with the reed switch open,the voltage polarity of the coil changes/inverts as the magnetic field collapses around the primary coil. So now the bottom of the primary coil is the positive potential,and the top of the primary coil becomes the negative potential. So which way will the current be flowing now?--thats right,from the positive potential to the negative potential. So the current is still flowing in the same direction through the coil,where it enters the top of the coil,and exits the bottom of the coil-->exactly the same way it was flowing through the coil when the reed switch was closed.
Now,before anyone gets started--yes,i know electron current cannot flow through a capacitor. But the current flow is still flowing through the coil due to the depletion of electrons on the positive plate of the capacitor,and the increase of electrons on the negative plate of the capacitor(true current electron flow used here to explain). The current flow through the circuit in this case is via displacement current flow through the capacitor.
So i hope that has cleared that up Synchro.
Current continues to flow in the same direction through the coil when the reed switch is open,and will continue to flow until such time as the magnetic field around the coil has totally collapsed.
Quote from: MileHigh on November 20, 2015, 02:36:58 AM
It should slow down the capacitor discharge curve. This is because the passing rotor magnet induces counter-EMF in the secondary coil as compared to the voltage that the capacitor is putting across the coil. So that counter-EMF is "stealing" or "eating" some of the voltage put across the coil by the charged capacitor. Therefore there is less net voltage available to push current through the coil so the capacitor discharge slows down. It seems counter-intuitive when you first think about it.
In the previous posting to this posting you make reference to something that Laurent did that is apparently opposite to what I am saying. All that I can say is earlier in the thread I made reference to something similar that Laurent did in his clip that supports what I am saying.
The following comments are about the drive coil and you see it in the first(?) clip: He is running the motor normally and the capacitor gets charged to say 60 volts. Then he moves the drive coil away from the spinning rotor, and you notice that now the capacitor gets charged to say 80 volts. So to repeat the logic: With the drive coil in place, the passing rotor magnets induce a CEMF in the coil that acts against the applied voltage from the power supply. So that CEMF "steals" some of the power supply voltage and that reduces the final current flow in the coil before the reed switch opens -> cap charges to 60 volts. When the drive coil is not in place, there is no CEMF "inside" the drive coil, and therefore there is a "normal" level of current flow when the reed switch opens -> cap charges to 80 volts.
@Milehigh,
Based on your analysis increasing rotor magnet strength would not improve the "Flyback Motor". What effect do you think placing magnets on the auxiliary coil's ferrite U core would have?
Quote from: tinman on November 20, 2015, 05:17:53 AM
Synchro-your just not getting it,so i will try again one last time.
Picture 1 below shows the current flowing through the primary coil when the reed is closed,and the coil is getting it's current supplied via the battery. The red arrows show the current flow path--this is conventional current flow,as if we use true current flow,then that would lead only to more confusion.
Picture one shows the current flow from the positive of the battery,through the primary coil,and to the negative side of the battery. Current is flowing from the positive potential to the negative potential via the primary coil.
Picture two.
Picture two is when the reed switch opens,and current flow from the battery is interrupted.
So in picture one,we see the current flowing into the top of the primary coil,and exiting out through the bottom. Now with the reed switch open,the voltage polarity of the coil changes/inverts as the magnetic field collapses around the primary coil. So now the bottom of the primary coil is the positive potential,and the top of the primary coil becomes the negative potential. So which way will the current be flowing now?--thats right,from the positive potential to the negative potential. So the current is still flowing in the same direction through the coil,where it enters the top of the coil,and exits the bottom of the coil-->exactly the same way it was flowing through the coil when the reed switch was closed.
Now,before anyone gets started--yes,i know electron current cannot flow through a capacitor. But the current flow is still flowing through the coil due to the depletion of electrons on the positive plate of the capacitor,and the increase of electrons on the negative plate of the capacitor(true current electron flow used here to explain). The current flow through the circuit in this case is via displacement current flow through the capacitor.
So i hope that has cleared that up Synchro.
Current continues to flow in the same direction through the coil when the reed switch is open,and will continue to flow until such time as the magnetic field around the coil has totally collapsed.
@Tinman,
How much time do you imagine it takes for the magnetic field to collapse? This is a function of the rate at which the Reed switch is opened, which is proportional to the rotor R.P.M. times the number of rotor magnets divided by seconds. So lets say the frequency is 75 hertz! You believe that's perhaps too a short an interval for the collapse to fully occur?
The way you have the schematic drawn now the flyback would simply jump the Reed switch gap to get back to the positive electrode creating a nice blue spark and not follow any path you drew! The "Flyback" has a very Hi-voltage potential compared to the original current. That's why it's not going to follow your direction. I can't believe you're persisting in trying to prove this absurdity.
synchro,
There is nothing absurd about what Tinman is saying. I have a few questions for you if you don't mind.
Are you willing to learn or are you going to stick to your misconceptions no matter what? If you are not willing to learn then just ignore the rest of this post.
If you are willing to learn then please answer the following questions:
Can you do the simple test I proposed earlier with a 2 channel scope?
What direction does the current flow internally in a battery?
What is the definition of an inductor?
My apologies to luc for the side track of this thread. Luc, if you want we can take this discussion to a new thread if synchro wants to continue this discussion.
Respectfully,
Carroll
Quote from: citfta on November 20, 2015, 06:43:14 AM
synchro,
There is nothing absurd about what Tinman is saying. I have a few questions for you if you don't mind.
Are you willing to learn or are you going to stick to your misconceptions no matter what? If you are not willing to learn then just ignore the rest of this post.
If you are willing to learn then please answer the following questions:
Can you do the simple test I proposed earlier with a 2 channel scope?
What direction does the current flow internally in a battery?
What is the definition of an inductor?
My apologies to luc for the side track of this thread. Luc, if you want we can take this discussion to a new thread if synchro wants to continue this discussion.
Respectfully,
Carroll
@Citfta,
Start a new thread. Where did you come up with these theories from? I've never encountered anything like this before. Can you direct me to any kind of legitimate source material whatsoever to support your claims?
I have started a new thread. It is called "Inductive Kickback" http://overunity.com/16203/inductive-kickback/msg466238/#msg466238 (http://overunity.com/16203/inductive-kickback/msg466238/#msg466238)
This information is standard textbook information about inductors. See you on the new thread.
Carroll
ADDED BY GOTOLUC
Please note that Carroll has started a new topic (link above) to discuss or debate anything related to "Inductive Kickback"
Normal discussions are acceptable in this topic, however, debates are not and your posts could get deleted.
Luc
Hi all
I have tried the 10 ohm scr between the entry of the cap and the assistant coil (X4), the trace is much better but is at no place at zéro volt, so i think perhaps my probes or my scope should be recalibrated, so i stop measuring so small current with my scope to avoid junk datas. If somebody has a better equipment, feel free to follow MH recommendations for the measurement.
So this morning i have tried to get some feeling with my hands, and i decided to place a big ventilator fan on the rotor to significantly increase the mechanical load. So i decreased the voltage to 1.9 volts and i noticed that i had to increase a lot the cap in the flyback circuit (from 0.3 up to 10 uF ) to get the best rotation speed. For info, in this case the " parent " pulse duration in the main coil, is almost the same as the "child" pulse duration in the assistant coil.
So some results on this setup
1- when the main coil and the assistant coil are working together, the input voltage is 1.9 V and the average current is 0.110 A. that is 0.21 Watts and the fan spins at 500 rpm
2- i disconnected completely the assistant coil and flyback circuitery and put away the assistant coil in order to not influence the rotor magnetism.
- i put a freewheeling diode across the main coil and put the power on. 1.9 volts and average current at 0.12 A that is 0.23Watts to get only 450 rpm.
- than same setup and i disconnected the diode so the reed switch is strongly arcing so 1.9 volt at around 0.12 A , that is 0.23 Watts and only 400 rpm
3- I remounted the flyback circuitery but put away the assistant coil from the rotor - in open magnetic assistant C core = 1.9 V and 0.130 A that is 0.247 Watts with 420 rpm
-and than i magnetically closed the C core and 1.9 V and 0.125 A that is 0.24 Watts and 416 rpm
So it seems that the assistant coil bring an strong torque addition in comparison with the main coil alone and for the same or less input power, and it is what is important to me at this stage.
Just for info
Laurent
Quote from: woopy on November 20, 2015, 09:45:11 AM
Hi all
I have tried the 10 ohm scr between the entry of the cap and the assistant coil (X4), the trace is much better but is at no place at zéro volt, so i think perhaps my probes or my scope should be recalibrated, so i stop measuring so small current with my scope to avoid junk datas. If somebody has a better equipment, feel free to follow MH recommendations for the measurement.
So this morning i have tried to get some feeling with my hands, and i decided to place a big ventilator fan on the rotor to significantly increase the mechanical load. So i decreased the voltage to 1.9 volts and i noticed that i had to increase a lot the cap in the flyback circuit (from 0.3 up to 10 uF ) to get the best rotation speed. For info, in this case the " parent " pulse duration in the main coil, is almost the same as the "child" pulse duration in the assistant coil.
So some results on this setup
1- when the main coil and the assistant coil are working together, the input voltage is 1.9 V and the average current is 0.110 A. that is 0.21 Watts and the fan spins at 500 rpm
2- i disconnected completely the assistant coil and flyback circuitery and put away the assistant coil in order to not influence the rotor magnetism.
- i put a freewheeling diode across the main coil and put the power on. 1.9 volts and average current at 0.12 A that is 0.23Watts to get only 450 rpm.
- than same setup and i disconnected the diode so the reed switch is strongly arcing so 1.9 volt at around 0.12 A , that is 0.23 Watts and only 400 rpm
3- I remounted the flyback circuitery but put away the assistant coil from the rotor - in open magnetic assistant C core = 1.9 V and 0.130 A that is 0.247 Watts with 420 rpm
-and than i magnetically closed the C core and 1.9 V and 0.125 A that is 0.24 Watts and 416 rpm
So it seems that the assistant coil bring an strong torque addition in comparison with the main coil alone and for the same or less input power, and it is what is important to me at this stage.
Just for info
Laurent
@Woopjump,
Another video would be appreciated.
Hi Laurent,
Thanks for all your efforts and sharing the results.
Gyula
Hi everyone,
here is a new demo video of the GTL Flux Gate with a better comparative test then my first attempt which I deleted.
The test starts with the scale showing the pull force of 2.6Kg from the permanent magnets which are imbedded in the MOT (GTL Flux Gate)
Test 1: I power only the Low impedance coil (0.4 Ohms) with pure DC to establish a baseline to have a comparative for the 2nd test.
Results are: with an input of 2.20vdc @ 4.3a = 9.5 watts, the low impedance coil can alleviate 2Kg of the PM pull force.
Test 2: I power both the low impedance and high impedance coil connected in series (93 Ohms) with the flyback diode connected across the series coils.
Results are: with the pulse circuit at 235Hz / 50% duty cycle with 60vdc @ 0.151a = 9 watts the high impedance coils can alleviate the same 2Kg of the PM pull force.
The results are not as spectacular as I would of hoped for but how can we explain Test 2 with such a high coil resistance to perform the same task with 1/2 a watt less input by adding components that represents only losses. pulse switch, high impedance coil and diode?
Link to video: https://www.youtube.com/watch?v=2k3iGi9VPCU (https://www.youtube.com/watch?v=2k3iGi9VPCU)
Luc
@ synchro1 and anyone else wanting to debate inductive-kickback (http://overunity.com/16203/inductive-kickback/msg466238/#msg466238) related effects.
please stop posting your debates in this topic. citfta has started a topic for anyone interested to do that: http://overunity.com/16203/inductive-kickback/msg466238/#msg466238 (http://overunity.com/16203/inductive-kickback/msg466238/#msg466238)
I'm deleting all the new posts that were added after citfta started the topic relating to this debate.
It's filling up the topic with too many posts that will make it difficult for future replicators.
Luc
Quote from: synchro1 on November 20, 2015, 05:58:00 AM
@Milehigh,
Based on your analysis increasing rotor magnet strength would not improve the "Flyback Motor". What effect do you think placing magnets on the auxiliary coil's ferrite U core would have?
The vast majority of experiments where people place magnets in magnetic circuits as part of a pulse motor setup with the belief that they will do something special or add to the output power are simply wrong. In the majority of cases they don't test their setup with and then without the magnets in place so they never know one way or the other. You are also subjecting the magnets to a lot of changing flux and for some magnets you will start to demagnetize them.
Quote from: gotoluc on November 20, 2015, 10:28:42 AM
Hi everyone,
here is a new demo video of the GTL Flux Gate with a better comparative test then my first attempt which I deleted.
The test starts with the scale showing the pull force of 2.6Kg from the permanent magnets which are imbedded in the MOT (GTL Flux Gate)
Test 1: I power only the Low impedance coil (0.4 Ohms) with pure DC to establish a baseline to have a comparative for the 2nd test.
Results are: with an input of 2.20vdc @ 4.3a = 9.5 watts, the low impedance coil can alleviate 2Kg of the PM pull force.
Test 2: I power both the low impedance and high impedance coil connected in series (93 Ohms) with the flyback diode connected across the series coils.
Results are: with the pulse circuit at 235Hz / 50% duty cycle with 60vdc @ 0.151a = 9 watts the high impedance coils can alleviate the same 2Kg of the PM pull force.
The results are not as spectacular as I would of hoped for but how can we explain Test 2 with such a high coil resistance to perform the same task with 1/2 a watt less input by adding components that represents only losses. pulse switch, high impedance coil and diode?
Link to video: https://www.youtube.com/watch?v=2k3iGi9VPCU (https://www.youtube.com/watch?v=2k3iGi9VPCU)
Luc
Hi Luc,
The only explanation to your question is the much more number of turns for the 93 Ohm MOT coil than the 0.4 Ohm thick wire MOT coil has. The high number of turns can insure a similar excitation for the core at a lower current level compared to a coil with low number of turns at a higher input current level.
The 93 Ohm coil resistance dissipates 2.1 W power (0.151A*0.151A*93).
The 0.4 Ohm coil resistance dissipates 7.4 W power (4.3A*4.3A*0.4).
This may seem surprising but shows 2 things: loss in a wire increases with the square of the current and high current can be circumvented by many number of turns in favor of getting less loss in a coil. Of course in any particular case a trade-off should be chosen on the number of turns and the wire diameter.
Gyula
Quote from: woopy on November 20, 2015, 09:45:11 AM
Hi all
I have tried the 10 ohm scr between the entry of the cap and the assistant coil (X4), the trace is much better but is at no place at zéro volt, so i think perhaps my probes or my scope should be recalibrated, so i stop measuring so small current with my scope to avoid junk datas. If somebody has a better equipment, feel free to follow MH recommendations for the measurement.
It's entirely possible that when you switch over to the 10-ohm resistor that you have a continuous current flow through the secondary coil. Therefore there may be no problem at all with your probes and your scope calibration. If you check the voltage across the 1 uF capacitor you may also notice that it never goes to zero volts.
The explanation for this is that the current pulses from the drive coil keep filling up the 1 uF capacitor with charge so that it never goes down to zero volts. Therefore there is always current flowing through the secondary coil.
Quote from: gyulasun on November 20, 2015, 12:02:15 PM
Hi Luc,
The only explanation to your question is the much more number of turns for the 93 Ohm MOT coil than the 0.4 Ohm thick wire MOT coil has. The high number of turns can insure a similar excitation for the core at a lower current level compared to a coil with low number of turns at a higher input current level.
The 93 Ohm coil resistance dissipates 2.1 W power (0.151A*0.151A*93).
The 0.4 Ohm coil resistance dissipates 7.4 W power (4.3A*4.3A*0.4).
This may seem surprising but shows 2 things: loss in a wire increases with the square of the current and high current can be circumvented by many number of turns in favor of getting less loss in a coil. Of course in any particular case a trade-off should be chosen on the number of turns and the wire diameter.
Gyula
Thanks Gyula for keeping the explanation simple enough for me to understand.
Even though today's tests are not so bad with the GTL Gate, yesterdays tests reveled I could achieve 2Kg. for pull force from a MOT (without PM) using just the primary with only 3 watts of input. So for now the GTL Gate idea is shelved until something new come.
So I'm going back to my original idea of building a bucking MOT motor with the assisting flyback coil separated from the primary low impedance coil.
Luc
Running out of time here at lunch so dont have time to find the post by Brad I read earlier to quote.
Brad was talking about 1kv jumping across 1mm gap, suggesting that there wasnt enough to jump the reed at 60v. What is missing there is the fact that as the reed opens, the tiniest fraction of a mm, the spark starts and continues to flow across the gap as it goes to fully open. Like if you set up a standard relay to buzz, 12v in usually can give 90v out. But while it buzzes, there are sparks across that gap. The initial spark when the contacts just open ionizes the air allowing the spark to continue even when the gap is opened more to its fully open state, then dies out.
Not knocking you brad. Just providing info that I know on the subject. ;) ;D
Mags
Hi Luc,
In the meantime Hob showed a paper he and his colleague made on just electromagnets to increase their force, see attachment and then in his next post a video :
http://overunity.com/15796/elementary-physics-revisited/msg466245/#msg466245 (http://overunity.com/15796/elementary-physics-revisited/msg466245/#msg466245)
They introduced the force/power ratio, tested and compared 6 different coils and the conclusion is that
"The amount of copper in an electromagnet determines the force per power
ratio, not the number of turns or the wire thickness in the coil, the more copper the greater force."
Well, interesting approach, I have not seen such comparison before, and "to my rescue" I think the amount of copper is also increased when you increase the number of turns...(what I said) and of course there are other factors to be considered. 8)
Gyula
Thanks Gyula for bringing Hob's experiment to my attention.
It's all these small details we need to know and combine in or design that may lead us to the goal.
Luc
Quote from: Magluvin on November 20, 2015, 01:12:55 PM
... What is missing there is the fact that as the reed opens, the tiniest fraction of a mm, the spark starts and continues to flow across the gap as it goes to fully open. Like if you set up a standard relay to buzz, 12v in usually can give 90v out. But while it buzzes, there are sparks across that gap. The initial spark when the contacts just open ionizes the air allowing the spark to continue even when the gap is opened more to its fully open state, then dies out...
Good points. Newman's book discusses ways to achieve fast opens with less sparking. Maybe try parallel reed switches.
MagnaMoRo
Quote from: gyulasun on November 20, 2015, 02:12:59 PM
Hi Luc,
In the meantime Hob showed a paper he and his colleague made on just electromagnets to increase their force, see attachment and then in his next post a video :
http://overunity.com/15796/elementary-physics-revisited/msg466245/#msg466245 (http://overunity.com/15796/elementary-physics-revisited/msg466245/#msg466245)
They introduced the force/power ratio, tested and compared 6 different coils and the conclusion is that
"The amount of copper in an electromagnet determines the force per power
ratio, not the number of turns or the wire thickness in the coil, the more copper the greater force."
Well, interesting approach, I have not seen such comparison before, and "to my rescue" I think the amount of copper is also increased when you increase the number of turns...(what I said) and of course there are other factors to be considered. 8)
Gyula
But it is also true that by decreasing the number of turns while increasing the size of the wire,can also result in more copper used. The advantage of this is less resistive losses-ohms law rules.
The ultimate inductor would use square copper wire to reduce the air gaps between windings ;)
I am half way through my Woopy replication-should be up later tonight.
Brad
Quote from: MoRo on November 20, 2015, 11:35:12 PM
Good points. Newman's book discusses ways to achieve fast opens with less sparking. Maybe try parallel reed switches.
MagnaMoRo
Hey MoRo
Thanks
That would be series on the reeds I believe. 'If' they open at the same time, the total gap possible is twice the distance as just 1 reed, to fully open in the same time frame. So the total gap gains distance 2 times as fast. Would probably be best to make a spinning commutator switch in which many opening series contacts open at once, crating a super fast close to fully open situation. ;) ;) Seems like a Tesla thing. Have to look that up.
The reason for the commutator version is that reeds are not all exactly the same even from the same batch. There will be issues of magnets(of many on a rotor) not having the exact same field strength and or orientation as the others. best to use tiny ones to avoid pole position discrepancies and just move the weaker ones out further to make up for field strength differences. But I would be way more confident in a well designed commutator with say 8 series switch points than 8 series reeds. Dont get me wrong. Reads a very good. But they also can only work up to certain freq as resonance of the reed contact arms does interfere with keeping control of the on off times. Got a vid of that i think where the rotor slows in acceleration till it gets past that resonant point then surges further in rpm. Also happens at harmonics of the reeds resonant freq. So you get more than 1 resonance break through, as I remember I got at least 3 on my setup.
Good points MoRo. Never thought of the increase in speed of the contacts gaining distance from each other, only thought of the fact that the distance would be more. Now that interests me more that the speed of the distance is increased with multiple switches in series. Thanks for bringing that up. Was probably about 6 to 7 years ago i played with the series reeds but didnt consider the speed till you brought it up.
Should probably make a thread on the subject. ;)
Mags
Quote from: tinman on November 21, 2015, 12:21:22 AM
But it is also true that by decreasing the number of turns while increasing the size of the wire,can also result in more copper used. The advantage of this is less resistive losses-ohms law rules.
The ultimate inductor would use square copper wire to reduce the air gaps between windings ;)
I am half way through my Woopy replication-should be up later tonight.
Brad
Hey Brad
I suppose the idea of having 2 coils of the same dimensions and volume of copper but each having 2 diff sizes of wire, that they could produce the same field strength as long as the (P)ower dissipated into each is the same, where the smaller wire coil gets higher voltage input to overcome higher resistance and the larger wire coil gets less voltage to calculate the power as the same. Makes sense.
The 24 coils on my lasersaber motor are in series. each are 42 awg 3200 turns at 15k ohm all in series. Seems like a large inductance would be had having a high rise time, but due to the high resistance the rise time seems almost instantaneous. Strange to experience.
They do make square enameled magnet wire. Would probably do wonders for the capacitance of a bifi coil. ;D
Mags
Quote from: Magluvin on November 21, 2015, 12:45:46 AM
Hey Brad
The 24 coils on my lasersaber motor are in series. each are 42 awg 3200 turns at 15k ohm all in series. Seems like a large inductance would be had having a high rise time, but due to the high resistance the rise time seems almost instantaneous. Strange to experience.
They do make square enameled magnet wire. Would probably do wonders for the capacitance of a bifi coil. ;D
Mags
QuoteI suppose the idea of having 2 coils of the same dimensions and volume of copper but each having 2 diff sizes of wire, that they could produce the same field strength as long as the (P)ower dissipated into each is the same, where the smaller wire coil gets higher voltage input to overcome higher resistance and the larger wire coil gets less voltage to calculate the power as the same. Makes sense.
Yes,it dose in that situation. But we dont have that situation with Woopy's DUT. The load resistance determines the voltage,current,and time base for the available flyback energy. There is also the matter of the power being generated by the rotating magnets.
Anyway,i will have my motor up and running by tonight. I have mounted my rotor onto a PM stator motor that i will be using as a generator i can load to variable resistive loads.
Brad
Well the motor is up and running.
First i will show a scope shot of both the voltage across the primary coil,and the current flowing through it. The flyback current is being sent to a resistive load ATM. As you can see,the current flow from the flyback is not quite as linear as Woopy's is,but not to bad.
Dont know if this thread has died out or not. Been thinking on it.
First thought is we have shown a couple ways to accomplish what is stated in the topic. But they only show that there can be an advantage to using bemf, as most call it.
So from there one needs to build a motor that is very efficient to begin with. The output of the mechanical part is not in any of the equations so far. Rotors are not loaded and just free spinning.
When building a motor that is very efficient, meaning input compared to loaded output, the time frame for capturing collapse and reusing it has to be instilled into the design. Pwm drive comes to mind where the coils are pulsed multiple times during a push and pull cycle and bemf is captured multiple times.
When we look at a chart curve of a coil building a field and taking on more current, as the curve just starts to level out, then we begin to input more and more as the curve flattens and the bemf becomes less and less as compared to the input total. So in my opinion it is best to cut the input while the current and field build is still on a strong vertical climb. Thats when the bemf will be closest to equaling the input of the cycle. For example, you could leave the input on for min or even hours, but still only get a very short period of bemf that does not build over that period of time.
My other thoughts on this are that the cap across the second coil design will be freq dependent in a much stronger way than using a switch or reed to fire the second coil from a precharged bank of caps charged by bemf. Just from studying the circuit in circuit sim with different freq inputs.
I have an electric bike that has an asymmetrical brushless motor where only 1 coil is switched to opposite polarity at a time, while all the others are on push or pull mode. A very smooth near zero cogging operation. The controller uses pwm to power the coils with varying pulses depending on the throttle input. The coils of that motor have thicker wire and less turns than the coils shown in my earlier vid. Possibly a virtual farm for capturing bemf.
If we were to try and capture bemf from say a standard dc, 2 brush, multiple commutator/winding motor, most of the windings for each com switch are in a steady on state and only one winding 'might' produce some bemf out. And a study of that is that there may not be much due to when the brushes are making a switch, they are contacting 2 commutator tabs at the same time, thus actually shorting a winding while the motor is running, causing large inefficiency in the motor to begin with. Big loss there, especially when the shorted windings are in the middle of the stator pole field, where pole switching generally happens.
So a particular motor design will have to be decided on in order to take the most advantage of bemf in my honest opinion. And if the initial motor design is up in the 90%, while being able to secure high amounts of bemf to recycle back into the motor function, I believe it can go over 100%
Mags
Quote from: Magluvin on November 22, 2015, 05:18:28 PM
Pwm drive comes to mind where the coils are pulsed multiple times during a push and pull cycle and bemf is captured multiple times.
When we look at a chart curve of a coil building a field and taking on more current, as the curve just starts to level out, then we begin to input more and more as the curve flattens and the bemf becomes less and less as compared to the input total. So in my opinion it is best to cut the input while the current and field build is still on a strong vertical climb. Thats when the bemf will be closest to equaling the input of the cycle. For example, you could leave the input on for min or even hours, but still only get a very short period of bemf that does not build over that period of time.
Magluvin,
All of your points are spot on! But, I really like the point above. It is well within the point Newman stated in his book on firing a coil multiple times.
Of importance too, and I don't know if this was mentioned with regard to this possible build, but, we should keep in mind that Newman started with 'higher voltages' to begin with. It becomes obviously important when you consider the advantages we receive due to the kick-back being of such high voltage.
Regarding the point mentioned earlier on same-time-frame-fired series reed switches. I understand the possible inconsistencies. Yet, a commutator is a terrible source of friction. So, maybe a an optical switch (for even greater efficiency) can act as a relay to fire the series switches that open and close in the correct time frame.
MagnaMoRo
I just enjoy siting back and thinking about this stuff!
Quote from: MoRo on November 22, 2015, 07:20:52 PM
Magluvin,
All of your points are spot on! But, I really like the point above. It is well within the point Newman stated in his book on firing a coil multiple times.
Of importance too, and I don't know if this was mentioned with regard to this possible build, but, we should keep in mind that Newman started with 'higher voltages' to begin with. It becomes obviously important when you consider the advantages we receive due to the kick-back being of such high voltage.
Regarding the point mentioned earlier on same-time-frame-fired series reed switches. I understand the possible inconsistencies. Yet, a commutator is a terrible source of friction. So, maybe a an optical switch (for even greater efficiency) can act as a relay to fire the series switches that open and close in the correct time frame.
MagnaMoRo
I just enjoy siting back and thinking about this stuff!
Thanks MoRo
The point of the series mechanical switches is the possibility of opening the switch fast enough to beat the collapse and resist possibly thousands of volts of potential crossing the gap. Like making an ideal high voltage switch, when it comes to the off cycle.
Can transistors be set up in series to do that? Havnt looked yet to see if that is being done out there.
As for the friction, if we are dealing with higher voltage and lower currents, the brushes dont need to be as tight against the commutator as a high current motor would. ;)
Mags
Quote from: Magluvin on November 22, 2015, 05:18:28 PM
Dont know if this thread has died out or not. Been thinking on it.
First thought is we have shown a couple ways to accomplish what is stated in the topic. But they only show that there can be an advantage to using bemf, as most call it.
So from there one needs to build a motor that is very efficient to begin with. The output of the mechanical part is not in any of the equations so far. Rotors are not loaded and just free spinning.
When building a motor that is very efficient, meaning input compared to loaded output, the time frame for capturing collapse and reusing it has to be instilled into the design. Pwm drive comes to mind where the coils are pulsed multiple times during a push and pull cycle and bemf is captured multiple times.
When we look at a chart curve of a coil building a field and taking on more current, as the curve just starts to level out, then we begin to input more and more as the curve flattens and the bemf becomes less and less as compared to the input total. So in my opinion it is best to cut the input while the current and field build is still on a strong vertical climb. Thats when the bemf will be closest to equaling the input of the cycle. For example, you could leave the input on for min or even hours, but still only get a very short period of bemf that does not build over that period of time.
My other thoughts on this are that the cap across the second coil design will be freq dependent in a much stronger way than using a switch or reed to fire the second coil from a precharged bank of caps charged by bemf. Just from studying the circuit in circuit sim with different freq inputs.
I have an electric bike that has an asymmetrical brushless motor where only 1 coil is switched to opposite polarity at a time, while all the others are on push or pull mode. A very smooth near zero cogging operation. The controller uses pwm to power the coils with varying pulses depending on the throttle input. The coils of that motor have thicker wire and less turns than the coils shown in my earlier vid. Possibly a virtual farm for capturing bemf.
If we were to try and capture bemf from say a standard dc, 2 brush, multiple commutator/winding motor, most of the windings for each com switch are in a steady on state and only one winding 'might' produce some bemf out. And a study of that is that there may not be much due to when the brushes are making a switch, they are contacting 2 commutator tabs at the same time, thus actually shorting a winding while the motor is running, causing large inefficiency in the motor to begin with. Big loss there, especially when the shorted windings are in the middle of the stator pole field, where pole switching generally happens.
So a particular motor design will have to be decided on in order to take the most advantage of bemf in my honest opinion. And if the initial motor design is up in the 90%, while being able to secure high amounts of bemf to recycle back into the motor function, I believe it can go over 100%
Mags
As far as I'm concerned the topic is still going. It's a little quiet now since I asked the guys to go to the flyback topic to debate things there.
I'm sure they'll come back once they've reached a consensuses.
This summer I played around with multiple on off time (per commutator segment) on a universal motor: https://www.youtube.com/watch?v=pib2fpHRsaM
Luc
I'm going through reed switches like a knife through butter. Current draw down at 10ma on 12VDC. Simplest bloody pulse motor I've ever made. I'm measuring the flyback at several hundred volts though so it's probably that.
Quote from: Jimboot on November 23, 2015, 01:51:42 AM
I'm going through reed switches like a knife through butter. Current draw down at 10ma on 12VDC. Simplest bloody pulse motor I've ever made. I'm measuring the flyback at several hundred volts though so it's probably that.
The average current may be 10mA Jimboot,but the instantaneous current will be much higher i would think. Lets say your coil has a resistance of 10 ohms,and your running 12 volt's-thats 1.2 amp's of current flowing through your reed switch. If your coil is 50 ohms,that is 240mA through the reed switch. In order for there to be only 10mA through the reed switch,your coil would have to have a resistance of 1200 ohm's.
Brad
Hi everyone and Laurent (woopy)
I just watched Laurent's newest video: https://www.youtube.com/watch?v=TAx7Y0UIyHA (https://www.youtube.com/watch?v=TAx7Y0UIyHA)
and decided to share my latest ideas and experiments with TinMan's RT : https://www.youtube.com/watch?v=n3IxLmdR9O0 (https://www.youtube.com/watch?v=n3IxLmdR9O0)
Laurent (if you wish) please feel free to post your experiments in this topic as we may find this is all related.
I think there is two things going on when you bring the tungsten rod forward, one is the rotor coils resistance is less, so the flyback is more and the other could be the rod is now on the side Laurent has every second commutator removed (drilled out) and maybe the plasma is ionizing across two (series) rotor coils on with the flyback pulse (only) which may amplify the effect.
Also, I have found that after an hour or so of playing with a universal motor with the modified rotor in this configuration the motor starts to lock up which may be due to the constant magnetization direction.
All very interesting
Thanks for sharing
Luc
Quote from: tinman on November 23, 2015, 08:39:45 AM
The average current may be 10mA Jimboot,but the instantaneous current will be much higher i would think. Lets say your coil has a resistance of 10 ohms,and your running 12 volt's-thats 1.2 amp's of current flowing through your reed switch. If your coil is 50 ohms,that is 240mA through the reed switch. In order for there to be only 10mA through the reed switch,your coil would have to have a resistance of 1200 ohm's.
Brad
Aaah, thanks brad.
Quote from: Jimboot on November 23, 2015, 01:51:42 AM
I'm going through reed switches like a knife through butter. Current draw down at 10ma on 12VDC. Simplest bloody pulse motor I've ever made. I'm measuring the flyback at several hundred volts though so it's probably that.
Here is a vid back when I bought these 3in reads. Can find on ebay like I did. They are not as fast at switching as the tiny ones and have resonance issues at higher freq and harmonics around that point. But will work well for what you are working on.
The second vid is showing the pulse motor slowing acceleration when approaching resonance 'points' then accelerating to the next level.
https://www.youtube.com/watch?v=lkxRNhPG2sE (https://www.youtube.com/watch?v=lkxRNhPG2sE)
https://www.youtube.com/watch?v=b7PR8JEVp7A (https://www.youtube.com/watch?v=b7PR8JEVp7A) Lol I burn a slot in the tip of my finger accidentally touching the rotor.
The amount the speed jumps isnt from just adjusting. This also shows bemf being sent to a 5w resistor and gets very hot to touch. I think I say in the vid the ohm. If I remember it was 5?
Mags
Hi everyone,
here is the Bucking Reluctance Motor I Designed
Link to video: https://www.youtube.com/watch?v=WL3kO4n2XsE
If someone can help confirm if CEMF or Lenz is negated in my test or suggest a test that would be appreciated.
Thanks
Luc
Thanks for sharing
Quote from: gotoluc on November 24, 2015, 12:28:17 AM
Hi everyone,
here is the Bucking Reluctance Motor I Designed
Link to video: https://www.youtube.com/watch?v=WL3kO4n2XsE
If someone can help confirm if CEMF or Lenz is negated in my test or suggest a test that would be appreciated.
Luc
Something is not right with your CVR,be it the value of .1 ohm,or the ability of it to carry the current.
I am leaning toward the second,as the CVR seems far to small to be able to do the job. If your blue trace is across the CVR,and the VPD on your scope for that channel is 5 volt's,then i estimate the average voltage to be about 7 volts across that CVR during the on time. This is an average current of 70 amps,and the average voltage across the two coils seems to be about 8 volt's,as your VPD are 500mV and your probe is set at 10x<-- not sure why you have done it this way,as that is still 5 volts per division.
So using the screen capture below,i see around 560 watts per pulse,and that CVR looks like a 1 watt CVR ?. :D
Quote from: tinman on November 24, 2015, 05:19:06 AM
Luc
Something is not right with your CVR,be it the value of .1 ohm,or the ability of it to carry the current.
I am leaning toward the second,as the CVR seems far to small to be able to do the job. If your blue trace is across the CVR,and the VPD on your scope for that channel is 5 volt's,then i estimate the average voltage to be about 7 volts across that CVR during the on time. This is an average current of 70 amps,and the average voltage across the two coils seems to be about 8 volt's,as your VPD are 500mV and your probe is set at 10x<-- not sure why you have done it this way,as that is still 5 volts per division.
So using the screen capture below,i see around 560 watts per pulse,and that CVR looks like a 1 watt CVR ?. :D
I think saw <black,brown,black> on the resistor so that would make it a one-ohm CVR. So that would translate into an average current of about 7 amps during the pulse which would make it seem reasonable. The voltage pulse starts at 11 volts, which is the battery voltage.
You should say, "around 560 watts
during the pulse" so as to not confuse power with energy. You quote energy
per pulse.
The battery voltage is about 11 volts, and I am going to guess that it is a lead-acid battery that would normally be at 12.6 volts. So I believe that gives two options, the battery is relatively new but quite discharged, or the battery is quite old.
For the sake of argument let's assume that it is a relatively new battery that is quite discharged. That means that it's internal impedance is rather high, and that's why you see the huge voltage droop from the battery as it starts to supply current to the two coils. The battery starts to croak from the current demands being made by the coils. The moral of the story is to try to do your experiments with fresh batteries, especially if the load will be drawing heavier currents. Also, when currents are in the range of 5 amps and above I personally would switch from alligator clips to terminal block type connections.
I am not a fan of this "bucking coils" business, for the most part I view it as doing something "alternative" just for the sake of "being alternative." In this case when the slab of metal is between the bucking coils, the magnetic fields are mostly cancelling each other out. Even when the coils have air between them, they are cancelling each other out.
If I can make a suggestion to you Luc it's this: Just wire a switch to turn the coils on and off manually. Move the slab of metal through the gap between the two coils step-by-step and feel the pull on the slab when you switch the coils on. Very low tech, just feel what is going on with your hands. Then reverse the wiring in one of the coils so they are no longer bucking and repeat the same procedure.
I am willing to bet you that you will get a much stronger pull force when the coils are "normally" wired and not in a "bucking" configuration.
Finally, what do you mean by "counter-EMF trying to negate Lenz" for your design?
Hi Luc,
I think your question on the Lenz issue could be answered by giving a certain mechanical load to the spinning wheel when it already established a constant speed from the input pulses and then monitor the input current. In order to do this, you would need to insure a more stable current measurement method to filter out fluctuation.
It is sure your bucking coils draw several Ampers of peak current what the power supply cannot readily insure, this is seen from the decreasing top part of the voltage across the coil (voltage is lower at the switch-off moments with respect to the instanteneous ON moment value).
You could make the power supply more happy by using a choke coil of some ten mH (or higher) and with low DC resistance in series with the positive output of the power supply, preceeding the RC filter you have with the input DC voltage and current meters. A choke like an original MOT primary coil with an open core like you use for the bucking coils or similar would serve. The series inductance would smooth out the huge current demand the ON moments represent and current meter fluctuations would be much less. More filter capacitors would also help but a series choke is better.
Brad addresses problems you need to check of course. MileHigh, I think Luc uses the power supply for the bucking coils and the battery is used for the pulse generating circuit on his PCB board.
Gyula
Quote from: gyulasun on November 24, 2015, 06:49:10 AM
MileHigh, I think Luc uses the power supply for the bucking coils and the battery is used for the pulse generating circuit on his PCB board.
Gyula
Thanks, I missed that in the clip. So the power supply is croaking from the load put on it.
Quote from: MileHigh on November 24, 2015, 06:19:53 AM
So that would translate into an average current of about 7 amps during the pulse which would make it seem reasonable. The voltage pulse starts at 11 volts, which is the battery voltage.
The battery voltage is about 11 volts, and I am going to guess that it is a lead-acid battery that would normally be at 12.6 volts. So I believe that gives two options, the battery is relatively new but quite discharged, or the battery is quite old.
For the sake of argument let's assume that it is a relatively new battery that is quite discharged. That means that it's internal impedance is rather high, and that's why you see the huge voltage droop from the battery as it starts to supply current to the two coils. The battery starts to croak from the current demands being made by the coils. The moral of the story is to try to do your experiments with fresh batteries, especially if the load will be drawing heavier currents. Also, when currents are in the range of 5 amps and above I personally would switch from alligator clips to terminal block type connections.
I am not a fan of this "bucking coils" business, for the most part I view it as doing something "alternative" just for the sake of "being alternative." In this case when the slab of metal is between the bucking coils, the magnetic fields are mostly cancelling each other out. Even when the coils have air between them, they are cancelling each other out.
If I can make a suggestion to you Luc it's this: Just wire a switch to turn the coils on and off manually. Move the slab of metal through the gap between the two coils step-by-step and feel the pull on the slab when you switch the coils on. Very low tech, just feel what is going on with your hands. Then reverse the wiring in one of the coils so they are no longer bucking and repeat the same procedure.
I am willing to bet you that you will get a much stronger pull force when the coils are "normally" wired and not in a "bucking" configuration.
Finally, what do you mean by "counter-EMF trying to negate Lenz" for your design?
QuoteYou should say, "around 560 watts during the pulse" so as to not confuse power with energy. You quote energy per pulse.
Where did i quote that MH ?.
I quoted-->So using the screen capture below,i see around 560 watts per pulse,and that CVR looks like a 1 watt CVR
QuoteI think saw <black,brown,black> on the resistor so that would make it a one-ohm CVR.
I only used what Luc quoted the value as,as i cant make heads or tails of what the colors are on the resistor-->see screen capture below.
It looks like a 5-band resistor with colours black (0) - brown (1) - black (0) - silver (x0.01) - brown (1%)
making it a 0.1 Ohm resistor like Luc said, see http://nearbus.net/wiki/index.php?title=File:Resistor_color_codes.jpg
I think there is more a probe x1 or x10 mixup in play here.
Itsu
Brad:
"i see around 560 watts per pulse" - that is the wrong terminology and doesn't make sense, please see my posting again. You should quote energy per pulse to use the terminology of electronics properly.
Itsu:
Thanks, it's been so long for me that I got it wrong and I thought it was <digit, digit, multiplier, tolerance>. I realize now how I missed a digit. There were some neurons in my head that haven't been fired up in a long time.
Quote from: itsu on November 24, 2015, 08:13:27 AM
It looks like a 5-band resistor with colours black (0) - brown (1) - black (0) - silver (x0.01) - brown (1%)
making it a 0.1 Ohm resistor like Luc said, see http://nearbus.net/wiki/index.php?title=File:Resistor_color_codes.jpg
I think there is more a probe x1 or x10 mixup in play here.
Itsu
I think it is the CVR not being able to handle the current,and so it is showing a way higher voltage across it than should be there. Looking at the video,i can see the yellow channel is set to 500mV per division. The yellow trace is the voltage across the bucking coils. Luc says the probe is set at 10x for the yellow channel(channel 1) If the probe is set to 1x on channel 1,then he has only about 1.2 volts across his coils. Now,if channel 2 is set to 10x on the probe,and 5 VPD on his scope,then he actually has only 700mV across the .1 ohm resistor. This being the case,that is still 7 amps at around 12 volts,which is still 84 watts. That CVR is definitely not rated at 84 watts. I would say 1,maybe 2 watts at most.
Quote from: MileHigh on November 24, 2015, 08:23:39 AM
Brad:
"i see around 560 watts per pulse" - that is the wrong terminology and doesn't make sense, please see my posting again. You should quote energy per pulse to use the terminology of electronics properly.
Itsu:
Thanks, it's been so long for me that I got it wrong and I thought it was <digit, digit, multiplier, tolerance>. I realize now how I missed a digit. There were some neurons in my head that haven't been fired up in a long time.
560 watts of power flowing through the CVR per/during each pulse.
Whats wrong with that?
Quote from: tinman on November 24, 2015, 08:26:32 AM
I think it is the CVR not being able to handle the current,and so it is showing a way higher voltage across it than should be there. Looking at the video,i can see the yellow channel is set to 500mV per division. The yellow trace is the voltage across the bucking coils. Luc says the probe is set at 10x for the yellow channel(channel 1) If the probe is set to 1x on channel 1,then he has only about 1.2 volts across his coils. Now,if channel 2 is set to 10x on the probe,and 5 VPD on his scope,then he actually has only 700mV across the .1 ohm resistor. This being the case,that is still 7 amps at around 12 volts,which is still 84 watts. That CVR is definitely not rated at 84 watts. I would say 1,maybe 2 watts at most.
Brad,
84W yes, but only for a very short time, so we have a low Pulse repetition Frequency.
See also this post: http://overunity.com/12736/kapanadze-cousin-dally-free-energy/msg466397/#msg466397
There i use a ½W 50 Ohm resistor which easily can handle the 28.8KW for the 13ns pulse every 2.3KHz.
Itsu
Quote from: itsu on November 24, 2015, 08:41:05 AM
Brad,
84W yes, but only for a very short time, so we have a low Pulse repetition Frequency.
See also this post: http://overunity.com/12736/kapanadze-cousin-dally-free-energy/msg466397/#msg466397
There i use a ½W 50 Ohm resistor which easily can handle the 28.8KW for the 13ns pulse every 2.3KHz.
Itsu
1200 volts at 24 amp's through a 1/2 watt resistor Itsu :o
I really cannot see a 1/2 watt resistor doing that. There is also a big difference in time on between yours and Luc's system. I have found that it looks like you have that current flowing through,as you see the voltage across it. But the thing is,when a resistors power value is way under the required power value,you see a higher voltage across it when there is less current flowing through it.
The resistors power rating should always be above that of the peak power flowing through it.
Quote from: MileHigh on November 24, 2015, 06:19:53 AM
...
I am not a fan of this "bucking coils" business, for the most part I view it as doing something "alternative" just for the sake of "being alternative." In this case when the slab of metal is between the bucking coils, the magnetic fields are mostly cancelling each other out. Even when the coils have air between them, they are cancelling each other out.
....
Hi MileHigh,
I would like to understand what you mean by magnetic fields cancelling each other out when the coils have air between them?
I think that bucking or repelling magnetic fields try to push away each other, towards sideways. The closer to each other the repel poles get,
the more compressed (squeezed) the fields become and they come out from the gap towards sideways and then spread out again.
Long time ago I downloaded from a now defunct yahoo group some measured flux data (collected by a flux meter probe) on the surface of
ceramic magnets that included the repel mode between them too, see attached. (I cannot recall who uploaded it back then, unfortunately.)
In those measurements the magnets were fully pressed to touch each other surfaces when they were in repel mode. It is interesting that
the squeezed-out North fields have more than twice measured field values on the touching surfaces than the sum of the
two North field values of the individual magnets.
When there is a small air gap left between the facing surfaces, (still no entering slab of metal), the flux from both magnets should be there
(albeit in a less squeezed form) and repel force would still be there of course between the facing cores as per the gap size would control it.
When the slab of metal approaches the two magnetically bucking cores, the bucking fields from both cores will attract the slab
(more or less equally if their fields are pretty close to each other in strength).
So how do you mean the magnetic fields cancelling each other out when the coils have air gap between them?
Thanks
Gyula
Hi everyone,
sorry for the delay on the replies. Had other things to attend to today.
Below is a new test 2 video of my Bucking Field Reluctance Motor which should address the CSR concerns and issues.
It was late last night (1 am) when I made the first video and today I found the probe X settings in the scope menu were reversed.
Link to test 2 video: https://www.youtube.com/watch?v=TeO9iM2-29o
Post your comments or test suggestions (within reason) of what could be further tested on this (limited) proof of concept.
Luc
Hi Luc,
Thanks for the new video, very good, albeit I have two puzzles. The scope still shows a 2.6V drop in the pulse amplitude level across the bucking coils and I wonder why it happens?
My other puzzle is that in case the analog meter shows a 140-150 mA current draw for the 'unloaded' wheel and you mentioned it went up to 200 mA when you loaded the wheel with your hand. Is this not Lenz effect? albeit in a small degree for sure (a normal pulse motor would probably have higher current draw increase than that). Question is how the input current increases if you load the rotor further down to say 30-40 RPM?
EDIT: Of course the wider pulse (longer ON time of the switch) can also increase the input current.
I think the voltage drop in the pulse amplitude may still come from the supply voltage fluctuation when the MOSFET is ON, the voltage drop may mainly be caused by the added up series inner resistances of the current meters. There may be other cause for it too. If we could add up all the DC resistances which are in series with the coil (MOSFET rDS, current meters, shunt, secunder coil of the AC step down transformer), that would help to estimate whether the drop is simply due to by those accumulated drops.
Gyula
Quote from: gyulasun on November 24, 2015, 04:15:21 PM
Hi Luc,
Thanks for the new video, very good, albeit I have two puzzles. The scope still shows a 2.6V drop in the pulse amplitude level across the bucking coils and I wonder why it happens?
Yes I agree!... I'm quite sure it's the Analogue Amp meter, since as soon as I introduced it in series with the positive input terminal its internal resistance caused a decrease in RPM, so I had to incease the input voltage to keep the rotor turning at the same speed.
Quote from: gyulasun on November 24, 2015, 04:15:21 PM
My other puzzle is that in case the analog meter shows a 140-150 mA current draw for the 'unloaded' wheel and you mentioned it went up to 200 mA when you loaded the wheel with your hand. Is this not Lenz effect? albeit in a small degree for sure (a normal pulse motor would probably have higher current draw increase than that). Question is how the input current increases if you load the rotor further down to say 30-40 RPM?
I don't think it is so. When you slow down the rotor the pulse on time gets wider so naturally the current of an analogue (moving coil) meter will have more time to get to its real level, however, the pulses now are less frequent so it should all come to the same vs. many shorter on times. If we could have a continual reading I'm quite sure we would see the input power stays the same no matter what RPM. It's unfortunate I can't raise the RPM of this prototype. The next build won't have this limitation and will also have High Impedance assist coils powered by the flyback of the Low Impedance coils.
Quote from: gyulasun on November 24, 2015, 04:15:21 PM
EDIT: Of course the wider pulse (longer ON time of the switch) can also increase the input current.
Yes, correct and what I tried to explain above.
Quote from: gyulasun on November 24, 2015, 04:15:21 PM
I think the voltage drop in the pulse amplitude may still come from the supply voltage fluctuation when the MOSFET is ON, the voltage drop may mainly be caused by the added up series inner resistances of the current meters. There may be other cause for it too. If we could add up all the DC resistances which are in series with the coil (MOSFET rDS, current meters, shunt, secunder coil of the AC step down transformer), that would help to estimate whether the drop is simply due to by those accumulated drops.
Gyula
Yes, also correct!... the Amp meter is the biggest cause in this case.
Thanks for your input
Luc
Quote from: gotoluc on November 24, 2015, 04:58:23 PM
Yes I agree, I quite sure it the Analogue Amp meter I introduced in series across the positive terminal that is causing the voltage drop across its internal resistance. As soon as I connected it I had to incease the input voltage to keep the rotor turning at the same speed.
....
Luc, does it mean that in case you short cicuit the analogue Amp meter, the top of the voltage pulse becomes more or less a straight
line instead of the curved sloping one?
Gyula
Quote from: gyulasun on November 24, 2015, 05:17:17 PM
Luc, does it mean that in case you short cicuit the analogue Amp meter, the top of the voltage pulse becomes more or less a straight
line instead of the curved sloping one?
Gyula
It helped but there is still something else. First scope shot is with Amp meter and motor at 108 RPM and second scope shot is removing the Amp meter and lowering the voltage to maintain the same 108 RPM.
I wonder if the switch could be part of the losses?... it's a dual mosfet (IRFP4310Z) arrangement so current can go either direction?
Let me know what you think
Thanks for your help
Luc
The MOSFET data sheet says the ON resistance is max 6 milliOhm (at 10 Vgs and at 75A drain current), so in case you have two such FETs in series, the 12 mOhm at the cca 8.5 A peak current would cause only 102 mV drop across the FETs and we still have at least a cca 2.3V drop.
As the current increases from zero to about 7-8 Amper, the voltage drops from 10 V to cca 8 V in about 2 millisecond, this what the scopeshot says.
One test would be to further reduce the supply voltage to say 4-5 V DC and see how the curve on the top changes if any.
Gyula
Quote from: gyulasun on November 24, 2015, 06:30:49 PM
The MOSFET data sheet says the ON resistance is max 6 milliOhm (at 10 Vgs and at 75A drain current), so in case you have two such FETs in series, the 12 mOhm at the cca 8.5 A peak current would cause only 102 mV drop across the FETs and we still have at least a cca 2.3V drop.
As the current increases from zero to about 7-8 Amper, the voltage drops from 10 V to cca 8 V in about 4 millisecond, this what the scopeshot says.
One test would be to further reduce the supply voltage to say 4-5 V DC and see how the curve on the top changes if any.
Gyula
Okay Gyula,
the below scope shots are 2 different circuit scenarios using the minimum voltage the motor needs to rotate. Keep in mind the air gap between E's and I cores is large and not idealized since the rotor flexes. This was built as a proof of concept.
I've eliminated potential resistance in the circuit and obviously the Amp meter is not connect at all in the below scope shots.
First scope shot is, removing the cheap Watt meter I have in top of my dual capacitor bank meters and next scope shot is removing my dual cap bank meters and connecting directly in the large 100,000uf source capacitor.
The only other resistance would be the switch, alligator clips and the 0.4 Ohm of the parallel MOT primary coils have.
Does it look any better? or should I need to use heavier wire leads to replace all the alligator clips?
Thanks for your input
Luc
Quote from: tinman on November 24, 2015, 08:26:32 AM
I think it is the CVR not being able to handle the current,and so it is showing a way higher voltage across it than should be there. Looking at the video,i can see the yellow channel is set to 500mV per division. The yellow trace is the voltage across the bucking coils. Luc says the probe is set at 10x for the yellow channel(channel 1) If the probe is set to 1x on channel 1,then he has only about 1.2 volts across his coils. Now,if channel 2 is set to 10x on the probe,and 5 VPD on his scope,then he actually has only 700mV across the .1 ohm resistor. This being the case,that is still 7 amps at around 12 volts,which is still 84 watts. That CVR is definitely not rated at 84 watts. I would say 1,maybe 2 watts at most.
Tinman,
Just FYI, the power dissipated by a resistor is equal to the current flowing thru the resistor times the voltage across that same resistor.
For example, assuming DC current, a CVR with 7amps flowing thru it and 700mV measured across it is dissipating 4.9 watts (not 84).
PW
Here is the best scope shot after removing all the alligator clip and replacing it with no. 12 AWG.
I also added two 0.1 Ohms in parallel as CSR, so it's now a 0.05 Ohms CSR.
It makes quite the difference to reduce accumulated resistive losses. I've also included the original scope shot (last one) of before I started to reduce circuit losses so you can compare the difference of before and after.
Now the most resistive losses in the circuit now the two primary coils which are 0.4 Ohms connected in parallel bucking fields.
The parallel Inductance is 1.05mH with no I core, 1.18mH with I core at switch on position, 2.38mH with I core at switch off position and 4.85 with I core fully in between E cores.
Luc
Quote from: picowatt on November 25, 2015, 12:04:46 AM
Tinman,
Just FYI, the power dissipated by a resistor is equal to the current flowing thru the resistor times the voltage across that same resistor.
For example, assuming DC current, a CVR with 7amps flowing thru it and 700mV measured across it is dissipating 4.9 watts (not 84).
PW
yes-this i know PW.
But i was talking about the power flowing through the resistor,not the power being dissipated by the resistor. The power being dissipated by the resistor as heat is directly determined by the power being delivered to the load through that resistor. If the resistor is rated at 1 watt,then trying to dissipate 4.9 watts of power from a resistor rated at 1 watt is going to lead to wrong calculations.
Remember,my first calculations were made from Luc's first video where he said that channel 2(the one across the .1 ohm CVR) was set at 5 VPD and the probe at 1x.(see post 148) This gave us 7 volts across a .1 ohm resistor--that 1 or 2 watt resistor is now dissipating 490 watts of power during the pulse.
When Luc corrected 10x on the scope channel for the CVR,we then had an average voltage of about 700mV across the .1 ohm CVR during the pulse,and so now that 1 or 2 watt CVR is dissipating your 4.9 watts every pulse.
If the pulse is long enough,and the resistor is rated well below the power it is trying to dissipate,then incorrect(higher) voltages can be seen across that resistor,but the current flowing through it will not be as high as ohms law says it should be.
Brad
Quote from: tinman on November 25, 2015, 04:20:21 AM
yes-this i know PW.
But i was talking about the power flowing through the resistor
"Current" flows thru the resistor....
Ok I'm doing something wrong here but I'm not sure what. I must have butchered woopys circuit some how http://youtu.be/GCE84y7gIrs
Quote from: picowatt on November 25, 2015, 04:29:07 AM
"Current" flows thru the resistor....
Are you sure it is just current that flows through a resistor in all circumstances?.
Quote from: tinman on November 25, 2015, 04:59:39 AM
Are you sure it is just current that flows through a resistor in all circumstances?.
As opposed to what?
Consider a .01R CVR in series with a 10R load being fed 100VDC.
Load dissipates very close to 1000 watts. CVR dissipates very close to 1 watt.
Surely you don't think the CVR must be a 1000 watt resistor.
The 84 watts you mentioned in the post I commented on had nothing to do with CVR requirements.
PW
Quote from: gotoluc on November 25, 2015, 03:09:39 AM
Here is the best scope shot after removing all the alligator clip and replacing it with no. 12 AWG.
I also added two 0.1 Ohms in parallel as CSR, so it's now a 0.05 Ohms CSR.
It makes quite the difference to reduce accumulated resistive losses. I've also included the original scope shot (last one) of before I started to reduce circuit losses so you can compare the difference of before and after.
Now the most resistive losses in the circuit now the two primary coils which are 0.4 Ohms connected in parallel bucking fields.
The parallel Inductance is 1.05mH with no I core, 1.18mH with I core at switch on position, 2.38mH with I core at switch off position and 4.85 with I core fully in between E cores.
Luc
great work Luc i have watched a lot of yours and TM work - for us now maybe time is out for some humans.
great jobs you have done, hopefully in the future you keep working on this.
at that time you will have much more help.
regards.
What you need is probably the square root of the wattage
divided by the resistance. HAND!
Quote from: picowatt on November 25, 2015, 05:26:19 AM
As opposed to what?
Consider a .01R CVR in series with a 10R load being fed 100VDC.
Load dissipates very close to 1000 watts. CVR dissipates very close to 1 watt.
Surely you don't think the CVR must be a 1000 watt resistor.
The 84 watts you mentioned in the post I commented on had nothing to do with CVR requirements.
PW
I love how you guys re arrange things so as they suit your need to look right.
The fact is,the example you have above has nothing at all to do with the calculations i made with the posted data by Luc--and my power calculations of the power being dissipated by the .1 ohm resistor are correct in both cases,and where in this case,the power being dissipated in Luc's setup was dependent on the power rating of the used resistor.
It would be great if you guru's could use the examples that relate to the test being carried out,in stead of going off in a different direction to show us all how you !!can!! be right in other situation.
We all know this,but how about sticking with the subject and DUT being investigated.
Brad
Quote from: Jimboot on November 25, 2015, 04:41:09 AM
Ok I'm doing something wrong here but I'm not sure what. I must have butchered woopys circuit some how http://youtu.be/GCE84y7gIrs (http://youtu.be/GCE84y7gIrs)
Hi Jimboot,
One question: why is your multimeter in the AC voltage mode??
It should be in DC voltage mode, no?
Gyula
Quote from: tinman on November 25, 2015, 06:43:24 AM
I love how you guys re arrange things so as they suit your need to look right.
The fact is,the example you have above has nothing at all to do with the calculations i made with the posted data by Luc--and my power calculations of the power being dissipated by the .1 ohm resistor are correct in both cases,and where in this case,the power being dissipated in Luc's setup was dependent on the power rating of the used resistor.
It would be great if you guru's could use the examples that relate to the test being carried out,in stead of going off in a different direction to show us all how you !!can!! be right in other situation.
We all know this,but how about sticking with the subject and DUT being investigated.
Brad
Timman,
I was commenting on this statement you made:
Quote
Now,if channel 2 is set to 10x on the probe,and 5 VPD on his scope,then he actually has only 700mV across the .1 ohm resistor. This being the case,that is still 7 amps at around 12 volts,which is still 84 watts. That CVR is definitely not rated at 84 watts.
I "rearranged" nothing. The 84 watts you mentioned has nothing to do with CVR requirements as you appear to suggest.
PW
Quote from: picowatt on November 25, 2015, 06:54:51 AM
Timman,
I was commenting on this statement you made:
I "rearranged" nothing. The 84 watts you mentioned has nothing to do with CVR requirements as you appear to suggest.
PW
It has quite a lot to do with it as far as Lus DUT go's.
Do you know the resistance of the two drive coils?
Dose he have them hooked in series or parallel ?.
Did you read my reply to you a few post back?
QuoteRemember,my first calculations were made from Luc's first video where he said that channel 2(the one across the .1 ohm CVR) was set at 5 VPD and the probe at 1x.(see post 148) This gave us 7 volts across a .1 ohm resistor--that 1 or 2 watt resistor is now dissipating 490 watts of power during the pulse.
Did i get that part right?.
QuoteWhen Luc corrected 10x on the scope channel for the CVR,we then had an average voltage of about 700mV across the .1 ohm CVR during the pulse,and so now that 1 or 2 watt CVR is dissipating your 4.9 watts every pulse.
Did i get that part right?.
QuoteIf the pulse is long enough,and the resistor is rated well below the power it is trying to dissipate,then incorrect(higher) voltages can be seen across that resistor,but the current flowing through it will not be as high as ohms law says it should be.
And do you believe the above to be correct?
If it's a yes to all three,why are we having this conversation?,as i have also stated and corrected my wording in saying that i was referring to the 84 watts as being the power consumption of the device during the on time. Knowing all this,we can also get a pretty accurate measurement of what the resistance of the two drive coils are.
I think you guys spend more time going through post looking for mistakes others make(which are more often corrected later on than not),than you do looking for ways to help improve on what we are trying to achieve.
Brad
Brad
Quote from: gyulasun on November 25, 2015, 06:43:46 AM
Hi Jimboot,
One question: why is your multimeter in the AC voltage mode??
It should be in DC voltage mode, no?
Gyula
Also i seen a 30uF cap-->is that not to large?,as i think woopy was using only a 1uF cap from a microwave oven ?.
Quote from: tinman on November 25, 2015, 07:28:39 AM
It has quite a lot to do with it as far as Lus DUT go's.
Do you know the resistance of the two drive coils?
Dose he have them hooked in series or parallel ?.
Regarding the CVR wattage rating, it doesn't matter.
Only the resistance of the CVR, the voltage across the CVR, and the applied duty cycle, determine the wattage requirement of the CVR, which was and remains the point I was making.
Quote
Did you read my reply to you a few post back?
Yes, I read your posts, and although you originally had it right, you drifted off onto a wrong track and started equating power dissipated somewhere else in the circuit as somehow determining the required CVR wattage rating, hence my original post regarding your "84 watts" comment.
PW
Quote from: tinman on November 25, 2015, 07:30:00 AM
Also i seen a 30uF cap-->is that not to large?,as i think woopy was using only a 1uF cap from a microwave oven ?.
Yes, it may be a large value. Woopy used first a 1 uF, then tested 2 uF but finally found 0.3 uF as the best fit for giving highest RPM for his 5 mH (and 0.5 Ohm) drive coil. Jim may have to test several cap values for his drive coil (which may have higher inductance and definetely higher DC resistance) plus for his 'assistant' coil.
Gyula
Hi Luc,
Good "detective" work to explore the resistive losses in the circuit. Now the voltage pulse across the bucking coils has an almost straight top line, albeit it slopes towards the switch-off edge of the pulse but then the voltage across the coils at the moment of switch-off goes below zero to a negative value (cca to -2V), and then in cca 15 ms time goes up to zero (I judged this from earlier zoom-out time base settings), well before the next ON-pulse comes.
I think this negative amplitude 'excursion' is due to the flyback diode in parallel with the paralleled bucking coils. It is interesting the parallel coils extend the field collapse time beyond to the ON time of the input pulse. Also, the long collapse time is helped maintained by the increasing inductance value of the paralleled coils as the rotor I core moves across them.
Thanks for showing this interesting test.
Gyula
Quote from: gotoluc on November 25, 2015, 03:09:39 AM
Here is the best scope shot after removing all the alligator clip and replacing it with no. 12 AWG.
I also added two 0.1 Ohms in parallel as CSR, so it's now a 0.05 Ohms CSR.
It makes quite the difference to reduce accumulated resistive losses. I've also included the original scope shot (last one) of before I started to reduce circuit losses so you can compare the difference of before and after.
Now the most resistive losses in the circuit now the two primary coils which are 0.4 Ohms connected in parallel bucking fields.
The parallel Inductance is 1.05mH with no I core, 1.18mH with I core at switch on position, 2.38mH with I core at switch off position and 4.85 with I core fully in between E cores.
Luc
When you think about all of these ideas about using the flyback pulse, perhaps the simplest way of using the flyback pulse would be the best. That would be to have your decreasing recirculating current during the field collapse be used as part of the initial propulsion pulse. In other words, no secondary coil at all.
One more time, this is all just an exercise in the proper timing of the energizing of the drive pulse, and the proper timing of the cutting off of the voltage to the drive coil so that the field collapse is part of the drive pulse itself. Let's call that a "unified drive pulse."
Note, there would be no excess energy from the drive coil discharge going into a charging battery, all of the energy from the drive coil discharge would be used to push on the rotor.
Now, the start and stop timing of the unified drive pulse could be put into a sweet spot where you get the maximum possible usable torque on the rotor to give you the maximum possible energy impulse to make the rotor turn faster.
Nothing is stopping you from measuring the usable torque that this timing system would generate also: Using a separate pickup-coil and load resistor, measure the electrical power that you can draw off of the spinning rotor and convert that into a torque measurement.
Will the strategy of using a unified drive pulse give you a higher RPM rotor for the same input power as compared to any other strategy, as in using a secondary drive coil? I am willing to bet you it will.
All of this would be an interesting investigation doing precise measurements on your favourite pulse motor. However, you need completely flexible timing and that rules out the good old extremely limited pickup coil wound coaxial with the drive coil. Either go with the MHOP pickup coil design or at least have a separate conventional pickup coil that is movable, or use an optical timing system so that you have the flexibility you need to play with the pulse timing.
In my opinion, trying to measure the theoretical usable torque from your pulse motor makes all the difference in the world in terms of getting some tangible satisfaction from what you are building. You can finally make a real-world efficiency measurement: rotary mechanical power out vs. electrical power in. Just like a real motor in the real world!
Just watching a pulse motor spin and measuring the RPM gets boring after a while. When you do that you are looking at a device with no useful output at all. People might not like this way of looking at it but a pulse motor just spinning there and doing nothing with no useful output is just a glorified resistor. That's why adding the separate pickup coil and load resistor and then converting the electrical dissipation into mechanical power and torque is so much more interesting.
Tinman and Luc -
have you viewed this video ?
he displays in this video multiple devices using 'BEMF' and conducts measurements of input and output.
there could be something here of value, he did however design a custom rotor.
https://www.youtube.com/watch?v=HK3JOlY0V8Y
also is displayed a solid state device which is of course based on similar principals.
runs 53 minutes, but is pretty much display the whole time.
demonstrates COP>3 and above I believe.
regards.
Yes digitalindustry, TinMan and I know of Jim Murray.
I don't usually follow or replicate the work of others and mostly if they keep things secret.
There is still more to come from this topic and my motor design. There will be a new build that I'll be able to do high rpm tests coming out soon. This will allow to verify if the input power changes under different loads. If that build is successful then there will be a large 2hp version built next.
So, at this time I'm not looking for more ideas but more time to do the work.
Regards
Luc
Quote from: Jimboot on November 25, 2015, 04:41:09 AM
Ok I'm doing something wrong here but I'm not sure what. I must have butchered woopys circuit some how http://youtu.be/GCE84y7gIrs (http://youtu.be/GCE84y7gIrs)
Jim mate, it looks to me your motor coil is a fine wire (HV) MOT secondary. Is this correct?
Luc
@woopy
Great job. Do you need any other ideas?
@gotoluc
I read the whole thread so good on your topic and holding it and also good for all.
About the flux wheel with the two MOT primaries, I guess "bucking" is because the two pri's are in series. But to me that's not bucking. Bucking is when the two primaries are on the same core. Right now those two MOTs (usually 120 or 220 v driven) primary coils are driven with 10 volts. So in series that second primary coil should be dead. Just scope both. I would put them in parallel and add a third MOT primary off the wheel in series to the two in parallel. The third MOT should be on the side of the coils that is always connected, that is the non-pulsed side of the now two parallel primaries. Also, the pulsed side of the two primaries should be at the core.
Yes you can leave them in series and add the third MOT as above and you will see better performance but in this type of situation where you want both primaries to impart their maximum effect to the wheel plates, bucking is not the answer in my perspective.
Put it this way. If you just run it without connecting the second MOT, what happens? Chances are a 25% drop max. (Due to classic half coil syndrome - yep it's everywhere man).
But ultimately, if this design is to remain a 10 volt driven toy, you may consider unwinding the primaries until you start seeing a drop in effect. I guess 2/3rds of the windings should come out and then this will show the actual optimum with your idea of maximum core surface area.
Scope shots are often misleading. The first vertical voltage rise (1) indicates the average rise across the coil but this average means some parts are higher then others and overrrrrrallllllll this is what it looks like. hehehe The voltage rise is your pulse creating a copper atom streamer (influence of one copper nuclei to the next and to the next etc.) inside the coil wire that will penetrate to a certain depth into the wire length. That streamer starts agitating because itself it is as unstable as a spark gap, always wobbling around along the length of wire and making more and more atoms agitated. This shows up in your current waveform rise (4) and then steady (5). The drop in voltage (3) on the scope is because the 10 volts is not enough to keep all the copper atoms of that thick primary wire busy so it gets, well call it natures rms'ing of the line since all the other copper atoms around the stream will either do one of three things, turn to the new influence, not turn because it is already in the angle of influence, or do nothing since it is be too far to be influenced.
Small analogy of amperage variations. If you look at a spark gap that is 2 inches long with a very thin spark stream, then double the amperage, the stream will be thicker, if you double the voltage, the stream will go 4 inches long. In a wire it's the same thing man. (I could get into that later but this is relative to hydro plants outputing 750000 volts to stream along hundreds and hundreds of miles of wire.).
I don't know how your circuit is set up but if the positive is being pulsed, when the pulse opens, right away the next thing you have is the negative pushing back into the coil to reclaim it again and that is your flyback moment on the scope shot, You need to zoom in on that moment. Is it up first then down? hahaha If the negative is being pulsed then the above happens in reverse.
After that big spiking (6) it trails off just below the zero (7) and goes up to zero before the next pulse (8). That trailing off needs to be lower then where it is right now, since that is a sign of how deep the pulse entered the coil wire length. Right now its saying to me, "not too deep".
One other never considered key also is the flat line (8) just before the pulse in your scope shot. That flat line time is just perfect to introduce a high frequency (well if high is 5khz to 50khz) signal into the primary as a tickle pulse designed to realign the copper atoms before the next pulse. The tickle should stop right when the pulse starts so this way the tickle will keep the polarity side that is already connected (permanent side) busy so when the pulse occurs and the tickle stops, both the positive and negative sides will claim a freshly reset zone and this should increase the primary effect since the more copper atoms that are away from their potentialized (maybe not a word "yet") alignment the more change is created and the more magnetic polarization is produced.
But this could be future stuff. Or a second reed for @woopy. Accumulate the flyback
and release it at this time only back into the primary.
Hope this is not off topic. Keep go'in.
wattsup
PS: If you added a wattmeter before your variac and deduct its zero consumption loss maybe that will enable a more precise reading. But please do not get overly obsessive about input output since a this stage it seems rather premature to spend so much attention on that. Trust your instincts on when to consider taking measurements. You will know.
Quote from: wattsup on November 26, 2015, 09:02:42 AM
About the flux wheel with the two MOT primaries, I guess "bucking" is because the two pri's are in series. But to me that's not bucking. Bucking is when the two primaries are on the same core. Right now those two MOTs (usually 120 or 220 v driven) primary coils are driven with 10 volts. So in series that second primary coil should be dead. Just scope both. I would put them in parallel and add a third MOT primary off the wheel in series to the two in parallel. The third MOT should be on the side of the coils that is always connected, that is the non-pulsed side of the now two parallel primaries. Also, the pulsed side of the two primaries should be at the core.
Yes you can leave them in series and add the third MOT as above and you will see better performance but in this type of situation where you want both primaries to impart their maximum effect to the wheel plates, bucking is not the answer in my perspective.
Hi wattsup,
just to let you know, the MOT primaries are and have always been connected in parallel. I have never tried connecting them in series and was not planing to. I want minimal coil resistance for maximum flyback power.
Thanks
Luc
Quote from: gotoluc on November 26, 2015, 09:19:08 AM
Hi wattsup,
just to let you know, the MOT primaries are and have always been connected in parallel. I have never tried connecting them in series and was not planing to. I want minimal coil resistance for maximum flyback power.
Thanks
Luc
@gotoluc
Great, but is the pulsed side going to the inside start of the winds nearest the core? Sorry to ask so many questions that I could not figure out with the vids. If it is, then just consider trying the added third MOT primary in series between the always connected side of those two primaries. But again, 10 volts in those coils is like playing ping ping in a Wallmarts.
I understand the "minimal coil resistance" but I think 10 volts on that wire is way to low to see some of the better effects that should only be starting to scintillate. Like exactly the flyback. Adding the third MOT will increase the flyback and increase how far the now 10v pulse goes into the two primaries.
So where and how does the "bucking" come in?
Always a pleasure to see your works man.
@woopy that goes for you as well. Sorry if I did not respond to anything about your build here because I am always afraid that the topic is now advancing further along and would not want to cause a topical regression. But I can if you want, hehehe
wattsup
Quote from: gotoluc on November 26, 2015, 08:32:21 AM
Yes digitalindustry, TinMan and I know of Jim Murray.
I don't usually follow or replicate the work of others and mostly if they keep things secret.
There is still more to come from this topic and my motor design. There will be a new build that I'll be able to do high rpm tests coming out soon. This will allow to verify if the input power changes under different loads. If that build is successful then there will be a large 2hp version built next.
So, at this time I'm not looking for more ideas but more time to do the work.
Regards
Luc
ahh i see thank you
indeed time is a valuable resource always appearing so anyhow.
great work i'll watch in future and hope to contribute maybe, 'in time'
Quote from: wattsup on November 26, 2015, 10:57:40 AM
Great, but is the pulsed side going to the inside start of the winds nearest the core?
Not sure what you mean by "the pulse side"... the way I have
both MOT primaries connected is, the positive is connected to the beginning of the windings nearest to the core. Both MOT's are a mirror of each other. I'm sure it could be connected the other way around but I doubt it would make a difference. The important thing is for the magnetic fields to be repulsing each other
and that it is.
Quote from: wattsup on November 26, 2015, 10:57:40 AM
Sorry to ask so many questions that I could not figure out with the vids. If it is, then just consider trying the added third MOT primary in series between the always connected side of those two primaries. But again, 10 volts in those coils is like playing ping ping in a Wallmarts.
Again, not sure what you mean by always connected side? Both primaries are always connected in parallel.
What would the series coil in between the two MOT's do?... would it be an air coil or a fully closed MOT?
Anything over 10vdc will cause the present (proof of concept) I core rotor to rub on the E-cores.
I'm presently building another prototype that should be able to handle more power and RPM.
Quote from: wattsup on November 26, 2015, 10:57:40 AM
I understand the "minimal coil resistance" but I think 10 volts on that wire is way to low to see some of the better effects that should only be starting to scintillate.
I agree but it already shows it to be a promising design idea.
Quote from: wattsup on November 26, 2015, 10:57:40 AM
So where and how does the "bucking" come in?
Point two identical speakers, coils or magnets and the will be neutral point no matter the strength of the magnetic field or sound level. Obviously it gets more interesting as the strength goes up but it's all the same effect.
Back in 2009 Gyula had suggested to me to consider building a all repelling motor using two identical transformers. I considered it but after this test: https://www.youtube.com/watch?v=wAYsAN5QPnA (https://www.youtube.com/watch?v=wAYsAN5QPnA) I thought it may not have much torque so I never built it.
But using my most recent idea of a I core rotor in between the bucking transformers it turns it back to attraction force as long as you don't over saturate the I core rotor. It's kind of brilliant if I may say so ;) ... also, there is an increase in inductance (double) from the time the power is turned on to the time it's turned off: Inductance is 1.05mH no I core, 1.18mH with I core at switch on position, 2.38mH with I core at switch off position and 4.85mH with full core in position. So if a gain of inductance can help the flyback to have more kick (to be proven), then this design will help with that also.
Anyways, the correct I core rotor saturation level will be important. The I core thickness may need to be twice the thickness then the ones I'm using to allow the full MOT's flux potential but for now it will be good enough for proof of concept.
Hope this helps you and others better understand where I'm going with this. Also, don't forget, the primary coil flyback will be going to the high impedance assist coils to further boost the torque of this non CEMF reluctance motor.
Luc
Quote from: gotoluc on November 26, 2015, 08:48:26 AM
Jim mate, it looks to me your motor coil is a fine wire (HV) MOT secondary. Is this correct?
Luc
Yep MOT secondary Luc. I'm having better results now with a smaller cap. Thank Brad & Gyula. I have a condition called dickheadedness which manifests when I build circuits hence my DMM setting Gyula :)
@gotoluc
OK, thanks for the bucking explain. I see that you consider this as a bucking effect and also understand that there is no relation between this and the standard bucking coils that are generally wound on a same core. Great.
Coils are wound on a core so you have the core layer that starts the winding and the outer layer that ends the winding.
A) The "pulse side" is the side of your parallel primaries that gets the make break, what ever it is - positive or negative of a source, applied to the coil at the core or outer layer.
B) The "always connected side" is the side of your parallel primaries that is always connected to source - positive or negative, applied to the coil at the core or outer layer.
So what I am trying to explain is that you add a third MOT primary in series to the two in parallel on the always connected side. This third one is just there sitting on the table and does not have to be mounted in any way on your wheel. Depending on which side is pulsed, these are the two possible ways;
====Primary 1====
---- pulsed on positive ---F-- ====Primary 3==== ----- constant negative------
====Primary 2====
or,
====Primary 1====
---- pulsed on negative --F--- ====Primary 3==== ----- constant positive------
====Primary 2====
OK, so where are you connecting your flyback diode? Using the above scheme, the flyback diode should come off the line where you see the letter F, that is between the pulse and the two parallel primaries.
So here are two questions for you guys about flyback.
1) If you run an electric motor or transformer with AC, A) is there flyback, or, B) is flyback strictly present after a DC pulse only?
2) If #1 is A or B or both, why?
wattsup
Quote from: Jimboot on November 27, 2015, 05:08:49 AM
Yep MOT secondary Luc. I'm having better results now with a smaller cap. Thank Brad & Gyula. I have a condition called dickheadedness which manifests when I build circuits hence my DMM setting Gyula :)
In that case you don't have a replication of woopy's setup or what I've been recommending. Your motor coil should be the MOT's primary and send its flyback to the MOT secondary as the assist coil. Doing it the way you have it will have next to no effect to the assist coil as the high resistance of a MOT secondary as motor coil will give you a weak flyback.
What I've been saying is, start with low impedance (mot primary) and send it's flyback to a high impedance (mot secondary) to assist the motor.
All the best mate
Luc
Quote from: wattsup on November 27, 2015, 09:35:06 AM
@gotoluc
OK, thanks for the bucking explain. I see that you consider this as a bucking effect and also understand that there is no relation between this and the standard bucking coils that are generally wound on a same core. Great.
Coils are wound on a core so you have the core layer that starts the winding and the outer layer that ends the winding.
A) The "pulse side" is the side of your parallel primaries that gets the make break, what ever it is - positive or negative of a source, applied to the coil at the core or outer layer.
B) The "always connected side" is the side of your parallel primaries that is always connected to source - positive or negative, applied to the coil at the core or outer layer.
So what I am trying to explain is that you add a third MOT primary in series to the two in parallel on the always connected side. This third one is just there sitting on the table and does not have to be mounted in any way on your wheel. Depending on which side is pulsed, these are the two possible ways;
====Primary 1====
---- pulsed on positive ---F-- ====Primary 3==== ----- constant negative------
====Primary 2====
or,
====Primary 1====
---- pulsed on negative --F--- ====Primary 3==== ----- constant positive------
====Primary 2====
OK, so where are you connecting your flyback diode? Using the above scheme, the flyback diode should come off the line where you see the letter F, that is between the pulse and the two parallel primaries.
So here are two questions for you guys about flyback.
1) If you run an electric motor or transformer with AC, A) is there flyback, or, B) is flyback strictly present after a DC pulse only?
2) If #1 is A or B or both, why?
wattsup
I'll give it a test at some point and let you know. You should see the result in about a week, if I forget please remind me.
Thanks for sharing
Luc
Quote from: gotoluc on November 26, 2015, 01:17:29 PM
I agree but it already shows it to be a promising design idea.
Point two identical speakers, coils or magnets and the will be neutral point no matter the strength of the magnetic field or sound level. Obviously it gets more interesting as the strength goes up but it's all the same effect.
Back in 2009 Gyula had suggested to me to consider building a all repelling motor using two identical transformers. I considered it but after this test: https://www.youtube.com/watch?v=wAYsAN5QPnA (https://www.youtube.com/watch?v=wAYsAN5QPnA) I thought it may not have much torque so I never built it.
But using my most recent idea of a I core rotor in between the bucking transformers it turns it back to attraction force as long as you don't over saturate the I core rotor. It's kind of brilliant if I may say so ;) ... also, there is an increase in inductance (double) from the time the power is turned on to the time it's turned off: Inductance is 1.05mH no I core, 1.18mH with I core at switch on position, 2.38mH with I core at switch off position and 4.85mH with full core in position. So if a gain of inductance can help the flyback to have more kick (to be proven), then this design will help with that also.
Anyways, the correct I core rotor saturation level will be important. The I core thickness may need to be twice the thickness then the ones I'm using to allow the full MOT's flux potential but for now it will be good enough for proof of concept.
Hope this helps you and others better understand where I'm going with this. Also, don't forget, the primary coil flyback will be going to the high impedance assist coils to further boost the torque of this non CEMF reluctance motor.
Luc
@gotoluc
If your 2009 video was redone today, the first thing I would have wanted to see is while the coils are energized in series as you have them, show that each one can lift up a separate piece of laminate. That is the first thing, are both coils working because I am sure the second coil in series would not have picked up anything so it always remained a piece of metal. That's why in your repulsion mode it was not repulsing because one coil had a polarity while the other was just metal sticking to a polarity. Even at first when you had it in attraction mode, I am sure one coil was attracting the others metal but not the others polarity because series coils cannot be equal. Which also confirms that metal to polarity can only work under attraction, so great again. You can pull in and release but never repulse. I guess if you added some magnet on the trailing side of the wheel plates and rigged a second pulse as the plate leaves the coils confines, it could also work under repulsion as a secondary motivation.
wattsup
Hi wattsup,
My "2 cents" answers on your questions. I repeat here your questions:
"1) If you run an electric motor or transformer with AC, A) is there flyback, or, B) is flyback strictly present after a DC pulse only?
2) If #1 is A or B or both, why?"
Answers:
1) I do not think there is flyback with AC input so my aswer is B
2) only the B because with a normal (50 or 60 Hz or whatever frequency) sinusoidal input voltage there is no interruption of the input current (zero crossing yes but it is not sudden). Flyback pulse is strictly associated with a switch-off event under which no input current is drawn from the the input source which normally provides the input current during the ON times. Again, the zero crossing event is not equivalent at all with a switch-off event by a dedicated switch.
Regarding your suggestion on using a 3rd primary MOT coil in series with the paralleled ones: the use of this coil increases the overall impedance of the whole setup and as such you have to increase a little the input voltage to compensate for the reduced input current what this increased impedance causes. It is okay that you would receive flyback energy from the 3rd coil too, question is would this give more juice than what the increased input cost? needs testing.
You wrote earlier that the 10 V or so input voltage level is rather low with respect to the mains 110-120 V these MOT primary coils were designed for originally. Here comes again that Luc drives these coils with pulsed DC and not AC, notice the pulsed DC involves already several Amper peak currents when the voltage level is at the 10 V area, while normal AC drive at 110-120 V input would result only in some hundred mA maximum or less input current levels (assuming no load condition).
One more thing: these "transformers" have open magnetic cores, rendering the original primary coil inductances much less than a closed core insures, and add to this the bucking magnetic coupling which further reduces the resulting inductance. The I core entering the gap between the two open cores increases this lowered coil inductance as we learnt from Luc inductance measurements, this is good.
Gyula
Quote from: gotoluc on November 27, 2015, 10:02:49 AM
In that case you don't have a replication of woopy's setup or what I've been recommending. Your motor coil should be the MOT's primary and send its flyback to the MOT secondary as the assist coil. Doing it the way you have it will have next to no effect to the assist coil as the high resistance of a MOT secondary as motor coil will give you a weak flyback.
What I've been saying is, start with low impedance (mot primary) and send it's flyback to a high impedance (mot secondary) to assist the motor.
All the best mate
Luc
Yeah thanks I have been playing with both. I think I've got the right setup now so I'll switch back to the primary. Finally stopped the plasma in the switches with the right cap.
Quote from: gyulasun on November 24, 2015, 12:14:27 PM
Hi MileHigh,
I would like to understand what you mean by magnetic fields cancelling each other out when the coils have air between them?
I think that bucking or repelling magnetic fields try to push away each other, towards sideways. The closer to each other the repel poles get,
the more compressed (squeezed) the fields become and they come out from the gap towards sideways and then spread out again.
Long time ago I downloaded from a now defunct yahoo group some measured flux data (collected by a flux meter probe) on the surface of
ceramic magnets that included the repel mode between them too, see attached. (I cannot recall who uploaded it back then, unfortunately.)
In those measurements the magnets were fully pressed to touch each other surfaces when they were in repel mode. It is interesting that
the squeezed-out North fields have more than twice measured field values on the touching surfaces than the sum of the
two North field values of the individual magnets.
When there is a small air gap left between the facing surfaces, (still no entering slab of metal), the flux from both magnets should be there
(albeit in a less squeezed form) and repel force would still be there of course between the facing cores as per the gap size would control it.
When the slab of metal approaches the two magnetically bucking cores, the bucking fields from both cores will attract the slab
(more or less equally if their fields are pretty close to each other in strength).
So how do you mean the magnetic fields cancelling each other out when the coils have air gap between them?
Thanks
Gyula
Gyula,
I think that you are just forgetting the basics for a momentary lapse. The only thing that two separate magnetic fields do is vector addition. So for those that may not understand that means the two magnetic fields act like the other field doesn't even exist, the two fields just pass right through each other. You just add the two fields together to get the net magnetic field.
So, if you have a magnetic field going horizontally left to right, and another magnetic field going horizontally right to left, you just add them together. Because of their respective directions, they cancel each other out.
Therefore, there is no squeezing or compressing going on like you are alluding to. The people in the Yahoo group led themselves down the proverbial garden path.
Okay, in the next posting let me deconstruct the image you posted from the Yahoo group.
MileHigh
When you look at the measurements for the individual magnets, what is seriously missing is the orientation of the Hall sensor. That is a major mistake but we can live with it. We are going to assume that the Hall sensor is always placed parallel to the surface of the magnets.
Looking at the rectangular shape of the magnets, you can see increasing field strength going towards the corners of the magnets. However, they fail to make any measurements at the corners themselves for the individual magnet fields, but effectively they do do corner measurements when making the combined magnet field measurements. However, we can make some reasonable extrapolations with the data and arrive at a satisfactory conclusion.
For the individual magnet fields, we are pretty certain that the strength is the strongest at the corners, and we are pretty certain that the angle of the magnetic field emanating at the corners is roughly 45 degrees. Let's assume for the sake of argument that a Hall sensor placed normal to the 45-degree-angle field at the corners will measure 1760 units of Gauss. It seems perfectly reasonable because the measurements close to the corners are are between 1000 and 1100 Gauss and we can assume that the Hall sensor was placed parallel the surface of the magnet. If the Hall sensor in that position was actually normal to the magnetic field you would measure a higher Gauss value, perhaps around 1400 units of Gauss. So it is perfectly reasonable to assume about 1760 units of Gauss with the Hall sensor normal to the magnetic field.
So now let's look at the junction of the two magnets where the Hall sensor measures about 2500 units of Gauss. Again, we will assume that the Hall sensor is parallel to that surface.
You have two magnets with 1760 units of Gauss at the corners at a roughly 45 degree angle. Therefore you divide each measurement by the square root of two to get the individual components of the magnetic fields that are normal to the Hall sensor. Therefore each magnet contributes about 1245 units of Gauss normal to the Hall sensor. There are two magnets and therefore you get 2490 units of Gauss normal to the surface of the magnets. This is also normal to the Hall sensor that is assumed to be parallel to the surface of the magnets at that position. That is in line with the measured values of 2486 and 2680 units of Gauss in the Yahoo data.
So you can see it all works out. By looking at the Yahoo data and making a few inferences and reasonable assumptions everything checks out as it should.
Quote from: gyulasun on November 24, 2015, 12:14:27 PM
Hi MileHigh,
I would like to understand what you mean by magnetic fields cancelling each other out when the coils have air between them?
When there is a small air gap left between the facing surfaces, (still no entering slab of metal), the flux from both magnets should be there
(albeit in a less squeezed form) and repel force would still be there of course between the facing cores as per the gap size would control it.
When the slab of metal approaches the two magnetically bucking cores, the bucking fields from both cores will attract the slab
(more or less equally if their fields are pretty close to each other in strength).
So how do you mean the magnetic fields cancelling each other out when the coils have air gap between them?
Thanks
Gyula
Let's do a thought experiment that anybody can replicate in real life. You have two long and thin bar magnets, where one is slightly stronger than the other.
You set up the magnets like this: [S============N] [N=============S]
As the two magnets are brought together you feel increasing repulsion. Then at a certain close distance from each other you feel nothing, no repulsion or attraction. That is where the opposing north fields have cancelled each other out. Then as you bring the two magnets closer together they are lightly attracted and the two north ends stick together. This is where the stronger magnet has dominated over the weaker magnet and there is a small net magnetic field and a small attraction.
From the side view in Luc's setup, looking at a cross-section, you have two "E" cores facing each other.
If the metal slab is not between the cores, you have a wide air gap between the cores. In this case much less flux produced by one core crosses the air gap and goes into the facing core. Nonetheless, some flux will cross over into the facing core. Whatever flux produced by the left core that goes into the right core will cause cancellation in the flux produced by the right core. Likewise, whatever flux produced by the right core that goes into the right core will cause cancellation in the flux produced by the left core.
When the metal slab is in place, there is much less of a reluctance gap between the two "E" cores and therefore there will be correspondingly more flux going from the left core into the right core, and vice versa. Hence, less overall inductance, less attraction for the slab, and more of the wire proportionally in the pair of coils that only functions like a resistor and not like an inductor.
If the two "E" cores produce additive magnetic fields, then one must assume that the attraction of the slab will be that much stronger and presumably balanced enough so that the slab does not scrape against one of the "E" cores. In effect, for this setup, and for the bucking setup, the slab is dragged down into the bottom of a magnetic potential well, and then at TDC (or just before TDC) you stop energizing the coils to make the magnetic potential well disappear. If the two coils are additive instead of bucking, one assumes that the slab will fall into a much deeper magnetic potential well and have much more kinetic energy as a result of the steeper fall into the well.
I asked Luc to try an additive instead of a bucking configuration but I don't think the question was acknowledged. I would be very surprised if I missed something in my analysis. From what I can see bucking coils in this configuration will work, but not be nearly as well as additive coils. The wild card is that with the stronger magnetic forces, will the slab hit one of the "E" cores or not?
MileHigh
Dear MileHigh,
I have been busy for a while but will answer to you tonight or tomorrow morning, please bare with me till then.
Thanks,
Gyula
@gyulasun
Thanks for your comments and I will answer them in bold.
Quote from: gyulasun on November 27, 2015, 07:16:01 PM
Hi wattsup,
My "2 cents" answers on your questions. I repeat here your questions:
"1) If you run an electric motor or transformer with AC, A) is there flyback, or, B) is flyback strictly present after a DC pulse only?
2) If #1 is A or B or both, why?"
Answers:
1) I do not think there is flyback with AC input so my answer is B
2) only the B because with a normal (50 or 60 Hz or whatever frequency) sinusoidal input voltage there is no interruption of the input current (zero crossing yes but it is not sudden). Flyback pulse is strictly associated with a switch-off event under which no input current is drawn from the the input source which normally provides the input current during the ON times. Again, the zero crossing event is not equivalent at all with a switch-off event by a dedicated switch.
Thank you very much for your answers which where exactly as one would hope to have in our present days.
My answer is written but is too long and I do not want to clutter @gotolucs' thread. haha
I will post in my own thread when it's time with your answers and my response.
Regarding your suggestion on using a 3rd primary MOT coil in series with the paralleled ones: the use of this coil increases the overall impedance of the whole setup and as such you have to increase a little the input voltage to compensate for the reduced input current what this increased impedance causes. It is okay that you would receive flyback energy from the 3rd coil too, question is would this give more juice than what the increased input cost? needs testing.
You wrote earlier that the 10 V or so input voltage level is rather low with respect to the mains 110-120 V these MOT primary coils were designed for originally. Here comes again that Luc drives these coils with pulsed DC and not AC, notice the pulsed DC involves already several Amper peak currents when the voltage level is at the 10 V area, while normal AC drive at 110-120 V input would result only in some hundred mA maximum or less input current levels (assuming no load condition).
One more thing: these "transformers" have open magnetic cores, rendering the original primary coil inductances much less than a closed core insures, and add to this the bucking magnetic coupling which further reduces the resulting inductance. The I core entering the gap between the two open cores increases this lowered coil inductance as we learnt from Luc inductance measurements, this is good.
Gyula
The added impedance is peanuts compared to what the added primary will do for the two working primaries on the wheel. As I had shown in my Half Coil Syndrome videos, by adding a second coil in series, the first coil displays a much higher action through the coil where the first coil will display a much larger single polarity. This single polarity would be more advantageous in @gotolucs case where his two primaries are on "open cores" as you so well defined. Right now the open core has a primary coil that has two polarities hitting it with no where to go and I suspect there is a good portion of the combined polarities that are simply cancelling each other out in relation to the passing plate. So by adding the third primary, this pushes one of the polarities out of the first two and into the third, thus this will lower cancellation potential in the working coils and increase the attraction of the passing plate.
Also, at pulse open, the side that is always connected (third coil) will rebias the first two primaries much more fully thus preparing the coil and core for the next complete change in polarity. I know you guys stand by the never proven, but always can I say "blindly" accepted, notion that the two primaries will produce a magnetic field and it is this field that will attract the passing plate, but in Spin Conveyance theory, this is not the case. It is the iron and copper atoms that do all the work. All that is required is to produce a change in the atomic alignments and the more change that is produced the more attraction it will produce simply mass to mass without any field requirement. I will explain this further soon because I do not want to use @gotolucs thread on anything off topic.
wattsup
Quote from: wattsup on November 28, 2015, 10:42:09 AM
@gyulasun
QuoteAs I had shown in my Half Coil Syndrome videos
Thanks for your comments and I will answer them in bold.
wattsup
Im not seeing this half coil syndrome wattsup ?.
In fact,everything seems to be as it should be.
https://www.youtube.com/watch?v=_YFBs7DVqok
Hi MileHigh,
I have done a test with permanent magnets in repel configuration, the attached picture shows what kind of magnets I used. The single half crescent-shaped magnets are magnetised through their thickness, so these make up for two long "bar" magnets with the poles at the ends, I hope you accept this as if they were two long and thin bar magnets.
First I held the two "bar" magnets in my left and right hands and started to force them closer and closer together with facing repel poles and maintaining their lengthwise axis in exact alignment. I clearly felt an increasing repel force all the way till the facing surfaces actually touched each other.
These magnets are ALNICO types and I managed to fully force them together. I have some cylinder shaped Neodymium magnets but they are so strong I cannot force them fully together with facing repel poles in my hands, a gap of about 1.5-2mm remains between their facing surfaces. But as I force them closer and closer, I do feel an increasing repel force as the gap reduces between them till I reach the 1.5-2mm gap I cannot defeat by my hands.
In the pictures I show the two Alnico "bar" magnets in repel, with the help from a family member. Top picture shows what distance they let themselves place in repel, it was about 29-30mm on the table (of course this is influenced by friction). The middle and bottom pictures show as we forced the "bar" magnets closer and closer while sensing the repel force by our fingers continuously. Any time we stopped forcing them together and let the bars toss each other out to the left and right, they always snapped backwards immediately. There was no any distance my helper or I found where the repel force would have been weaker, nor we found a position any close to each other where we could feel a no repulsion, no attraction situation.
The above text is my answer to this part of your post:
QuoteLet's do a thought experiment that anybody can replicate in real life. You have two long and thin bar magnets, where one is slightly stronger than the other.
You set up the magnets like this: [S============N] [N=============S]
As the two magnets are brought together you feel increasing repulsion. Then at a certain close distance from each other you feel nothing, no repulsion or attraction. That is where the opposing north fields have cancelled each other out. Then as you bring the two magnets closer together they are lightly attracted and the two north ends stick together. This is where the stronger magnet has dominated over the weaker magnet and there is a small net magnetic field and a small attraction.
On the web there are such pieces of information like this:
"The point X is called a neutral point. The forces due to both magnets cancel each other, i.e. there is no net force, at X."
See the next attachment with the drawing belonging to that text, from this link:
http://www.animatedscience.co.uk/ks5_physics/general/Electricity%20&%20Magnetism/Magnetic%20Fields.htm (http://www.animatedscience.co.uk/ks5_physics/general/Electricity%20&%20Magnetism/Magnetic%20Fields.htm)
Well, I can accept that in a tiny volume like some pin-heads would represent in the middle area between the facing repel magnets, the flux lines leaving both magnets are so much diverted sideways by the mutual repulsion that there would remain only a tiny neutral point or volume as shown. However, this must be so small force cancellation that cannot affect much the resulting main repel force which comes about from flux lines leaving the magnets elsewhwere on the facing areas. At least we have not sensed any such reducement, let alone small attraction force.
I can also accept that in case the two repel magnets are different in strength like in the case of a ferrite magnet and a rare earth magnet interaction, here the stronger magnet may change the weaker magnet's original properties by remagnetizing it (changes the original poles for instance) temporarily or for a longer time.
However this may not happen with two ceramic magnets, for instance which may have but a small difference in strength.
I did not mean to consider these latter situations, especially not in the case of two repelling electromagnets which was the initial situation when I asked what you had meant by magnetic field cancellation when the coils had only air between them.
Let me copy some characteristics of magnetic lines of force from this link http://www.tpub.com/neets/book1/chapter1/1i.htm (http://www.tpub.com/neets/book1/chapter1/1i.htm)
1. Magnetic lines of force are continuous and will always form closed loops.
2. Magnetic lines of force will never cross one another.
3. Parallel magnetic lines of force traveling in the same direction repel one another. Parallel magnetic lines of force traveling in opposite directions tend to unite with each other and form into single lines traveling in a direction determined by the magnetic poles creating the lines of force.
4. Magnetic lines of force tend to shorten themselves. Therefore, the magnetic lines of force existing between two unlike poles cause the poles to be pulled together.
5. Magnetic lines of force pass through all materials, both magnetic and nonmagnetic.
6. Magnetic lines of force always enter or leave a magnetic material at right angles to the surface.
Can you agree with these points?
Gyula
PS Next time I will try to answer some of your other statements / opinions made in your previous two posts.
Quote from: gyulasun on November 28, 2015, 07:00:31 PM
Hi MileHigh,
I have done a test with permanent magnets in repel configuration, the attached picture shows what kind of magnets I used. The single half crescent-shaped magnets are magnetised through their thickness, so these make up for two long "bar" magnets with the poles at the ends, I hope you accept this as if they were two long and thin bar magnets.
First I held the two "bar" magnets in my left and right hands and started to force them closer and closer together with facing repel poles and maintaining their lengthwise axis in exact alignment. I clearly felt an increasing repel force all the way till the facing surfaces actually touched each other.
These magnets are ALNICO types and I managed to fully force them together. I have some cylinder shaped Neodymium magnets but they are so strong I cannot force them fully together with facing repel poles in my hands, a gap of about 1.5-2mm remains between their facing surfaces. But as I force them closer and closer, I do feel an increasing repel force as the gap reduces between them till I reach the 1.5-2mm gap I cannot defeat by my hands.
In the pictures I show the two Alnico "bar" magnets in repel, with the help from a family member. Top picture shows what distance they let themselves place in repel, it was about 29-30mm on the table (of course this is influenced by friction). The middle and bottom pictures show as we forced the "bar" magnets closer and closer while sensing the repel force by our fingers continuously. Any time we stopped forcing them together and let the bars toss each other out to the left and right, they always snapped backwards immediately. There was no any distance my helper or I found where the repel force would have been weaker, nor we found a position any close to each other where we could feel a no repulsion, no attraction situation.
Gyula,
Okay, I can see your test apparently refutes my claim, which made me realize that I have to further fine tune my statements and description. For starters, the phenomenon that I described is something that I have observed myself several times in the past. It was so long ago and I was just casually playing with magnets that I can't remember any specifics.
The reason I suggested long and thin bar-type magnets was to reduce the gradient in the magnetic field strength at the end of each magnet. That way presumably you have more opportunity for the phenomenon I described to take place.
I believe what I missed in my description is that the magnets can't be such that the ferrite or other material is fully saturated. There has to be some remaining headroom for supporting the flow of increased magnetic flux. Without that property I don't think it will work. So I am going to assume that your "bar" magnets made with the Alnico material are nearly or are fully saturated. So that means from the perspective of the opposing magnet, there is no conduit for magnetic flux, and the relative permeability of the material is close to one.
So, if you can imagine the two approaching bar magnets, north approaching north, each magnet "sees" an approaching opposite pole (repulsion) and it "sees" an approaching conduit for magnetic flux (attraction). When the magnets cross the "zero line" threshold, the stronger magnet nullifies the magnetic field of the weaker magnet, and the remaining net magnetic field and potential for increasing flux will cause attraction between the two magnets. The net magnetic field gradient is working "for attraction" in this case as more net flux continues to flow between the two magnets as they get closer together.
This is all partially shown in Luc's clip: https://www.youtube.com/watch?v=wAYsAN5QPnA (https://www.youtube.com/watch?v=wAYsAN5QPnA)
Nothing happens when he energizes the two core assemblies that are facing each other in repulsion mode. If you assume the same modified transformers were used, you have the same number of turns and they are connected in series for the same number of ampere-turns. Both cores are only partially magnetized so they both have extra headroom for magnetic flux to flow through them in the "opposite" direction.
The net result of all of this is near-perfect flux cancellation and essentially no net magnetic field, so the force between the two transformer cores is near zero. There is a gradient, but with no net magnetic field when the two cores are so close together, that you can do nothing with it. If the experiment was done so that you had 10% more current flowing in one of the modified transformers, then the gradient has something to work with, and then the two transformer cores would be attracted to each other, as per my suggested experiment.
Likewise, if in the clip Luc had moved the two transformer cores about one centimeter away from each other, then you would have had less flux flowing into the "opposite" core because of the big air gap. At the same time, the repulsive magnetic field gradient would have taken over and the two transformer cores would have pushed away from each other from the now-manifesting magnetic repulsion.
So, if you agree with this, Luc's clip is a partial demonstration of the phenomenon that I am referring to.
MileHigh
Quote from: gyulasun on November 28, 2015, 07:00:31 PM
1. Magnetic lines of force are continuous and will always form closed loops.
2. Magnetic lines of force will never cross one another.
3. Parallel magnetic lines of force traveling in the same direction repel one another. Parallel magnetic lines of force traveling in opposite directions tend to unite with each other and form into single lines traveling in a direction determined by the magnetic poles creating the lines of force.
4. Magnetic lines of force tend to shorten themselves. Therefore, the magnetic lines of force existing between two unlike poles cause the poles to be pulled together.
5. Magnetic lines of force pass through all materials, both magnetic and nonmagnetic.
6. Magnetic lines of force always enter or leave a magnetic material at right angles to the surface.
Can you agree with these points?
Gyula
#6 is wrong. It looks like the author is mixed up and was actually thinking about static electric field lines leaving a conductive surface. In that case the electric field lines have to be at right angles to the surface.
I moved away from the reeds as I ran out of them :) Back using my hall effect circuit (designed by woopy circa 2006) Anyway interesting effects. I'll hook the scope up tonight. Also I could not replicate in this vid but when I had the cap hooked up in parallel with the drive coil and I was trying to hook up a load I accidentally created a spark gap and the light shone very brightly. I'll try to replicate that tomorrow night with the scope attached. In the meantime I thought you may find this interesting. https://www.youtube.com/watch?v=WeJkyMNeTzE (edit changed URL)
Quote from: MileHigh on November 29, 2015, 01:44:09 PM
Gyula,
Okay, I can see your test apparently refutes my claim, which made me realize that I have to further fine tune my statements and description. For starters, the phenomenon that I described is something that I have observed myself several times in the past. It was so long ago and I was just casually playing with magnets that I can't remember any specifics.
The reason I suggested long and thin bar-type magnets was to reduce the gradient in the magnetic field strength at the end of each magnet. That way presumably you have more opportunity for the phenomenon I described to take place.
I believe what I missed in my description is that the magnets can't be such that the ferrite or other material is fully saturated. There has to be some remaining headroom for supporting the flow of increased magnetic flux. Without that property I don't think it will work. So I am going to assume that your "bar" magnets made with the Alnico material are nearly or are fully saturated. So that means from the perspective of the opposing magnet, there is no conduit for magnetic flux, and the relative permeability of the material is close to one.
So, if you can imagine the two approaching bar magnets, north approaching north, each magnet "sees" an approaching opposite pole (repulsion) and it "sees" an approaching conduit for magnetic flux (attraction). When the magnets cross the "zero line" threshold, the stronger magnet nullifies the magnetic field of the weaker magnet, and the remaining net magnetic field and potential for increasing flux will cause attraction between the two magnets. The net magnetic field gradient is working "for attraction" in this case as more net flux continues to flow between the two magnets as they get closer together.
This is all partially shown in Luc's clip: https://www.youtube.com/watch?v=wAYsAN5QPnA (https://www.youtube.com/watch?v=wAYsAN5QPnA)
Nothing happens when he energizes the two core assemblies that are facing each other in repulsion mode. If you assume the same modified transformers were used, you have the same number of turns and they are connected in series for the same number of ampere-turns. Both cores are only partially magnetized so they both have extra headroom for magnetic flux to flow through them in the "opposite" direction.
The net result of all of this is near-perfect flux cancellation and essentially no net magnetic field, so the force between the two transformer cores is near zero. There is a gradient, but with no net magnetic field when the two cores are so close together, that you can do nothing with it. If the experiment was done so that you had 10% more current flowing in one of the modified transformers, then the gradient has something to work with, and then the two transformer cores would be attracted to each other, as per my suggested experiment.
Likewise, if in the clip Luc had moved the two transformer cores about one centimeter away from each other, then you would have had less flux flowing into the "opposite" core because of the big air gap. At the same time, the repulsive magnetic field gradient would have taken over and the two transformer cores would have pushed away from each other from the now-manifesting magnetic repulsion.
So, if you agree with this, Luc's clip is a partial demonstration of the phenomenon that I am referring to.
MileHigh
Hi MH,
your test and results can be achieved even with neodymium if you use of piece of steel between them. You must tune to correct thickness and you will get exactly what you described. Did it many times and can works with any magnets.
I do agree it can also be done with a weak ceramic magnets that has space to store flux of the overtaking magnet.
Luc
Quote from: webby1 on November 29, 2015, 07:41:05 AM
....
But I digress,,
Hi webby1,
Yes I understand how the use of a soft steel between the repel magnets changes the situation versus the case when
there is air gap between the two repel magnets. I agree that with a soft iron in-between, several possibilities come
up, depending on the properties of the iron piece (thickness, length, permeability etc) and as you mention, repel or
attract modes can be attained, including a shoot away situation for the soft iron piece.
But my question was not involved with an iron piece placed between two repel poles but was involved with how MileHigh
meant the "cancellation" of magnetic repel fields between two facing electromagnets with the air gap in between.
Thanks
Gyula
Quote from: MileHigh on November 29, 2015, 01:44:09 PM
...
I believe what I missed in my description is that the magnets can't be such that the ferrite or other material is fully saturated. There has to be some remaining headroom for supporting the flow of increased magnetic flux. Without that property I don't think it will work. So I am going to assume that your "bar" magnets made with the Alnico material are nearly or are fully saturated. So that means from the perspective of the opposing magnet, there is no conduit for magnetic flux, and the relative permeability of the material is close to one.
...
Hi MileHigh,
Yes, permanent magnets are fully or almost fully saturated materials. It is a fact that any permanent magnet (which is not abused and indeed a magnet) has got a relative permeability of very near to 1. We already agreed on this here:
http://overunity.com/13993/the-magneformer-lenzless-transformer/msg377008/#msg377008 (http://overunity.com/13993/the-magneformer-lenzless-transformer/msg377008/#msg377008)
and here is data on relative permeability of Neodymium and Samarium-Cobalt magnets:
https://en.wikipedia.org/wiki/Neodymium_magnet#Physical_and_mechanical_properties (https://en.wikipedia.org/wiki/Neodymium_magnet#Physical_and_mechanical_properties)
You also wrote:
QuoteSo, if you can imagine the two approaching bar magnets, north approaching north, each magnet "sees" an approaching opposite pole (repulsion) and it "sees" an approaching conduit for magnetic flux (attraction). When the magnets cross the "zero line" threshold, the stronger magnet nullifies the magnetic field of the weaker magnet, and the remaining net magnetic field and potential for increasing flux will cause attraction between the two magnets. The net magnetic field gradient is working "for attraction" in this case as more net flux continues to flow between the two magnets as they get closer together.
If you accept that permanent magnets have a magnetic permeability very close to one, then why would any of the magnets see an approaching conduit for flux in the other approaching magnet?
For, if you accept that magnetic lines of flux coming from two like poles would always repel each other, then this would mean that no or very little part of the flux lines from one magnet could reach the body of the other magnet so any attraction force would be amply defeated by the ruling repel forces.
If you accept these, then the question is why would the magnetic lines of flux (that repel each other) cancel each other?
My answer is they cannot cancel each other in repel but they both get diverted (sideways) and compressed (I refer to your two bar magnets in repel example). If your answer is the fields get cancelled, that is fine with me but at least I tried and I see it differently.
Regarding the measurements on individual and combined magnet fields I uploaded as the FluxStrengthOfMagPoles.jpg file from an old yahoo group, the flux numbers clearly show what happens to combined fields of magnets when they are in repel or in attract mode.
Sorry but I disagree with the conclusions you wrote here (for reasons I wrote above) :
http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg466910/#msg466910 (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg466910/#msg466910)
Regarding Luc's video clip on the two E cored coils in attract and repel modes, https://www.youtube.com/watch?v=wAYsAN5QPnA (https://www.youtube.com/watch?v=wAYsAN5QPnA) I will return to it later tomorrow or a day after.
Gyula
Quote from: gyulasun on November 30, 2015, 04:22:17 PM
If you accept that permanent magnets have a magnetic permeability very close to one, then why would any of the magnets see an approaching conduit for flux in the other approaching magnet?
For, if you accept that magnetic lines of flux coming from two like poles would always repel each other, then this would mean that no or very little part of the flux lines from one magnet could reach the body of the other magnet so any attraction force would be amply defeated by the ruling repel forces.
If you accept these, then the question is why would the magnetic lines of flux (that repel each other) cancel each other?
My answer is they cannot cancel each other in repel but they both get diverted (sideways) and compressed (I refer to your two bar magnets in repel example). If your answer is the fields get cancelled, that is fine with me but at least I tried and I see it differently.
I am making the assumption that some permanent magnets are manufactured so as to not be too strong. So in these magnets not all of the magnetic domains are strongly aligned and there is indeed available headroom for more magnetic flux to flow. I am not an expert but I think the phenomenon of a stronger magnet overcoming the repelling force of a weaker magnet is something that many people have observed. Think about inducing magnetism into the tip of a screwdriver. Then of course the magnetic material itself is something that can come in many different formulations.
I don't accept you notion of compression and diversion. I will just repeat that the magnetic fields from the two north poles of the two magnets just pass right through each other. The net magnetic field, the vector addition of these two separate field sources only appears to show "compression." It's simply an illusion. "Areas of high compression" are often areas where the magnetic field is actually quite weak, simply because the vector addition of the two fields because of the direction component is actually a subtraction. If you sketched out two north fields facing each other as if the opposing field did not exist, and then you sketched out the net field from the vector addition, you would clearly see that it looks exactly what the "compression field" looks like. So indeed, the magnetic north field from one magnet can pass directly into the north end of a facing magnet. If the facing magnet has some permeability headroom to offer to the other field then there will be more flux flow from the facing magnet.
Yes, I agree that in most cases when you push two magnet north faces together you feel repulsion. However, you know that magnetic material that has a soft plastic feel to it? I am not talking about kitchen cabinet magnets, which I believe are ceramic. The material I am talking about is related to those rubbery and pliable magnetic materials that might be used say for a magnetic clasp to say to keep a woman's purse closed or to keep an iPad case closed. They are some kind of a rubber/magnetic composite material. When you take that type of magnetic material, say in the shape of two rectangular blocks, and you bring opposite poles together, you feel a dramatic decrease in the repulsive force when the opposite pole faces are brought together. Some food for thought.
MileHigh
would anyone here be willing to post a summary , even two sentences, on what is going on with this bucking motor? For newbies who don't understand this clearly (like me) Please forgive my ignorance , seriously, I dont' want to take away anything from what you have all worked so hard to learn and achieve, by my coming here and not understanding this.
I have read through the thread and tried to take as much in as I can understand with my limited (0) knowledge of electricity and electronics. I gather that these microwave transformers are somehow capturing and storing previously wasted power in an innovative new way, and the task at hand is now to find ways to use it to propel the rotor in an innovative new way.
If so it is a promising new development, hoping to understand a bit more via plain English
about what is going on, and the potential applications for this, potential ways to achieve this, etc.
Quote from: Over Goat on November 30, 2015, 07:38:02 PM
would anyone here be willing to post a summary , even two sentences, on what is going on with this bucking motor? For newbies who don't understand this clearly (like me) Please forgive my ignorance , seriously, I dont' want to take away anything from what you have all worked so hard to learn and achieve, by my coming here and not understanding this.
I have read through the thread and tried to take as much in as I can understand with my limited (0) knowledge of electricity and electronics. I gather that these microwave transformers are somehow capturing and storing previously wasted power in an innovative new way, and the task at hand is now to find ways to use it to propel the rotor in an innovative new way.
If so it is a promising new development, hoping to understand a bit more via plain English
about what is going on, and the potential applications for this, potential ways to achieve this, etc.
The best Electric motor can be up to 95% efficient. The established Science says it can never be 100% efficient since a motor, transformer or other electromagnetic device have a counter reactive force which build up (Lenz law) that opposes the initial action (input power). Call it
action - reaction if you wish.
A very small number of us here at this forum do actual experiments to see if we can find a combination that may negate this counter opposing force.
It is well known in Science that when two poles (N_N or S-S) of the same polarity (aka bucking magnetic field) come together this counter opposing force is not present. However, in this situation the established Science says no useful work can be achieved from such an arrangement.
My motor design is an attempt to see if this is correct or not.
The use of microwave oven transformer (MOT) is not because they are special in any way. They are used for convenience as they are easy to open and offer you the two extremes in coils. One of heavy wire which can take current and one with fine wire which can make a strong magnetic field with high voltage. Each have their own use.
There is something else that I'm trying to use to assist a motor known as Inductive kickback (aka flyback).
This is the reverse effect of what happens to a coil when you cut off (switch off) the power (input current)
Even though your coil input had much current, everything changes when the coil is switched off... the counter effect is it kicks back a super fast and super high voltage spike.
So coils are interesting creatures, they have the ability to transform high current to high voltage. It is this fast high voltage I'm also trying to use to assist the motor.
Here is how... DC motors are switched on and off and have been designed to cancel this flyback spike from the motor coils since it causes many issues, mainly arcing of the switch contacts. So motor designers have been taught to just short out this flyback spike and problem solved.
I'm proposing to use this flyback and send it to a coil which would be more suited to use this high voltage, hence the fine wire high voltage coil.
By placing this flyback coil at the appropriate position (timing) in the motor I believe it could further assist the motor at no cost to the input. So if a motor can be made to be up to 95% efficient without using flyback, we only need it to make up for a 5% additional assistance and now we have cacaos in the established Science :o
Keep in mind I'm not just looking to use flyback to assist a motor but to also use bucking fields. So my motor design would be considered quite controversial to Science as I'm proposing to use two things they say cannot do work.
Hope this helps explain it in a simple way?
Luc
Quote from: gotoluc on November 30, 2015, 09:06:24 PM
DC motors are switched on and off and have been designed to cancel this flyback spike from the motor coils since it causes many issues, mainly arcing of the switch contacts. So motor designers have been taught to just short out this flyback spike and problem solved.
I'm proposing to use this flyback and send it to a coil which would be more suited to use this high voltage, hence the fine wire high voltage coil.
This "spike" represents the remaining energy, which is still stored in the motor winding. Wasting it, indeed is a waste.
Why not send this energy into a capacitor ?
Capacitors can hold this recovered energy efficiently and almost indefinitely, unlike most inductors, which continually suffer I
2R losses.
More
here (http://www.overunityresearch.com/index.php?topic=2684.msg43698#msg43698)
Hi Luc, You said;
"There is something else that I'm trying to use to assist a motor known as Inductive kickback (aka flyback).
This is the reverse effect of what happens to a coil when you cut off (switch off) the power (input current)
Even though your coil input had much current, everything changes when the coil is switched off... the counter effect is it kicks back a super fast and super high voltage spike"
I have seen the spike also by shorting the coil ends together but when being used as a generating coil.
Have you seen this or am I off track?
If you use blocking diodes at the coil leads with capacitors you automatically catch the spike and it is now stored power you can use.
artv
Quote from: shylo on December 01, 2015, 04:34:26 AM
Hi Luc, You said;
"There is something else that I'm trying to use to assist a motor known as Inductive kickback (aka flyback).
This is the reverse effect of what happens to a coil when you cut off (switch off) the power (input current)
Even though your coil input had much current, everything changes when the coil is switched off... the counter effect is it kicks back a super fast and super high voltage spike"
I have seen the spike also by shorting the coil ends together but when being used as a generating coil.
Have you seen this or am I off track?
If you use blocking diodes at the coil leads with capacitors you automatically catch the spike and it is now stored power you can use.
artv
I actually used that config and added a load via an inaccurate sparkgap and saw some very interesting things. Still utilising the flyback in a more direct way though. Probably wont be able to film until Thursdy
Quote from: verpies on December 01, 2015, 02:34:52 AM
This "spike" represents the remaining energy, which is still stored in the motor winding. Wasting it, indeed is a waste.
Why not send this energy into a capacitor ?
Capacitors can hold this recovered energy efficiently and almost indefinitely, unlike most inductors, which continually suffer I2R losses.
More here (http://www.overunityresearch.com/index.php?topic=2684.msg43698#msg43698)
Hi verpies,
thanks for your post and suggestion. I am trying to avoid storing in capacitors as from what I understand this would represent a conversion loss. Let me explain.
Please correct me if I am wrong but the way I see and understand it is, a coil (inductor) is truly a voltage device but current is needed to create a magnetic field. Capacitors are the opposite, they are current devices and are not known to create a magnetic field.
So if you want to make a motor you first have to start supplying current in your coil to build a strong magnetic field for the motor to do work. Then once the current is switched off the coil does something rather interesting as we all have observed, it naturally converted the current to a very fast and high voltage spike. That to me is telling us something.
Now if you try to capture that fast high voltage spike in a capacitor, will there not be a fight (delay) = heat losses? as capacitors don't accept fast high voltage spikes, so why try to do this, is it not a waste? why not send it to another coil which would perfectly accept this high voltage spike without any fight (losses) other then the coils wire resistance. The idea here is, the flyback is only a portion of the power we first imputed to the coil. Let's say we can get 50% of the power back from the flyback and we chose to send it to another coil which has double the inductance of the the first coil and this coil is strategically placed in the motor to further assist the rotor. Do you not think this would be a more efficient way to use flyback then to try to store it in a capacitor or a battery?
If you disagree please post your argument (in plain English) as to why you believe it not to be so.
Thanks for sharing
Luc
Quote from: shylo on December 01, 2015, 04:34:26 AM
Hi Luc, You said;
"There is something else that I'm trying to use to assist a motor known as Inductive kickback (aka flyback).
This is the reverse effect of what happens to a coil when you cut off (switch off) the power (input current)
Even though your coil input had much current, everything changes when the coil is switched off... the counter effect is it kicks back a super fast and super high voltage spike"
I have seen the spike also by shorting the coil ends together but when being used as a generating coil.
Have you seen this or am I off track?
If you use blocking diodes at the coil leads with capacitors you automatically catch the spike and it is now stored power you can use.
artv
Yes Shylo, I have experimented with coil shorting. The results are interesting and when you think of it mimics the similar effect of when a coil is powered and switched off, no?
Thanks
Luc
Quote from: gotoluc on November 30, 2015, 09:06:24 PM
The best Electric motor can be up to 95% efficient. The established Science says it can never be 100% efficient since a motor, transformer or other electromagnetic device have a counter reactive force which build up (Lenz law) that opposes the initial action (input power). Call it action - reaction if you wish.
A very small number of us here at this forum do actual experiments to see if we can find a combination that may negate this counter opposing force.
It is well known in Science that when two poles (N_N or S-S) of the same polarity (aka bucking magnetic field) come together this counter opposing force is not present. However, in this situation the established Science says no useful work can be achieved from such an arrangement.
My motor design is an attempt to see if this is correct or not.
The use of microwave oven transformer (MOT) is not because they are special in any way. They are used for convenience as they are easy to open and offer you the two extremes in coils. One of heavy wire which can take current and one with fine wire which can make a strong magnetic field with high voltage. Each have their own use.
There is something else that I'm trying to use to assist a motor known as Inductive kickback (aka flyback).
This is the reverse effect of what happens to a coil when you cut off (switch off) the power (input current)
Even though your coil input had much current, everything changes when the coil is switched off... the counter effect is it kicks back a super fast and super high voltage spike.
So coils are interesting creatures, they have the ability to transform high current to high voltage. It is this fast high voltage I'm also trying to use to assist the motor.
Here is how... DC motors are switched on and off and have been designed to cancel this flyback spike from the motor coils since it causes many issues, mainly arcing of the switch contacts. So motor designers have been taught to just short out this flyback spike and problem solved.
I'm proposing to use this flyback and send it to a coil which would be more suited to use this high voltage, hence the fine wire high voltage coil.
By placing this flyback coil at the appropriate position (timing) in the motor I believe it could further assist the motor at no cost to the input. So if a motor can be made to be up to 95% efficient without using flyback, we only need it to make up for a 5% additional assistance and now we have cacaos in the established Science :o
Keep in mind I'm not just looking to use flyback to assist a motor but to also use bucking fields. So my motor design would be considered quite controversial to Science as I'm proposing to use two things they say cannot do work.
Hope this helps explain it in a simple way?
Luc
yes, thanks, I know it must be frustrating to have to simplify it such, as 'lost in translation' is probably significant information, thank you for your willingness to do this.
This motor will re-use previously wasted energy, and also use the stored energy of magnets to push/pull or impel/propel itself and consume almost no more energy than that which was used to set in motion to begin with.
I am very new to this, this is my true curiosity, I am not trying to restate this or put words into anyone's mouth, it is my interpretation of what is being done here and trying not to waste anyone's time
the combination of the two is the key to this, because Luc has finally found a way to have the extra energy needed to access the stored energy of the magnets for mechanical work.
now the task at hand is to find the best magnets and the best coils or capacitors or any other way to capture, store and recycle the energy.
so do we google all the latest capacitor or transformer research and latest magnet research to try to find if someone already made new things which could be applicable, and try to get these things involved? I found new Cerium magnets which might be available relatively soon (they are light and cheap replacements of neodymium)
I hope this helps, the more people searching out what's already been recently done might help. Not to distract from Luc's concerted effort of course. thanks again for the reply, Luc
Quote from: webby1 on December 01, 2015, 10:56:24 AM
Would the second coil then not impede the current flow from the first
Yes, the connection of an empty second coil in series with the first, charged coil, would represent a huge current discontinuity, that that charged coil would not tolerate.
The EM pulse during the switch-over would radiate a lot of energy away and with practical components the resulting arcs would dissipate most of the energy stored in the first coil. The stress on the switch would be extreme, leading to its breakdown :(
Quote from: gotoluc on December 01, 2015, 09:17:42 AM
I am trying to avoid storing in capacitors as from what I understand this would represent a conversion loss.
I thought that you chose an Inductor --> Inductor energy transfer for other reasons than efficiency, because such transfer is less energy efficient than Inductor --> Capacitor.
Quote from: gotoluc on December 01, 2015, 09:17:42 AM
Let me explain. Please correct me if I am wrong but the way I see and understand it is, a coil (inductor) is truly a voltage device but current is needed to create a magnetic field. Capacitors are the opposite, they are current devices and are not known to create a magnetic field.
I can't agree with that.
Capacitors store energy as electric field, which is proportional to the voltage across them.
Inductors store energy as magnetic field, which is proportional to the electric current flowing through them.
This is not to be confused with what happens when their energy levels are changed:- An electric current must flow through a capacitor when it is charged (or discharged).
- A voltage must appear across an inductor when it is charged (or discharged).
Quote from: gotoluc on December 01, 2015, 09:17:42 AM
So if you want to make a motor you first have to start supplying current in your coil to build a strong magnetic field for the motor to do work.
Yes
Quote from: gotoluc on December 01, 2015, 09:17:42 AM
Then once the current is switched off the coil does something rather interesting as we all have observed, it naturally converted the current to a very fast and high voltage spike.
That voltage spike represents a rapid discharge of energy that was stored in the coil in the form of magnetic field.
That discharge does not have to be quick, but it can be when a high resistance, e.g. 100MΩ, is switched in series with the coil. With lower resistance the discharge will be slower and voltage will be lower, too.
While the coil is discharging, it acts as a current source.
Conversely, if a low resistance, e.g. 0.01Ω, is switched in parallel with a charged capacitor, then an analogous effect happens - the energy is quickly discharged as a high current spike.
While the capacitor is discharging, it acts as a voltage source.
If I were you, in case of the discharging capacitor, I would have written: "it naturally converted the voltage to a very fast and high current spike"
Quote from: gotoluc on December 01, 2015, 09:17:42 AM
Now if you try to capture that fast high voltage spike in a capacitor, will there not be a fight (delay) = heat losses?
There will be a delay proportional to the capacitance, but this delay will be negligible for small capacitances.
Quote from: gotoluc on December 01, 2015, 09:17:42 AM
...as capacitors don't accept fast high voltage spikes, so why try to do this, is it not a waste?
Yes, capacitors don't tolerate voltage discontinuities across them, but that is not applicable in this situation.
You seem to be assuming that an empty capacitor will experience a high voltage spike from a charged inductor, while the opposite is true, since the initial impedance of an empty capacitor presented to an inductor is 0Ω (short circuit) and that means that at the moment the cap is connected, the current flowing through it will be at maximum and the voltage across it will be at the minimum (zero).
Quote from: gotoluc on December 01, 2015, 09:17:42 AM
why not send it to another coil which would perfectly accept this high voltage spike without any fight (losses) .
Because switching an empty inductor, in series with a charged inductor represents a huge current discontinuity, which inductors don't tolerate...just like capacitors don't tolerate voltage discontinuities.
The resulting EM radiation would prevent a 100% efficient energy transfer even in theory with ideal components, while a 100% energy transfer is possible with an ideal coil and capacitor.
Also, the current in high inductance coil (many turns) changes much slower than current in low-inductance coil when all other conditions are being equal.
Quote from: gotoluc on December 01, 2015, 09:17:42 AM
...other then the coils wire resistance
Resistive losses exist in both the Inductor --> Inductor energy transfer and Inductor --> Capacitor energy transfer. Actually, the former has higher resistive losses when non-ideal contemporary components are considered.
Quote from: gotoluc on December 01, 2015, 09:17:42 AM
The idea here is, the flyback is only a portion of the power we first imputed to the coil. Let's say we can get 50% of the power back from the flyback and we chose to send it to another coil which has double the inductance of the the first coil and this coil is strategically placed in the motor to further assist the rotor. Do you not think this would be a more efficient way to use flyback then to try to store it in a capacitor or a battery?
No.
You seem to be under the impression that an inductor is better suited to receive energy from another inductor, compared to a capacitor, while the opposite is true. In fact Inductor --> Inductor, as well as, Capacitor --> Capacitor are the worst combinations for efficient energy transfers.
The mechanical analogy would be two coaxial flywheels (one spinning and the other stationary) suddenly slammed together ...such as in "dumping the clutch".
The best efficiencies are achieved by conjugate components, such as Inductor --> Capacitor, as well as, Capacitor --> Inductor.
The mechanical analogy would be a spinning flywheel being stopped by a suddenly connected (but relaxed) spiral torsion spring.
This preference for conjugacy manifests itself in the common life, too, ...such as: male & female pipe fittings ;)
Quote from: verpies on December 01, 2015, 11:22:51 AM
Yes, the connection of an empty second coil in series with the first, charged coil, would represent a huge current discontinuity, that that charged coil would not tolerate.
The EM pulse during the switch-over would radiate a lot of energy away and with practical components the resulting arcs would dissipate most of the energy stored in the first coil. The stress on the switch would be extreme, leading to its breakdown :(
Do you think this would happen with a diode is between the two coils so only the flyback (collapsing field) can go through?
Luc
Quote from: gotoluc on December 01, 2015, 12:45:08 PM
Do you think this would happen with a diode is between the two coils so only the flyback (collapsing field) can go through?
First draw a schematic depicting these two coils, power supply, diode and the switch.
Quote from: verpies on December 01, 2015, 12:54:44 PM
First draw a schematic depicting these two coils, power supply, diode and the switch.
Here at the first page of this topic was a post with video demo done by woopy exactly following my instructions. You will see a schematic in his video and the device operating. Please note that later on woopy confirmed a 0.3uf capacitor give best rpm results. I have also suggested the ideal capacitor value to be chosen with the motor under load as more torque may be favorable over rpm which would require a little more cap uf value.
Post: http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg465964/#msg465964 (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg465964/#msg465964)
Link to woppy's video: https://youtu.be/tag5OlvPi54 (https://youtu.be/tag5OlvPi54)
Then, yes - the reed switch will fail pretty quickly if the 1μF capacitor is deleted and most of the energy will be radiated away in the resulting arcs.
That capacitor has almost 0Ω impedance when the switch opens, while the second (high inductance assistance coil) has ∞ impedance when the switch opens. So this capacitor makes a world of difference. That's why the system slows down when the capacitor is disconnected.
Of course after the time equal to π(LC)½ the capacitor will spontaneously charge the 2nd coil and after another period π(LC)½ that coil will recharge the capacitor back (albeit with an opposite voltage polarity), and so on ...over and over - in the typical sinusoidal self-oscillation manner of an LC circuit.
If the rotor magnets approach and depart that "assistance coil" at the same frequency as the natural self-oscillating frequency of the LC circuit (119Hz for Woopy), then a second resonance (mechanical-LC) will appear and the entire efficiency of the motor will increase even further.
It would be better to move the 2nd coil behind a 2nd switch and energize it from the charged capacitor at a proper time and with greater timing control.
The voltage to which the capacitor becomes charged to (before the 2nd switch closes) will be a measure of the recovered energy according to the formula E=½CV2
The largest energy loss in this circuit is caused by the diode voltage drop (~0.6V) and that's why a synchronous MOSFET rectifier would increase the Inductor --> Capacitor energy transfer efficiency, even further.
About two weeks ago I spent a couple of days explaining more or less the same stuff about inductors and capacitors and I gave a half a dozen coil-flywheel explanations in great detail and at the expenditure of a lot of effort.
I clearly got almost nowhere with that thankless endeavour and I am almost certain that I will never make the considerable effort and invest the time to do it again.
Quote from: MileHigh on December 01, 2015, 02:00:15 PM
About two weeks ago I spent a couple of days explaining more or less the same stuff about inductors and capacitors and I gave a half a dozen coil-flywheel explanations in great detail and at the expenditure of a lot of effort.
I noticed and there will always be some individuals that this is wasted on. I just treat it as community service.
There seems to be this intuitive conviction that
the same type of components are the best energy receivers for each other, despite that life is full of examples demonstrating that opposites complement each other the best....like hand in glove.
BTW: I monkeyed off the flywheel examples from you. Grumage could probably even make instructional videos out of them if he did not have to pay for materials.
Hi MH,
I liked those analogies , I found them very helpful, Thank-you.
You also mentioned something about ; how the more components you add to a circuit the more loss you will suffer, is this correct?
I have to disagree.
artv
Quote from: MileHigh on December 01, 2015, 02:00:15 PMAbout two weeks ago I spent a couple of days explaining more or less the same stuff about inductors and capacitors and I gave a half a dozen coil-flywheel explanations in great detail and at the expenditure of a lot of effort.
I clearly got almost nowhere with that thankless endeavour and I am almost certain that I will never make the considerable effort and invest the time to do it again.
Quote from: verpies on December 01, 2015, 02:07:32 PMI noticed and there will always be some individuals that this is wasted on. I just treat it as community service.
There seems to be this intuitive conviction that inductors are the best energy receivers for inductors, despite that life is full of examples demonstrating that opposites complement each other the best....like hand in glove.
BTW: I monkeyed off the flywheel examples from you. Grumage could probably even make instructional videos out of them if he did not have to pay for materials.
You guys create your own situations with your know it all attitudes, then claim no one bothers listening to me :'( and look at all the time I've wasted when you actually have
not contributed anything that will improve what is being tested or worked on.
Do you for a minute think about how much time I've contributed in building and testing ideas because you know it all are convinced, or should I say brain washed there is no alternative?
Run along now and find someone else to play your gratifying games.
Don't bother posting a justification as they will be deleted. However, if you wish to contribute something that will improve what I'm working on, by all means do. Then you may get as many gratifications as I get for building and taking the time to share the real results ::)
Regards
Luc
Quote from: MileHigh on December 01, 2015, 02:00:15 PM
About two weeks ago I spent a couple of days explaining more or less the same stuff about inductors and capacitors and I gave a half a dozen coil-flywheel explanations in great detail and at the expenditure of a lot of effort.
I clearly got almost nowhere with that thankless endeavour and I am almost certain that I will never make the considerable effort and invest the time to do it again.
Hi MileHigh,
No harm intended but then what shall I say in connection with You? Because I already wrote to you on the permeability of permanent magnets and still you did not recall it at once, only later when I mentioned it. And also I mention this debate on the magnetic lines of flux for the facing like poles case, you disregard measured experience.
Here are two links on the repel pole bar magnets for your consideration. I consider the meaning of the texts as correct and they confirm other people's and my own experience.
http://www.homofaciens.com/technics-electrical-engineering-magnets_en_navion.htm (http://www.homofaciens.com/technics-electrical-engineering-magnets_en_navion.htm)
http://my.execpc.com/~rhoadley/field04.htm (http://my.execpc.com/~rhoadley/field04.htm)
I do not think I continue this "repel flux cancellation debate" with you, I clearly desribed my stance on this in my recent posts, it is up to you what you accept or do not accept of course, I understand this and also that you make assumptions and insist on examples which compare a weaker magnet with a stronger one: it is obvious that the stronger may change the weaker magnet's strength, poles etc. Why don't you consider testing some real magnets instead of doing false thought experiements? Again, no harm intended.
Gyula
Quote from: gyulasun on December 01, 2015, 04:58:23 PM
Hi MileHigh,
No harm intended but then what shall I say in connection with You? Because I already wrote to you on the permeability of permanent magnets and still you did not recall it at once, only later when I mentioned it. And also I mention this debate on the magnetic lines of flux for the facing like poles case, you disregard measured experience.
Here are two links on the repel pole bar magnets for your consideration. I consider the meaning of the texts as correct and they confirm other people's and my own experience.
http://www.homofaciens.com/technics-electrical-engineering-magnets_en_navion.htm (http://www.homofaciens.com/technics-electrical-engineering-magnets_en_navion.htm)
http://my.execpc.com/~rhoadley/field04.htm (http://my.execpc.com/~rhoadley/field04.htm)
I do not think I continue this "repel flux cancellation debate" with you, I clearly desribed my stance on this in my recent posts, it is up to you what you accept or do not accept of course, I understand this and also that you make assumptions and insist on examples which compare a weaker magnet with a stronger one: it is obvious that the stronger may change the weaker magnet's strength, poles etc. Why don't you consider testing some real magnets instead of doing false thought experiements? Again, no harm intended.
Gyula
Gyula & Luc
You are both correct-there is no cancellation of magnetic fields when like poles are pointed toward each other. In fact,something else happens-the opposite of cancellation. The two combined fields actually create a situation where the pull force from that bucking field is around 3 times that of what you would get from a single magnet alone !! but you have to be careful with what your seeing here in regards to how your test setup is setup !!. For example-if a single magnet was able to create a pull force !of say 100 grams! on a piece of magnetic material at a set distance of say 10mm,then by adding a second magnet of equal size and strength directly opposite to the first so as like poles face each other,then the pull force on that same piece of magnetic material will be 3x that given by the single magnet alone.
But what if the two magnets are placed in the same positions under the same test,but where as the two magnets now have the opposite poles facing each other?. Will the two magnets now still apply a pull force on the magnetic material that is 3x that of a single magnet alone ?. The answer will tell you if this bucking field setup is any better or more efficient than a !!standard!! setup.
I have done these test,but i will let everyone place there bets first.
Some of you will be surprised,and some will see nothing out of the ordinary.
Then there is the second thing to consider.
If we have two identical electromagnets/inductors as in Luc's setup,dose having those two inductors in bucking mode effect the amount of inductive kickback energy that can be gained from those two inductors,as apposed from that if they were in a standard (not bucking) mode,where unlike poles are formed when facing each other ?.
There are also things to consider when looking at Luc's setup--is there really a bucking field taking place-even though his two primary coils produce magnetic fields that are like poles facing each other?. It is my belief that with the E core configuration that there is no bucking fields in the true sense of the meaning-that being the two like fields distorting each other in the shape of a squashed balloon. I picture the fields to be that of the diagram bellow,where the fields must form loops due to the E core configuration,and the size of the stator block on the wheel.
Hi Brad,
Thanks and my two "cents" on your first question: it depends on what position the iron piece has with respect to the stator cores in the moment when the current to the coils is switched on and then what position it has when the current is off? When the iron piece starts approaching and entering into the gap between the facing cores say from the bottom and moves upward, first it sounds good if both flux poles can close via the iron because pulling from two unlike poles gives stronger pull in general versus a single pole pull BUT if current is still ON when the leading edge of the iron piece is about to leave the center column of the E core then a cogging possibility may occur between say the lower and middle prongs if the iron entered at 6 o'clock and moves upward.
In the position you drew the iron piece in your attached drawing, a second cogging point may occur if the current is still ON, and the flux could close via the iron piece at all the 4 routes you indicated. This would not be good to let it happen.
So while I agree with the quasi 3 times more force with the 2 like poles presence (when no opposite poles are present) but when the latter are present and the iron piece is not yet started to enter the gap then the 3 times more pull force may change to less. Thanks for pointing to the presence of the opposite pole.
I will try to answer the rest tomorrow, I have to finish now.
Gyula
Thanks Brad for taking the time to post your opinion and test experience.
I do agree that my design may not necessarily represent a true bucking coil configuration. Only real tests will prove if it can work in this configuration or not.
I've just finished building the higher RPM 2nd test device and presently testing and fine tuning.
A video demo should be released in the next day or two.
Stay tuned for the Bucking Field truth 8)
Luc
Gyula,
From your first link:
QuoteThe field strength around two magnets arranged in parallel is boosted if the north poles are pointing into the same direction. In an aintiparallel arrangement (the north pole of one magnet is pointing into the direction of the south pole of the second magnet) the field strength is weakened (nearly to zero at greater distances).
What do you think "boosted" and "weakened" mean in the above quote? They are just euphemisms for vector addition and vector subtraction. Your link is at too much of an entry level to use the proper terminology. Look at the three screen captures I attached.
So, what happens when two like poles face each other? Does common sense get thrown out the window and instead the magnetic fields are "diverted" and "compressed?" Or is it something like as long as the poles are not facing each other, then there is vector addition, but when the poles are facing each other there is "diversion and compression?" If that's true, when does that transition from vector addition to "diversion and compression" take place? At what angle between the facing poles? At what distance? At what strength?
Do you see the trap that you seem to be falling into? There is vector addition -
period. Like I already told you the net magnetic field after the vector addition is only creating the
appearance, or let's say "
illusion" of "diversion and compression."
MileHigh
Hi Brad,
Here is my answer to your second question, I qoute your question:
"Then there is the second thing to consider.
If we have two identical electromagnets/inductors as in Luc's setup, does having those two inductors in bucking mode effect the amount of inductive kickback energy that can be gained from those two inductors, as apposed from that if they were in a standard (not bucking) mode, where unlike poles are formed when facing each other ?"
Answer: The paralled bucking coils have a residual inductance value which is less than any of the individual coils and additionally depends on the air gap between the facing cores too (the smaller the gap, the lower the resulting inductance is).
So the residual inductance as Luc measured was changing roughly between 1.2 mH and 2.4 mH as it moves in the gap between the ON and OFF moments. The latter value (2.4 mH) is to be considered for estimating the inductive kickback energy I think.
In the not bucking mode, even if the gap is the same as in the bucking mode, the resulting inductance of the coils would still be higher than the 2.4 mH at switch-off, so higher inductance would result in higher stored hence collapsing energy (assuming identical input currents for both cases).
So in brief, the inductors in bucking mode affect (reduce) the amount of inductive kickback versus the non bucking inductors I think.
Gyula
Quote from: gyulasun on December 02, 2015, 04:38:52 PM
Hi Brad,
Here is my answer to your second question, I qoute your question:
"Then there is the second thing to consider.
If we have two identical electromagnets/inductors as in Luc's setup, does having those two inductors in bucking mode effect the amount of inductive kickback energy that can be gained from those two inductors, as apposed from that if they were in a standard (not bucking) mode, where unlike poles are formed when facing each other ?"
Answer: The paralled bucking coils have a residual inductance value which is less than any of the individual coils and additionally depends on the air gap between the facing cores too (the smaller the gap, the lower the resulting inductance is).
So the residual inductance as Luc measured was changing roughly between 1.2 mH and 2.4 mH as it moves in the gap between the ON and OFF moments. The latter value (2.4 mH) is to be considered for estimating the inductive kickback energy I think.
In the not bucking mode, even if the gap is the same as in the bucking mode, the resulting inductance of the coils would still be higher than the 2.4 mH at switch-off, so higher inductance would result in higher stored hence collapsing energy (assuming identical input currents for both cases).
So in brief, the inductors in bucking mode affect (reduce) the amount of inductive kickback versus the non bucking inductors I think.
Gyula
OK,so piece by piece we put this together.
If we have put in the same amount of energy in each case,but we capture less electrical energy out from the bucking coil's-where did that lost energy go?-it cant just disappear. I also stated that two magnets in a bucking configuration triples the pull force of that of 1 magnet on a magnetic material that is close by. Is it conceivable that this lost inductive kickback energy from the bucking coils was transformed into mechanical energy by way of a larger/stronger magnetic field produced by the bucking coils?.
The good news is,i have all the equipment to conduct a controlled test,where as we can put to rest once and for all if a bucking magnetic field is more efficient than that of a standard magnetic field.
My only concern with such a setup where soft iron laminates are used,is the chance that the core will slowly become magnetized it self. So throughout the test, i will switch polarities every so often to try and reduce the residual magnetism in the cores.
My setup will be smaller than Luc's MOT core setup,as i do not have two MOT's the same size. I do how ever have bulk amount of surplus transformers that have the bridged E core transformer inside them. It will be an all day project,but set up right,we will be able to carry out many controlled tests to see which gives us our best bang for buck-so to speak. I guess the best way to do this would be to design some sort of pulse motor based around the E core transformer setup,and use a small generator as the bearing for the rotor. This way we can apply a known load across the generator,and carry out the many tests to obtain the best results.
Ok,enough said--time to go build.
Brad
Get rid of all steel
artv
Quote from: shylo on December 02, 2015, 07:00:34 PM
Get rid of all steel
artv
Silicon steel/soft iron is the best material to use in low frequency applications.
Quote from: tinman on December 02, 2015, 06:23:54 PM
...
If we have put in the same amount of energy in each case,but we capture less electrical energy out from the bucking coil's-where did that lost energy go?-it cant just disappear. I also stated that two magnets in a bucking configuration triples the pull force of that of 1 magnet on a magnetic material that is close by. Is it conceivable that this lost inductive kickback energy from the bucking coils was transformed into mechanical energy by way of a larger/stronger magnetic field produced by the bucking coils?.
....
Hi Brad,
Unfortunately I forgot to consider in my answer yesterday that if in the not bucking coils mode you asked, when the resulting inductance would be say indeed higher than in the bucking mode, then the coils AC impedance would also be higher, thus to insure the same input current, the input voltage should be increased and this would involve higher input energy. So we need to consider this when we wish to compare the bucking coils with the not bucking coils performance and answer your question.
I do not think that transformer laminates you would use for the tests will become magnetized in time (unless very strong peak currents would be involved on the long run).
Gyula
Hi everyone,
as promised here is v.2 of the Bucking Field Reluctance Motor. You may be surprised at what happened.
Link to video: https://www.youtube.com/watch?v=5f9TQ8AXglg
Luc
Quote from: gotoluc on December 03, 2015, 05:01:41 PM
Hi everyone,
as promised here is v.2 of the Bucking Field Reluctance Motor. You may be surprised at what happened.
Link to video: https://www.youtube.com/watch?v=5f9TQ8AXglg (https://www.youtube.com/watch?v=5f9TQ8AXglg)
Luc
Good link bait headline! I'll be watching tonight.
Here is the first part of the bucking field motor test bench build.
Sorry about the small product review lol.
https://www.youtube.com/watch?v=Wj_Ecj2NZj0
looks like a have a playlist for tonight
Hey Luc
Nice setup. Try something here....
Use a full wave bridge rectifier for the flyback. At first it might seem odd to do because that would mean that the input to the coil would be transferred to the bemf cap. But once charged the caps voltage is mostly higher than the input. Supposedly this is what Bedini did and that was the explanation I understood.
It may help ease up on all that oscillation after switch off also. There may be an oscillation in the coils after shutoff that are messing with the transistor circuits.
But when you get it back together, try the bridge and see what happens. It may help get more into the cap and possibly help kill off the oscillations.
Mags
Quote from: tinman on December 03, 2015, 05:45:12 PM
Here is the first part of the bucking field motor test bench build.
Sorry about the small product review lol.
https://www.youtube.com/watch?v=Wj_Ecj2NZj0 (https://www.youtube.com/watch?v=Wj_Ecj2NZj0)
Great job mate!
Keeping the two inductors identical is important. Good to have isolated flyback coils also.
Believe it or not, I had the same idea to use two identical dc motors as bearings generators to supports for the rotor if I was to build v.3
I have not demonstrated the direction the I cores are in my v.2 build. It's 180 degrees from v.1
The I core have more then twice the attraction pull then from v.1
See attached drawing I made for you.
Luc
Quote from: shylo on December 02, 2015, 07:00:34 PM
Get rid of all steel
That would leave only air coils in the stator and a ferrite in the rotor.
Quote from: gyulasun on December 02, 2015, 04:38:52 PM
So the residual inductance as Luc measured was changing roughly between 1.2 mH and 2.4 mH as it moves in the gap between the ON and OFF moments.
That's only 1:2
inductance ratio between these "ON and OFF moments".
With air coils, that inductance ratio would be ~100x larger.
Indeed, that would be very good for efficiency.
Quote from: Magluvin on December 03, 2015, 06:42:22 PM
Hey Luc
Nice setup. Try something here....
Use a full wave bridge rectifier for the flyback. At first it might seem odd to do because that would mean that the input to the coil would be transferred to the bemf cap. But once charged the caps voltage is mostly higher than the input. Supposedly this is what Bedini did and that was the explanation I understood.
It may help ease up on all that oscillation after switch off also. There may be an oscillation in the coils after shutoff that are messing with the transistor circuits.
But when you get it back together, try the bridge and see what happens. It may help get more into the cap and possibly help kill off the oscillations.
Mags
Humm :-\ ... that is interesting. Too bad there's nothing left to test ;D
I'm not rebuilding this v.2, it served its purpose of observing an even power input.
Thanks for sharing the idea.
Luc
Quote from: verpies on December 03, 2015, 06:51:02 PM
That would leave only air coils in the stator and a ferrite in the rotor.
That's only 1:2 inductance ratio between these "ON and OFF moments".
With air coils, that inductance ratio would be ~100x larger.
Indeed, that would be very good for efficiency.
Really ??? ... Bucking Air Coils would make this more efficient?
That I have a hard time visualizing
Please share an example
Thanks
Luc
Quote from: gotoluc on December 03, 2015, 07:04:43 PM
Really ??? ... Bucking Air Coils would make this more efficient?
More efficient means more recovered electric energy relative to the electric input energy.
It does not mean that the motor would have a higher torque. In fact the torque would be quite smaller.
It's all about maximizing that inductance ratio. If you have an inductance meter, notice how large this ratio becomes for an air coil.
Just divide the inductance without the rotor nearby and with the rotor nearby.
Quote from: gotoluc on December 03, 2015, 05:01:41 PM
as promised here is v.2 of the Bucking Field Reluctance Motor. You may be surprised at what happened.
Link to video: https://www.youtube.com/watch?v=5f9TQ8AXglg (https://www.youtube.com/watch?v=5f9TQ8AXglg)
That MOSFET is oscillating because that HV inductive spike from the motor winding is turning it ON when it is supposed to turn off. This effect is known as Miller Turn-on or Miller Oscillation and happens when the voltage on the drain is changing too quickly for the MOSFET to handle.
There are two ways to solve it:
1) The brute force method: Drive the gate of the MOSFET with a strong driver that can sink many amps of current from the gate, such as the UCC27511.
2) The elegant method: Empty the capacitor before each HV spike arrives from the motor winding.
Quote from: verpies on December 03, 2015, 07:06:14 PM
More efficient means more recovered electric energy relative to the electric input energy.
It does not mean that the motor would have a higher torque. In fact the torque would be quite smaller.
It's all about maximizing that inductance ratio. If you have an inductance meter, notice how large this ratio becomes for an air coil.
Just divide the inductance without the rotor nearby and with the rotor nearby.
I understand but I didn't know for sure that an Inductance boost during on time will make the flyback have more power. This is what Steorn Orbo was claiming but I never saw any proof to it. Is this what you're sawing?
Luc
Quote from: verpies on December 03, 2015, 07:11:06 PM
That MOSFET is oscillating because that HV inductive spike from the motor winding is turning it ON when it is supposed to turn off. This effect is known as Miller Turn-on or Miller Oscillation and happens when the voltage on the drain is changing too quickly for the MOSFET to handle.
There are two ways to solve it:
1) The brute force method: Drive the gate of the MOSFET with a strong driver that can sink many amps of current from the gate, such as the UCC27511.
2) The elegant method: Empty the capacitor before each HV spike arrives from the motor winding.
I'll test "The elegant method" next time this issue comes up.
Good to know! thanks for sharing
Luc
Just another idea to increase the efficiency of this experiment.
the electromagnet has to be polarized as per diagram .drilling the
hole is ease .place the core in a vise and tighten the vise as much
as you can then drill the hole or holes..then place the cerramic
magnet as diagram. you can use you scale to check if there is a magnetic
pull increase accordingly . Good luck. LUCKY LUKE..
Quote from: seychelles on December 03, 2015, 07:25:19 PM
Just another idea to increase the efficiency of this experiment.
the electromagnet has to be polarized as per diagram .drilling the
hole is ease .place the core in a vise and tighten the vise as much
as you can then drill the hole or holes..then place the cerramic
magnet as diagram. you can use you scale to check if there is a magnetic
pull increase accordingly . Good luck. LUCKY LUKE..
Thanks seychelles,
I don't know if I missing the benefit of your idea but as soon as you add magnets you are now adding potential of this device to generate electricity (generator effect) no?
Please explain if I'm not understanding
Luc
Quote from: gotoluc on December 03, 2015, 07:20:46 PM
I understand but I didn't know for sure that an Inductance boost during on time will make the flyback have more power. This is what Steorn Orbo was claiming but I never saw any proof to it. Is this what you're sawing?
Yes, but you can leave that improvement for the last supper.
Quote from: gotoluc on December 03, 2015, 07:24:42 PM
I'll test "The elegant method" next time this issue comes up.
If you can get four of these HEF4047B cheap chips (CD4047) then I could draw you a schematic that will allow you to trigger on the
edge of that white tape and have an adjustable ON pulse width (independent of the tape width) as well as a signal for emptying that capacitor later.
All I would need to know is whether that pulse from your opto is positive going or negative going....
Quote from: verpies on December 03, 2015, 08:13:10 PM
Yes, but you can leave that improvement for the last supper.
Okay, then how about you share on the results you saw in my last video demo before it broke apart. Do you think this design idea has any value?... what changes would you make if it needs any? etc.
Thanks
Luc
Quote from: verpies on December 03, 2015, 08:20:07 PM
If you can get four of these HEF4047B cheap chips (CD4047) then I could draw you a schematic that will allow you to trigger on the edge of that white tape and have an adjustable ON pulse width (independent of the tape width) as well as a signal for emptying that capacitor later.
All I would need to know is whether that pulse from your opto is positive going or negative going....
I have some MC14013b, LM3915, IR2103 on hand, any good?
The opto is a positive pulse.
Thanks
Luc
The answer to you answer is below.
An increased in magnetic field = increase in torque and BEMF..
Hi everyone,
I think the limitation we have is no one really knows for sure the exact behavior of a non CEMF motor. What we have heard most often is, the current shouldn't change no matter the load. However, mine does change a little, so does this mean it's not a non CEMF motor?
This was bothering me so I decided to take an off the shelf DC motor and observe its behavior when set to as close to the same conditions as my v.2 bucking motor.
Well, to my surprise I can't see much of a difference once everything is setup to the same.
You be the judge: https://www.youtube.com/watch?v=SozrH_IyPXc (https://www.youtube.com/watch?v=SozrH_IyPXc)
Let me know if you think I'm missing something
Thanks
Luc
Quote from: gotoluc on December 03, 2015, 08:29:18 PM
Okay, then how about you share on the results you saw in my last video demo before it broke apart. Do you think this design idea has any value?... what changes would you make if it needs any? etc.
The build is fine, except for the rotor design.
The electric current profile and its covariance with the supply voltage is nothing unusual and it is accordance with
this analysis (http://www.overunityresearch.com/index.php?topic=2684.msg43698#msg43698) but it has potential to be interesting because the operating principle is different than a common DC motor.
My changes to the mechanical design:
- I would embed the I-cores in a solid wheel by routing out sockets for them (stronger and less windage),
My changes to the electronic design:
- Fully electronic adjustment of the driving pulse position instead of the mechanic adjustment
- Fully electronic adjustment of the driving pulse width
- Deleting that inductive load resistor across the capacitor and substituting it with a reed relay in series with an automotive light bulb for emptying the capacitor (could be a different resistor but the bulb is more fun)
- Judging how much energy is recovered from the motor winding by measuring the peak voltage to which the capacitor gets charged, as this is very precise and dependable method if the capacitor is good, meaning: it does not leak or soak.
- not sharing the same power supply between the motor winding and the electronics...or at least a separate voltage regulator.
- Using a real MOSFET gate driver.
...and much more refinements later (such as closing flux paths).
Quote from: gotoluc on December 03, 2015, 08:35:28 PM
I have some MC14013b, LM3915, IR2103 on hand, any good?
If I turn on my McGyver mode, then I can put sth together with these parts.
Quote from: verpies on December 04, 2015, 04:43:38 AM
The build is fine, except for the rotor design.
Agree, I knew that was the week part and why I could predict it could brake apart at any moment. It was a quick build so not to invest too much time and material. If successful I would take the time to embed the cores in a rotor just as you described below.
Quote from: verpies on December 04, 2015, 04:43:38 AM
The electric current profile and its covariance with the supply voltage is nothing unusual and it is accordance with this analysis (http://www.overunityresearch.com/index.php?topic=2684.msg43698#msg43698) but it has potential to be interesting because the operating principle is different than a common DC motor.
Also agree with you here. However, I thought it would of had more potential. Maybe because I changed the I cores entry by rotating the E core by 180 degrees to favor torque. could that of changed the results?
Quote from: verpies on December 04, 2015, 04:43:38 AM
My changes to the mechanical design:
- I would embed the I-cores in a solid wheel by routing out sockets for them (stronger and less windage),
Yes, agree and exactly what I would of done for v.3
Quote from: verpies on December 04, 2015, 04:43:38 AM
My changes to the electronic design:
- Fully electronic adjustment of the driving pulse position instead of the mechanic adjustment
- Fully electronic adjustment of the driving pulse width
- Deleting that inductive load resistor across the capacitor and substituting it with a reed relay in series with an automotive light bulb for emptying the capacitor (could be a different resistor but the bulb is more fun)
- Judging how much energy is recovered from the motor winding by measuring the peak voltage to which the capacitor gets charged, as this is very precise and dependable method if the capacitor is good, meaning: it does not leak or soak.
- not sharing the same power supply between the motor winding and the electronics...or at least a separate voltage regulator.
- Using a real MOSFET gate driver.
...and much more refinements later (such as closing flux paths).
I would like to try all the above but I'll need help on the circuit side which is not part of my skills.
Quote from: verpies on December 04, 2015, 06:26:41 AM
If I turn on my McGyver mode, then I can put sth together with these parts.
Okay McGyver, do what you can and I'll try the circuit out. I've got lots of IRF3710 as low R switches.
Thanks for your help
Luc
Do you have any small transistors?
The LM3915 has constant current outputs which makes it hard to make a sequencer out of it while using a timing capacitor being charged at different rates.
Quote from: verpies on December 04, 2015, 02:15:04 PM
Do you have any small transistors?
The LM3915 has constant current outputs which makes it hard to make a sequencer out of it while using a timing capacitor being charged at different rates.
Yes, I have 4 of MPS8599, 4 of MPSA06, 4 of MPS222A, 4 of 2N4401 and 3 of 2N3904
I also found 1 of NTE4049 1 of CD4043BCN and 1 of SN74ACT08N
I have some opto 4 of 4N35 and 2 of CNY17-3
and found lots of mosfet drivers MIC4451
Hope it helps?
Luc
The below comment was posted youtube on my Bucking Field Reluctance Motor vs off the shelf DC Motor (https://www.youtube.com/watch?v=SozrH_IyPXc) test.
I think he has a point. I was looking mostly at the current but it is true that The bucking field motor has a perfectly stable and flat voltage.
What do you guys think?
Luc
YouRogga (https://www.youtube.com/user/YouRogga) Hmm, but the voltage drops down about 20% during the pulse as the current increases with the standard DC motor. Your motor shows a constant voltage during the pulse, despite the increase of current. Isn't that significant? I don't understand why the voltage is constant in your motor, I think it should drop when the current increases but it doesn't. Why?
Quote from: gotoluc on December 04, 2015, 05:28:13 PM
The below comment was posted youtube on my Bucking Field Reluctance Motor vs off the shelf DC Motor (https://www.youtube.com/watch?v=SozrH_IyPXc) test.
I think he has a point. I was looking mostly at the current but it is true that The bucking field motor has a perfectly stable and flat voltage.
What do you guys think?
Luc
YouRogga (https://www.youtube.com/user/YouRogga) Hmm, but the voltage drops down about 20% during the pulse as the current increases with the standard DC motor. Your motor shows a constant voltage during the pulse, despite the increase of current. Isn't that significant? I don't understand why the voltage is constant in your motor, I think it should drop when the current increases but it doesn't. Why?
All will be answered soon Luc-just need to finish my test bed(today hopefully).
But for now,ask yourself this-->what is different between the two motors? What dose one have that the other dose not?
You will then have your answer. ;)
Brad
Quote from: tinman on December 04, 2015, 06:37:02 PM
All will be answered soon Luc-just need to finish my test bed(today hopefully).
But for now,ask yourself this-->what is different between the two motors? What dose one have that the other dose not?
You will then have your answer. ;)
Brad
Magnets?
Hi guys, Verpies said;
"Deleting that inductive load resistor across the capacitor and substituting it with a reed relay in series with an automotive light bulb for emptying the capacitor (could be a different resistor but the bulb is more fun)"
Instead of the bulb use it as the input supply.
And the magnets are a must.
artv
Quote from: shylo on December 05, 2015, 03:43:28 AM
Instead of the bulb use it as the input supply.
Of course, but that is a next stage of the project.
Quote from: shylo on December 05, 2015, 03:43:28 AM
And the magnets are a must.
Getting rid of a movable soft ferromagnetics and replacing them all with magnets would alter motor's Modus Operandi: the variable reluctance and variable inductance ...effectively converting it into the conventional DC motor we all know.
Hi all
I am trying to answer my own questions of the beginning of this thread. That is, is a magnetic field dissipated in the work it does on a passing magnet?
https://youtu.be/vLiUkno3tSY
So it seems in the video that the magnetic fiels needs power to be created, but once it is created, the work it does or not seems to have no need of more or less power. But when it collapses we can recover a part of the energy which was dedicated to its creation.
So is a created magnetic field really almost free lunch on the torque it produces?
Not sure to be very clear here.
Laurent
Quote from: woopy on December 05, 2015, 12:03:55 PM
Hi all
I am trying to answer my own questions of the beginning of this thread. That is, is a magnetic field dissipated in the work it does on a passing magnet?
https://youtu.be/vLiUkno3tSY (https://youtu.be/vLiUkno3tSY)
So it seems in the video that the magnetic fiels needs power to be created, but once it is created, the work it does or not seems to have no need of more or less power. But when it collapses we can recover a part of the energy which was dedicated to its creation.
So is a created magnetic field really almost free lunch on the torque it produces?
Not sure to be very clear here.
Laurent
Hey Woopy
Once the rotor is up to speed, the swinging magnet is pulled up toward the coil but it isnt moving anymore. Its probably vibrating and may be difficult to measure that energy of the vibrations.
When I read the post I anticipated that the test was to see if the drive coil input would be affected by the rotor magnets as it spins, and wondered how you would make the comparison. But Im not sure the swinging magnet test would affect the motor input much in this case because the secondary coil in that circuit is supposedly only getting the bemf from the drive coil. Maybe this test is too far down the chain to say its conclusive. What would be better is to measure the capacitor that is dumping into the secondary coil and see if the swinging magnet makes changes there, being that cap is the primary source for that coil.
Mags
What came to mind on how the test might be before I saw the vid was, hmm, how?
I thought, well the coil needs to be operating whether the rotor is there or not. So one would have to have a timed input circuit that emulates the input to the coil as the motor would produce when present and in motion, then compare measurements. But it gets deeper. Now do measurements with the rotor with a load where the rpms are lower and we are actually doing work instead of the rotor running itself at max rpm. Then do the input emulation again simulating the reed switch timing. If there is no change or a change for the better, then this could be very interesting.
I really liked the idea Luc found with pulsing the motor instead of straight dc didnt change the input when loaded. That is crazy impressive. To me anyway. May be interesting to do that test with one measurement of the motor free wheeling and one measurement with the motor shaft locked in place. Use lower input to not burn the coils, just measure the differences.
Mags
Hi Laurent,
My thoughts:
Of course the movement of the permanent magnet in front of the input coil does some work by being attracted towards the coil while it is drawn back by its elastic suspension and partly by gravitation. But this work done seems to be very little I agree with Magluvin.
1) I wonder whether the input current would change when you use a piece of ferrite or laminated soft iron core instead of the permanent magnet. In this case the inductance of the input coil should change as the core would "wobble" in front of it. (it is possible that the magnet may influence the permeability of the I core too what you mention to have in the input coil but this may also be only a small change) However, with a ferrite piece in front of the coil instead of the perm. magnet, the wobbling of this core would possible 'average out' to be a tiny trembling, so what we could sense would possibly be whether the presence of such core would make a difference or not
2) I wonder how the input current would change if at all, when you firmly hold the permanent magnet in your fingers (or fix it mechanically and do not let it move at all) but the magnet would still under a strong attraction (very close to the coil). This way the work done by the fields should increase I suppose and this may or may not have influence on the input current. I think though that this may only have the same effect as the case when the magnet is let trembling with a little 'averaged' amplitude against its elastic suspension.
3) I wonder how the input current would change if you use a much stronger permanent magnet (if you have such at hand, that is).
Thanks for showing this test.
Gyula
Quote from: woopy on December 05, 2015, 12:03:55 PM
Hi all
I am trying to answer my own questions of the beginning of this thread. That is, is a magnetic field dissipated in the work it does on a passing magnet?
https://youtu.be/vLiUkno3tSY (https://youtu.be/vLiUkno3tSY)
So it seems in the video that the magnetic fiels needs power to be created, but once it is created, the work it does or not seems to have no need of more or less power. But when it collapses we can recover a part of the energy which was dedicated to its creation.
So is a created magnetic field really almost free lunch on the torque it produces?
Not sure to be very clear here.
Laurent
Merci Laurent pour continuer les tests.
Vos questions sont très valide!
Ceci est exactement ce que je propose depuis le premier jour. Une fois le champ magnétique est créé dans la bobine et le commutateur est ouvert, nous pouvons utiliser le champ magnétique (flyback) pour créer un deuxième champ magnétique pour faire plus de travail.
Beaucoup diront, la flyback ne contient pas autant d'énergie que l'entrée initiale à la bobine et fera un champ magnétique faible. Cela est vrai
si, le second champ magnétique entre une bobine de la même valeur (inductance) que la première. Cependant, en utilisant une bobine d'impédance plus élevé, le fyback peut créer un champ magnétique aussi forte que la première bobine en raison que c'est l'inductance qui va dicter la force du champ magnétique dans la seconde bobine.
Il doit y avoir un équilibre idéal dans le choix de valeur d'inductance pour la deuxième bobine. Exemple, si le flyback est seulement 33% de la puissance que nous avons commencé avec la première bobine, est ce que la deuxième bobine aura besoin de 3 fois l'inductance pour faire un champ magnétique aussi forts? et si nous avon un commutateur a la deuxième bobine, pourrions-nous envoyer à nouveau son flyback à une troisième bobine et ainsi de suite?
Comme on le voit, il ya encore beaucoup de questions? je suis heureux que vous recherchez ces options.
Ce que Magluvin et Gyula essaient de vous dire est, l'aimant ne se déplace pas assez pour faire votre test concluant. Peut-être que si vous attachez un élastique de sorte que l'aimant a plus de mouvement.
Merci pour le partage. Je suis toujours heureux de voir vos tests vidéo.
Luc
Thanks Laurent for continuing the tests.
Your questions are very valid!
This is exactly what I've been proposing since day one. Once a magnetic field is created in the coil and the switch is opened, we can use it's collapsing magnetic field (flyback) to create a second magnetic field to do more work.
Many will say, the flyback does not contain as much energy as the initial input to the coil and therefore will make a weak magnetic field. This is true,
if the second magnetic field enters a coil of the same value as the first. However, by using a higher impedance coil it can create as strong of a magnetic field as the first coil did since inductance is what will dictate the force of the magnetic field in the second coil.
There must be an ideal balance in choosing the ideal assisting coil inductance value. lets say the flyback from the first coil is only 33% of the power we imputed, does this mean the second coil need to be 3 times the inductance of the primary coil to make as strong of a magnetic field? and if we switched off the second coil at its peak, could we again send its flyback to a third coil and so on?
As we see there are still many unanswered questions.
What Magluvin ans Gyula are trying to tell you is, the magnet is not moving enough to make your test conclusive. Maybe if you attached an elastic so the magnet has more of an in and out movement.
I am always happy to see your video tests.
Thanks for sharing and looking forward to your next tests.
Luc
Hi Woopy, Nice work, put a way to interupt the supply to your coil so field rises and collapses ,(on &off) also put another rotor with all the same poles facing out, to replace the swinging magnet, that rotor will turn if timed properly.
Right now it's just being a "on always electro-magnet"
The on & off will replace mister hand.
artv
Quote from: woopy on December 05, 2015, 12:03:55 PM
So it seems in the video that the magnetic field needs power to be created,
Yes, but more precisely: an inductor needs energy to build up the current and magnetic flux.
Quote from: woopy on December 05, 2015, 12:03:55 PM
but once it is created, the work it does or not seems to have no need of more or less power.
Yes, the energy does not escape the coil if it is shorted by a jumper or a C.V. power supply.
Any resistance in the coil's circuit (including the resistance of its windings) causes the energy in the coil to leak out (as heat).
The more resistance, the faster the energy leaks out. In a superconductive coil, the energy never leaks out and the coil acts as a permanent magnet. See
this video (https://www.youtube.com/watch?v=uL4pfisCX14).
Quote from: woopy on December 05, 2015, 12:03:55 PM
But when it collapses we can recover a part of the energy which was dedicated to its creation.
Yes, but if the coil's circuit had no resistance then we could recover 100% of the energy that we had put in.
If the coil's circuit has resistance (as most coils do) then we can recover more energy than gets dissipated in the resistance, only when we charge the coil for a shorter time than 0.5757 Tau (assuming equal discharge rate).
This article (http://www.overunityresearch.com/index.php?topic=2684.msg43692#msg43692) explains why.
Quote from: woopy on December 05, 2015, 12:03:55 PM
So is a created magnetic field really almost free lunch on the torque it produces?
Not when a coil attracts a permanent magnet, as in your video.
This is because an approaching magnet increases the flux through the coil and this induces current in the coil, according to Mr. Lenz.
This current gets subtracted from the current that was already circulating in the coil. Your video does not show this well because you are continually replenishing the current by the power supply.
If you connect a current sensor (e.g. a CSR) to an empty coil and approach a magnet to that coil, then you will observe a current flowing through that coil. This is the same current that would get subtracted from a coil that was not empty (already energized earlier).
If you did your experiment in proper isolation, that is:
- energized the coil
- prevented any energy leaks and external energy inputs (isolation).
- attracted a magnet (doing mechanical work)
- recovered energy from the coil
...then the recovered electric energy would be smaller than the initial input energy.
When a coil does not attract a hard magnet but a piece of soft steel or ferrite instead, then the behavior is different, because a magnet does not change the inductance of an air coil and soft steel, does. This is described in detail
here (http://www.overunityresearch.com/index.php?topic=2684.msg43698#msg43698).
Quote from: verpies on December 06, 2015, 04:18:56 AM
This is because an approaching magnet increases the flux through the coil and this induces current in the coil. This current gets subtracted from the current that was already circulating in the coil. Your video does not show this well because you are continually replenishing the current by the power supply.
Thanks Verpies, probably a dumb question, but why does the current that is being created by the moving magnet "gets subtracted" . I guess I'm asking what happens to the current created by a moving magnet in the coil that is being pulsed by DC?
Quote from: Jimboot on December 06, 2015, 04:31:55 AM
...but why does the current that is being created by the moving magnet "gets subtracted".
Because the ultimate "goal" of any coil is to keep the magnetic flux constant (the flux that penetrates the coil).
If the coil is ideal (like in
this video (https://www.youtube.com/watch?v=uL4pfisCX14)) then the coil always succeeds in keeping the flux constant, but if the coil's circuit is resistive then the energy will eventually leak out and the coil will not be able to maintain the constant flux. This is shown in
this video (https://www.youtube.com/watch?v=wUaqXk6axOo).
As the magnet is attracted to a preenergized coil, the magnet wants to add its flux to the flux of the coil. This addition would represent flux increase. An isolated coil tries to keep the sum of the fluxes constant and it accomplishes this by decreasing its own flux...and current.
Quote from: Jimboot on December 06, 2015, 04:31:55 AM
I guess I'm asking what happens to the current created by a moving magnet in the coil that is being pulsed by DC?
If the magnet is attracted by a coil (as in Woopy's video) then the magnet's flux wants to add to the coil's flux and current induced by the magnet's motion subtracts from any preexisting current in the coil, to compensate.
If the magnet is repelled by a coil then the magnet's flux wants to subtract from the coil's flux and the current induced by the magnet's motion adds to any preexisting current in the coil, to compensate.
...but in both of these cases, the flux penetrating the coil stays constant, if there are no energy leaks or replenishments by an external power supply.
If an external power supply replenishes the lost energy faster than the leak, then the energy content of the coil increases, if not - then it decreases.
A constant voltage power supply can pump up the coil's energy and current only up to the
V/R limit (http://www.overunityresearch.com/index.php?topic=2684.msg43692#msg43692).
Quote from: verpies on December 06, 2015, 07:02:44 AM
Because the ultimate "goal" of any coil is to keep the magnetic flux constant (the flux that penetrates the coil).
If the coil is ideal (like in this video (https://www.youtube.com/watch?v=uL4pfisCX14)) then the coil always succeeds in keeping the flux constant, but if the coil's circuit is resistive then the energy will eventually leak out and the coil will not be able to maintain the constant flux. This is shown in this video (https://www.youtube.com/watch?v=wUaqXk6axOo).
As the magnet is attracted to a preenergized coil, the magnet wants to add its flux to the flux of the coil. This addition would represent flux increase. An isolated coil tries to keep the sum of the fluxes constant and it accomplishes this by decreasing its own flux...and current.
If the magnet is attracted by a coil (as in Woopy's video) then the magnet's flux wants to add to the coil's flux and current induced by the magnet's motion subtracts from any preexisting current in the coil, to compensate.
If the magnet is repelled by a coil then the magnet's flux wants to subtract from the coil's flux and the current induced by the magnet's motion adds to any preexisting current in the coil, to compensate.
...but in both of these cases, the flux penetrating the coil stays constant, if there are no energy leaks or replenishments by an external power supply.
If an external power supply replenishes the lost energy faster than the leak, then the energy content of the coil increases, if not - then it decreases.
A constant voltage power supply can pump up the coil's energy and current only up to the V/R limit (http://www.overunityresearch.com/index.php?topic=2684.msg43692#msg43692).
Thanks Verpies.
Salut Luc
Merci pour ta réponse en français, ça fait toujours plaisir. Et on semble être d'accord sur beaucoup de points sur ce fil.
Hi Mag and Gyula
just for info i have tried much bigger magnets and also inserting magnets inside the air core coils and the current seems not to be very much affected even as i feel a strong permanent and wobling attraction ??
Hi Shylo
yes i have tried to place a diamagnetic round cylinder magnets , and when it spin at the right speed, it Kicks in and rotates as a rotor. When it rotates , the current is very slightly lower.
Hi Verpies
Thank's for your info, just a question, could you please explain precisely what happens when the magnetic field collapse, i mean how is the process of the creation of the strong negative voltage in the wires. Perhaps some videos ?
I ask this because i am not satisfied with the flywheel explanation by Milehigh, i need to "see" the "mecanics " as you did in your graph for the creation of the magnetic field.
I also ask it because today i have replaced the reed by a hall sensor and a transistor so i can get a much shorter "on time " for the main coil, in order to stay as near as the 0.58 TAU. And it seems that by drawing far less current , the flyback spike stays almost at the same amplitude (all this to be checked of course)
Thank's to all of you for your suport
laurent
Hello everyone, Hello Gotoluc!
I've been reading this thread from day 1, and after seeing this video of yours: http://www.youtube.com/watch?v=7dmKEOWOhQA
...I asked myself why did you not do a comparision between your last test in that video and a direct driving of the HV coil, with the same energy input (The signal - voltage and on time, should be as close as possible to the kind of kickback spike that you get from the low voltage coil).
OK, i realize that it would not be easy to get that kind of signal, and measurements would be difficult also, but if done right, we could see if the low voltage coil is giving us any benefit, or is it just acting as a loosy low voltage to high voltage converter.
I just thought to mention this, hopefully it brings a new view to the thread.
Wish you luck and thanks for doing all the experimentation and video sharing.
Quote from: woopy on December 06, 2015, 04:10:08 PM
could you please explain precisely what happens when the magnetic field collapse, I mean how is the process of the creation of the strong negative voltage in the wires. Perhaps some videos ?
Does
this animation (http://overunity.com/14378/akula0083-30-watt-self-running-generator/dlattach/attach/135807/) help ?
Notice that the wheel keeps turning for a short time AFTER the valve is closed.
Coils just do not like to have their current interrupted suddenly and they try to fight any current interruption by increasing the voltage in such manner that the current keeps flowing through the coil
in the same direction and with the same magnitude...even if it means ionizing the air to keeping the current flowing through it.
This is the same behavior as that of a constant current power supply.
Quote from: woopy on December 06, 2015, 04:10:08 PM
I also ask it because today i have replaced the reed by a hall sensor and a transistor so i can get a much shorter "on time " for the main coil, in order to stay as near as the 0.58 TAU. And it seems that by drawing far less current , the flyback spike stays almost at the same amplitude...
Remember that the 0.58 Tau is the "break even point" - not the "sweet spot".
Only 1 thing to say
!! FAIL !!
https://www.youtube.com/watch?v=ALg_9FR6egY
Quote from: madddann on December 06, 2015, 04:20:05 PM
Hello everyone, Hello Gotoluc!
I've been reading this thread from day 1, and after seeing this video of yours: http://www.youtube.com/watch?v=7dmKEOWOhQA (http://www.youtube.com/watch?v=7dmKEOWOhQA)
...I asked myself why did you not do a comparision between your last test in that video and a direct driving of the HV coil, with the same energy input (The signal - voltage and on time, should be as close as possible to the kind of kickback spike that you get from the low voltage coil).
OK, i realize that it would not be easy to get that kind of signal, and measurements would be difficult also, but if done right, we could see if the low voltage coil is giving us any benefit, or is it just acting as a loosy low voltage to high voltage converter.
I just thought to mention this, hopefully it brings a new view to the thread.
Wish you luck and thanks for doing all the experimentation and video sharing.
Hi madddann, thanks for your interest.
There are reasons for doing things a certain way and not enough time to go over it all in the videos.
The MOT's low voltage coil has 0.8 Ohm of resistance compared to the high voltage coil has over 100 Ohms.
If I was to connect the same input to the HV coil and expect the rotor to turn at the same speed it wouldn't, unless I raised the input voltage quite considerably to get anywhere close to the speed I had. The other problem the high resistance causes is, it would needs much more time for the magnetic field to build compared to the LV coil since . Also, the flyback from a HV coil would not have much energy and again because of high resistance.
So by far a low resistance (low impedance) coil is best for what I want to do.
I'm not sure if this answers your question?
Luc
Quote from: tinman on December 06, 2015, 05:50:40 PM
Only 1 thing to say
!! FAIL !!
https://www.youtube.com/watch?v=ALg_9FR6egY (https://www.youtube.com/watch?v=ALg_9FR6egY)
Sorry it's giving you a hard time mate!
I did make a drawing to show you how to get the best torque but for some reason you didn't build it that way :-\
I guess something else is now being tested ;D
Thanks for your time with this
Luc
Quote from: gotoluc on December 06, 2015, 06:51:19 PM
Sorry it's giving you a hard time mate!
I did make a drawing to show you how to get the best torque but for some reason you didn't build it that way :-\
I guess something else is now being tested ;D
Thanks for your time with this
Luc
Well its much the same-just on a smaller scale. My laminated blocks are to small, but if I swap the stepper motor out for some free running bearings, then it spins up just fine-like yours. But this design has no torque-that much is true.
If I had of taken a little time to think about it, then I already have the most efficient motor of this type-that being the RT. The only difference between the two is your design has a lump ofstell laminates being pulled toward an electromagnet, and the RT has an electromagnet being pulled toward a lump of steel laminates. So the two are one in the same. But as we have seen, the RT is capable of far more torque, and using the flyback increases the efficiency by about 80%.
So in all reality, I have already done extensive testing on this type of motor
Brad.
Quote from: tinman on December 06, 2015, 11:17:48 PM
... the RT is capable of far more torque, and using the flyback increases the efficiency by about 80%....
80% total or 80% over what ever % the motor was before being turned into a RT? I'm guessing the latter.
Quote from: tinman on December 06, 2015, 05:50:40 PM
Only 1 thing to say
!! FAIL !!
https://www.youtube.com/watch?v=ALg_9FR6egY (https://www.youtube.com/watch?v=ALg_9FR6egY)
How large is the yellow gap in your design ?
Could you be closing the flux path through the yellow gap as depicted with the red flux lines ?
Are you using ferrite, solid steel, or laminated steel? If it is the latter, then which way are the laminations oriented ?
I know you have 2 coils in your C-core. Does their flux add or subtract? ...in other words if the coils were free to slide left & right, would they attract or repel?
Are you using ferrite, solid steel, or laminated steel? If it is the latter, then which way are the laminations oriented ?
Verpies I've been trying to get my head around what you said. Below is a screenshot from 2010 when I was working on the ossie motor. I assumed at the time the spikes were pulses from the reeds and the sine was from the spinning magnets. I have this wrong it seems? I thought that in a pulse situation a current would induced in the coil by the magnets when the dc pulse was not present. From memory that build had 4 coils but I'd have to go back and check.
Quote from: Jimboot on December 07, 2015, 04:24:44 AM
Verpies I've been trying to get my head around what you said. Below is a screenshot from 2010 when I was working on the ossie motor. I assumed at the time the spikes were pulses from the reeds and the sine was from the spinning magnets. I have this wrong it seems? I thought that in a pulse situation a current would induced in the coil by the magnets when the dc pulse was not present. From memory that build had 4 coils but I'd have to go back and check.
Quote from: verpies on December 07, 2015, 03:48:02 AM
QuoteHow large is the yellow gap in your design ?
10mm
QuoteCould you be closing the flux path through the yellow gap as depicted with the red flux lines ?
Only when the laminated blocks on the rotor are further than 10mm away from the C core inductor.
QuoteAre you using ferrite, solid steel, or laminated steel? If it is the latter, then which way are the laminations oriented ?
Laminated transformer core from the transformers that the C core inductor is made from. They are orientated the same way as the inductors core.
QuoteI know you have 2 coils in your C-core. Does their flux add or subtract? ...in other words if the coils were free to slide left & right, would they attract or repel?
Have tried both ways-found attraction mode to be better than bucking mode.
Quote from: tinman on December 06, 2015, 11:17:48 PM
Well its much the same-just on a smaller scale. My laminated blocks are to small, but if I swap the stepper motor out for some free running bearings, then it spins up just fine-like yours. But this design has no torque-that much is true.
If I had of taken a little time to think about it, then I already have the most efficient motor of this type-that being the RT. The only difference between the two is your design has a lump ofstell laminates being pulled toward an electromagnet, and the RT has an electromagnet being pulled toward a lump of steel laminates. So the two are one in the same. But as we have seen, the RT is capable of far more torque, and using the flyback increases the efficiency by about 80%.
So in all reality, I have already done extensive testing on this type of motor
Brad.
I agree and have been working on my next version
https://www.youtube.com/watch?v=LMWQwRt1KNc
Luc
Hi all,
i've been exploring this concept from a slightly different angle.
It's a fact that, given the same power input to two inductors of the same resistance, the inductor with twice the length of wire (and half the per-metre resistance) will give you a stronger magnetic field. For the same power input.
This is what Joseph Newman's motor was based on.
He used ten watts input to a very large coil, and the magnetic field it created was strong enough to rotate a 600LB magnet.
He then designed an elegant mechanical commutator to motor and generate at the right times and for the right durations.
Looking at how people rewired various Rotoverter incarnations, i think they were sinking the energy from the collapsing fields back in to aid rotation.
Looking at other inventions, it may be the basis for many of them.
The high impedance coils, as Gyulasun said, inhibit the maximum RPM, so we have to keep the resistance low, but this results in larger coils.
It's an effect that can be exploited with varied geometries and i look forward to exploring it here.
Y.
Thanks brad that is what I thought, I must have misunderstood verpies thanks. Btw nice little "fail" very nice demo.
Quote from: gotoluc on December 07, 2015, 09:48:17 AM
I agree and have been working on my next version
https://www.youtube.com/watch?v=LMWQwRt1KNc
Luc
Nice-very nice.
It is good to see there are some that still have hand skill's,and dont rely on machines to do all the work.
This should be a very interesting build.
Brad
Quote from: gotoluc on December 07, 2015, 09:48:17 AM
I agree and have been working on my next version
https://www.youtube.com/watch?v=LMWQwRt1KNc (https://www.youtube.com/watch?v=LMWQwRt1KNc)
Luc
That is some sweet work Luc. What are those shaded poles made from? They don't look laminated.
Quote from: tinman on December 07, 2015, 06:15:59 PM
Nice-very nice.
It is good to see there are some that still have hand skill's,and dont rely on machines to do all the work.
This should be a very interesting build.
Brad
Thanks Brad,
I know of all people,
you can appreciate the work and the flexibility of tests this design can achieve.
Quote from: Jimboot on December 07, 2015, 06:30:03 PM
That is some sweet work Luc. What are those shaded poles made from? They don't look laminated.
They are form shaded pole motors (pic below) and they are laminations but when you fine finish the surface to remove shorts you can't tell it's laminations any longer.
#notworthy
I found this one pretty cool. And in line with the discussion.
https://www.youtube.com/watch?v=41cfHcb7qd8 (https://www.youtube.com/watch?v=41cfHcb7qd8)
Quote from: MoRo on December 07, 2015, 08:41:19 PM
I found this one pretty cool. And in line with the discussion.
https://www.youtube.com/watch?v=41cfHcb7qd8 (https://www.youtube.com/watch?v=41cfHcb7qd8)
Yes MoRo, it's definitely relevant!... and it's not coincidence because Cool Joule is user minoly who was here at the beginning of the topic preaching about JB: http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg466014/#msg466014 (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg466014/#msg466014)
He hasn't posted since I was strait forward with his assumptions, some pages later but I'm glad he is experimenting with the concept which apparently is JB's ???
Anyways, it's for everyone to experiment with and I wish him well and he is always welcome to post his results here if he wishes as I don't hold grudges.
Thanks for sharing and please continue to post his experiments if he chooses not to do so himself.
Luc
Quote from: Jimboot on December 07, 2015, 04:24:44 AM
Verpies I've been trying to get my head around what you said. Below is a screenshot from 2010 when I was working on the ossie motor. I assumed at the time the spikes were pulses from the reeds and the sine was from the spinning magnets. I have this wrong it seems? I thought that in a pulse situation a current would induced in the coil by the magnets when the dc pulse was not present. From memory that build had 4 coils but I'd have to go back and check.
Your scopeshot seems to be showing a voltage across a coil, so it is difficult to conclude what the current flowing through that coil is when that coil is closed by a circuit. Anyway, with this voltage oscillogram, the spikes appear to be occurring when the coil's current is interrupted (the reed switch opens) and the sinewave could be the emf induced by moving magnets (twice per reed closure) or an LC interaction with some capacitance.
When a coil is connected across a C.V. power supply then the coil is effectively shorted and its current can flow uninterrupted. Thus the
externally induced current can flow in a coil regardless whether it is shorted by a jumper or by a C.V. power supply ...in a "pulse situation".
Quote from: gotoluc on December 07, 2015, 09:04:20 PM
Yes MoRo, it's definitely relevant!... and it's not coincidence because Cool Joule is user minoly who was here at the beginning of the topic preaching about JB: http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg466014/#msg466014 (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg466014/#msg466014)
He hasn't posted since I was strait forward with his assumptions, some pages later but I'm glad he is experimenting with the concept which apparently is JB's ???
Anyways, it's for everyone to experiment with and I wish him well and he is always welcome to post his results here if he wishes as I don't hold grudges.
Thanks for sharing and please continue to post his experiments if he chooses not to do so himself.
Luc
Yes, a visual on how the joule thief works, only repelling a magnet with the kickback in stead of lighting an LED. How ever, believing that the kickback has more energy than that supplied to the primary , just because the primary dosnt move the magnet-is wrong.
Brad
Quote from: verpies on December 07, 2015, 10:11:50 PM
Your scopeshot seems to be showing a voltage across a coil, so it is difficult to conclude what the current flowing through that coil is when that coil is closed by a circuit. Anyway, with this voltage oscillogram, the spikes appear to be occurring when the coil's current is interrupted (the reed switch opens) and the sinewave could be the emf induced by moving magnets (twice per reed closure) or an LC interaction with some capacitance.
When a coil is connected across a C.V. power supply then the coil is effectively shorted and its current can flow uninterrupted. Thus the externally induced current can flow in a coil regardless whether it is shorted by a jumper or by a C.V. power supply ...in a "pulse situation".
Thanks mate, yeah I should have explained that the probes are simply across the coil. The sine wave was definitely being created by the spinning magnet so idea of how much current was there at the time. I assumed (probably wrongly) that the efficiency of this motor was due to that power recirculating in the circuit. I believe the general consensus at the time tho was the capturing of the flyback was what made it so efficient. The Ossie motor is basically a couple of reeds and a FWBR. Certainly no caps, I probably still have the inductance numbers somewhere. Thanks
Quote from: Yttrium on December 07, 2015, 02:43:39 PM
It's a fact that, given the same power input to two inductors of the same resistance, the inductor with twice the length of wire (and half the per-metre resistance) will give you a stronger magnetic field. For the same power input.
What about the difference in inductances of these two coils ?
If the inductances are different then how do you propose to keep the input power the same? Alas, the input power vs. time, greatly depends on the inductance.
Quote from: Jimboot on December 08, 2015, 12:40:06 AM
I believe the general consensus at the time tho was the capturing of the flyback was what made it so efficient.
It is still true that "capturing of the flyback" makes a motor more efficient.
Quote from: Jimboot on December 08, 2015, 12:40:06 AM
The Ossie motor is basically a couple of reeds and a FWBR. Certainly no caps, ...
The lack of caps is a problem, because the most efficient way to transfer energy between two coils is:
Coil1 --> Cap -->Coil2
Transferring directly Coil1 --> Coil2 is a most inefficient scheme.
Before somebody asks whether S2 is absolutely necessary to maintain high efficiency:
The answer is "no".
S2 can be deleted but then L2 will be energized soon after S1 opens and this will constitute the loss of precise control over L2 timing.
Also, the purpose of D2 is to prevent the energy in L2 from returning back to C1 and L1.
If D2 is deleted then energy will slosh back and forth between L2 and C1 (as a decaying sinewave oscillation with the max p-p amplitude equal to the forward voltage drop of D3, which is not much for a Schottky diode) and the L2 energy will return to L1 if the forward voltage drop of D3 is larger than forward voltage drop of D1.
If S2 is opened as soon as the voltage across C1 falls to zero, then D2 becomes completely unnecessary and can be deleted, leading to increased circuit efficiency.
The purpose of D3 is to maintain current flowing through L2 once it becomes energized. Without it, the energy will escape from L2 very quickly (as soon as C1 discharges to zero).
BTW: All diodes should be Synchronous Rectifiers or Schottky diodes with low forward voltage drops (that means no SiC diodes!) because these voltage drops cause the majority of electric energy losses.
The capacitor does not have to be bipolar, but it should have a low ESR - unfortunately electrolytics are not known for their low ESR.
P.S.
Since capacitors store energy as voltage and coils store energy as current, if you are interested in measuring their energy content, then capacitors should have their voltages measured across them and coils should have their currents measured through them.
When the caps are full just discharge them into the drive? No?
artv
Quote from: shylo on December 08, 2015, 06:18:25 PM
When the caps are full just discharge them into the drive? No?
artv
If by "drive" coil you mean L2, then yes.
...but remember, that if you do not have D3 (and D2 possibly) then the energy will go from C1 to L2 and then later it will return from L2 to C1 and L1. Would you want that ?
Quote from: verpies on December 08, 2015, 07:06:29 AM
Before somebody asks whether S2 is absolutely necessary to maintain high efficiency:
The answer is "no".
S2 can be deleted but then L2 will be energized soon after S1 opens and this will constitute the loss of precise control over L2 timing.
Also, the purpose of D2 is to prevent the energy in L2 from returning back to C1 and L1.
If D2 is deleted then energy will slosh back and forth between L2 and C1 (as a decaying sinewave oscillation with the max p-p amplitude equal to the forward voltage drop of D3, which is not much for a Schottky diode) and the L2 energy will return to L1 if the forward voltage drop of D3 is larger than forward voltage drop of D1.
If S2 is opened as soon as the voltage across C1 falls to zero, then D2 becomes completely unnecessary and can be deleted, leading to increased circuit efficiency.
The purpose of D3 is to maintain current flowing through L2 once it becomes energized. Without it, the energy will escape from L2 very quickly (as soon as C1 discharges to zero).
BTW: All diodes should be Synchronous Rectifiers or Schottky diodes with low forward voltage drops (that means no SiC diodes!) because these voltage drops cause the majority of electric energy losses.
The capacitor does not have to be bipolar, but it should have a low ESR - unfortunately electrolytics are not known for their low ESR.
P.S.
Since capacitors store energy as voltage and coils store energy as current, if you are interested in measuring their energy content, then capacitors should have their voltages measured across them and coils should have their currents measured through them.
If you use a lot of little electrolytic caps in parallel the esr is lowered considerably. Plus, smaller caps have lower esr compared to larger ones, so the gain of a lower esr is even better than just the paralleling of larger caps.
Mags
Quote from: Magluvin on December 08, 2015, 09:38:34 PM
If you use a lot of little electrolytic caps in parallel the esr is lowered considerably. Plus, smaller caps have lower esr compared to larger ones, so the gain of a lower esr is even better than just the paralleling of larger caps.
You you can use various techniques to adapt to the peculiarities of electrolytic caps, but don't loose the site of the goal in pursuit of these techniques.
...and the goal is to transfer all of the energy from L1 to L2 as efficiently as possible. This usually means as quickly as possible, too....so the current through D1 flows as little time as possible.
Note, that with a large C1, the time to transfer energy from L1 to C1 is large too, [ precisely t=π(L1*C1)
½ ] and during that long time, the forward voltage drop of D1 can waste a lot of energy.
So, a larger C1 does not mean a more efficient operation ...and the electrolytic caps just lost their biggest advantage (large capacitance).
BTW: C1 must be able to tolerate voltages of at least V = i
L1*(L1 / C1)
½ , thus the smaller the C1, the larger voltage it must tolerate to hold the same energy.
This capacitor voltage also defines the blocking voltages of the switches S1 and S2 and reverse voltage of diode D1 and D3. This is an issue with Schottky diodes which usually have low reverse blocking voltages.
Sorry guys I should clear things up, I'm using a mag rotor for generating the spikes , storing them in caps ,and the trying to dump them back into the generating coils to serve as motor coils.
I have 7 coils in place which gives the possibilty of 56 storage caps, currently I only have 36 wired in (it's a nightmare) room for 20 more.
The problem is trying to dump the caps ,since I have zero electronic skills, I have been trying to design a mechanical switch.
Luc if this too far off topic I apologize and will understand if you delete.
Btw what does esr mean?
Quote from: shylo on December 09, 2015, 05:34:53 AM
Btw what does esr mean?
https://en.wikipedia.org/wiki/Equivalent_series_resistance
Thanks Tagor, I just measured the resistance of my 7 coils wired in series to be 334 ohms.
I tried to measure a cap but just got an error reading.
But as soon as the mag rotor moves the coil readings just bounce all over the place, so is this esr really that important?
artv
Quote from: shylo on December 09, 2015, 06:08:22 AM
I tried to measure a cap but just got an error reading.
It is not possible to measure ESR of a capacitor with an ohmmeter.
Quote from: shylo on December 09, 2015, 05:34:53 AM
I'm using a mag rotor for generating the spikes , storing them in caps ,and the trying to dump them back into the generating coils to serve as motor coils.
I have 7 coils in place which gives the possibility of 56 storage caps, ...
Please draw a schematic of your coils, caps, diodes and switches.
Quote from: shylo on December 09, 2015, 05:34:53 AM
Luc if this too far off topic I apologize and will understand if you delete.
No, it's fine and relevant
Please continue sharing
Luc
I don't think I posted Hob Nilre experiments which would be relevant to some of what we are doing.
Here are a few video demos he has done:
https://www.youtube.com/watch?v=IIeRR6NjMPQ
https://www.youtube.com/watch?v=zu_v7jMs-ms
and for those who are into technical papers, he has 5 of them here: https://sites.google.com/site/nilrehob/home/elementary-physics
Luc
Hi Verpies, I don't have any switches yet , I am just collecting inductive kickback in the caps.
This kickback is not the same as the spike that can be collected due to shorting a generating coil, motoring coils also create a spike,
Which would mean 168 caps to deal with,
I need a switch that can handle all the operations in a sequenced order, 1 cap or perhaps 3 or 6 , I don't know , what ever it takes to sustain rotation.
I have more charged caps than I know what to do with.
artv
Hi to all,
Can this design be any good using magnetic cancellation and fly back recovery?
Mihai
Quote from: mihai.isteniuc on December 10, 2015, 04:02:40 AM
Hi to all,
Can this design be any good using magnetic cancellation and fly back recovery?
Mihai
I'll take a shot
Those ferromag core will flip polarity and so will be in full attraction mode.
Hi Jim and ty for taking time to reply.
The way i see it it's like this: energize the left coil to cancel magnetic flux from permanent magnets from left. The central magnets then are attracted by ferromagnetic core from left and repelled by the magnetic flux on the right. Ergo the central magnets are moving to the left. The process is then repeated by the other half of the system, so central magnets are gonna move to the right, inducing some electromagnetic power in central coil. This is one way to look at it ( and TL494 is a good IC to make this type of oscillator). Of course we recover fly-back energy from left coil, put some energy from the power supply and energize then the right coil, recover again fly-back and repeat the process.
The other possible way is to connect the two coils in series. In this configuration with the left coil we do magnetic cancellation while the right one is doing magnetic amplification, then reverse polarity. Again recover the fly-back pulse. (BTW the idea with magnetic cancellation/amplification of a magnetic field came from the www.gap-power.com (http://www.gap-power.com)).
The real problem is: it is possible to make it work with a very short pulse of energy? As Verpies said (and my tests confirm) the pulse must have an optimum time duration. After that duration depletes we have to disconnect the power supply and collect the fly-back. My tests shows then 8-10% pulse width can lead to more then 70% of energy recover from fly-back pulse. Question: is this low time duration pulse able to do any useful magnetic work in the system (except of course fly-back pulse)?
Independent question about a coil; we have the following energy's when we send pulsed power to a coil:
- input energy=A
- losses energy (all of them)=B
- fly-back energy=C
- magnetic energy (if we consider magnetic field doing some useful work)=D
What is the mathematical relation between this energy's? A=?
mihai
I tried to harvest the energy in a fly back spike by help of a transformer (see the attached circuit diagram).
There is no motor. The drive coil (if there were a motor) is switched with a IRF840 transistor and a function generator.
It takes less than 720 mW to drive the circuit (i.e. the drive coil) and one can harvest about 15 mW from the fly back spike of the drive coil (via a transformer).
This test shows, that there is very little energy in the fly back spike. May be one has to use a better technique (not a transformer) to get more energy out of the fly back spike.
The set up provided the best output at 270 Hz. The drive coil is driven with a 50% duty cycle. At less than 40% duty cycle the output starts to drop.
I had to use a rather small capacitor of 47 nF. With a bigger capacitor ( e.g. 1 µF or 10 µF) the output dropped.
Greetings, Conrad
Quote from: conradelektro on December 10, 2015, 12:15:14 PM
It takes less than 720 mW to drive the circuit (i.e. the drive coil) and one can harvest about 15 mW from the fly back spike of the drive coil (via a transformer).
This test shows, that there is very little energy in the fly back spike. May be one has to use a better technique (not a transformer) to get more energy out of the fly back spike.
Greetings, Conrad
Hi Conrad,
thanks for taking the time to post your results.
I would agree that the power of your flyback would be very little as your coils DC resistance is very high.
Please find a MOT or other transformer that the primary or secondary is below 1 ohm and redo your test and post the difference.
Looking forward to see 8) the difference in the results
Thanks for sharing
Luc
Quote from: gotoluc on December 10, 2015, 12:32:20 PM
Hi Conrad,
thanks for taking the time to post your results.
I would agree that the power of your flyback would be very little as your coils DC resistance is very high.
Please find a MOT or other transformer that the primary or secondary is below 1 ohm and redo your test and post the difference.
Looking forward to see 8) the difference in the results
Thanks for sharing
Luc
Hi Luc!
May be you have misunderstood my circuit.
The drive coil has 90 Ohm DC resistance (a 12 V relays coil). What you and some people call the "high voltage coil" to regain something from the fly back spike is in my circuit the primary of a transformer (this primary has a DC resistance of 3000 Ohm). I changed my circuit diagram to indicate the transformer with a box around it.
I use a transformer, because it allows to decouple the energy in the fly back spike. On the secondary of this transformer one has then some sort of strange alternating current (see scope shot at pos 3 in the circuit) which could be used for something else.
You try to use the magnetic force in the second coil (the so called high voltage coil) to make the motor stronger. I try to get "electricity" out of the fly back spikes (at the secondary of a transformer).
Of course I could use a drive coil with only about 1 Ohm DC resistance (instead of my drive coil with 90 Ohm DC resistance). And then a transformer with a primary of about 30 Ohm DC resistance for getting "electricity" out of the fly back spikes. The input power will be much higher and also the output power at the 10 Ohm resistor. But the result will be similar (only about 2% of the input power can be harvested from the fly back spikes). More input results in more output, but the relative gain (about 2%) will be about the same.
What one could try, is a better technique than a transformer to decouple the energy in the fly back spikes. May be it is easier to regain magnetic force (as you try) and not "electricity" (as I try).
Greetings, Conrad
Quote from: conradelektro on December 10, 2015, 12:56:32 PM
Hi Luc!
May be you have misunderstood my circuit.
The drive coil has 90 Ohm DC resistance (a 12 V relays coil).
"high voltage coil" has a DC resistance of 3000 Ohm
That is how I understood it.
Quote from: conradelektro on December 10, 2015, 12:56:32 PM
Of course I could use a drive coil with only about 1 Ohm DC resistance (instead of my drive coil with 90 Ohm DC resistance).
Yes, that's what I was recommending and for your drive coil to be even lower than 1 Ohm will be better.
The flyback coil also should be as low resistance as possible but 5 to 10 time the inductance of the drive coil.
Still looking forward to your new test
Luc
Quote from: gotoluc on December 10, 2015, 02:44:42 PM
That is how I understood it.
Yes, that's what I was recommending and for your drive coil to be even lower than 1 Ohm will be better.
The flyback coil also should be as low resistance as possible but 5 to 10 time the inductance of the drive coil.
Still looking forward to your new test
Luc
I got these two FEROXCUBE transformer cores (see the attached photo) and I can rewind them according to your specifications.
One will be the drive coil with about 0.5 Ohm DC and about 20 mH.
And the fly back coil (in my case the primary of a step down transformer) will have about 200 mH (and about 5 Ohm DC). The secondary will step down 10:1, just to have a low Voltage output (because the fly back spike will have several hundred Volts).
It will take some time.
To be honest, I do not expect a big difference. The input will be several Watts and the output again about 2% of the input power. Just to play with higher power does not necessarily give a a higher relative output.
Many people underestimate what a power input of e.g. 50 Watt can produce. 2% of 50 Watt will be 1 Watt, which looks like a lot if taken by it self. But in comparison to the 50 Watt spent, it is not really a big deal. Electric motors have a loss of up to 40%, really good ones loose 20% of the input power.
So, a loss of 2% in the fly back spikes is not much in reality.
Greetings, Conrad
Quote from: conradelektro on December 10, 2015, 12:15:14 PM
This test shows, that there is very little energy in the fly back spike. May be one has to use a better technique (not a transformer) to get more energy out of the fly back spike.
Yes, you can disconnect the transformer altogether and measure the voltage to which the capacitor gets charged. The recovered energy will be E=½CV
2The only caveat is that you need to empty the capacitor before each flyback pulse.
The capacitor should be able to withstand the voltage V = i
L*(L / C)
½, where i
L is the current flowing through the coil when the MOSFET opens.
...and this voltage should not be higher than the blocking voltage of the MOSFET and the reverse breakdown voltage of the diode (non-SiC Schottky preferable)
And with "non-SiC Schottky" you mean something like this:
Gallium Arsenide Schottky Rectifier http://pdf1.alldatasheet.com/datasheet-pdf/view/166395/IXYS/DGS20-025A.html (250V)?
or this: http://ixdev.ixys.com/DataSheet/L652.pdf (600V)?
Thanks, itsu
Quote from: verpies on December 10, 2015, 04:32:40 PM
Yes, you can disconnect the transformer altogether and measure the voltage to which the capacitor gets charged. The recovered energy will be E=½CV2
The only caveat is that you need to empty the capacitor before each flyback pulse.
The capacitor should be able to withstand the voltage V = iL*(L / C)½, where iL is the current flowing through the coil when the MOSFET opens.
...and this voltage should not be higher than the blocking voltage of the MOSFET and the reverse breakdown voltage of the diode (non-SiC Schottky preferable)
The "capacitor alone" is a better and cleaner approach for a theoretical power calculation in the spikes. Practically it will be difficult to empty the capacitor after each spike in a clean way (without effecting the storage of the spike energy).
I like the idea of harvesting the spike energy with a coil. After all we are dealing with a magnetic effect and why not try to get back a magnetic force. Putting a coil in the fly back path is a new idea, which I have not seen before (or which I have not recognised before). Whenever I tried to measure the energy which is fed back to the power supply with a diode alone, I failed. It never seemed that less power had to be supplied to a pulse motor with a fly back diode in place than without it. May be I am to stupid to measure that, but I also never saw a convincing measurement.
And the next step for me is to see the spike coil as the primary of a step down transformer, because the spikes have very high Voltage and one wants a low Voltage output (for practical use). The capacitor and the hysteresis of the step down transformer "transform" the very narrow fly back spike into a more sine wave like signal, which is more useful than a spike.
But I doubt that there is much energy in the spikes. And the energy seen as "magnetic force" or seen as "electricity" will be about the same. So, either to speed up the rotor (magnetic force) or to gain some electricity (which could be fed back to the power supply as an output from the spikes), it will not be much (only in the order of a few percent of the input power to the circuit or motor).
May be Luc could tell us where exactly he is looking for "more energy"? I tried to measure the energy in the spikes (more an estimate than an exact measurement, but it gives an order of magnitude) and it was not much.
I am not criticising Luc, I just want to understand and to measure the energy in the fly back spikes. Where is the magic? Or better said, where does Luc hope to find the magic? Any speculation is accepted and I will try to measure it in an "equivalent circuit". It can not be, that a motor is absolutely necessary? Or does it have to be a motor? I also accept that the magic is in the motor, but where in the motor, for sure not in the spikes.
Greetings, Conrad
Some thoughts about the capacitor in parallel with the "fly back coil":
During my tests I got the impression that the capacitor has to be chosen according to the energy in the spikes.
If one uses a low DC resistance and low impedance drive coil (like Luc or woopy), the capacitor should be 1 µF to 10 µF (because it has to store a considerable amount of energy, e.g. there is about 1 Watt in the spikes if about 50 Watts are fed into the motor or circuit).
If one uses a high DC resistance and high impedance drive coil (like I do in my equivalent circuit), the capacitor has to be very small (e.g. 47 nF), because the energy in the spikes is only about 10 mW if about 500 mW are fed into the circuit.
No capacitor is bad, because the fly back spikes are not stored (and not flattened). Too high a capacitance is also bad, because the spikes are flattened too much. In my case I use a fly back coil (the primary of a step down transformer) which would like 220 Volt (because it is a step down transformer from 200 V to 6 V). Therefore the spikes should be flattened to about 220 V in the ideal case. (Remark: I only manage 80 Volt in my circuit).
Just some very empiric considerations.
Greetings, Conrad
Quote from: itsu on December 10, 2015, 04:48:24 PM
And with "non-SiC Schottky" you mean something like this:
Gallium Arsenide Schottky Rectifier
http://pdf1.alldatasheet.com/datasheet-pdf/view/166395/IXYS/DGS20-025A.html (http://pdf1.alldatasheet.com/datasheet-pdf/view/166395/IXYS/DGS20-025A.html) (250V)?
or this: http://ixdev.ixys.com/DataSheet/L652.pdf (http://ixdev.ixys.com/DataSheet/L652.pdf) (600V)?
No, because the forward voltage drop (V
F) of these diodes is 1.2V or more :(
A good Schottky diode for this purpose would have a low voltage drop around 0.3V
Quote from: conradelektro on December 10, 2015, 05:14:56 PM
The "capacitor alone" is a better and cleaner approach for a theoretical power calculation in the spikes. Practically it will be difficult to empty the capacitor after each spike in a clean way (without effecting the storage of the spike energy).
Not really - e.g. Itsu did that once with a relay ;)
However, the simplest way to empty the capacitor is to connect a 1MΩ resistor in parallel with the cap and give it a very long time between the "flyback" pulses, so this resistor has the time to discharge the cap meanwhile.
Quote from: conradelektro on December 10, 2015, 05:14:56 PM
I like the idea of harvesting the spike energy with a coil. After all we are dealing with a magnetic effect and why not try to get back a magnetic force. Putting a coil in the fly back path is a new idea, which I have not seen before (or which I have not recognized before).
I also like the idea of recovering the remaining energy stored in inductor (L1) and transferring it into a second inductor (L2) for doing extra mechanical work.
...but using a direct connection between L1 and L2 is the least efficient method for transferring this energy. You are not doing this fortunately, but your transformer idea is only a little better since it does not maximize the time that the current flows in the second inductor.
You actually do not even have a 2nd inductor (L2) that could attract a piece of a rotor in a motor (a transformer is not an inductor, unless the flux coupling coefficient is below unity). Maybe you were planning to put the 2nd inductor in place of that 10Ω resistor but that would change the behavior of the circuit, entirely and render the intermediate transformer extraneous.
If the goal is for L2 to do as much mechanical work as possible, then the circuit below is the most efficient way to do it ...while observing
these notes (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg467827/#msg467827).
The ultimate efficiency is realized with Synchronous Rectifiers ...and second best - with low voltage drop Schottky diodes.
Quote from: conradelektro on December 10, 2015, 05:31:54 PM
During my tests I got the impression that the capacitor has to be chosen according to the energy in the spikes.
If one uses a low DC resistance and low impedance drive coil (like Luc or woopy), the capacitor should be 1 µF to 10 µF (because it has to store a considerable amount of energy, e.g. there is about 1 Watt in the spikes if about 50 Watts are fed into the motor or circuit).
If one uses a high DC resistance and high impedance drive coil (like I do in my equivalent circuit), the capacitor has to be very small (e.g. 47 nF), because the energy in the spikes is only about 10 mW if about 500 mW are fed into the circuit.
Actually the resistance of the coil just wastes energy and has little bearing on the capacitor parameters, unless it wastes so much of it that the capacitor barely charges up. The final current and inductance matter much more.
The ideal match of the capacitor to the coil is defined by the equation ½Li
2 = ½CV
2, where i is the current flowing in the coil when the MOSFET opens.
Quote from: conradelektro on December 10, 2015, 05:31:54 PM
No capacitor is bad, because the fly back spikes are not stored (and not flattened).
In your circuit, the energy contained in these "flyback" pulses is not stored but burned up in that transformer, but if you deleted that transformer, then this energy would be stored in that capacitor with utmost efficiency.
Actually, there is such a thing as a "bad capacitor" in this circuit. For example a capacitor that has a voltage rating lower than V=i
L*(L/C)
½, where i
L is the current flowing through the coil when the MOSFET opens.
Quote from: conradelektro on December 10, 2015, 05:31:54 PM
Too high a capacitance is also bad, because the spikes are flattened too much.
There is some truth in that when the capacitor is too big, because when the transfer current flows too long ( time is equal to t=π(LC)
½ ) then it gets more time to be converted to heat by the diode's forward voltage drop and the circuit's resistance (including coil's resistance).
Quote from: conradelektro on December 10, 2015, 05:14:56 PM
May be Luc could tell us where exactly he is looking for "more energy"?
I never claimed more energy!
Quote from: conradelektro on December 10, 2015, 05:14:56 PM
I am not criticising Luc, I just want to understand and to measure the energy in the fly back spikes. Where is the magic? Or better said, where does Luc hope to find the magic?
I never claimed there was magic either.
Better move along as I don't think this is for you. No magic or more energy in what is suggested.
Regards
Luc
Quote from: conradelektro on December 10, 2015, 05:14:56 PM
May be Luc could tell us where exactly he is looking for "more energy"? I tried to measure the energy in the spikes (more an estimate than an exact measurement, but it gives an order of magnitude) and it was not much.
I am not criticising Luc, I just want to understand and to measure the energy in the fly back spikes. Where is the magic? Or better said, where does Luc hope to find the magic? Any speculation is accepted and I will try to measure it in an "equivalent circuit". It can not be, that a motor is absolutely necessary? Or does it have to be a motor? I also accept that the magic is in the motor, but where in the motor, for sure not in the spikes.
Greetings, Conrad
It doesnt have to be just a spike when it comes to bemf capture. Look at a switching supply. They are generally very efficient. So we have an in and an out, but we also want to turn a rotor with the magnetic field developed in the supply. If we can do that and not have the addition of the rotor affect the original in out of the switching circuit, then we may have something good.
Mags
Another way to use the bemf to create more rotor energy is to just put a diode across the drive coil and shorten the input on time. The recycled collapse current will temporarily accomplish more push or pull time on the rotor after the switch is open. To see the effect well the drive coil should be very low ohm and say 2mh and up. Will need a diode of decent current capability.
Some relay data sheets will describe the time delay of opening of the contacts when using a snubber diode to recirculate field collapse currents. In a relay application, it wouldnt be considered an energy saving advantage as relays are not really designed to be high frequency devices. More just on now, off later, etc.
Mags
Quote from: conradelektro on December 10, 2015, 05:14:56 PM
I am not criticizing Luc, I just want to understand and to measure the energy in the fly back spikes. Where is the magic? Or better said, where does Luc hope to find the magic?
I don't think he ever told us to find "magic" in the "flyback" spike. He just made an experiment that showed that energy in this spike can perform useful mechanical work.
I, however, think that there is no "magic" in the interaction of an ideal coil with a permanent magnet.
If there is any "magic" to be found, then I would expect it to be in the motor interaction between a coil and a
soft ferromagnetic, because its domain alignment energy can be recovered in the "flyback" pulse at TDC. With a permanent magnet - it cannot.
Quote from: Magluvin on December 10, 2015, 10:59:46 PM
It doesnt have to be just a spike when it comes to bemf capture. Look at a switching supply. They are generally very efficient.
Yes and the most efficient ones use synchronous rectifiers or Schottky diodes with low voltage drops.
Quote from: Magluvin on December 10, 2015, 10:59:46 PM
So we have an in and an out, but we also want to turn a rotor with the magnetic field developed in the supply.
To maximize the distance x mechanical force acting on the rotor by the "assistance coil" (L2), the current through L2 should flow as long as possible...not for several microseconds of a typical "flyback" pulse. The L2 and diode D3 makes this possible.
Quote from: verpies on December 11, 2015, 05:26:20 AM
I don't think he ever told us to find "magic" in the "flyback" spike. He just made an experiment that showed that energy in this spike can perform useful mechanical work.
I, however, think that there is no "magic" in the interaction of an ideal coil with a permanent magnet.
With a permanent magnet - it cannot.
The opposite is actually true.
Having permanent magnets on the rotor actually increases the efficiency of the motor in two way's.
First,the magnets actually generate a current within the drive coil,and this is additive to the inductive kickback output energy-so the inductive kickback output will rise.
Second,the power consumption (P/in) will also drop when a rotor with magnets is used,as apposed to there being no rotor at all.
This is of course using a non ideal coil,as we cannot make an ideal coil.
QuoteIf there is any "magic" to be found, then I would expect it to be in the motor interaction between a coil and a soft ferromagnetic, because its domain alignment energy can be recovered in the "flyback" pulse at TDC.
The problem here is that it takes energy to align those domains in the first place,so there will be no gain to be had this way.
Brad
Quote from: Erfinder on December 11, 2015, 11:27:34 AM
It sad that more experimenters working with pulsed machines don't know realize this!
It's well established that CEMF limits the input current. When CEMF and its generating mechanism is "truly" comprehended, we will then be in the position to reverse the effect CEMF has on the input current, specifically, we should note current increase with increasing RPM and CEMF, versus the well known decrease in current with increasing RPM and CEMF. This would constitute something akin to a phase shift, far removed from the text book definition of a phase shift. This is the condition I establish in my machines.
Regard
Yes Erfinder.
But we do our best to show what we can.
Is a magnet doing useful work when it alone is responsible for increasing the efficiency of a device or system ?.
The screen shots you see below are across the emitter/collector.
Enjoy.
https://www.youtube.com/watch?v=tVNABy8fSlI
Brad
Quote from: tinman on December 11, 2015, 10:15:45 AM
The opposite is actually true.
Having permanent magnets on the rotor actually increases the efficiency of the motor in two way's.
I know that there is more torque when the rotor is made out of magnets, but torque is a different concept than O/I efficiency.
These types of motors can have two output energies: the mechanical output and recovered electric energy.
Quote from: tinman on December 11, 2015, 10:15:45 AM
First,the magnets actually generate a current within the drive coil,and this is additive to the inductive kickback output energy-so the inductive kickback output will rise.
I think that the approaching attracted magnets generate current that is subtractive to the "inductive kickback output energy".
In other words: the current generated by the approaching attracted magnets, subtracts from the external current that made the coil an attracting electromagnet in the first place.
If I am wrong about this point then you've already won the OU prize.
Quote from: tinman on December 11, 2015, 10:15:45 AM
Second,the power consumption (P/in) will also drop when a rotor with magnets is used,as apposed to there being no rotor at all.
That's to be expected when the
total current flowing through the coil decreases.
Quote from: tinman on December 11, 2015, 10:15:45 AM
This is of course using a non ideal coil,as we cannot make an ideal coil.
Note that a ferro-cored coil is less ideal than an air coil, so there are different grades of non-ideality.
Quote from: tinman on December 11, 2015, 10:15:45 AM
The problem here is that it takes energy to align those domains in the first place,so there will be no gain to be had this way.
Of course, but if the entire energy required to align those domains can be recovered back at TDC (even in theory) and in the meanwhile the attraction to these aligned domains has performed mechanical work, then there is gain to be had.
Quote from: verpies on December 11, 2015, 11:42:21 AM
I know that there is more torque when the rotor is made out of magnets, but torque is a different concept than O/I efficiency.
These types of motors can have two output energies: the mechanical attraction and recovered electric energy.
I think that the approaching attracted magnets generate current that is subtractive to the "inductive kickback output energy".
In other words: the current generated by the approaching attracted magnets, subtracts from the external current that made the coil an attracting electromagnet in the first place.
If I am wrong about this point then you've already won the OU prize.
That's to be expected when the total current flowing through the coil decreases.
This is of course using a non ideal coil,as we cannot make an ideal coil.
Of course, but if the energy required to align those domains can be recovered back at TDC and in the meanwhile the attraction to these aligned domains performed mechanical work, then there is gain to be had.
QuoteI think that the approaching attracted magnets generate current that is subtractive to the "inductive kickback output energy".
Verpies
Watch the video i just posted a few post back. Here is the link again.
What happens when we have alternating magnets?. ;)
https://www.youtube.com/watch?v=tVNABy8fSlI
Brad
Quote from: tinman on December 11, 2015, 11:47:53 AM
Verpies
Watch the video i just posted a few post back. Here is the link again.
What happens when we have alternating magnets?. ;)
When we have magnets with N poles sticking out, alternating with magnets with S poles sticking out, while the coil is supplied with unipolar Pulsating DC, then half of the time a repelling magnet is approaching the coil and in such case it is my opinion that the flux of this magnet induces current that
adds to the current already circulating in the coil.
The other half of the time, when an attracting magnet is approaching the coil, the flux of this magnet induces current that subtracts from the current already circulating in the coil.
Quote from: tinman on December 11, 2015, 11:47:53 AM
https://www.youtube.com/watch?v=tVNABy8fSlI (https://www.youtube.com/watch?v=tVNABy8fSlI)
I can't put the schematic together in my mind. Unlike Hoppy, I am not very good in analyzing wired connections in videos :( .
Without knowing the schematic, the only objection that I can make now is that you are measuring a voltage across a capacitor (behind a diode) and if that voltage is supposed to represent the energy recovered from the coil, then that is an accurate measure only without that 100Ω resistor across it and when starting with an empty capacitor.
Quote from: verpies on December 11, 2015, 01:43:14 PM
When we have magnets with N poles sticking out, alternating with magnets with S poles sticking out, while the coil is supplied with unipolar Pulsating DC, then half of the time a repelling magnet is approaching the coil and in such case it is my opinion that the flux of this magnet induces current that adds to the current already circulating in the coil.
The other half of the time, when an attracting magnet is approaching the coil, the flux of this magnet induces current that subtracts from the current already circulating in the coil.
Without knowing the schematic, the only objection that I can make now is that you are measuring a voltage across a capacitor (behind a diode) and if that voltage is supposed to represent the energy recovered from the coil, then that is an accurate measure only without that 100Ω resistor across it and when starting with an empty capacitor.
QuoteI can't put the schematic together in my mind. Unlike Hoppy, I am not very good in analyzing wired connections in videos :( .
Circuit below.
Inductive kickback dissipated across R1.
Quote from: tinman on December 11, 2015, 11:37:02 AM
Is a magnet doing useful work when it alone is responsible for increasing the efficiency of a device or system ?.
The screen shots you see below are across the emitter/collector.
Enjoy.
https://www.youtube.com/watch?v=tVNABy8fSlI (https://www.youtube.com/watch?v=tVNABy8fSlI)
Brad
I agree with you Brad and believe it or not I was working on that a few months back.
What I found is, there's a special way we can arrange the magnets to make what I call a mono induced wave form (first scope shot)
When the timing is right, this special arrangement of magnets actually assists to get out of the attraction when the coil is energized (second scope shot), so it takes very power as you can see from the yellow current trace as most of the field in the coil has been induced by the
powerful attraction entry of the magnet arrangement. It's still a work in progress but I'm back refining it. My wish is to get the current to swing to the other side a little more ;D
I'm sure Erfinder will enjoy these scope shots ;)
Luc
PS, if anyone is needing to know, the probes and scope are on x1 and the CSR is 1% 0.05 Ohms
Quote from: Erfinder on December 12, 2015, 03:01:15 PM
Welcome to the party.....If I am not mistaken, the wave you are showing is not the one you want. I don't know how you are generating it, and I don't need to know. I have my own methods for doing it at least 12 different ways. Out of all those options, only one gives me what I am looking for.
Here is a short video showing the waves, the one you definitely don't want and the one you want. There are at least 4 which look like the one you want, you are in for some serious work if you chase this one, knowing exactly what you want helps you get through the maze. Good luck....and have fun!
https://vimeo.com/148736377 (https://vimeo.com/148736377)
Regards
Thank you Erfinder for taking the time to make a video for me demonstrating what you have found for some time.
I did say it needed improving as what I used to make the wave was not symmetrical.
I'm presently making the needed modifications and will re-post a new scope shot once satisfied.
Again, thanks for confirming what I believed to be the direction to go.
Regards
Luc
PS please feel free to give me some pointers when you see necessary ;)
When the induction stops , everything after that is free?
The collapsing field, the reverse magnetic polarity, and the opposite current, or are they all the same?
Were fools if we let all that go to waste.
Just saying artv
A video update on the recent build and test device
Link to video: https://www.youtube.com/watch?v=yN_VJ5YnoX0
Thanks for your interest
Luc
Here is a new video which will explain how the above scope shots are achieved and where the research is at this time.
Link to video: https://www.youtube.com/watch?v=tCUpmYLPdj0
Thanks for your interest.
Luc
I've started this other topic for anyone interested: http://overunity.com/16263/billions-in-change-alternative-electric-power-solution/new/#new
Thanks
Luc
Quote from: gotoluc on December 14, 2015, 01:53:44 PM
A video update on the recent build and test device
Link to video: https://www.youtube.com/watch?v=yN_VJ5YnoX0
Thanks for your interest
Luc
good video, thanks for update, interested to see results.
Quote from: gotoluc on December 14, 2015, 03:02:06 PM
Here is a new video which will explain how the above scope shots are achieved and where the research is at this time.
Link to video: https://www.youtube.com/watch?v=tCUpmYLPdj0 (https://www.youtube.com/watch?v=tCUpmYLPdj0)
Well made video.
I am missing the current waveform when the coil is shorted and the wheel is spinning (without external electric energy input)
Also, what is the inductance of that 0.3Ω coil ?
Quote from: verpies on December 14, 2015, 08:18:20 PM
Well made video.
I am missing the current waveform when the coil is shorted and the wheel is spinning (without external electric energy input)
The coil is not shorted, all I do is turn off the switch so you can see only the coils induced waveform.
Luc
Quote from: verpies on December 14, 2015, 08:18:20 PM
Also, what is the inductance of that 0.3Ω coil ?
The Inductance is 2.39mH
Luc
Quote from: gotoluc on December 14, 2015, 08:47:12 PM
The coil is not shorted, all I do is turn off the switch so you can see only the coils induced waveform.
I know, that's why I am missing the current waveform
Later, the coil becomes shorted by the low impedance of the Constant Voltage Source (your pulsing driver), so it would be nice to compare that to a case when the pulsing driver is not enabled but the coil is shorted through the CSR only.
Quote from: verpies on December 14, 2015, 09:11:15 PM
I know, that's why I am missing the current waveform
Later, the coil becomes shorted by the low impedance of the Constant Voltage Source (your pulsing driver), so it would be nice to compare that to a case when the pulsing driver is not enabled but the coil is shorted through the CSR.
Okay, I see where you're going. I'll do the test for you. One scope shot of switch on, one open coil and one coil shorted through 0.05 Ohm CSR
Will that be all for today?
Luc
Quote from: gotoluc on December 14, 2015, 09:32:19 PM
...and one for coil shorted through 0.05 Ohm CSR
Please make that scopeshot while the wheel is spinning and that means that you'll need to put some mechanical energy input into the wheel, because it will experience a braking action from the coil.
Quote from: gotoluc on December 14, 2015, 09:32:19 PM
Will that be all for today?
Yes :)
On another note.
It was brought to my attention that I'm receiving thumbs down on most of all my YouTube videos (see bellow) and I've been keeping an eye on it.
It doesn't really bother me but thought of bring it up as it's a member from here that's doing it as it happened to videos that are only listed at this forum.
The other noticeable pattern is it eventually get to 4 or 6 on most of the videos over a week. It's like it's the same person. However, in order for them to do that they would have to have 4 or 6 YouTube accounts. I know it's possible to get that many accounts but it boggles my mind to think a single person would go through the trouble of doing such a crazy thing :o ... and to think he is among us at this forum.
I wonder who hates me that much :'(
Just thought I would share the bad as well ;D
Luc
Quote from: gotoluc on December 14, 2015, 09:53:50 PM
On another note.
It was brought to my attention that I'm receiving thumbs down on most of all my YouTube videos (see bellow) and I've been keeping an eye on it.
It doesn't really bother me but thought of bring it up as it's a member from here that's doing it as it happened to videos that are only listed at this forum.
The other noticeable pattern is it eventually get to 4 or 6 on most of the videos over a week. It's like it's the same person. However, in order for them to do that they would have to have 4 or 6 YouTube accounts. I know it's possible to get that many accounts but it boggles my mind to think a single person would go through the trouble of doing such a crazy thing :o ... and to think he is among us at this forum.
I wonder who hates me that much :'(
Just thought I would share the bad as well ;D
Luc
Hi Luc,
I also have had the same thing, anywhere from 2 - 6. Some are out to try to stop this, at all costs! My private videos, that I have only shared with my close friends have only had 1. So yes some are deliberately Thumbs downing the video no matter how good the content is! Maybe this is a guide, the more thunbs down the closer you are? Certainly I know this is the case for me!!! 32 is the most for my best video and I know for sure this is my best, most detailed Video!!! Hahaha Oh well, its something that I guess spins some peoples wheels...
TinselKoala has also mentioned it also. It really is sad, some people must be very unhappy souls!!!
Keep up the good work mate, you're doing a good job!
Chris Sykes
hyiq.org
Thanks Chris for letting me know how the same thing is going on your channel.
Luc
Quote from: verpies on December 14, 2015, 09:40:43 PM
Please make that scopeshot while the wheel is spinning and that means that you'll need to put some mechanical energy input into the wheel, because it will experience a braking action from the coil.
Yes :)
Here you go verpies,
the first scope shot is with a 2vdc input to the 0.3 Ohm, 2.39mH motor coil which rotates the 2 pole rotor at 55 RPM
the second scope shot is the first capture of the voltage across the coil when I disconnect the switch
and the third scope shot is the first capture of the voltage and current across the 0.05 Ohm CSR when I disconnect the switch.
I did not rig up a motor to the rotor but I had the rotor RPM at 55.5 so it was close enough and I got the shot from the first shorting but it would loose momentum fast and you may be able to see it.
Hope this is good enough as setting up a motor to this wheel is not an easy task.
Anyways, in the next few days I'll be building another prototype which the shaft and bearing to the rotor will be an excellent quality DC motor, so we can get all the performance data we would like to know.
Luc
Quote from: gotoluc on December 14, 2015, 09:32:19 PM
Okay, I see where you're going. I'll do the test for you. One scope shot of switch on, one open coil and one coil shorted through 0.05 Ohm CSR
Will that be all for today?
Luc
this made me laugh.
: D
how long before the 'doers' ignore that 'testers' ? to me unless you have video of a device it's hard to call yourself a 'peer' (i include myself in that presently)
history shows that your paper qualification are worthless and that's what i think they are indeed.
it's good to get measurements and learn but better to experiment and make something work and then work backwards to learn, or simply build another device and see the different.
this makes me think this is why all the great innovations have been discovered by 'doers'
take Hendershot as a good example.
Luc it is just my (non peer) humble opinion, focus on symmetry and geometry in building just as much as 'ohms' 'volts' and 'currents' life is much more simple than the 'paperwork' guys make out.
It is my opinion that their aim is to tie you guys down in numbers and bullshitthink about this...
how much time would you spend arguing around the why of numbers with lot so these guys, as opposed to just building a whole new device with a variation on an original device that say didn't function as you expected.
my guess is at a point of tinkering you will see the simplicity in geometry, and this is
all related to the reward of innovationsome people here what their aim to do is to tie you practical guys down in numbers,
and take the reward of innovation away.i.e it is their aim to try to make this shit boring and waste your time. when it is clear, it is not boring you and Tinman (and others) are breaking new ground each day.
so just keep doing what you are doing, and hopefully more doers will participate.
Quote from: digitalindustry on December 15, 2015, 12:17:45 AM
this made me laugh.
: D
how long before the 'doers' ignore that 'testers' ?
to me unless you have video of a device it's hard to call yourself a 'peer' (i include myself in that presently)
history shows that your paper qualification are worthless and that's what i think they are indeed.
it's good to get measurements and learn but better to experiment and make something work and then work backwards to learn, or simply build another device and see the different.
this makes me think this is why all the great innovations have been discovered by 'doers'
take Hendershot as a good example.
Luc it is just my (non peer) humble opinion, focus on symmetry and geometry in building just as much as 'ohms' 'volts' and 'currents' life is much more simple than the 'paperwork' guys make out.
It is my opinion that their aim is to tie you guys down in numbers and bullshit
think about this...
how much time would you spend arguing around the why of numbers with lot so these guys, as opposed to just building a whole new device with a variation on an original device that say didn't function as you expected.
my guess is at a point of tinkering you will see the simplicity in geometry, and this is all related to the reward of innovation
some people here what their aim to do is to tie you practical guys down in numbers, and take the reward of innovation away.
i.e it is their aim to try to make this shit boring and waste your time.
when it is clear, it is not boring you and Tinman (and others) are breaking new ground each day.
so just keep doing what you are doing, and hopefully more doers will participate.
I had a Giggle at this - It is so true.
Digital, you're so right!!!
Nice post!
Chris Sykes
hyiq.org
Quote from: gotoluc on December 15, 2015, 12:13:49 AM
Here you go verpies,
the first scope shot is with a 2vdc input to the 0.3 Ohm, 2.39mH motor coil which rotates the 2 pole rotor at 55 RPM
the second scope shot is the first capture of the voltage across the coil when I disconnect the switch
and the third scope shot is the first capture of the voltage and current across the 0.05 Ohm CSR when I disconnect the switch.
I did not rig up a motor to the rotor but I had the rotor RPM at 55.5 so it was close enough and I got the shot from the first shorting but it would loose momentum fast and you may be able to see it.
Hope this is good enough as setting up a motor to this wheel is not an easy task.
Anyways, in the next few days I'll be building another prototype which the shaft and bearing to the rotor will be an excellent quality DC motor, so we can get all the performance data we would like to know.
Luc
I have seen this waveform before. But it was produced by turning the coil(with core) sideways.
With the coil end facing the pole of the rotor mag, the mag first cuts the approaching side of the coil producing say a positive output of the coil, and when the mag passes tdc, its field cuts the departing side of the coil producing a negative out from the coil.
When the coil is turned sideways, the rotor magnet only cuts one side of the coil producing a mostly all positive, or negative output from the coil, depending on magnet polarity and direction.
Mags
@verpies,
since I was so kind to take the time to do a test for you and provide you with the data I would like in return you to take a minute and write what you had in mind and how it compared to the test data I provided.
Thanks for your time and sharing.
Luc
I am still thinking.
Please verify that this is the circuit that generates these waveforms (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg468532/#msg468532).
Okay verpies,
I took the liberty to modify your circuit and re-post it in your above post so not to confuse anyone.
So it is now correct.
Please give us an update once you have come to a conclusions
Thanks for sharing
Luc
Excellent.
Now it is clear to me that the voltage drop across the CSR is added to the voltage appearing across the coil.
This lack of I&V signals separation was giving me problems because they are normally measured with both of the ground clips at point B (Current at C ...and voltage at A).
In case some of you may of missed it, I made a new video using my most recent build to support TinMan's topic on how magnets boost the efficiency of a pulse motors drive coil.
Here is TinMan's topic: http://overunity.com/16261/rotating-magnetic-fields-and-inductors/#.VnC_YEb_rIU
Here is my the video demo which demonstrates a 10x efficiency boost using the magnet rotor of my most recent build.
Link to test video: https://www.youtube.com/watch?v=TwKd7UG1Wb8
Luc
Quote from: Erfinder on December 15, 2015, 01:22:57 AM
It's unfortunate that the scope is still playing a central role here. It was refreshing to see that you see how well this orthogonal magnet almost completely neutralizes cogging. You correctly note the relation between the size of the stator and rotor magnet cross section, where the latter should be larger than the prior. You experienced the almost pulse like nature of the magnet and stator pole piece interaction, extremely important. There are giants in this area, you would be wise to identify them, one in particular. That particular individual has mastered this concept, and perfected it to a very high degree. When looking at his device, at first glance you don't even see the orthogonal relationship! This principle is not limited to spinning the magnet 90°! It took me what feels like ages to understand that.
If you can, reduce your rotor diameter. Find a better means for locking those magnets in place. Unfortunately, your desire to make this a strong motor will not pan out, especially when you use solenoids like the one you are demonstrating. Decrease your on time so that you are only pulsing during the peak of the wave, allowing current to flow only long enough for you to assist the pulse that you get for free. It should be clear that the pulse is at TDC, should also be clear that the voltage node is peaked at TDC, this is what makes this topology interesting, as under normal circumstances, TDC is the zero crossing. This concept lends itself for building an electric turbine, a properly built machine will be extremely fast, but will produce very little torque. It can and take the skin off your hand if the rotor has enough mass, and is spinning fast enough, however, owing to the very short duty, consumption drops when the shaft is loaded. The device performs best when duty is short. A torque machine based on concepts learned from spinning the magnets 90° uses "conventional" alternating magnets, these magnets are configured so as to induce orthogonal fields in a properly configured stator.
I spent years investigating this. Don't make the mistake most make and just throw something together, and broadcast it like its the end all. That will motivate many to look, and "replicate", but they will not think! They need to think about whats going on. They need to try and connect this concept with other concepts which have been demonstrated over the years, find the common links. Its not enough to throw something together for the sake of just doing it. Your desire to help others is noble, but you of all folk should know that its best to show a man how to fish so that he can feed himself. You aren't showing anyone how to fish by just throwing them the bone of turning the magnet sideways...
Regards
Thanks erfinder for taking the time to offer the above.
I'm sure I'll do honor to your Jewel
Luc
Quote from: gotoluc on December 15, 2015, 09:20:31 PM
In case some of you may of missed it...
I did miss it so thanks for pointing it out. Brilliant idea using the magnet at 90 degrees. Little gems of info like that really help us not spend time reinventing the wheel over and over and help move progress forward in the right direction. Video up voted and credit will be given where credit is due should I use that idea of yours.
Quote from: gotoluc on December 15, 2015, 08:25:31 PM
Please give us an update once you have come to a conclusions
I don't have hard conclusions yet because these probes are not connected how I'd like them to be and I don't have any scopeshots of your system when the rotor is not moving.
...but I strongly suspect that the conclusion is
Case #3 (http://overunity.com/16261/rotating-magnetic-fields-and-inductors/msg468683/#msg468683)
P.S.
I am also confused how the blue trace can be non-zero when doing a scopeshot with a moving rotor and
this circuit (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/153869/).
Quote from: verpies on December 16, 2015, 04:59:30 PM
I don't have hard conclusions yet because these probes are not connected how I'd like them to be and I don't have any scopeshots of your system when the rotor is not moving.
...but I strongly suspect that the conclusion is Case #3 (http://overunity.com/16261/rotating-magnetic-fields-and-inductors/msg468683/#msg468683)
P.S.
I am also confused how the blue trace can be non-zero when doing a scopeshot with a moving rotor and this circuit (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/153869/).
Thanks for the update and please find the scopeshots of the coil with the 2 pole rotor not moving.
There is 1.57vdc across the 100 Ohm flyback tank load.
Let me know if you need anything else or have any other conclusions.
Thanks
Luc
OK, that shows me classical inverted exponential current waveform for an LR circuit.
Was the magnet at the coil's core when this scopeshot was taken...or was it away from the core? - I am asking because of saturation concerns.
Also, the voltage should be constant while the transistors are ON. Since it is not (it is sloping down), I must assume that RDS_ON of these MOSFETS (or wire resistance) is allowing for this voltage drop.
I expected most ringing when the transistor shuts OFF and interupts the current, but that is not the case because D1 takes over, starts conducting and keeps the current flowing trough L1.
The CSR does not show this current because it is not in the L1, D1, C2 circuit.
So the question remains: What starts ringing after a delay and what determines this delay. D1 stopping its conduction ? Drift Step recovery?
Anyway, when this current waveform (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/153930/) (when the rotor is stationary) is superposed on the current waveform obtained in this circuit (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/153869/) (while the rotor is moving), the result should yield the same current waveform as when the stator is powered externally (while the rotor is moving ...at the same speed).
If it does not, then you have an anomaly.
Are you open to moving the CSR and the scope probes according to the schematic below, to clear these questions?
i hooked up my 8 magnet (all north up) one coil pulse motor like in this diagram (12V from a battery):
http://overunity.com/16261/rotating-magnetic-fields-and-inductors/msg468654/#msg468654
Difference is that i have 1K across the flyback cap (1000uF) and using a MOSFET (IRF530)
I use my SG to switch the MOSFET (50 Hz) and measure the signals as seen below in screenshot 1
Blue is the gate signal referenced to ground
green is the current (probe) at the junction coil / drain 500ma/div.
purple is the drain signal referenced to ground
When using my SG to drive the MOSFET, i cannot hook up my probe ground to point C in the above mentioned diagram.
I have 22.42V in the flyback cap.
There is no difference when stopping the rotor, see screenshot 2.
Itsu
Quote from: itsu on December 17, 2015, 05:15:33 PM
i hooked up my 8 magnet (all north up) one coil pulse motor like in this diagram (12V from a battery):
http://overunity.com/16261/rotating-magnetic-fields-and-inductors/msg468654/#msg468654 (http://overunity.com/16261/rotating-magnetic-fields-and-inductors/msg468654/#msg468654)
I was trying to adapt to Tinman's design when drawing that diagram, thus it was not optimal.
This circuit (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/153957/)
* and probe hookups are much more preferable because:
- it allows the potential of the scope's ground to vary only minimally.
- the CSR also measures the current in the L1,D1,C2 circuit
- it separates I & V signals (Itsu's setup does too)
Anyway, your green current waveform is not showing strong signs of flattening out, so I judge that the pulse width is around 2*Tau. That's better than most people do but still beyond the
break-even point (http://overunity.com/16261/rotating-magnetic-fields-and-inductors/msg468599/#msg468599) shown
here (http://overunity.com/16261/rotating-magnetic-fields-and-inductors/dlattach/attach/153925/).
Also, what is the current waveform induced in the coil measured in
this circuit (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/153869/) ...when the rotor is spun by hand (or motor)?
Quote from: itsu on December 17, 2015, 05:15:33 PM
I have 22.42V in the flyback cap.
The 22.4V in the recovery cap is about correct. The Source-Drain voltage waveform shows a peak close to 40V and with 12V power supply that gives 40V - 12V = 28V and the difference from 22.4V is probably wasted between pulses in that resistor in parallel with the cap.
Cheers
* iso-drivers and double MOSFETs are optional
Quote from: verpies on December 17, 2015, 01:21:43 PM
OK, that shows me classical inverted exponential current waveform for an LR circuit.
Was the magnet attracted to the coil's core when this was taken...or was it away from the core? - I am asking because of saturation concerns.
The magnet was far away from the core.
Quote from: verpies on December 17, 2015, 01:21:43 PM
Also, the voltage should be constant while the transistors are ON. Since it is not (it is sloping down), I must assume that RDS_ON of these MOSFETS (or wire resistance) is allowing for this voltage drop.
Part of the cause of the voltage drop on the above scope shot is the power supply is limited to 3 amps. Even though I have a 100,000uf tank before the switch my DMM displays a voltage drop when the switch closes. So I added a second power supply and it helped a little. See below new scope shot. However, the current is so demanding
(without the magnet rotor assisting) that I have nothing else available that can mantain the 2vdc when the switch closes.
Quote from: verpies on December 17, 2015, 01:21:43 PM
I expected most ringing when the transistor shuts OFF, but that is not the case because D1 starts conducting and keeps the current flowing trough L1.
The CSR does not show this current because it is not in the D1, C1 circuit.
So the question remains: When does it start ringing after a delay and what determines this delay. D1 stopping conduction.
Anyway, when this current waveform (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/153930/) is superposed on the current waveform obtained in this circuit (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/153869/) while the rotor is moving, the result should yield the current waveform when the stator is powered externally, while the rotor is moving.
If it does not, then you have an anomaly.
I'll look at this in time.
Quote from: verpies on December 17, 2015, 01:21:43 PM
Are you open to moving the CSR and the scope probes according to the schematic below, to clear these questions?
Yes I would be but unfortunatly I've already built another test device and removed the coil and 1 inch cube magnets from the rotor.
Lets see what the next test device does with the new CSR.
Btw where would the scope grounds be located?
Thanks
Luc
The scope grounds are marked on the schematic with little letters G on black background (photo negative)
Quote from: itsu on December 17, 2015, 05:15:33 PM
i hooked up my 8 magnet (all north up) one coil pulse motor like in this diagram (12V from a battery):
http://overunity.com/16261/rotating-magnetic-fields-and-inductors/msg468654/#msg468654
Difference is that i have 1K across the flyback cap (1000uF) and using a MOSFET (IRF530)
I use my SG to switch the MOSFET (50 Hz) and measure the signals as seen below in screenshot 1
Blue is the gate signal referenced to ground
green is the current (probe) at the junction coil / drain 500ma/div.
purple is the drain signal referenced to ground
When using my SG to drive the MOSFET, i cannot hook up my probe ground to point C in the above mentioned diagram.
I have 22.42V in the flyback cap.
There is no difference when stopping the rotor, see screenshot 2.
Itsu
Itsu
Is it hard to change every second magnet in your rotor to south up ?.
It is good to see that even with the all north up rotor,there was no energy cost to spin your rotor as MH seems to think there should be.
Brad
Quote from: gotoluc on December 17, 2015, 06:03:41 PM
Part of the cause of the voltage drop on the above scope shot is the power supply is limited to 3 amps. Even though I have a 100,000uf tank before the switch my DMM displays a voltage drop when the switch closes. So I added a second power supply and it helped a little. See below new scope shot. However, the current is so demanding (without the magnet rotor assisting) that I have nothing else available that can mantain the 2vdc when the switch closes.
That 100mF cap should not discharge that much in several milliseconds, but I can live with the voltage drop if I know where it comes from.
Anyway, the good way to combat the excessive input current is to decrease the pulse width and increase the voltage. Note my statement about the break even point that I've just mentioned to Itsu and the graph below.
Practically, if you are not seeing sawtooth or triangular current waveforms flowing through that inductor during energizing it by a constant voltage source and a resistance, then you are wasting energy in that resistance.
Look at the dashed yellow curve: cut that curve with a vertical line at 0.5 Tau and you get a triangle (sawtooth wave) - that's good.
...but cut that curve at 2 Tau and you don't get a triangle any more ...more like a rounded triangle - that's bad.
Quote from: tinman on December 17, 2015, 06:11:42 PM
Itsu
It is good to see that even with the all north up rotor,there was no energy cost to spin your rotor as MH seems to think there should be.
Brad
You have zero data to state that there was no energy cost for Itsu to spin his rotor. What is readily apparent when looking at his waveforms is that the current draw increases when he removes his rotor, just like it happens in your setup.
I do not see any difference in Itsu's current waveforms (green) ...maybe, just maybe more amplitude and more flattening in the stopped case, but I would not swear by it.
Looking at the pair of waveforms in the center of the screen, look where each one crosses relative to the (+1, +1) major division point. The one with the rotor in place passes more or less right through the point, while for the one without the rotor in place, the lower limit of the noise band clears the point. The area under the no-rotor waveform is clearly larger.
I'm having trouble getting decent spikes.
The first thing i wanted to do was ascertain the difference in the magnetic fields of two inductors when supplied with the same input.
I know it's physics fact but i like to hold it in my hands.
Inductor 1 is 0.25mm wire and very close to 40 ohms, weighing ~50g.
Inductor 2 is 0.355mm wire and very close to 40 ohms, weighing ~200g.
My Henry meter is broken so no idea there, but the important thing is the comparison, if we charge coil 1, cut the power, dump to cap, how many volts do we have, then same for coil 2 ... coil 3 is 800g ...
I made a small pulse motor, saw regular 60V spikes, normal RMS was around 9.
But i don't like Bedini motors, Fakient spikes and the rest so tried it solid-state.
I replaced the trigger coil with a pulse-train, and tested coils 1 and 2 in place of the normal power coil.
The spikes are shit, i can only assume it's because there isn't a Neo flying overhead and inducing into the power as well as the trigger coil.
Perhaps the larger spikes are due to opposing, collapsing fields, as mentioned everywhere in the literature, since simply cutting juice doesn't give a large collapse field.
Y.
Quote from: Yttrium on December 17, 2015, 07:39:44 PM
if we charge coil 1, cut the power, dump to cap, how many volts do we have
V=i
L*(L/C)
½, where i
L is the current flowing through the coil when the switch (S1) opens.
L is the inductance of the coil and C is the capacitance of the recovery capacitor and V is the final voltage across that capacitor (providing you do not have any resistor connected across it).
In real world you have to subtract the diode's voltage drop from V, too (usually 0.6V)
...and the time (t) that this transfer of energy from the coil (L) to the capacitor (C) takes, is:
t=π(LC)
½
Here is my new test device which uses the coils flyback to do work.
Link to video demo: https://www.youtube.com/watch?v=K8kwdrHCyig
Same observation, it's more efficient with the coil and magnet rotor.
Luc
Quote from: gotoluc on December 18, 2015, 12:08:05 AM
Here is my new test device which uses the coils flyback to do work.
Link to video demo: https://www.youtube.com/watch?v=K8kwdrHCyig (https://www.youtube.com/watch?v=K8kwdrHCyig)
The video is marked as "Private" and I cannot watch it.
Which circuit and probe positions were used to make these scopeshots? The current flowing through a coil cannot suddenly jump up like on that annotated scopeshot below !
Also, is Ch2 inverted?
Quote from: verpies on December 18, 2015, 01:39:18 AM
The video is marked as "Private" and I cannot watch it.
Which circuit and probe positions were used to make these scopeshots? The current flowing through a coil cannot suddenly jump up like on that annotated scopeshot below !
Also, is Ch2 inverted?
Yes-odd how the current shot up as soon as the voltage inverted,which would seem like the switch off time.
Verpies.
If we wound an inductor with say 400 turn's,but at the 200 turn mark we place a 1 ohm CVR in series with the next 200 turns. So we have a series as such-->200 turns of wire on the core,then bring the wire outside the former,through a 1 ohm CVR,then bring the wire back into the former,and continue to wind on our last 200 turns. Will this CVR now still measure the current flowing through the coil,or will it be seeing the inverted voltage across the winding's when the transistor opens ?.
Im thinking that there should be no voltage reading at the middle point of the windings,but from that mid point to either end of the winding's,we should see a voltage that is 1/2 that of the supplied voltage?.
Brad
Quote from: tinman on December 18, 2015, 05:00:07 AM
If we wound an inductor with say 400 turn's,but at the 200 turn mark we place a 1 ohm CVR in series with the next 200 turns. So we have a series as such-->200 turns of wire on the core,then bring the wire outside the former,through a 1 ohm CVR,then bring the wire back into the former,and continue to wind on our last 200 turns. Will this CVR now still measure the current flowing through the coil,or will it be seeing the inverted voltage across the winding's when the transistor opens ?.
It will read the current flowing through both coils if you drive it end-to-end.
The midpoint current might only show a difference when the coil's length is comparable to 1/2 of its driving wavelength and standing waves form -
read this (http://www.teslatechnologyresearch.com/corum/index.htm).
At frequencies appearing in motors - this is a non-issue as they can be treated as lumped-element systems, in which the current is distributed uniformly along the coil.
Quote from: tinman on December 18, 2015, 05:00:07 AM
Yes-odd how the current shot up as soon as the voltage inverted,which would seem like the switch off time.
Yes, it does seem like the switch-off time.
I suspect this odd current waveform is caused by improper scope probe positions.
Can you take a look at
this schematic (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/153957/) and let me know if you can spot the ground clips of the scope probes, appearing on this schematic around point B ? ...or are they too small?
Quote from: verpies on December 18, 2015, 01:39:18 AM
The video is marked as "Private" and I cannot watch it.
Which circuit and probe positions were used to make these scopeshots? The current flowing through a coil cannot suddenly jump up like on that annotated scopeshot below !
Also, is Ch2 inverted?
The video is now public, sorry about that.
Link to video: https://www.youtube.com/watch?v=K8kwdrHCyig (https://www.youtube.com/watch?v=K8kwdrHCyig)
The circuit is as below.
I inverted Ch2 to keep the scope shots looking like the previous or else the voltage would be inverted to the current. Do you want to see the voltage inverted to the current?
Luc
Quote from: verpies on December 18, 2015, 01:39:18 AM
The current flowing through a coil cannot suddenly jump up like on that annotated scopeshot below !
That's just noise from the mosfet having a hard time switching off. See close up below.
I also included a non inverted scope shot with magnet rotor to coil.
Luc
Quote from: gotoluc on December 18, 2015, 09:35:56 AM
The circuit is as below.
I inverted Ch2 to keep the scope shots looking like the previous or else the voltage would be inverted to the current. Do you want to see the voltage inverted to the current?
Inverting Ch2 was a nice touch. I knew it had to be done to keep up the trace appearances with these probe placements, but I did not want to overwhelm you with suggestions.
Quote from: verpies on December 18, 2015, 08:48:43 AM
The midpoint current might only show a difference when the coil's length is comparable to 1/2 of its driving wavelength and standing waves form - read this (http://www.teslatechnologyresearch.com/corum/index.htm).
At frequencies appearing in motors - this is a non-issue as they can be treated as lumped-element systems, in which the current is distributed uniformly along the coil.
QuoteIt will read the current flowing through both coils if you drive it end-to-end.
Rather than say both coil's,lets just see it as one coil where a 1 ohm CVR is tapped into the mid point of the turn number. Anyway,lets say the DUT is my circuit on the other thread-the SSG circuit. In this case,the CVR that is in the center of the coil windings should read a higher average current flowing through the coil than that which is flowing into the coil--where that current is also viewed by way of another 1 ohm CVR.
Correct?.
Brad.
Quote from: gotoluc on December 18, 2015, 10:47:51 AM
That's just noise from the mosfet having a hard time switching off. See close up below.
That should not be happening because D1 should maintain the current at the same level as before.
Of course, when C2 is charged up then not much current can flow through D1.
Please disconnect R2 and try to manually discharge C2 and catch it on the scope when C2 is charging up from 0V, for the first time. The upward current ramp should be followed by a downward ramp and there should be no oscillations. That downward ramp can tell us a lot about the maximum efficiency of the energy recovery possible.
Quote from: verpies on December 17, 2015, 05:27:34 PM
I was trying to adapt to Tinman's design when drawing that diagram, thus it was not optimal.
This circuit (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/153957/)* and probe hookups are much more preferable because:
- it allows the potential of the scope's ground to vary only minimally.
- the CSR also measures the current in the L1,D1,C2 circuit
- it separates I & V signals (Itsu's setup does too)
Anyway, your green current waveform is not showing strong signs of flattening out, so I judge that the pulse width is around 2*Tau. That's better than most people do but still beyond the break-even point (http://overunity.com/16261/rotating-magnetic-fields-and-inductors/msg468599/#msg468599) shown here (http://overunity.com/16261/rotating-magnetic-fields-and-inductors/dlattach/attach/153925/).
Also, what is the current waveform induced in the coil measured in this circuit (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/153869/) ...when the rotor is spun by hand (or motor)?
The 22.4V in the recovery cap is about correct. The Source-Drain voltage waveform shows a peak close to 40V and with 12V power supply that gives 40V - 12V = 28V and the difference from 22.4V is probably wasted between pulses in that resistor in parallel with the cap.
Cheers
* iso-drivers and double MOSFETs are optional
verpies,
i will change my circuit to the one which is preferable, including a 555 timer for the 50 Hz signal as to be able to place my scopes ground probes anywhere in the circuit.
The pulse width was 31%, see earlier blue gate signal, lowering it even 1% maked the rotor loose sync with the SG.
The current waveform in that circuit is as in the screenshot below, where blue is the signal across a 0.1 ohm 1% csr, and green my current probe signal (100mA/div.).
(unfortunatly the current probe offset is toasted appearently (due to kacher activities), so never mind the green trace offset value)
The coil measures 11 Ohm and 37.8mH @ 100Hz.
Wire is AWG 24 and core is made of cut-up welding rods
Tinman
QuoteItsu
Is it hard to change every second magnet in your rotor to south up ?.
It is good to see that even with the all north up rotor,there was no energy cost to spin your rotor as MH seems to think there should be.
Brad
the magnets are superglued in, so i doubt it, but i will try gently
Itsu
Quote from: verpies on December 18, 2015, 11:15:44 AM
That should not be happening because D1 should maintain the current at the same level as before.
Of course, when C2 is charged up then not much current can flow through D2.
Please disconnect R2 and try to manually discharge C2 and catch it on the scope when C2 is charging up from 0V, for the first time. The upward current ramp should be followed by a downward ramp and there should be no oscillations. That downward ramp can tell us a lot about the maximum efficiency of the energy recovery possible.
Verpies.
Watching the video,i see Luc was running a brushed DC motor from the flyback- storage cap. Could that noise that is being seen be brush noise from the DC motor ?.
Brad
Quote from: tinman on December 18, 2015, 11:27:14 AM
Watching the video,i see Luc was running a brushed DC motor from the flyback- storage cap. Could that noise that is being seen be brush noise from the DC motor ?.
No, no!
That noise is definitely caused by lack of D1 conduction after switch-off and if C2 is emptied before the switch-off then the t
OFF time will lengthen and the noise will disappear.
The current waveform, when the coil is not under the influence of the rotor, should look like this:
Quote from: Erfinder on December 18, 2015, 06:04:22 AM
I see you have identified the MW waveform.
Thanks again for sharing some details.
Let's have a closer look at the MW waveform you have identified that my current 2 pole magnet rotor is producing.
It looks quite different then the one in your picture, so I was wondering if that has to do that mine is produced using strictly a magnet and yours may be a magnet and iron rotor combination?
Thanks for sharing
Luc
Quote from: verpies on December 18, 2015, 11:15:44 AM
That should not be happening because D1 should maintain the current at the same level as before.
Of course, when C2 is charged up then not much current can flow through D2.
You must mean "not much current can flow through D1" and not D2 as you wrote above right?
Quote from: verpies on December 18, 2015, 11:15:44 AM
Please disconnect R2 and try to manually discharge C2 and catch it on the scope when C2 is charging up from 0V, for the first time. The upward current ramp should be followed by a downward ramp and there should be no oscillations. That downward ramp can tell us a lot about the maximum efficiency of the energy recovery possible.
Would a low value resistor across C2 not do the same?
Luc
Quote from: itsu on December 18, 2015, 11:24:41 AM
I will change my circuit to the one which is preferable, including a 555 timer for the 50 Hz signal as to be able to place my scopes ground probes anywhere in the circuit.
If you have four of these cheap HEF4047B chips (CD4047) then a universal motor sequencer can be built that will have the capability of positioning the ON-pulse anywhere and have an additional output for emptying C2 before each "flyback" pulse, for accurate recovered energy measurement..
Quote from: itsu on December 18, 2015, 11:24:41 AM
The pulse width was 31%, see earlier blue gate signal, lowering it even 1% made the rotor loose sync with the SG.
Yup, you have to compensate lower pulse width with higher amplitude (supply voltage).
Quote from: itsu on December 18, 2015, 11:24:41 AM
The current waveform in that circuit is as in the screenshot below, where blue is the signal across a 0.1 ohm 1% CSR, and green my current probe signal (100mA/div.).
(unfortunately the current probe offset is toasted apparently (due to Kacher activities), so never mind the green trace offset value)
Was that current waveform obtained from
this (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/153869/) trivial circuit?
It is the induced current waveform shape that matters the most anyway - with alternating N, S magnet poles (like Gotoluc suggests) this waveform will appear significantly different.
Quote from: gotoluc on December 18, 2015, 11:52:51 AM
You must mean "not much current can flow through D1" and not D2 as you wrote above right?
Right. That's what happens when I'm writing in a hurry.
I corrected it already.
Quote from: gotoluc on December 18, 2015, 11:52:51 AM
Would a low value resistor across C2 not do the same?
No, because the transfer of energy from a coil to a cap is much more efficient than from a coil to the resistor.
With a low value resistor the T
OFF time will be very long. With a small capacitor it will be short and the final voltage across the capacitor will be an accurate measure of the recovered energy (minus diode losses)
...but you can try a low value resistor (or even a 0Ω short) to see that the noise at switch-OFF really disappears. If it does, then emptying C2 will accomplish the same feat and more.
Quote from: verpies on December 18, 2015, 12:01:43 PM
with alternating N, S magnet poles (like Gotoluc suggests) this waveform will appear significantly different.
Sorry verpies but I would like to clear up the way you wrote the above.
I don't consider my rotor to be Alternating magnets. The magnets are rotated 90 degrees from the conventional way and all oriented the same way.
So each magnets give you instantaneous N/S which produces the unusual waveform below.
You simply power (TDC) at the peaks
Luc
Quote from: gotoluc on December 18, 2015, 12:16:03 PM
Sorry verpies but I would like to clear up the way you wrote the above.
Don't be sorry. The devil is in the details.
With your rotor polarization the coil will see bidirectional flux. In Tinman's case the flux will be bidirectional, too but the flux amplitude should be larger.
With Itsu's rotor polarization the coil will see unidirectional flux (unipolar) with the smallest flux amplitude of the three cases.
Quote from: tinman on December 18, 2015, 11:11:14 AM
Rather than say both coil's,lets just see it as one coil where a 1 ohm CVR is tapped into the mid point of the turn number. Anyway,lets say the DUT is my circuit on the other thread-the SSG circuit. In this case,the CVR that is in the center of the coil windings should read a higher average current flowing through the coil than that which is flowing into the coil--where that current is also viewed by way of another 1 ohm CVR.
Correct?.
No, unless the wavelength of the driving waveform is close to the length of the coil as described in
this article (http://www.teslatechnologyresearch.com/corum/index.htm).
Kirchhoff's Current Law is the one that describes the conservation of currents in circuit branches, unless significant EM radiation occurs according to Maxwell at higher frequencies. That's how radio antennas work.
Quote from: verpies on December 18, 2015, 12:01:43 PM
If you have four of these cheap HEF4047B chips (CD4047) then a universal motor sequencer can be built that will have the capability of positioning the ON-pulse anywhere and have an additional output for emptying C2 before each "flyback" pulse, for accurate recovered energy measurement..
Yup, you have to compensate lower pulse width with higher amplitude (supply voltage).
Was that current waveform obtained from this (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/153869/) trivial circuit?
It is the induced current waveform shape that matters the most anyway - with alternating N, S magnet poles (like Gotoluc suggests) this waveform will appear significantly different.
No 4047's here.
Yes, that current waveform was obtained by that trivial circuit.
Itsu
Quote from: verpies on December 18, 2015, 12:11:15 PM
...but you can try a low value resistor (or even a 0Ω short) to see that the noise at switch-OFF really disappears. If it does, then emptying C2 will accomplish the same feat and more.
I tried it with a 1 Ohm and a 0.1 Ohm on C2 (see shots) and there is a difference but it does not fix it. Even a short does no fix it.
I found this gets worse as I raise the voltage. I'll scope the opto and gate of the switch to see what is going on there.
Luc
There is no spiking noise on your waveform for the current - that's impossible. It's just spiking noise on the voltage waveform channel cross-coupling to the current waveform channel. The way you have the probe cables arranged may have something to do with it.
Quote from: gotoluc on December 18, 2015, 01:38:13 PM
I tried it with a 1 Ohm and a 0.1 Ohm on C2 (see shots) and there is a difference but it does not fix it. Even a short does no fix it.
So there is something seriously wrong.
Are you sure that you have the
point A (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/153986/) connected to the negative terminal of C2 and the CSR ?
Quote from: verpies on December 18, 2015, 02:29:12 PM
So there is something seriously wrong.
Are you sure that you have the point A (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/153986/) connected to the negative terminal of C2 ?
I think I see the problem now. Point A was connected to C1 as per you schematic, but the negative leg of C2 was on the other side of the CSR. However, this means the flyback will be going though 0.05 Ohms of resistance and waste a little ;) .
I'm just not use to looking or using schematics :P
I'll try it now and post the new shots
Luc
Quote from: gotoluc on December 18, 2015, 02:42:52 PM
However, this means the flyback will be going though 0.05 Ohms of resistance and waste a little ;) .
Yes, but negligibly at 0.05Ω. The up side is that this configuration allows you observe the current pulse charging the recovery capacitor C2.
Later, when all the measurements are done, you can remove the CSR from the L1, D1, C2 circuit ...or remove it altogether.
Okay, sorry about the schematic screw up.
Please find the new shots. First is with the rotor then the next is without and below is the same but with math.
Thanks for your help and interest
Luc
It looks much better now. The downward ramp, that illustrates the current flowing into the recovery capacitor (C2) is clearly visible. If the capacitor was empty, then this current would be a half of a sinewave.
There is still some oscillation at switch-off but that must be caused by the by the lack of starpoint wire connections.
Now it would be good to scope be voltage across the capacitor as it is rising from 0V (even after manual discharge), and if you have more than 2 channels on your scope, then that could scope that voltage simultaneously between points B and D (ground clip on B just like the other channels) with the downside of the small CSR's voltage drop getting added. Pulsing the MOSFETs only once will make C2 measurements from 0V, easier.
I setup the circuit as in the below diagram, only i still have all 8 magnets facing north to the coil.
I use a 555 timer to sync the rotor with, presently running at 50Hz @ 31% duty cycle (12V input).
So i put the scope ground points as mentioned in the diagram (point B)
See screenshot 1 for a running situation, screenshot 2 with the rotor stopped (coil inbetween 2 magnets).
Blue is the CSR voltage (0.1 Ohm @ 1%) point A to ref. B
Green is my current probe set at 500mA/div. taken next to the CSR
Purple is the drain voltage point C to ref. B
yellow is the cap voltage point D to ref. B (not shortend/emptied)
The DMM across the cap showed 5.8V when running, and 5.9 when stopped.
Regards Itsu
Fantastic work like usual Itsu. If I can suggest something it would be to not bother reversing some of the magnets in your rotor. They are glued in place and there is nothing to be gained and nothing to be learned by doing all of the work to reverse them.
For you scope traces, this time after taking a one-minute look, I cannot see any appreciable differences between the two waveforms. However, your multimeter is showing slightly higher voltage across the capacitor which is telling us that slightly more current is being drawn by the drive coil when the rotor is stopped. This is to be expected because there is no influence from any counter-EMF in the drive coil from the moving magnets in the spinning rotor.
With respect to attempting to measure the added power that has to be pumped into the drive coil to make the rotor spin, that could be done. However, that would require some thought and careful preparation and developing the right measurement regime to detect it. Right now what is happening is that when the rotor is in place and spinning, the reduced current draw due to the counter-EMF induced into the drive coil is overshadowing the extra power that is added to make the rotor spin. In other words, the preliminary analysis is that more current draw is reduced due to the counter-EMF than the extra current draw that is required to make the rotor spin. There are two opposite effects that are happening at the same time with respect to the current draw and it is not necessarily that easy to separate them from each other.
Quote from: verpies on December 18, 2015, 04:53:50 PM
It looks much better now. The downward ramp, that illustrates the current flowing into the recovery capacitor (C2) is clearly visible. If the capacitor was empty, then this current would be a half of a sinewave.
There is still some oscillation at switch-off but that must be caused by the by the lack of starpoint wire connections.
Now it would be good to scope be voltage across the capacitor as it is rising from 0V (even after manual discharge), and if you have more than 2 channels on your scope, then that could scope that voltage simultaneously between points B and D (ground clip on B just like the other channels) with the downside of the small CSR's voltage drop getting added. Pulsing the MOSFETs only once will make C2 measurements from 0V, easier.
Okay verpies, you should be happy with the below scope shots as C2 is drained to 0v before the next on time.
C2 is now a 60uf motor run cap and has a 47 Ohm 10w resistor across it.
The DC motor is not connected (assisting) in this test.
Input is still at 20vdc
Also, the switch off is much cleaner, see last shot.
Thanks for your help
Luc
Quote from: MileHigh on December 18, 2015, 05:35:50 PM
Fantastic work like usual Itsu.
For you scope traces, this time after taking a one-minute look, I cannot see any appreciable differences between the two waveforms. However, your multimeter is showing slightly higher voltage across the capacitor which is telling us that slightly more current is being drawn by the drive coil when the rotor is stopped. This is to be expected because there is no influence from any counter-EMF in the drive coil from the moving magnets in the spinning rotor.
With respect to attempting to measure the added power that has to be pumped into the drive coil to make the rotor spin, that could be done. However, that would require some thought and careful preparation and developing the right measurement regime to detect it. Right now what is happening is that when the rotor is in place and spinning, the reduced current draw due to the counter-EMF induced into the drive coil is overshadowing the extra power that is added to make the rotor spin. In other words, the preliminary analysis is that more current draw is reduced due to the counter-EMF than the extra current draw that is required to make the rotor spin. There are two opposite effects that are happening at the same time with respect to the current draw and it is not necessarily that easy to separate them from each other.
MH,
Did you read what you wrote? You just posted that the magnets are generating more power back into the coil than the extra power it takes to turn the rotor. That has to mean the system is more efficient with the rotor than without. I am amazed you can't see that. Every single person that has done this test for you has shown you that either the rotor made no difference (only one as I recall) or in all other cases the input power went down. I just don't understand how you can keep saying adding a rotor is not making these systems more efficient.
And by the way you never did comment about why so many industrial motors now have permanent magnets in them since you claim magnets can't do any work.
Respectfully,
Carroll
Carroll:
I appreciate your comments but I think that they will be more appropriately addressed after Tinman posts his test results in his thread, presumably in a day or two. Then I will gladly pick up the discussion with you in that thread.
MileHigh
Quote from: MileHigh on December 18, 2015, 06:30:27 PM
Carroll:
I appreciate your comments but I think that they will be more appropriately addressed after Tinman posts his test results in his thread, presumably in a day or two. Then I will gladly pick up the discussion with you in that thread.
MileHigh
Thanks MH for taking that discussion somewhere else.
Regards
Luc
Quote from: itsu on December 18, 2015, 05:20:30 PM
I setup the circuit as in the below diagram, only i still have all 8 magnets facing north to the coil.
I use a 555 timer to sync the rotor with, presently running at 50Hz @ 31% duty cycle (12V input).
Very clean waveforms - you should collaborate with Gotoluc so he can obtain the same.
I have extended the light blue current ramp-up on your scopeshot without rotor and came to the conclusion that your ON-pulse lasts for approximately 1.6Tau and that means you have burned 1.5x more energy in the resistance than the energy you had built up in the coil as the magnetic field.
If you decrease the ON time to 0.5Tau than this ratio will fall to 0.37x. Of course narrower pulse needs to be compensated with higher supply voltage.
The voltage across C2 (yellow Ch1) varies between 4V and 15V so the energy gain of this capacitor is ½C*( (15V)
2 - (4V)
2 ). This is how many Joules are recovered per 1 pulse.
At "switch-OFF" (dark blue vertical line) approximately 1.1A flows through the coil, so ½L*(1.1A)
2 Joules of energy is stored in the coil at that time.
I encourage you to plug the inductance of your coil and the capacitance of your recovery capacitor into these formulas to see how many Joules of energy each one of them represents. (I would do this but I am not sure what L & C are).
Finally, divide these two energies to see what your recovery efficiency is.
If you experiment further you will find out that this efficiency increases if C2 is completely discharged before the "flyback" pulse and no resistor is connected across C2 to burn the recovered energy.
BTW:
The red curve is
this curve (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/153963/) and the black vertical line is at the break-even-point in time (1.15Tau)
Quote from: gotoluc on December 18, 2015, 06:12:21 PM
Okay verpies, you should be happy with the below scope shots as C2 is drained to 0v before the next on time.
C2 is now a 60uf motor run cap and has a 47 Ohm 10w resistor across it.
Oh, yes I am very happy.
The current down ramp down has a nice partial sinusoidal shape (voltage waveform too) which is indicative of an efficient resonant power transfer from L1 to C2.
C2 gets charged from 0V to almost 90V peak, which represents ½*60uf*(90V)
2 = 243mJ of energy recovered from 1 "flyback" pulse. Energy burned up by R2 is not accounted for. Actually, you can delete R2 if measuring 1 pulse from the drained C2.
R2 is not needed to measure the recovered energy (only the voltage across C2 is).
At switch-off time there is close to 17A flowing through the coil so that represents ½*L*(17A)
2 Joules of energy.
I am not sure what the inductance of your coil (L) is now but if you plug it into the formula then you will obtain how much energy is stored in the coil at switch-off. Dividing this energy, by the energy recovered by C2, will give you the efficiency of your recovery process alone. An interesting number, albeit not the entire story...
P.S.
Your current ramp-up is flattening out similarly to Itsu's (it is not a nice triangle) but the reason for this is different.
I have analyzed Itsu's scopeshot without the rotor but your scopeshot is with the rotor. Consequently Itsu's current ramp-up is flattening out because he is getting too close to the V/R limit, but your pulse is flattening out because the current induced by the moving magnet in the rotor is getting superimposed on the LR current ramp. Sorry for comparing apples with oranges, but you gave me only a scopeshot with the rotor's influence.
Quote from: MileHigh on December 18, 2015, 05:35:50 PM
Fantastic work like usual Itsu. If I can suggest something it would be to not bother reversing some of the magnets in your rotor. They are glued in place and there is nothing to be gained and nothing to be learned by doing all of the work to reverse them.
For you scope traces, this time after taking a one-minute look, I cannot see any appreciable differences between the two waveforms. However, your multimeter is showing slightly higher voltage across the capacitor which is telling us that slightly more current is being drawn by the drive coil when the rotor is stopped. This is to be expected because there is no influence from any counter-EMF in the drive coil from the moving magnets in the spinning rotor.
With respect to attempting to measure the added power that has to be pumped into the drive coil to make the rotor spin, that could be done. However, that would require some thought and careful preparation and developing the right measurement regime to detect it. Right now what is happening is that when the rotor is in place and spinning, the reduced current draw due to the counter-EMF induced into the drive coil is overshadowing the extra power that is added to make the rotor spin. In other words, the preliminary analysis is that more current draw is reduced due to the counter-EMF than the extra current draw that is required to make the rotor spin. There are two opposite effects that are happening at the same time with respect to the current draw and it is not necessarily that easy to separate them from each other.
Thanks MileHigh,
i will try to do some power calculations on the voltage and current using the math function, but i do see some strange variations i cannot yet explain,
like the first screenshot (run) blue trace figure says Pk-Pk = 116mV while the second screenshot (stopped) blue trace figure says 113mV, but the second
screenshot blue trace display clearly shows a higher (off the screen) peak.
It probably are some spikes that cause that while running, but i would like to see them.
Itsu
Quote from: verpies on December 18, 2015, 07:38:54 PM
Very clean waveforms - you should collaborate with Gotoluc so he can obtain the same.
I have extended the light blue current ramp-up on your scopeshot without rotor and came to the conclusion that your ON-pulse lasts for approximately 1.6Tau and that means you have burned 1.5x more energy in the resistance than the energy you had built up in the coil as the magnetic field.
If you decrease the ON time to 0.5Tau than this ratio will fall to 0.37x. Of course narrower pulse needs to be compensated with higher supply voltage.
The voltage across C2 (yellow Ch1) varies between 4V and 15V so the energy gain of this capacitor is ½C*( (15V)2 - (4V)2 ). This is how many Joules are recovered per 1 pulse.
At "switch-OFF" (dark blue vertical line) approximately 1.1A flows through the coil, so ½L*(1.1A)2 Joules of energy is stored in the coil at that time.
I encourage you to plug the inductance of your coil and the capacitance of your recovery capacitor into these formulas to see how many Joules of energy each one of them represents. (I would do this but I am not sure what L & C are).
Finally, divide these two energies to see what your recovery efficiency is.
If you experiment further you will find out that this efficiency increases if C2 is completely discharged before the "flyback" pulse and no resistor is connected across C2 to burn the recovered energy.
BTW:
The red curve is this curve (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/153963/) and the black vertical line is at the break-even-point in time (1.15Tau)
verpies,
great analisys as always, many thanks.
I will shoot a video to show what i have running now, but the setups between GotoLuc and me are way different at the moment.
I will keep the 555 timer on 12V and "up" the voltage on the coil to 24V, hopefully i can narrow down the pulse to 0.5Tau.
C2 is 100uF, so ½C*( (15V)2 - (4V)2 ) = ½ 100 * 209 = 10450 (uJ?) which is way to low i think (see also http://www.calctool.org/CALC/eng/electronics/capacitor_energy)
L = 38mH, so ½L*(1.1A)² = 19 * 1,21 = 22.99 (mJ?)
Itsu
Hi Luc and all
Yes i have made a small rotor with 2 neomag oriented sideways.
After 2 days of playing with , i have stumbled upon some interesting things.
https://youtu.be/N7S_cbAYp6c
It seems that the timing is very important as well as the distance between the rotor and the core of the coil.
At some "sweet" point the current trace goes smoothly to zero at the end of the voltage pulse and so it seems that the flyback spike disappears completely.
I don't know if it helps, but this is very inteersting
Laurent
Quote from: itsu on December 19, 2015, 05:38:56 AM
C2 is 100uF, so ½C*( (15V)2 - (4V)2 ) = ½ 100 * 209 = 10450 (uJ?) which is way to low i think
½ * 0.0001 * (225 - 16) = ½ * 0.0001 * 209 = 0.01045 J = 10.45mJ
Quote from: itsu on December 19, 2015, 05:38:56 AM
(see also http://www.calctool.org/CALC/eng/electronics/capacitor_energy (http://www.calctool.org/CALC/eng/electronics/capacitor_energy))
The disparity comes from the fact that (15
2 - 4
2) <> (15 - 4)
2Quote from: itsu on December 19, 2015, 05:38:56 AM
L = 38mH, so ½L*(1.1A)² = 19 * 1,21 = 22.99 (mJ?)
That's correct
So at switch off you have 22.99mJ of energy stored in the coil, but after recovery you have only 10.45mJ of energy stored in the capacitor, which constitutes 45% recovery efficiency.
If you use a smaller C2 capacitor (e.g. 10μF 100V) and start the recovery from an empty capacitor and delete R2, then this efficiency number will increase a lot.
Cheers
P.S.
I have a 10μF 400V polypropylene cap sitting in front of my keyboard now, that would be ideal for this purpose because it is almost a lossless pulse cap. I wish I could teleport it to you.
Verpies,
thanks, i have some 10uF 630V poly caps here which i can use.
Itsu
Hi all
Just to add to my video , i placed a drum with winglets on my rotor to create an aerodynamic torque.
Pic 1 is the rotor with the drum and winglets
Pic 2 is the scope shot when the coil is at 20 mm distance from the rotor magnets
Pic 3 is the scope shot when the coil is at 4 mm at the better timing i could get.
It seems clearly that at 20 mm distance the rotor spins slowly but draws a lot of current. And there is a nice flyback spike.
At 4 mm distance the rotor spins faster (which means more torque for higher speed) and draws much less current. And there is also a nice flyback spike to recover.
With the aerodynamic load i can no more get the sweet spot where i got the current smoothing down to zero and no flyback spike as per the video with no load. And the best power is got when the current trace is as per Pic 3.
Food for thought
Laurent
Quote from: woopy on December 19, 2015, 11:11:25 AM
With the aerodynamic load i can no more get the sweet spot where i got the current smoothing down to zero and no flyback spike as per the video with no load. And the best power is got when the current trace is as per Pic 3.
Food for thought
Laurent
Excellent observation Laurent. I can share some additional thoughts. You clearly showed how with the aerodynamic load, even for a pulse motor, there is a price to pay for driving that load. As the aerodynamic load gets smaller and smaller, it gets more and more difficult to measure the decreasing amount of extra power consumption required to drive that load. There may be other secondary effects that mask that increased power consumption, especially if those secondary effects reduce the power consumption.
Now, it's possible to start making your measurements where the load is already quite small, and the secondary effects are already present. In that case, it becomes very difficult to figure out what is going on.
Quote from: woopy on December 19, 2015, 07:08:25 AM
Hi Luc and all
Yes i have made a small rotor with 2 neomag oriented sideways.
After 2 days of playing with , i have stumbled upon some interesting things.
https://youtu.be/N7S_cbAYp6c (https://youtu.be/N7S_cbAYp6c)
It seems that the timing is very important as well as the distance between the rotor and the core of the coil.
At some "sweet" point the current trace goes smoothly to zero at the end of the voltage pulse and so it seems that the flyback spike disappears completely.
I don't know if it helps, but this is very interesting
Laurent
Merci Laurent for taking the time to test and make a video of this interesting pulse motor magnet configuration.
You are right, at a certain timing you can eliminate flyback.
What I also found is if your pulse on time is centered in the induced EMF the input is the less and RPM the most with the coil at a certain distance from the magnets (sweet spot).
Hopefully you will find other interesting things to do with it.
Thanks for sharing
Luc
Quote from: woopy on December 19, 2015, 11:11:25 AM
Just to add to my video , i placed a drum with winglets on my rotor to create an aerodynamic torque.
Are your winglets metal or plastic?
Could you make a scopeshot when the rotor is spinning at 4mm with this
trivial circuit (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/153869/)? (with 2 pulses visible on the screen)
Could you make a scopeshots with the rotor stationary (with magnets far away and close by the coil's core)
I managed to swap every other magnet over, so i now have 8 magnets NSNSNSNS.
The trivial circuit below now shows the signals as in screenshot 1 when turning the rotor by hand
Blue is across 0.1 ohm 1% csr, green is the current probe.
I have changed to flyback cap to a 10uF 630V one in the "GotoLuc Schem1f" diagram
Then i managed to sync up the rotor to the 555 timer at 40Hz @ 18% duty cycle, see screenshot 2 (still at 12V input)
blue again the signal across the csr point A with reference to point B
purple the signal on point C with ref to point B
yellow the signal on point D with ref to point B
I think the purple and yellow channel needs to be inverted.
The DMM across the 120 Ohm R2 shows 1.57V
Screenshot 3 shows the same situation with the rotor stopped.
The DMM across the 120 Ohm R2 now shows 1.16V
Video here: https://www.youtube.com/watch?v=1r6a_hgJlko&feature=youtu.be
Itsu
Excellent test itsu
thanks for also making a video.
Well, for those who have done the tests we are all getting same results. The magnet rotor reduces the input and increases the output.
I like the explanation user citfta posted at Brad's topic:
Quote from: citfta on December 19, 2015, 02:25:57 PM
I think the real problem here is what people are defining as "work". A turbo does not do any "work". But when you add one to a gasoline or diesel engine it most certainly improves the efficiency. Maybe that is not "work" but is is most definitely beneficial. I see the rotor the same way. Maybe the magnets don't do "work" but they sure seem to be beneficial in almost all the tests that have been done on this forum. If you don't want to call that work that is fine with me. But until I see otherwise I have to believe the rotor with magnets is causing a beneficial effect.
Carroll
Are the magnets acting like a trubo?... maybe and who cares as long as we know the magnets can boost the efficiency like a turbo does to a ICE, then I say lets use it and move on and lets find a way to use it to our advantage.
I have some ideas, what about you guys?
Luc
ADDED: here is a video Lidmotor did back in 2009 which he just posted in Brad's topic that demonstrates the same effect: https://www.youtube.com/watch?v=afEWXadfpqY (https://www.youtube.com/watch?v=afEWXadfpqY)
Quote from: itsu on December 19, 2015, 04:23:27 PM
The DMM across the 120 Ohm R2 shows 1.57V
Very clean waveforms again and they appear how they are supposed to. No inversion changes necessary.
The double M waveform in the trivial circuit is caused by the fringing flux of the magnets. The 2 peaks occur when the magnet is half way eclipsing the coil's core and the dip between these peaks occurs at TDC.
The R2 is significantly distorting the voltage measurements across C2 !!! This happens because now you have higher voltage across C2 (~57V) and the peak discharge current increased to 57V/120Ω = 475mA ...and before it was 15V/120Ω=125mA, if the previous R2 was the same.
If you do not want to go to the trouble of making an active C2 discharger, at least increase R2 so much that the voltage across C2 falls down to 0V just before the MOSFET opens at the end of the
next energizing pulse....like I have drawn below.
If you do this, then the average DMM voltage reading will increase significantly, but it is the peak voltage calculated by the scope, that will be your true indicator of the recovered energy, according to E=½CV
2.
And if you zoom in and give me the rise time of the voltage across C2 I should be able to calculate how much higher the voltage peak would be if R2 was deleted altogether.
P.S.
If you had an active C2 discharger circuit, then the average voltage shown by the DMM would become equal to the peak voltage across C2, while the yellow waveform would become almost rectangular ...and the voltage peak itself would be higher, too.
Quote from: gotoluc on December 19, 2015, 04:46:22 PM
Are the magnets acting like a trubo?... maybe and who cares as long as we know the magnets can boost the efficiency like a turbo does to a ICE, then I say lets use it and move on and lets find a way to use it to our advantage.
I have some ideas, what about you guys?
Luc
You have to keep all options on the table. The spinning rotor will reduce the power draw. In Itsu's scope captures you can see that without the rotor the coil drew more power and the back spike also had more power. So for Itsu's setup, comparing power out vs. power in and rotor vs. no rotor does not look conclusive if you are looking at the power out vs. power in metric.
Let's just look at the power draw of the coil itself. When you add the spinning rotor you get reduced power draw. That is clear and has been replicated many times. But how about this: Instead of adding the rotor to reduce the power draw, why not forget about the rotor completely and just reduce the drive voltage for the coil? So that also gives you reduced power draw without adding a rotor. The back spike might still output the same amount or power. Do you see what I mean? This is highly academic because what can you do with a pulsing coil if you don't have a rotor?
Quote from: MileHigh on December 19, 2015, 05:33:07 PM
Instead of adding the rotor to reduce the power draw, why not forget about the rotor completely and just reduce the drive voltage for the coil? So that also gives you reduced power draw without adding a rotor. The back spike might still output the same amount or power. Do you see what I mean? This is highly academic because what can you do with a pulsing coil if you don't have a rotor?
That's a good suggestion MH
I will give it a try once my new smaller setup is ready for testing.
Itsu could beat me to it.
Thanks for sharing
Luc
Quote from: gotoluc on December 19, 2015, 04:46:22 PM
Excellent test itsu
thanks for also making a video.
Well, for those who have done the tests we are all getting same results. The magnet rotor reduces the input and increases the output.
I like the explanation user citfta posted at Brad's topic:
Are the magnets acting like a trubo?... maybe and who cares as long as we know the magnets can boost the efficiency like a turbo does to a ICE, then I say lets use it and move on and lets find a way to use it to our advantage.
I have some ideas, what about you guys?
Luc
ADDED: here is a video Lidmotor did back in 2009 which he just posted in Brad's topic that demonstrates the same effect: https://www.youtube.com/watch?v=afEWXadfpqY (https://www.youtube.com/watch?v=afEWXadfpqY)
Being a mechanic by trade,i can answer this one very accurately.
The problem with the turbo analogy is that when boost kicks in from the turbo,you also raise the fuel consumption of the motor,where as with the rotating magnet's,we actually decrease the fuel consumption of the motor--the fuel being of course our electrical P/in.
So i dont think the turbo analogy is right here.
Brad
Quote from: tinman on December 19, 2015, 09:23:39 PM
Being a mechanic by trade,i can answer this one very accurately.
The problem with the turbo analogy is that when boost kicks in from the turbo,you also raise the fuel consumption of the motor,where as with the rotating magnet's,we actually decrease the fuel consumption of the motor--the fuel being of course our electrical P/in.
So i dont think the turbo analogy is right here.
Brad
That's also a good point Brad... somehow I don't think we'll be able to satisfy everyone no matter how many tests we do. It's a little like religion, each will have their own take on it.
I've got a new build and should have the video up tomorrow.
Stay tuned
Luc
Quote from: gotoluc on December 20, 2015, 02:06:12 AM
That's also a good point Brad... somehow I don't think we'll be able to satisfy everyone no matter how many tests we do. It's a little like religion, each will have their own take on it.
I've got a new build and should have the video up tomorrow.
Stay tuned
Luc
This whole thing could be put to bed very easily Luc. All the EE guys have to do is tell me how to increase the P/in to P/out efficiency of my setup without the external alternating magnetic fields. It would then be game,set ,and match for them. But that will not happen. What will happen is they will want you to build a brand new coil,that in no way resembles the DUTs coil. And then who is to say that the same thing would not apply to that coil?.
We are being told this-
1-the P/in is providing the energy to spin the rotor
2-the rotor is inducing a CEMF in the coil
3-This CEMF is what is reducing the P/in-and increasing the overall P/in to P/out efficiency of the coil.
4- But the magnets are doing no useful work- ???
It is good to see that the bench work being done here is dismissing all there claim's so far against what we see.
Brad
Quote from: tinman on December 19, 2015, 09:23:39 PM
Being a mechanic by trade,i can answer this one very accurately.
The problem with the turbo analogy is that when boost kicks in from the turbo,you also raise the fuel consumption of the motor,where as with the rotating magnet's,we actually decrease the fuel consumption of the motor--the fuel being of course our electrical P/in.
So i dont think the turbo analogy is right here.
Brad
Hi Brad,
What you are saying is true. However, that is not all there is to it. Even when lightly spooled up the turbo helps the engine to burn the fuel more efficiently. It is well documented by Mercedes Benz and verified by Mercedes owners that the 5 cylinder diesel engine sold from the late 1970s until the mid 1980s got better overall fuel economy with the turbo. The non-turbo engine cars got on average 27-28 mpg on the open road and the turbo versions got 30-31 mpg on the open road. I probably could find a better analogy but I'll stand with my assertion that a turbo does improve efficiency.
Respectfully,
Carroll
Quote from: citfta on December 20, 2015, 06:46:51 AM
Hi Brad,
What you are saying is true. However, that is not all there is to it. Even when lightly spooled up the turbo helps the engine to burn the fuel more efficiently. It is well documented by Mercedes Benz and verified by Mercedes owners that the 5 cylinder diesel engine sold from the late 1970s until the mid 1980s got better overall fuel economy with the turbo. The non-turbo engine cars got on average 27-28 mpg on the open road and the turbo versions got 30-31 mpg on the open road. I probably could find a better analogy but I'll stand with my assertion that a turbo does improve efficiency.
Respectfully,
Carroll
If the compression ratio is raised in an engine,that will also improve efficiency,but you loose bottom end torque. Dont forget that diesel contains far more energy per ltr than petroleum ,and so MPG dont really tell the story of efficiency. There are many ways to increase the actual efficiency of an ICE,and a turbo is one that sits on the bottom of the list. But rather than clog up Lucs thread with this,i just put my point forward as you have,and will leave it at that.
During the week, i will be putting together a universal motor where one of the stator electromagnets can be pulled out,and a permanent magnet can take it's place.
Brad
Hi Brad,
Will you be starting another thread about the universal motor? I have just picked up a couple of small universal motors and have been playing around with just putting some ceramic magnets on the outside of the laminations and can see that just doing that makes a difference. Looking forward to what you are doing.
Carroll
Quote from: citfta on December 20, 2015, 07:54:29 AM
Hi Brad,
Will you be starting another thread about the universal motor? I have just picked up a couple of small universal motors and have been playing around with just putting some ceramic magnets on the outside of the laminations and can see that just doing that makes a difference. Looking forward to what you are doing.
Carroll
No-no new thread. I will just carry on on the thread i already have,as it is related to the topic at hand.
Brad
Quote from: tinman on December 20, 2015, 02:19:48 AM
This whole thing could be put to bed very easily Luc. All the EE guys have to do is tell me how to increase the P/in to P/out efficiency of my setup without the external alternating magnetic fields. It would then be game,set ,and match for them. But that will not happen. What will happen is they will want you to build a brand new coil,that in no way resembles the DUTs coil. And then who is to say that the same thing would not apply to that coil?.
We are being told this-
1-the P/in is providing the energy to spin the rotor
2-the rotor is inducing a CEMF in the coil
3-This CEMF is what is reducing the P/in-and increasing the overall P/in to P/out efficiency of the coil.
4- But the magnets are doing no useful work- ???
It is good to see that the bench work being done here is dismissing all there claim's so far against what we see.
Brad
they will simply never say you are correct because they want to draw out every point so that you keep spending time on videos running over tiny points and less time on innovation, trusting that if you self loop YT will pull the video, or it will be trolled as 'fake'
everyone can see what is obvious but every time you will be asked to do more videos, how many videos have these guys posted showing these concepts?
ha ha it's like they are sitting back and demanding you make videos for them, the 'EEE' = 'EE Elite'
ha ha
: D
on a serious note
i notice that creeping in here more and more is the geometry of the systems. with woopy video etc.
it's always good to take on new info by everyone but if it takes away your time spent innovating where you could test new 'out of the box' ideas is that time well spent?
As promised, here is the new test device and video.
I had to redo the pictures of the scope shots I took during the video since the scope data didn't display for some reason.
Link to video: https://www.youtube.com/watch?v=P7yf_J3dNNM (https://www.youtube.com/watch?v=P7yf_J3dNNM)
Coil Data:
Drive Coil is 0.3 Ohm and 2.39mH
Assist Coil is 8 Ohm and 105mH
Flyback Diode:
2x RHRG5060 Parellel give a 0.34 Conducting Forward Voltage
Luc
Quote from: verpies on December 19, 2015, 05:31:06 PM
Very clean waveforms again and they appear how they are supposed to. No inversion changes necessary.
The double M waveform in the trivial circuit is caused by the fringing flux of the magnets. The 2 peaks occur when the magnet is half way eclipsing the coil's core and the dip between these peaks occurs at TDC.
The R2 is significantly distorting the voltage measurements across C2 !!! This happens because now you have higher voltage across C2 (~57V) and the peak discharge current increased to 57V/120Ω = 475mA ...and before it was 15V/120Ω=125mA, if the previous R2 was the same.
If you do not want to go to the trouble of making an active C2 discharger, at least increase R2 so much that the voltage across C2 falls down to 0V just before the MOSFET opens at the end of the next energizing pulse....like I have drawn below.
If you do this, then the average DMM voltage reading will increase significantly, but it is the peak voltage calculated by the scope, that will be your true indicator of the recovered energy, according to E=½CV2.
And if you zoom in and give me the rise time of the voltage across C2 I should be able to calculate how much higher the voltage peak would be if R2 was deleted altogether.
P.S.
If you had an active C2 discharger circuit, then the average voltage shown by the DMM would become equal to the peak voltage across C2, while the yellow waveform would become almost rectangular ...and the voltage peak itself would be higher, too.
Ok, great.
I changed the flyback diode to a 0.4V forward voltage one and put a 5K potmeter as R2 and had to put it to the max. (5K), then the voltage across the C2 looks like what you had drawn, see screenshot 1.
The DDM now shows around 20V.
Screenshot 2 is somewhat zoomed in.
Both situations are with a running rotor.
Itsu
Quote from: itsu on December 20, 2015, 03:19:06 PM
I changed the flyback diode to a 0.4V forward voltage one and put a 5K potmeter as R2 and had to put it to the max. (5K), then the voltage across the C2 looks like what you had drawn, see screenshot 1.
The DDM now shows around 20V.
If L
1=38mH and peak i=880mA then E
1=½L
1i² = ½*0.038*0.880
2 = ½*0.038*0.7744 = 14.71mJ
If C
2=10μF and V
1=0V and peak V
2=106V then E
2=½C
2(V
22 - V
12) = ½ * 0.000010 * 106
2 - 0
2 = ½ * 0.000010 * 11236 = 56.18mJ
So the recovery efficiency calculates to: E
2/E
1 = 56.18mJ / 14.71mJ = 3.82 or 382% :o
...and that is before the C2's voltage sag due to the D1 and R2 wasting energy during the first 114.8μs, is compensated for.
All right, who divided by zero?
Hi Luc
Very impressive work as usual
Fantastic conclusion in your video
So if we use a main coil, the flyback of which is transfered to an assistant coil, so:
1 when the torque is increased on the rotor shaft, then the current in the main coil will increase
2 as the current is increased, the flyback spike is also increased
3 that increased flyback spike is recovered by the assistant coil .
4 the assistant coil is therefore more assisting the main coil to support the increased torque .
Seems very good all this stuff
Bravo and thank's for sharing
Laurent
Quote from: woopy on December 20, 2015, 05:13:12 PM
Hi Luc
Very impressive work as usual
Fantastic conclusion in your video
So if we use a main coil, the flyback of which is transfered to an assistant coil, so:
1 when the torque is increased on the rotor shaft, then the current in the main coil will increase
2 as the current is increased, the flyback spike is also increased
3 that increased flyback spike is recovered by the assistant coil .
4 the assistant coil is therefore more assisting the main coil to support the increased torque .
Seems very good all this stuff
Bravo and thank's for sharing
Laurent
Merci Laurent,
you have the same understanding as I do ;) ... I think feedback is the way to go for what we are looking for.
I did not demonstrate this but in my previous build it was much more apparent then in this one when using the high torque low rpm DC motor to assist instead of the assist coil. Something in the magnitude of around 3 times more torque (pressure on the rotor) to the input power to increase on the drive coil compared to just powering the DC motor directly with DC.
This gave me the idea of, what if we take an off the shelf brushless DC PM motor which are controlled by 3 hall effect sensors which tell the switches (mosfets or transistors) to turn on and off at the appropriate time ... I'm quite sure in these control circuits they do nothing with the flyback right?... so all we do is collect the flyback and with it power a second motor tied to the same shaft and there you go, a motor that once under load the assist motor will become stronger which should help maintain shaft torque.
I think Tesla had a device that would operate on this same principal.
What do you think of this instead of trying to re-invent a motor, we just add a turbo assist to it and we may see it go over 100%
Let me know what you think
Luc
Quoting myself:
Quote from: verpies on December 20, 2015, 05:08:22 PM
If L1=38mH and peak i=880mA then E1=½L1i² = ½*0.038*0.8802 = ½*0.038*0.7744 = 14.71mJ
We cannot trust the L
1 in the formula E
1=½L
1i² because L
1 might not be constant during the ON-pulse, when the rotor is influencing it.
However, if we use the fool-proof, albeit difficult method (V
inst *I
inst * t) of calculating the energy expended by the 12V power supply to bring L
1's current up to 880mA, we get approximately 45mJ (from ½ * 12V * 1A * 7.5ms, when we assume ideal sawtooth current fitted symmetrically into the real curved-down current waveform - see below).
A Math trace of Ch2*Ch3 integrated between cursors set to the width of the ON-pulse, would give us a more precise number.
We can trust that the capacitance of C2 is constant, so we can trust the formula E
2=½C
2V²
Nonetheless the energy recovery efficiency form L1 into C2 is still E
2/E
1 = 56.18mJ / 45mJ = 1.25 or 125% :o
So the first task for Itsu is to confirm, that the constant supply voltage applied to L1 through the MOSFET is really 12VDC and that PEAK current flowing through L1 is really 880mA and that C2=10μF and that the CSR = 0.1Ω and that scope probe attenuation is set correctly (e.g.: x10).
The second one is to do the Math trace Ch2*Ch3 and to integrate it for the duration of the ON-pulse, to confirm the energy delivered to L1 by the power supply.
Quote from: gotoluc on December 20, 2015, 05:41:54 PM
This gave me the idea of, what if we take an off the shelf brushless DC PM motor which are controlled by 3 hall effect sensors which tell the switches (mosfets or transistors) to turn on and off at the appropriate time ... I'm quite sure in these control circuits they do nothing with the flyback right?... so all we do is collect the flyback and with it power a second motor tied to the same shaft and there you go, a motor that once under load the assist motor will become stronger which should help maintain shaft torque.
Luc
This is a very good way to go in testing, Luc.
My thoughts were these:
* the in-line assist motor can be of a slightly larger diameter and have more copper winding (there by providing even greater torque).
* the in-line assist motor staters should be twist adjustable for timing purposes.
* if done right, the in-line assist motor could have its own kickback for a third in-line motor.
I found this video to be a good starting point for anyone wanting to make their own brush-less DC PM motor.
https://www.youtube.com/watch?v=vnjCrLTMGxQ
MagnaMoRo
Dear verpies,
could you please do the calculations on my below scope shot.
The input is exactly 2.00 Volts DC
The CSR is 1% 0.1 Ohm
The Capacitor is 9.92uf (measured)
The Load resistor is 1,001 Ohm (measured)
Many thanks
Luc
BTW I rearranged everything to look like what itsu provided so to keep it simple.
Dear verpies,
here is another one in case it looks better then the above.
The input is exactly 2.00 Volts DC
The CSR is 1% 0.1 Ohm
The Capacitor is 17.2uf (measured)
The Load resistor is 511 Ohm (measured)
Many thanks
Luc
At 1:40am here is the last update :P
Link to video: https://www.youtube.com/watch?v=Qt-eOxo6gjg
Thanks for your participation
Luc
Quote from: gotoluc on December 20, 2015, 11:00:35 PM
Dear verpies,
here is another one in case it looks better then the above.
The input is exactly 2.00 Volts DC
The CSR is 1% 0.1 Ohm
The Capacitor is 17.2uf (measured)
The Load resistor is 511 Ohm (measured)
Many thanks
Luc
Luc
Im guessing the question mark after ch2 and ch3 results are because the wave form continues outside of the screen,where as ch1 has no question mark after the resultant number ,as it is the only wave form that is within the screen of the scope. I am not sure how your scope calculates wave form value's,but if i have any of my wave forms continue out side the screen parameters,the calculated results are not correct.
Anyway,now it's time to go catch up on some of your video's ;)
Brad
Quote from: gotoluc on December 21, 2015, 01:42:08 AM
At 1:40am here is the last update :P
Link to video: https://www.youtube.com/watch?v=Qt-eOxo6gjg
Thanks for your participation
Luc
Ah,and i see in this scope shot,there are no question marks after the results,and all the wave forms are within the screen. :)
Quote from: gotoluc on December 21, 2015, 01:42:08 AM
At 1:40am here is the last update :P
Link to video: https://www.youtube.com/watch?v=Qt-eOxo6gjg
Thanks for your participation
Luc
Nice but why not go further and utilize the flyback from second coil and the next and next until you have no more room for coils?
Another thing to check is magnetic strengh of each coil before and after each one is added.
Another interesting thing to check would be heat differences.
Quote from: gotoluc on December 20, 2015, 10:10:52 PM
could you please do the calculations on my below scope shot.
The input is exactly 2.00 Volts DC
The CSR is 1% 0.1 Ohm
The Capacitor is 9.92uf (measured)
Your waveforms are pretty spikey due to long lead inductance and lack of starpoint connections, so I could not go by scope's automatic calculations and had to make my own.
600mV across a 0.1Ω resistor would mean that 6A is flowing through it.
So, the approximate energy delivered to L1 by the power supply is E
1 = 2V * 6A * 10ms * ½ = 120mJ ( *½ because it's a triangle ).
The peak energy recovered by C2 is E
2=½CV
2 = ½ * 0.00000992 * 50
2 = ½ * 0.00000992 * 2500 = 12.4mJ
Thus the "backspike" recovery efficiency is E
2/E
1 = 12.4mJ / 120mJ = 10.3%
Quote from: verpies on December 21, 2015, 05:03:18 AM
the "backspike" recovery efficiency is E2/E1 = 12.4mJ / 120mJ = 10.3%
is this 10% using just one coil, and could using ten coils you know what?
Quote from: verpies on December 20, 2015, 06:49:47 PM
Quoting myself:We cannot trust the L1 in the formula E1=½L1i² because L1 might not be constant during the ON-pulse, when the rotor is influencing it.
However, if we use the difficult, albeit fool-proof method (Vinst *Iinst * t) of calculating the energy expended by the 12V power supply to bring L1's current up to 880mA, we get approximately 45mJ (from ½ * 12V * 1A * 7.5ms, when we assume ideal sawtooth current fitted symmetrically into the real curved-down current waveform - see below).
A Math trace of Ch2*Ch3 integrated between cursors set to the width of the ON-pulse, would give us a more precise number.
We can trust that the capacitance of C2 is constant, so we can trust the formula E2=½C2V²
Nonetheless the energy recovery efficiency form L1 into C2 is still E2/E1 = 56.18mJ / 45mJ = 1.25 or 125% :o
So the first task for Itsu is to confirm, that the constant supply voltage applied to L1 through the MOSFET is really 12VDC and that PEAK current flowing through L1 is really 880mA and that C2=10μF and that the CSR = 0.1Ω and that scope probe attenuation is set correctly (e.g.: x10).
The second one is to do the Math trace Ch2*Ch3 and to integrate it for the duration of the ON-pulse, to confirm the energy delivered to L1 by the power supply.
Hmmm, i checked again the coil, and it measures 37.8mH @ 100Hz, the capacitor C2 measures 10.03uF @ 100Hz and the csr measures the mentioned 0.1 Ohm 1%, allthough
that is hard to confirm as my DDM is fluctuating in that low region, but after subtracting the leads resistance, it seems to be the correct 0.1 Ohm.
The battery voltage was/is 12.4V as can be seen on the screenshot below where i toke an input measurement of the battery voltage (12.4V yellow trace) and the
input current with my current probe set to 200mA/div. This screenshot was in a stopped state, i appearently forgot to save the same situation when running, but i
remember the wattage calculation to be 35mW, so a difference of about 10mA between running (35mW) and stopped (44.78mW).
(Note that the current controller was set to 200mA/div, so the green current value and the wattage values need to be taken x 20, so in the below screenshot
158.6mA RMS current and thus 895.6mW)
Taking more samples did not change the Math mean value.
By the way i have seperated the 555 timer circuit input voltage and the GotoLuc circuit input voltage by using a different 12V battery for both.
I will try to redo that input measurement test, but it is clear that the input wattage increases from 700mW (35mW * 20) when running to 895mW when stopped.
I will also try to come up with the data requested by verpies.
EDIT i have the input measurement when running on video which i will upload now. There you can see what happens with the input power when stopping the rotor.
Also it shows that i have changed the R2 resistor again, now for a 50K pot which gives better control over the cap trace.
Regards Itsu
Quote from: Over Goat on December 21, 2015, 05:18:12 AM
is this 10% using just one coil, and could using ten coils you know what?
Average of several 10% is still 10%
Long :( video here: https://www.youtube.com/watch?v=dANzzD6X75E&feature=youtu.be
The stopping of the rotor while measuring the input is shown around 5:50 minutes.
Itsu
Quote from: itsu on December 21, 2015, 05:21:11 AM
Hmmm, I checked again the coil, and it measures 37.8mH @ 100Hz,
That's good to know but the coil's inductance is not necessary to make the input energy calculation when using the fool-proof, albeit difficult method of integrating V
INST * i
INST for the length of the ON-period.
Quote from: itsu on December 21, 2015, 05:21:11 AM
the capacitor C2 measures 10.03uF @ 100Hz
So if the peak voltage across C2 is trustworthy then the energy recovered in C2 is trustworthy, too. All according to E
2=½C
2V
2Quote from: itsu on December 21, 2015, 05:21:11 AM
and the CSR measures the mentioned 0.1 Ohm 1%, although that is hard to confirm as my DMM is fluctuating in that low region, but after subtracting the leads resistance, it seems to be the correct 0.1 Ohm.
Yes, it is difficult to measure low resistances. The best way is to put exactly 2.000V across it and verify the the current flowing through it is exactly 20A since most current probes and multimeters can measure up to 20A.
Quote from: itsu on December 21, 2015, 05:21:11 AM
The battery voltage was/is 12.4V as can be seen on the screenshot below where i took an input measurement of the battery voltage (12.4V yellow trace)
OK, So my previous calculation of the input energy per 1 pulse would be now: ½ * 12.4V * 1A * 7.5ms = 46.5mJ
Quote from: itsu on December 21, 2015, 05:21:11 AM
the input current with my current probe set to 200mA/div.
so according to this, your latest scopeshot (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/154112/) shows 560mA peak current.
The RMS current calculated by the scope is pretty useless because it is
not calculated ONLY during the ON-pulse.
Quote from: itsu on December 21, 2015, 05:21:11 AM
This screenshot was in a stopped state, i apparently forgot to save the same situation when running,
Does this overunity "backspike" energy recovery efficiency also occur when the rotor is stopped?
Quote from: itsu on December 21, 2015, 05:21:11 AM
I remember the wattage calculation to be 35mW, so a difference of about 10mA between running (35mW) and stopped (44.78mW)
The average input power is pretty useless (just like the RMS current) unless it is measured only during the ON-pulse.
Does your scope have the ability to do measurements between cursors, or do you have to manipulate the display to show only the part of the waveform that you need to measure?
Quote from: itsu on December 21, 2015, 05:21:11 AM
(Note that the current controller was set to 200mA/div, so the green current value and the wattage values need to be taken x 20,
Yes, 200mA/div is pretty clear. That's how I got the 560mA peak current from your recent scopeshot (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/154112/).
It is also in the ballpark of the 880mA peak current that was shown on your previous scopeshot (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/154089/), so that confirms that this is not a x10 error or something silly like that.
Quote from: itsu on December 21, 2015, 05:21:11 AM
Taking more samples did not change the Math mean value.
Your i*V multiplication by the Math channel is set up correctly. Only the calculation of the mean value is not narrowed down to the ON-period like it is supposed to be.
Quote from: verpies on December 21, 2015, 05:03:18 AM
Thus the "backspike" recovery efficiency is E2/E1 = 12.4mJ / 120mJ = 10.3%
That kind of sucks :o
Thanks for taking the time to do what I can't do.
Luc
Quote from: gotoluc on December 21, 2015, 10:00:32 AM
That kind of sucks :o
Thanks for taking the time to do what I can't do.
I expected 80% "backspike" energy recovery efficiency.
Both 10% and 125% are surprises to me
Quote from: verpies on December 21, 2015, 10:29:04 AM
I expected 80% "backspike" energy recovery efficiency.
Both 10% and 125% are surprises to me
Me too!... I thought my big ass low resistance coil would do better. Apparently not ???
I'm going to have to change to hall sensors to clean up my switching spike problems.
Luc
QuoteDoes this overunity "backspike" energy recovery efficiency also occur when the rotor is stopped?
I think so as i do not see much difference when running or stopped, see this post:
http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg468976/#msg468976
QuoteThe average input power is pretty useless (just like the RMS current) unless it is measured only during the ON-pulse.
Does your scope have the ability to do measurements between cursors, or do you have to manipulate the display to show only the part of the waveform that you need to measure?
Ok, no no posibility to measure power between the cursors, so i manipulated the display to show only the "on" time, see screenshot 1 below.
Its negative as the purple trace is on the negative part, so i inverted the purple trace in screenshot 2
Both situations are when running at 40Hz @ 18% duty cycle (12V)
Remember, i use here the 0.1 Ohm csr
Thanks, regards Itsu
Quote from: gotoluc on December 21, 2015, 10:48:50 AM
Me too!... I thought my big ass low resistance coil would do better. Apparently not ???
I don't think that your big ass low resistance coil is the problem.
It is much more likely that your C2 or D1 are the problem as well as your wiring (no-starpoint).
Quote from: gotoluc on December 21, 2015, 10:48:50 AM
I'm going to have to change to hall sensors to clean up my switching spike problems.
A high R
DS-ON MOSFET transistor could be the culprit, although 6A is not a big deal for most MOSFETs today.
The gate timing seems to be fine, and Hall sensor will not improve it.
My McGyver circuit would improve it, but without these CD4047 chips it is very hard to get the LM3915 Vu-Meter to work as a sequencer...but not impossible
Quote from: itsu on December 21, 2015, 11:16:47 AM
Ok, no no posibility to measure power between the cursors, so i manipulated the display to show only the "on" time, see screenshot 1 below.
I remember that your previous ON-pulse (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/154101/) was almost 8ms wide but now the entire screen spans 4ms. Why?
It must be because of a lower speed and/or duty cycle.
This pulse motor when externally synced has a life of its own, sometimes it just won't run at the max 555 timer pots settings meaning 40Hz @ 18% dc, so i settle for a lower speed/dc to have some data taken.
I therefor try to mention the speed/dc each time.
Itsu
All right, so the mean power during the pulse is 4.478W
The scope shows 447.8mW mean power, but this is 10x too low because the current vertical scale is set to 50mA/div, while in fact 500mA flows through the 0.1Ω CSR when the scope trace is deflected 1 division ( IMPORTANT: please confirm this ).
So, if we multiply 4.478W * 4ms (the width of the ON-period), we obtain the input energy for 1 ON-pulse equal to 17.91mJ.
Let's see how that compares to the triangular/sawtooth method:
I estimate the interpolated peak current (blue Ch2) as approximately 650mA, so the input energy for 1 ON-pulse equals 12.4V * 650mA * 4ms * ½ = 16.12mJ
Both methods are pretty much in agreement.
P.S.
When setting up the Math trace to multiply 2 channels, make sure that the vertical scale/amplitude of these channels is set as high as possible, as it avoids inaccuracies due to multiplying two quantization errors introduced by the ADCs, which leads to large round-off errors.
Quote from: itsu on December 21, 2015, 11:16:47 AM
I think so as i do not see much difference when running or stopped, see this post:
http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg468976/#msg468976
Ok, no no posibility to measure power between the cursors, so i manipulated the display to show only the "on" time, see screenshot 1 below.
Its negative as the purple trace is on the negative part, so i inverted the purple trace in screenshot 2
Both situations are when running at 40Hz @ 18% duty cycle (12V)
Remember, i use here the 0.1 Ohm csr
Thanks, regards Itsu
Itsu,
Just browsing and ran across this thread and would like to make a comment in regards to your TDS3054 scope. I'm sure you are aware of this but the scope is capable of vertical or horizontal cursor measurements of any of it's functions. To access, press the front panel "Quick Menu" and then select desired cursor function in the "Cursor" screen menu.
If you are using the "sample" acquisition mode and are concerned about enough samples on the screen, select an "average" acquisition and set for "8" which I have found found to be more than adequate. You can now accurately measure one complete cycle of the periodic event captured in the screen with great resolution and view-ability.
Please disregard this if you have other reasons for not using the cursors and I apologize for interrupting!
partzman
Quote from: verpies on December 21, 2015, 12:23:54 PM
All right, so the mean power during the pulse is 4.478W
The scope shows 447.8mW mean power, but this is 10x too low because the current vertical scale is set to 50mA/div, while in fact 500mA flows through the 0.1Ω CSR when the scope trace is deflected 1 division ( IMPORTANT: please confirm this ).
So, if we multiply 4.478W * 4ms (the width of the ON-period), we obtain the input energy for 1 ON-pulse equal to 17.91mJ.
Let's see how that compares to the triangular/sawtooth method:
I estimate the interpolated peak current (blue Ch2) as approximately 650mA, so the input energy for 1 ON-pulse equals 12.4V * 650mA * 4ms * ½ = 16.12mJ
Both methods are pretty much in agreement.
P.S.
When setting up the Math trace to multiply 2 channels, make sure that the vertical scale/amplitude of these channels is set as high as possible, as it avoids inaccuracies due to multiplying two quantization errors introduced by the ADCs, which leads to large round-off errors.
Yes, thats confirmed.
Itsu
Quote from: partzman on December 21, 2015, 12:48:24 PM
Itsu,
Just browsing and ran across this thread and would like to make a comment in regards to your TDS3054 scope. I'm sure you are aware of this but the scope is capable of vertical or horizontal cursor measurements of any of it's functions. To access, press the front panel "Quick Menu" and then select desired cursor function in the "Cursor" screen menu.
If you are using the "sample" acquisition mode and are concerned about enough samples on the screen, select an "average" acquisition and set for "8" which I have found found to be more than adequate. You can now accurately measure one complete cycle of the periodic event captured in the screen with great resolution and view-ability.
Please disregard this if you have other reasons for not using the cursors and I apologize for interrupting!
partzman
Hi Partzman, thanks for your interruption, no problem.
I know the TDS3054 scope is capable of vertical or horizontal cursor measurements like amplitude and timing, but to my knowledge not for doing "power"
calculations using the MATH function, but i will take a look at your suggested settings.
Regards Itsu
Itsu,
Yes, it will do the "Math" channel in mean, rms, or any other function between vertical cursors. Use the Math menu to select the channels and function you wish.
I've attached a sample scope shot from my TDS3034 which has the Math set to CH1 x CH4. I have selected the result to be in "mean" watts which is the correct result for this particular measurement. In this example, the power is -125.8mvv mean for one cycle as seen. The result in 'mvv' which is really 'mw' because I am multiplying two voltages because I was using an AM503 current probe amplifier for this test.
Hope this helps!
partzman
Partzman,
my understanding is that your "red Math Mean -125.8mVV" value is the mean value from the total screen (so the 10 horizontal divisions).
So not the the Math mean value of the area between the 2 vertical cursors which is what the question was.
You presently have displayed the values of the area between the 2 cursors of the yellow trace as amplitude and timing (upper right corner).
If you now activate the red math trace, in that upper right corner you will see the data of that red math trace (wattage and timing) inbetween the 2 cursors, but
i doubt it will be the calculated power delta as it does not match my full screen calculations when trying, but i will double check.
This might be interesting for Luc also as he has a similar scope.
Itsu
Quote from: verpies on December 21, 2015, 11:21:38 AM
I don't think that your big ass low resistance coil is the problem.
It is much more likely that your C2 or D1 are the problem as well as your wiring (no-starpoint).
I don't know about the C2 resistance and why I gave you 2 different samples in case one is better then the other.
As for the diode, did you not see the quality of diode I'm using? Ultra fast RHRG5060 paired which gives a 0.34v forward voltage.
This is the second time you mention wiring starpoint!... I never heard of this? what is it and can you show me a picture of what it would look like.
Quote from: verpies on December 21, 2015, 11:21:38 AM
A high RDS-ON MOSFET transistor could be the culprit, although 6A is not a big deal for most MOSFETs today.
The gate timing seems to be fine, and Hall sensor will not improve it.
The mosfet's are IRFP4310Z 120A and RDS 8.4 milliohm resistance, so I don't think this is the problem.
Quote from: verpies on December 21, 2015, 11:21:38 AM
My McGyver circuit would improve it, but without these CD4047 chips it is very hard to get the LM3915 Vu-Meter to work as a sequencer...but not impossible
I can order some CD4047
Can you provide the complete circuit so I can make sure I order everything needed at the same time.
Could a on the fly pulse width adjustment capability be added to this?... and if so, what components are need to be ordered for that option?
I have 2 of the below outrunner motors. Do you think a circuit could be built to power it and collect the flyback to power the second one which would be attached to the same shaft?
Thanks for your time and help
Luc
Quote from: itsu on December 21, 2015, 02:57:37 PM
Partzman,
my understanding is that your "red Math Mean -125.8mVV" value is the mean value from the total screen (so the 10 horizontal divisions).
So not the the Math mean value of the area between the 2 vertical cursors which is what the question was.
You presently have displayed the values of the area between the 2 cursors of the yellow trace as amplitude and timing (upper right corner).
If you now activate the red math trace, in that upper right corner you will see the data of that red math trace (wattage and timing) inbetween the 2 cursors, but
i doubt it will be the calculated power delta as it does not match my full screen calculations when trying, but i will double check.
This might be interesting for Luc also as he has a similar scope.
Itsu
I don't know? the scope I have access to is a TDS 2024B which by the way has very low resolution screen shots captures compared to your and why I started taking pictures of the screen to show more details.
Luc
Hi itsu (and Hi to all people who do decent measurements like itsu):
To find out if the energy in the spikes can be "fed back" to save power, one should do the following measurement:
1) You do the input power measurement with the motor running like you do in your video https://www.youtube.com/watch?v=dANzzD6X75E&feature=youtu.be
2) But then you do not stop the motor, but you disconnect the path to the capacitor (by disconnecting the diode). This will prohibit that power is fed back from the spikes. And now you measure the input power. Is it less or more?
I tried that several times with different pulse motors and different drive circuits. I never could prove that input power to the motor was less if one tries to feed back power from the spikes (from the back EMF of the drive coil).
I think that the power in the spikes is forever lost.
Greetings, Conrad
Quote from: conradelektro on December 21, 2015, 03:17:08 PM
Hi itsu (and Hi to all people who do decent measurements like itsu)
So are you being selective to who you want to say hi to or what?
I've never seen such a written thing LOL
Luc
Quote from: gotoluc on December 21, 2015, 02:59:19 PM
I don't know about the C2 resistance and why I gave you 2 different samples in case one is better then the other.
I had in mind the quality of the capacitor - not only its capacitance. Itsu is using a polypropylene cap, which is an excellent pulse cap.
Quote from: gotoluc on December 21, 2015, 02:59:19 PM
As for the diode, did you not see the quality of diode I'm using? Ultra fast RHRG5060 paired which gives a 0.34v forward voltage.
I did not check its specs. However 0.34V is not very much, indeed.
Quote from: gotoluc on December 21, 2015, 02:59:19 PM
This is the second time you mention wiring starpoint!... I never heard of this? what is it and can you show me a picture of what it would look like.
Similar to this one (http://overunity.com/12736/kapanadze-cousin-dally-free-energy/msg340180/#msg340180)
Quote from: gotoluc on December 21, 2015, 02:59:19 PM
The MOSFETs are IRFP4310Z 120A and RDS 8.4 milliohm resistance, so I don't think this is the problem.
A good 100V MOSFET with low RDS_ON and large 6.8nF gate capacitance.
Let's see its gate-to-source voltage waveform (not simultaneously with the other waveforms because of scope ground hazard)
Quote from: gotoluc on December 21, 2015, 02:59:19 PM
I can order some CD4047
Can you provide the complete circuit so I can make sure I order everything needed at the same time.
Give me some time to build it so I don't make a mistake in a purely theoretical schematic.
But the other parts I expect will be non-electrolytic caps, multi-turn trim pots, which you probably already have.
Quote from: gotoluc on December 21, 2015, 02:59:19 PM
Could a on the fly pulse width adjustment capability be added to this?... and if so, what components are need to be ordered for that option?
Yes - that is half of the purpose of such circuit.
Quote from: gotoluc on December 21, 2015, 02:59:19 PM
I have 2 of the below outrunner motors. Do you think a circuit could be built to power it and collect the flyback to power the second one which would be attached to the same shaft?
I don't know offhand, I'd have to think about it.
The magnetic circuit can be source of large loses, too.
I don't remember what coil's core and magnets Itsu is using but if it is a ceramic ferrite and ceramic magnets then this would make a difference because NdFeB magnets are internally conductive and allow Eddy current to develop inside them just like in steel.
Quote from: gotoluc on December 21, 2015, 03:08:48 PM
I don't know? the scope I have access to is a TDS 2024B which by the way has very low resolution screen shots captures compared to your and why I started taking pictures of the screen to show more details.
Luc
Luc,
I use the Tektronix "OpenChoice Desktop" free application to capture / display my screenshots which can be found here:
http://www.tek.com/oscilloscope/tds210-software
I hooked up my scope via its ethernet connection, but you can do it with your USB connection too and get the same screenshots i do.
Itsu
Quote from: conradelektro on December 21, 2015, 03:17:08 PM
Hi itsu (and Hi to all people who do decent measurements like itsu):
To find out if the energy in the spikes can be "fed back" to save power, one should do the following measurement:
1) You do the input power measurement with the motor running like you do in your video https://www.youtube.com/watch?v=dANzzD6X75E&feature=youtu.be
2) But then you do not stop the motor, but you disconnect the path to the capacitor (by disconnecting the diode). This will prohibit that power is fed back from the spikes. And now you measure the input power. Is it less or more?
I tried that several times with different pulse motors and different drive circuits. I never could prove that input power to the motor was less if one tries to feed back power from the spikes (from the back EMF of the drive coil).
I think that the power in the spikes is forever lost.
Greetings, Conrad
Hi Conrad,
i did that test, but there was only a very small increase in input power after removing the diode (177mW to 180mW), but i don't think that the present circuit that i am using
has any capability to feed back any spikes.
Itsu
Quote from: itsu on December 21, 2015, 02:57:37 PM
Partzman,
my understanding is that your "red Math Mean -125.8mVV" value is the mean value from the total screen (so the 10 horizontal divisions).
So not the the Math mean value of the area between the 2 vertical cursors which is what the question was.
You presently have displayed the values of the area between the 2 cursors of the yellow trace as amplitude and timing (upper right corner).
If you now activate the red math trace, in that upper right corner you will see the data of that red math trace (wattage and timing) inbetween the 2 cursors, but
i doubt it will be the calculated power delta as it does not match my full screen calculations when trying, but i will double check.
This might be interesting for Luc also as he has a similar scope.
Itsu
Itsu,
OK, I may be misunderstanding you but the measurement shown on the previous scope shot is the mean or average of the Math waveform between the vertical cursors and not the full screen width. I've attached another scope pix taken from the same test but with the cursors re-positioned to show the mean of CH2 which is 234.8mv but also notice the Math mean of -82mvv.
You can use the vertical cursors to define any measurement window in time that you desire with any of the available measurement resources on your scope. It is really powerful IMO!
partzman
Quote from: conradelektro on December 21, 2015, 03:17:08 PM
2) But then you do not stop the motor, but you disconnect the path to the capacitor (by disconnecting the diode). This will prohibit that power is fed back from the spikes. And now you measure the input power. Is it less or more?
Please remember that we are measuring energy expended by the power supply only during the ON-pulse period and the "flyback" spike occurs
outside of the ON-pulse period, so anything we do outside of this period will have minimal influence on this energy measurement.
During the 100μs that the current flows from L1 to C2, the L1 remains an electromagnet and is capable of attracting the rotor.
Once all of the energy is transferred from L1 to C2, the electromagnetic influence on the rotor stops, after which the energy stored in C2 is simply wasted in R2.
I foresee, that once we make sure that Itsu's energy recovery efficiency numbers are not a result of some measuring error, then we will try to loop the energy stored in C2 back into the power supply (...or L1).
Quote from: partzman on December 21, 2015, 05:15:00 PM
I've attached another scope pix taken from the same test but with the cursors re-positioned to show the mean of CH2 which is 234.8mv but also notice the Math mean of -82mvv.
Show me the cursors re-positioned in such manner, that they encompass only the positive excursion of the red Math waveform and you will make a believer out of me if the Math Mean becomes all positive.
@Gotoluc
In case your scope cannot do a Mean of a Math channel that multiplies 2 other channels (a current and a voltage), here is how to eyeball the power contained in the pulse created by a switched constant voltage source, just by looking at its current waveform:
The real current pulse is blue in all four scopeshots below:
1) The area of the red triangle is much larger than the area of the blue pulse, thus the red triangle overestimates the energy of the blue pulse.
2) The area of the yellow triangle is much smaller than the area of the blue pulse, thus the yellow triangle underestimates the energy of the blue pulse.
3) The area of the green triangle is the same as the area of the blue pulse, thus the green triangle approximates the energy of the blue pulse, the best.
The image in the lower-right corner shows how to eyeball and draw this green triangle that approximates the blue pulse well - Namely, as you draw it, make sure that the orange area and the purple area are equal.
Once you have that green triangle drawn, its height will give you the interpolated peak current. Let's call it iPEAK
The width of the triangle is the period of the ON-pulse. Let's call it tON
If you multiply the constant supply Voltage * iPEAK * tON * 0.5 then the result will be the energy contained in this pulse expressed in Joules.
P.S.
If you want your result in Joules, then:
The Voltage must be expressed in Volts (e.g.: mV are not allowed) and must be constant during the entire ON-pulse.
The IPEAK must be expressed in Amps (e.g.: mA are not allowed)
The tON must be expressed in seconds (e.g.: ms or μs or ns are not allowed)
The tON is a little narrower than the blue pulse, because the remainder of the blue pulse's width is the short energy recovery period, when the energy is transferred from the coil to the cap.
Nice work Luc, others have investigated this area and probably still are, the infamous Romero diverted the spike into motoring coils, lots of info on his site, i can't remember the address.
I'll look for some example threads and remind myself of what's already been looked at in this area.
Y.
Quote from: gotoluc on December 21, 2015, 02:59:19 PM
I have 2 of the below outrunner motors. Do you think a circuit could be built to power it and collect the flyback to power the second one which would be attached to the same shaft?
Thanks for your time and help
Luc
My electric bike uses a 3 phase hub motor. I had posted about it in the rotating thread. I hooked up my scope to just 2 of the 3 wires going to the coils just to see if there is anything to grab when it comes to bemf. Surprisingly there doesnt seem to be much. When the motor is WOT wide open throttle, there is no pwm pulses. Just acts similar to a pulse motor. And when I throttle down some to where there is pulsing, im not seeing any great spikes. Like the whole pwm 3phase process is clamping the spikes. So the motor may have to be driven another way, possibly by even separating the windings and having individual drive circuits for each in order to avoid the clamping, or what ever is happening.
It is a 48v system. Scope seems to be showing more. Not sure what is up with that yet. Probably the little bit there is of spiking. I kinda expected some big spiking. Maybe the diodes on the fets putting out the flames. Was thinking to put a bridge rectifier across the coils in pairs where there would be 3 bridges to collect from. The bridge diodes would have to be lower conducting voltage than the fets diodes in order to get output, if the diodes in the fets are what is clamping the spikes.
The scope shots were of the hub motor wheel off the ground free wheeling. The pyramid shape of the trace turns into a clean sine when the throttle is turned to off showing the motor generating near what the wave forms show and dies out as the wheel slows to a stop.
Not trying to distract. Just noticed your 3 phase motor and posted this. Be nice if we could figure this out. Will be playing with this more.
Mags
Quote from: partzman on December 21, 2015, 05:15:00 PM
Itsu,
OK, I may be misunderstanding you but the measurement shown on the previous scope shot is the mean or average of the Math waveform between the vertical cursors and not the full screen width. I've attached another scope pix taken from the same test but with the cursors re-positioned to show the mean of CH2 which is 234.8mv but also notice the Math mean of -82mvv.
You can use the vertical cursors to define any measurement window in time that you desire with any of the available measurement resources on your scope. It is really powerful IMO!
partzman
Partzman,
ok, i see what you mean, i will make some further tests and see if it is useable, i would be indeed a powerfull function.
Thanks, itsu
verpies,
QuoteI don't remember what coil's core and magnets Itsu is using but if it is a ceramic ferrite and ceramic magnets then this would make a difference because NdFeB magnets are internally conductive and allow Eddy current to develop inside them just like in steel.
I use neo's as magnets and some cut up welding rods in a soft iron sleeve as a core.
QuoteI foresee, that once we make sure that Itsu's energy recovery efficiency numbers are not a result of some measuring error, then we will try to loop the energy stored in C2 back into the power supply (...or L1).
I will redo these tests, presently i am working on a coil shortener using a reed relais triggered by a hall sensor (+ transistor), hopefully the 15 or so mJ in the cap will not fuse together the reed relais contacts.
Itsu
I put up a simple test using my SG to provide a 5Vpp 40Hz sine wave signal to a 50 Ohm resistor with a 1 Ohm csr.
I setup the math function Ch1 x Ch2 to show the wattage across the 50 Ohm resistor.
The signals can be seen in the below screenshots, but as can be seen, both screenshots (with or without the cursors) show the same Math mean value.
So either Partzman has an extra feature (advanced analysis) in his scope or our models are behaving differently.
Video here: https://www.youtube.com/watch?v=tvnUN8u2ILk&feature=youtu.be
Regards Itsu
Quote from: MileHigh on December 19, 2015, 05:33:07 PM
So for Itsu's setup, comparing power out vs. power in and rotor vs. no rotor does not look conclusive if you are looking at the power out vs. power in metric.
Let's just look at the power draw of the coil itself. When you add the spinning rotor you get reduced power draw. That is clear and has been replicated many times. But how about this: Instead of adding the rotor to reduce the power draw, why not forget about the rotor completely and just reduce the drive voltage for the coil? So that also gives you reduced power draw without adding a rotor. The back spike might still output the same amount or power. Do you see what I mean? This is highly academic because what can you do with a pulsing coil if you don't have a rotor?
No MH
When the rotor was stopped,the input power went up,and the output power from the spike went down.
It is hard to keep up now Luc's thread seems to have drifted over to what my thread was about-the alternating external magnetic fields from a spinning rotor decreasing P/in and increasing P/out.
But it is good to se that Itsu is showing the same sort of results as i did with the alternating magnetic fields.
Brad
Quote from: Magluvin on December 21, 2015, 11:47:14 PM
My electric bike uses a 3 phase hub motor. I had posted about it in the rotating thread. I hooked up my scope to just 2 of the 3 wires going to the coils just to see if there is anything to grab when it comes to bemf. Surprisingly there doesnt seem to be much. When the motor is WOT wide open throttle, there is no pwm pulses. Just acts similar to a pulse motor. And when I throttle down some to where there is pulsing, im not seeing any great spikes. Like the whole pwm 3phase process is clamping the spikes. So the motor may have to be driven another way, possibly by even separating the windings and having individual drive circuits for each in order to avoid the clamping, or what ever is happening.
It is a 48v system. Scope seems to be showing more. Not sure what is up with that yet. Probably the little bit there is of spiking. I kinda expected some big spiking. Maybe the diodes on the fets putting out the flames. Was thinking to put a bridge rectifier across the coils in pairs where there would be 3 bridges to collect from. The bridge diodes would have to be lower conducting voltage than the fets diodes in order to get output, if the diodes in the fets are what is clamping the spikes.
The scope shots were of the hub motor wheel off the ground free wheeling. The pyramid shape of the trace turns into a clean sine when the throttle is turned to off showing the motor generating near what the wave forms show and dies out as the wheel slows to a stop.
Not trying to distract. Just noticed your 3 phase motor and posted this. Be nice if we could figure this out. Will be playing with this more.
Mags
It is a sure bet the inductive spikes are being clamped by a diode feeding into a cap. But at 100% duty cycle,you will not get any inductive kickback,as it is an AC cycle, and so the spike is being bombed with a reverse current flow.
Brad
Quote from: itsu on December 22, 2015, 06:12:48 AM
So either Partzman has an extra feature (advanced analysis) in his scope or our models are behaving differently.
...or Partzman has a different firmware
Quote from: itsu on December 22, 2015, 05:36:33 AM
I use neo's as magnets and some cut up welding rods in a soft iron sleeve as a core.
Could you show us a hires photos of these welding rods in a soft iron sleeve ?
Quote from: itsu on December 22, 2015, 05:36:33 AM
I will redo these tests, presently i am working on a coil shortener using a reed relay triggered by a hall sensor (+ transistor), hopefully the 15 or so mJ in the cap will not fuse together the reed relay's contacts.
I am afraid it will fuse the relay's contacts :(
But there is an easy remedy to this. Connect a choke to the relay's contacts in parallel with a second flyback diode (the diode must be oriented so it does not conduct when the relay's contacts initially close). Put a 100Ω resistor in series with that diode.
This way when the relay's contacts close, the current will raise gradually through the choke and when they open later, the diode and resistor will prevent any arc over. The resistor's function is to prevent the current from lingering in the choke, for too long.
P.S.
Also, you could drive the relay from a 2nd channel of your FG.
Quote from: itsu on December 22, 2015, 06:12:48 AM
I put up a simple test using my SG to provide a 5Vpp 40Hz sine wave signal to a 50 Ohm resistor with a 1 Ohm csr.
I setup the math function Ch1 x Ch2 to show the wattage across the 50 Ohm resistor.
The signals can be seen in the below screenshots, but as can be seen, both screenshots (with or without the cursors) show the same Math mean value.
So either Partzman has an extra feature (advanced analysis) in his scope or our models are behaving differently.
Video here: https://www.youtube.com/watch?v=tvnUN8u2ILk&feature=youtu.be
Regards Itsu
Itsu,
OK, you must have the cursor gating function "off". From the front panel menu, press "Measure" and then see what the bottom screen menu shows for the gating function (third from the left). It should say "Gating Off". On the right vertical menu select the soft key "Between Vertical Bar Cursors" and this will allow measurements between the cursors. The gating cursor function on the bottom screen menu should now say "Gating Cursors". This should do it for you- enjoy!
partzman
Quote from: verpies on December 22, 2015, 07:45:35 AM
Could you show us a hires photos of these welding rods in a soft iron sleeve ?
I am afraid it will fuse the relay's contacts :(
But there is an easy remedy to this. Connect a choke to the relay's contacts in parallel with a second flyback diode (the diode must be oriented so it does not conduct when the relay's contacts initially close). Put a 100Ω resistor in series with that diode.
This way when the relay's contacts close, the current will raise gradually through the choke and when they open later, the diode and resistor will prevent any arc over. The resistor's function is to prevent the current from lingering in the choke, for too long.
P.S.
Also, you could drive the relay from a 2nd channel of your FG.
Ok, hires photo below, the use of the slug or sleeve was suggested in an article about the Adam's motor which i was following years ago, see this link which explains why (eddy currents):
https://sites.google.com/site/ageechen/free-energy/adams-motor/hoptoads-articles/page-4
Thanks for the additional info in the reed relais setup, i know i can use my SG for that too, but i want to avoid using the SG for now.
Itsu
Quote from: partzman on December 22, 2015, 09:25:50 AM
Itsu,
OK, you must have the cursor gating function "off". From the front panel menu, press "Measure" and then see what the bottom screen menu shows for the gating function (third from the left). It should say "Gating Off". On the right vertical menu select the soft key "Between Vertical Bar Cursors" and this will allow measurements between the cursors. The gating cursor function on the bottom screen menu should now say "Gating Cursors". This should do it for you- enjoy!
partzman
Wow, yes, you are absolutly right, gating was off, so now it does calculate the power inbetween the cursors, many thanks, very usefull.
Itsu
Quote from: itsu on December 21, 2015, 04:35:46 PM
Luc,
I use the Tektronix "OpenChoice Desktop" free application to capture / display my screenshots which can be found here:
http://www.tek.com/oscilloscope/tds210-software
Itsu
Thanks Itsu, I give it a try.
Quote from: verpies on December 21, 2015, 06:04:03 PM
@Gotoluc
In case your scope cannot do a Mean of a Math channel that multiplies 2 other channels (a current and a voltage), here is how to eyeball the power contained in the pulse created by a switched constant voltage source, just by looking at its current waveform:
Thanks for the details verpies
Quote from: Magluvin on December 21, 2015, 11:47:14 PM
I hooked up my scope to just 2 of the 3 wires going to the coils just to see if there is anything to grab when it comes to bemf. Surprisingly there doesn't seem to be much.
Mags
Hi Mags, thanks for your post. You're right, the frequency drive controlled motors would not work for what I'm suggesting.
I think the motors I was thinking of is the BLDC which have hall sensors to trigger the 3 mosfets.
Thanks for all your help guys
Luc
Quote from: itsu on December 22, 2015, 05:36:33 AM
...presently i am working on a coil shortener using a reed relay triggered by a Hall sensor (+ transistor), hopefully the 15 or so mJ in the cap will not fuse together the reed relay's contacts.
If you find out that the reed relay is not fast enough, you could use a MOSFET to short out the C2 (see the 1
st schematic below)
Parts recommendations:
Q3 should be a 15A, N-ch MOSFET capable of blocking at least 150V.
Q4 should be a 1A, N-ch MOSFET, capable of blocking at least 50V.
L2 should be approximately a 1mH choke capable of withstanding 500mA of current.
ZD2 should be a 500mA Zener diode featuring Zener voltages from 12V to 18V.
Theory of operation:
The L2 acts as an inverting boost inductor, that provides a gate voltage for Q3, that is higher that +V
1A, because N-channel MOSFETs need a gate voltage that is higher than the Source, in order to turn on.
This higher voltage could've been delivered by a second external power supply, supplying a voltage that is +10V above +V
1A. However, in this circuit, L2 acts as such power supply, thus a second power supply is not needed.
When the "Discharge C2" signal is high, Q4 closes and the gate of Q3 is pulled low through R3. Consequently, Q3 opens and currents starts building up in L2 (up to the +V
1A/R3 limit).
When the "Discharge C2" signal goes low, Q4 opens and the L2 current is interrupted. Consequently an inverted voltage is generated at point E (this voltage is higher than the +V
1A supply rail). This voltage is clamped by ZD2 and D2, which act to maintain L2 current and protect the gate of Q3 from overvoltage. As a result of this, Q3 closes and shorts out C2 through R2.
Note, that the "Discharge C2" signal is active when LOW !
This circuit should not be operated from a supply voltage ( +V
1A) higher than 20V without alterations.
A MOSFET driver U2 and a diode D3 can be used instead of Q4 and R4 (see the 2
nd schematic) and if this driver has an inverting input, then the the "Discharge C2" signal becomes active when high.
P.S.
A P-ch MOSFET could not have been used for Q3 because its body diode would be always shorting out C2.
Great stuff verpies, i certainly have to try it.
Meanwhile i have redone my measurements as to see if the overunity cap recovery condition still exists.
I tried to incorporate my new knowledge about doing power measurements on selected parts (inbetween cursors) of a wave form, but it also
influence the display of the other channels, so for a running situation (40Hz @ 18% duty cycle) i have 2 screenshots, first one without the selected math,
second one with selected math (ontime only).
I have a 100K potmeter for R2 set for the C2 voltage to go to 0 when the next pulse comes (yellow trace)
Blue trace is the CSR (1 Ohm 1%), purple trace the drain voltage, all signals referenced to point B
Supply voltage is 12.4V from a battery.
The 2e screenshot has the cursors setup for the on time only to calculate Ch2 x Ch3 which gives a mean wattage of 404mW (x 10 because of the 0.1 csr = 4.04W).
If needed i can provide the same for a stopped condition.
Itsu
Here an expanded (zoomed in) version of the ontime with math:
Here the same situation as the one above, but with yellow cursors (and data in upper right corner) and with the math trace (red) in lower amplitude:
Well Itsu, I tried the software but I think it's the scopes limitation.
.99 looked into this scopes specs before and was surprised of the low resolution the capture side had.
Below is:
Capture from the Scopes USB port
Capture from Software Screen Capture Tab
Capture from Software Data Capture Tab
I like the what the Data Capture Tab best and you can see more details but no scope data is included.
Do you think a firmware update would help?
Anyone else can reply as well
Thanks
Luc
Luc,
i like the last one best, thats what i use i think, but you probably use a different tab on the "openchoice desktop" application.
I use the green tab, see below screenshot, i guess you are using the red circled tab.
Itsu
Quote from: itsu on December 22, 2015, 04:50:25 PM
Meanwhile i have redone my measurements as to see if the overunity cap recovery condition still exists.
It sure does.
Going by the zoomed scopeshot:
Input energy during the ON-pulse: E
1 = 4.070W * 4.36ms = 4.07 * 0.00436 = 0.0177452J = 17.7452mJ
Recovered energy in C2: E
2 = ½ * 10μF * (99V)
2 = ½ * 0.000010 * 9801 = 0.049005J = 49.005mJ
E
2 / E
1 = 49.005mJ / 17.7452mJ = 2.76159 = ~
276%
These welding rods is good stuff !
P.S.
The sawtooth current energy estimate yields: E
1 = 660mA * 12.4V *4.36ms * ½ = 0.660 * 12.4 * 0.00436 * ½ = 0.017841J = 17.841mJ ( pretty close ! )
Quote from: verpies on December 22, 2015, 05:27:03 PM
It sure does.
Going by the zoomed scopeshot:
Input energy during the ON-pulse: E1 = 4.070W * 4.36ms = 4.07 * 0.00436 = 0.0177452J = 17.7452mJ
Recovered energy in C2: E2 = ½ * 10μF * (99V)2 = ½ * 0.000010 * 9801 = 0.049005J = 49.005mJ
E2 / E1 = 49.005mJ / 17.7452mJ = 2.76159 = ~276%
These welding rods is good stuff !
P.S.
The sawtooth current energy estimate yields: E1 = 660mA * 12.4V *4.36ms * ½ = 0.660 * 12.4 * 0.00436 * ½ = 0.017841J = 17.841mJ ( pretty close ! )
Isnt it amazing what can be found when one has an open mind when it comes to these laws of physics. As i stated before Verpies,it would only be you of the EE guys that would take note of what others have to show when it go's against these known laws--oh and the books lol.
276%
No bad Itsu--not to bad at all ;)
Brad
Quote from: itsu on December 22, 2015, 05:26:08 PM
Luc,
i like the last one best, thats what i use i think, but you probably use a different tap on the "openchoice desktop" application.
I use the green tap, see below screenshot, i guess you are using the red circled tap.
Itsu
The one you use "green tab" give me the below. The red tab give me the next one with no data.
So by using the software I don't get anything better. It's the scopes limitation compared to yours which must be better.
Luc
Quote from: verpies on December 22, 2015, 05:27:03 PM
It sure does.
Going by the zoomed scopeshot:
Input energy during the ON-pulse: E1 = 4.070W * 4.36ms = 4.07 * 0.00436 = 0.0177452J = 17.7452mJ
Recovered energy in C2: E2 = ½ * 10μF * (99V)2 = ½ * 0.000010 * 9801 = 0.049005J = 49.005mJ
E2 / E1 = 49.005mJ / 17.7452mJ = 2.76159 = ~276%
These welding rods is good stuff !
P.S.
The sawtooth current energy estimate yields: E1 = 660mA * 12.4V *4.36ms * ½ = 0.660 * 12.4 * 0.00436 * ½ = 0.017841J = 17.841mJ ( pretty close ! )
WOW :o
Excellent results there Itsu!!!... congratulations
Thanks for your great tests and sharing
Luc
Quote from: gotoluc on December 22, 2015, 01:00:20 PM
Hi Mags, thanks for your post. You're right, the frequency drive controlled motors would not work for what I'm suggesting.
I think the motors I was thinking of is the BLDC which have hall sensors to trigger the 3 mosfets.
Luc
Hey Luc
Well the bike hub motor is a bldc and uses halls to detect timing. Just really getting a hang on how they really work. The outrunners use self detecting driver modules that dont need halls or other timing circuits from what I understand, like my bike motor could probably run on one of those driver circuits and eliminate the halls.
What I want to try is to separate the windings so they are not together at all. And possibly be able to capture what isnt there in their normal config by having separate drivers for each independent coil. This may involve only having the coils operate at 1 polarity and shut off while the others do their thing and so on. Like my Tidalforce bike motor, it has 7 stator coils, but none of them are connected like a typical 3 phase as shown earlier. They have their own separate drive circuits, kind of. Will explain later as I get into that.
Like Brad was saying, and Im thinking he is correct, when the transistors of my hub motor fire a coil, when the power is cut, the circuit immediately switches polarity to the coil, not giving bemf a chance to be captured. Thinking about it most of today, the whole pwm while the throttle is less than wide open, there are transistors that connect one coil to neg or gnd zero as it is 48 v dc in, and the plus side transistor does the pulsing. Then when time for a phase shift, another transistor grounds the other end of the coil and yet again another transistor applies plus to the other end of the coil. So there is a lot going on there in the 3 phase setup.
So what I want to try to do is change the circuitry to allow bemf capture. It may require a lessening of performance due to off periods described above while the others work. With all that pulsing going on during one phase, im seeing a basic switching supply that should have bemf or field collapse output capabilities, while driving the magnet end of the motor.
Its nice to work with well made consumer/commercial motors, if viable to changes I am suggesting, because we are enabled to already have a very good base to begin with, tolerances and all. Like Brads rt, it would take a bit of work to make the motor exactly like that, when it can be had readily for little investment. Plus, having nice tight tolerances for field paths should enhance the bemf output like a decent switching supply.
I have a 3 phase automotive cooling fan motor that has a bell rotor with 3 mags inside the bell and a stator that is inside the bell with 3 coils. As basic as it gets. Im going to disconnect the windings from each other ad see what I can come up with there as a test bed. Has heavy windings and has a small magnet wheel with alt poles to trigger 3 halls.
What you have shown with the pulsing and loading the motor without increase in has inspired me. It should be a goal to expand on that and test the limits and such. Like would there be an increase in if the motor were not allowed to turn at all while pulsing. If so then I am inspired more. ;D
Mags
Quote from: tinman on December 22, 2015, 06:39:42 PM
As i stated before Verpies,it would only be you of the EE guys that would take note of what others have to show when it go's against these known laws--oh and the books lol.
I would like some other EE to verify:
- my calculations,
- measurement methodology,
- conceptual errors.
Seeking a second opinion is good science. It not wise to rely on the expertise of only one person. I am only human.
Quote from: verpies on December 22, 2015, 05:27:03 PM
It sure does.
Going by the zoomed scopeshot:
Input energy during the ON-pulse: E1 = 4.070W * 4.36ms = 4.07 * 0.00436 = 0.0177452J = 17.7452mJ
Recovered energy in C2: E2 = ½ * 10μF * (99V)2 = ½ * 0.000010 * 9801 = 0.049005J = 49.005mJ
E2 / E1 = 49.005mJ / 17.7452mJ = 2.76159 = ~276%
These welding rods is good stuff !
P.S.
The sawtooth current energy estimate yields: E1 = 660mA * 12.4V *4.36ms * ½ = 0.660 * 12.4 * 0.00436 * ½ = 0.017841J = 17.841mJ ( pretty close ! )
Hmmm, so where is the flaw, as there must be something wrong, Right?
Guess i need to empty the C2 cap before each new spike capturering, so will try first with the reed, then with your
nice circuits using a MOSFET / driver if the reed does not work.
Itsu
Quote from: gotoluc on December 22, 2015, 06:51:05 PM
The one you use "green tab" give me the below. The red tab give me the next one with no data.
So by using the software I don't get anything better. It's the scopes limitation compared to yours which must be better.
Luc
Luc, seems strange to me as your scope is newer (i still have a 1.44MB floppy and 2007 Firmware).
Perhaps a scope Guru can step in and advice a different setup.
Itsu
Quote from: itsu on December 22, 2015, 07:23:43 PM
Hmmm, so where is the flaw, as there must be something wrong, Right?
So far I cannot detect any mistake of mine. That's why I'd like somebody smarter to check me. Cyril Smith (a.k.a. Smudge (http://www.overunityresearch.com/index.php?action=profile;u=639)) has not commented yet.
I have received one constructive criticism via PM, namely that the voltage drop across the CSR affects the C2 voltage as measured between the points B and D.
This is a valid observation, because you are not measuring this voltage directly across C2 (between +V
1A and point D), however at ~1A currents, this error is very small, approximately ~100mV for C2 peak voltages around ~100V ...so the error is only 1:1000 or 0.1%
I was aware of this quirk when I designed the scope probe placements.
Well, I know for a fact that I am not as smart or as skilled as Verpies, but I can off some suggestions for Itsu and the team to consider.
I see that C2 gets charged to a high voltage. How about trying another cap of the same value, or just trying another cap value completely, perhaps between 2X and 10X larger? You get the new peak voltages, do the numbers still crunch to "over unity?" I realize that you want to keep the cap voltages high to minimize the effect of the diode drop on the measurements.
Then there are two related mechanisms to check the output pulse energy. First of all, disconnect the MOSFT cap discharge circuit. Use a very large cap for C2 and charge it to say 12 volts at the start of the test. Then run the pulse motor with your time base set to very slow. Count motor pulses and record the gradual increase in voltage on C2. Then you can simply pick a pulse range and a voltage range and calculate the energy per pulse. So you have a certain advantage here with time-averaging helping you out.
The other thing is to use a large C2 and just put a known value of bleeder resistor across C2 and run the pulse motor. This is the "inflating a leaky tire" technique. Just run the pulse motor until the voltage across the big cap completely stabilizes. Take your scope and make sure that the voltage across the big cap is near-DC. Then with an accurate measurement of the RMS voltage across the cap with a good multimeter, and an accurate measurement of the resistance of the bleeder resistor, and an accurate measurement of the pulse rate, you just have to crunch the numbers to get the back spike energy per pulse.
Right now my assumption is for the sake of argument that something is amiss with the measurement of the energy in the back spike. Therefore I am suggesting a few alternative ways to measure that energy. If all of the measurement methods agree within five percent, my instincts would be telling me that the problem would be in the measuring the energizing of the drive pulse itself, even though at first glance that also looks pretty well done and quite straightforward and not prone to problems.
Anyway, I won't go into that for now, but the train of thought would be the same - come up with some alternative ways of measuring the energy going into the drive pulse.
MileHigh
Quote from: tinman on December 22, 2015, 06:39:42 PM
Isnt it amazing what can be found when one has an open mind when it comes to these laws of physics. As i stated before Verpies,it would only be you of the EE guys that would take note of what others have to show when it go's against these known laws--oh and the books lol.
276%
No bad Itsu--not to bad at all ;)
Brad
The problem is that you are jumping the gun again. Good science and good testing and measuring of experimental data is never achieved by jumping the gun.
Quote from: MileHigh on December 22, 2015, 09:18:05 PM
The problem is that you are jumping the gun again. Good science and good testing and measuring of experimental data is never achieved by jumping the gun.
I have been watching this very carefully MH, and have faith in Verpies measurement procedure and calculations. It is great to see Itsu taking it far enough to see the results he is getting. I would also ask if you really think an error of over 100% has been made here?
Brad.
Quote from: Magluvin on December 22, 2015, 07:16:11 PM
I have a 3 phase automotive cooling fan motor that has a bell rotor with 3 mags inside the bell and a stator that is inside the bell with 3 coils. As basic as it gets. Im going to disconnect the windings from each other ad see what I can come up with there as a test bed. Has heavy windings and has a small magnet wheel with alt poles to trigger 3 halls.
Mags
Sorry. 6 individual coils and 4 magnets. Dont know what I was thinking. ::)
Mags
Quote from: tinman on December 22, 2015, 11:08:24 PM
I have been watching this very carefully MH, and have faith in Verpies measurement procedure and calculations. It is great to see Itsu taking it far enough to see the results he is getting. I would also ask if you really think an error of over 100% has been made here?
Brad.
I have a lot of faith in Verpies and Itsu. Yes I think an error of over 100% has been made and I am willing to bet you that Verpies and Itsu also believe that an error has been made. My assumption right now is that they will continue this investigation and find out where the error was. Things like this have happened many times in the past and they are bound to happen again. I will just repeat that you are jumping the gun.
Let's see what happens.
MileHigh
Dear Itsu,
I have been testing all the coils I have and cores like metglass, steel laminations, welding rods and so on.
I have carefully tested them and noticed thing that I have noticed and said in the past.
Not surprisingly the better results were had with the coils of low resistance but with a super low duty cycle like 1% (if coil is below 1 Ohm) but use as high of a voltage as possible but obviously stop when the cap reaches max voltage and drop it maybe a volt or two.
If you can redo your test at 10Hz instead fo 40Hz which should be easier to sync your rotor and drop your duty cycle and increase your voltage as much as you can all while keeping the rotor turning, I would be prepared to say the results could be even better.
Thank you for your great work
Luc
Quote from: MileHigh on December 22, 2015, 11:37:17 PM
I have a lot of faith in Verpies and Itsu. Yes I think an error of over 100% has been made and I am willing to bet you that Verpies and Itsu also believe that an error has been made. My assumption right now is that they will continue this investigation and find out where the error was. Things like this have happened many times in the past and they are bound to happen again. I will just repeat that you are jumping the gun.
Let's see what happens.
MileHigh
I agree MH
There are two very bright people working on this, and yes-if there are errors, they will find them.
Itsu
Are these results still with the rotor in play within the system?
If so, what are the results without it in play?
Brad
Quote from: MileHigh on December 22, 2015, 09:16:28 PM
How about trying another cap of the same value, or just trying another cap value completely, perhaps between 2X and 10X larger? You get the new peak voltages, do the numbers still crunch to "over unity?" I realize that you want to keep the cap voltages high to minimize the effect of the diode drop on the measurements.
Not only. Larger cap also lengthens the energy recovery period and that has a negative correlation with the L1 discharge efficiency, which decreases with longer discharge periods (times that are larger fractions of its Tau (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/153963/)).
To accommodate 10x larger cap while keeping the efficiency, the inductance of L1 would have to be increased too. But I think we could get away with a 2x cap.
Quote from: MileHigh on December 22, 2015, 09:16:28 PM
First of all, disconnect the MOSFET cap discharge circuit.
I don't think the MOSFET cap discharge circuit has been implemented yet.
The latest scopeshots were made just with a 10K bleeder resistor in place of Q3, because Itsu found out empirically, that this resistance discharges C2 to 0V just before the next pulse begins.
I think Itsu's next version of the circuit will use a choked reed relay in place of Q3.
Quote from: MileHigh on December 22, 2015, 09:16:28 PM
...my instincts would be telling me that the problem would be in the measuring the energizing of the drive pulse itself, even though at first glance that also looks pretty well done and quite straightforward and not prone to problems.
Do you mean something like scope probe attenuation set incorrectly in the scope's menus ...or the probe's 10x switch not making a good contact?
Also, if the CSR had resistance of 0.01Ω instead of the expected 0.1Ω, then the ohmmeter would not catch the difference and that peak ~600mA L1 current reading would really mean ~6A.
...but the problem with that theory is that Itsu has used his Hall current probe and the current readings were comparable with those given by the 0.1Ω CSR.
If you think of a good method to triple-check the L1 current waveform - I am sure that Itsu will be open to it.
Quote from: gotoluc on December 22, 2015, 11:53:51 PM
Not surprisingly the better results were had with the coils of low resistance but with a super low duty cycle like 1% (if coil is below 1 Ohm)
Yes, and your coil appears to have superior L/R parameters compared to Itsu's, e.g. your coil's current is a straight up-ramp and it doesn't flatten like Itsu's current at the end.
Quote from: verpies on December 23, 2015, 02:52:54 AM
Yes, and your coil appears to have superior L/R parameters compared to Itsu's, e.g. your coil's current is a straight up-ramp and it doesn't flatten like Itsu's current at the end.
Verpies
Once the current ramp flattens out,this is the point of core saturation-is it not?
If so,then it is possible for Itsu to reduce the current flow time,and switch off before or just at the point of saturation,and gain even a higher efficiency.
Do i have that correct?
Brad
P.S
Do you think that these results have anything to your theory you told me about in an Email some time back?. I believe you were in hospital at the time.
Quote from: tinman on December 23, 2015, 03:28:12 AM
If so,then it is possible for Itsu to reduce the current flow time,
Yes, reducing current flow time increases the efficiency because of
this effect (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/153963/) but not because of saturation avoidance...at least in Itsu's current system.
Quote from: tinman on December 23, 2015, 03:28:12 AM
Do you think that these results have anything to your theory you told me about in an Email some time back?
My theory had to do with moving (or vibrating)
soft ferromagnets. Do Itsu's results have anything like that in them - I don't know yet.
Quote from: tinman on December 23, 2015, 03:28:12 AM
Once the current ramp flattens out, this is the point of core saturation - is it not?
No, the flattening out of current means hitting the V/R limit (usually caused by the current flowing too long).
On a scope, the saturation curves the coil's current up, that is in the opposite direction.
Quote from: verpies on December 23, 2015, 03:37:04 AM
Not.
On a scope, the saturation curves the coil's current up, that is in the opposite direction.
Ah-ok,learn something new every day. I was of the belief that once the current trace had flattened out,then maximum current flow had been achieved,and the magnetic field would not get any stronger or greater. Or maybe that part is correct,and the core saturation curve would only be seen when the current being supplied to the coil is much higher than we are delivering at the moment ?.
Brad
Quote from: tinman on December 23, 2015, 03:46:25 AM
Ah-ok,learn something new every day. I was of the belief that once the current trace had flattened out,then maximum current flow had been achieved,and the magnetic field would not get any stronger or greater. Or maybe that part is correct,
It is correct.
The magnetic field does not get any stronger or greater, because the circuit's resistance limits the current according to Ohm's Law i=V/R ,...not because of any magnetic or core effects.
Quote from: tinman on December 23, 2015, 03:46:25 AM
and the core saturation curve would only be seen when the current being supplied to the coil is much higher than we are delivering at the moment ?
Yes, but that can happen only if the V is high enough or the R is low enough, in the Ohm's equation i=V/R.
Bonus Rant:
What if both of these phenomena occur at the same time?
The V/R limit wins out eventually but for a while they can superimpose on each other and obscure each other.
The down curve (due to the V/R limit) usually has a different shape than the up curve (due to saturation) so they don't superimpose on each other perfectly and some intermediate current curve results for a short while.
Quote from: verpies on December 23, 2015, 03:44:09 AM
Yes, reducing current flow time increases the efficiency because of this effect (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/153963/) but not because of saturation avoidance...at least in Itsu's current system.
My theory had to do with moving (or vibrating) soft ferromagnets. Do Itsu's results have anything like that in them - I don't know yet.
I was just thinking about what Itsu has used for a core--an odd combination at best.
If i am correct,those welding rods are black iron welding rods used for fusion welding with a naked flame or tig welder(hot arc). The dina bolt sleeves have a slit in them from one end to the other,and so no full turn is complete--dose not act as a shorted turn.
I have gone and brought some mosfets-IRF 954,and will wind a coil in the same fashion as Itsu has,although i am not sure if he has left all the rods hanging out the back of his coil,or has trimed them back to the former length. I will do this on my thread,as this thread(Luc's) is about sending the flyback to a second inductor,and increasing motor efficiency,where as i am sending the fly back to a battery--but the cap seems to be a better way to measure P/in to P/out.
Brad
Quote from: verpies on December 23, 2015, 03:54:34 AM
It is correct.
The magnetic field does not get any stronger or greater, because the circuit's resistance limits the current according to Ohm's Law i=V/R ,...not because of any magnetic or core effects.
Yes, but that can happen only if the V is high enough or the R is low enough, in the Ohm's equation i=V/R.
OK-Thanks
Quote from: tinman on December 23, 2015, 03:57:37 AM
I have gone and brought some mosfets-IRF 954,and will wind a coil in the same fashion as Itsu has,
Do you know how he has wound his coil and with what wire ?
Quote from: tinman on December 23, 2015, 01:34:20 AM
I agree MH
There are two very bright people working on this, and yes-if there are errors, they will find them.
Itsu
Are these results still with the rotor in play within the system?
If so, what are the results without it in play?
Brad
Brad,
i think so, but i will check that later.
The core can be seen here:
http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg469253/#msg469253
so the sleeve and the welding rods (genuine bedini) are sticking out of the coil as can be seen there.
The coil is 37.8mH and 10 Ohm, wire is AWG 24 (0.5mm diameter) it was made some years ago for an Adams motor, see specs in the
earlier supplied link.
Its series bifilar, at least 4 wires where sticking out and i connected in to out getting the 10 Ohm.
I think the coil dimension should be as wide as thick.
But i don't think the core / coil is what is causing this effect, more likely some measuring error/ component screwup.
MOSFET in use is an IRF530
Will do some re-checking / tests later today.
Itsu
Quote from: itsu on December 23, 2015, 04:37:11 AM
But i don't think the core / coil is what is causing this effect, ...
You can confirm that by connecting a different coil in place of the present L1. That would be a negative experiment of sorts...
That different coil should have the same or larger inductance so it does nor screw up the current circuit.
Quote from: MileHigh on December 22, 2015, 09:16:28 PM
...come up with some alternative ways of measuring the energy going into the drive pulse.
Let me give you same data to chew on:
According to the
latest scopeshot (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg469297/#msg469297), the energy expended by the power supply to bring L1's current from 0 to ~600mA is 17.745mJ, however according to the standard formula for energy stored in a coil E=½Li
2, the magnetic energy stored in a 37.8mH coil conducting ~600mA of current is 6.804mJ.
That is a big difference between what was consumed and what was stored in L1
The missing ~11mJ is easily explained by the mechanical energy gain of the rotor as well as hysteresis and resistive losses, but maybe this calculated energy imbalance will point you to a better explanation, what is going on.
Quote from: verpies on December 23, 2015, 04:12:27 AM
Do you know how he has wound his coil and with what wire ?
I do now ;)
But 10 ohm's is quite high,so i think i will drop that down a bit-or a lot,and go with a lower resistance across the coil. I will start with 2 ohms i think-less resistive losses that way,and see how it looks. It would only take 15 minutes to rewind a coil anyway,so might as well see if we get the same sort of results using a different resistance in the coil.
As i only have a 2 channel scope,the measurements will have to be done in pieces-this may make things a little harder--see how we go.
Brad
Quote from: verpies on December 23, 2015, 05:30:29 AM
Let me give you same data to chew on:
According to the latest scopeshot (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg469297/#msg469297), the energy expended by the power supply to bring L1's current from 0 to ~600mA is 17.745mJ, however according to the standard formula for energy stored in a coil E=½Li2, the magnetic energy stored in a 37.8mH coil conducting ~600mA of current is 6.804mJ.
That is a big difference between what was consumed and what was stored in L1
The missing ~11mJ is easily explained by the mechanical energy gain of the rotor as well as hysteresis and resistive losses, but maybe this calculated energy imbalance will point you to a better explanation, what is going on.
I am not following the thread closely but I will assume that the timing is still coming from Itsu's signal generator and the rotor is slaving to that. I agree that the missing energy is most likely explained by the energy gain of the rotor.
To me that suggests an important test and I hope Itsu would consider doing it (if not already done): Just look at the coil traces when there is no rotor spinning and compare them with the case with the rotor spinnng, exactly the same pulse timing. I believe a lot can be learned from that and it will open up people's minds to a series of issues that many readers are probably not thinking about or aware of.
I am a little confused though because I thought that Itsu's measurements are showing more energy in the back spike than energy expended to energize the coil. Am I missing something? As a result I am not sure what the better explanation is about since things seem to have returned to normal? Sorry, I don't have the energy to read the entire thread.
MileHigh
Quote from: MileHigh on December 23, 2015, 06:44:40 AM
I am not following the thread closely but I will assume that the timing is still coming from Itsu's signal generator and the rotor is slaving to that. I agree that the missing energy is most likely explained by the energy gain of the rotor.
To me that suggests an important test and I hope Itsu would consider doing it (if not already done): Just look at the coil traces when there is no rotor spinning and compare them with the case with the rotor spinnng, exactly the same pulse timing. I believe a lot can be learned from that and it will open up people's minds to a series of issues that many readers are probably not thinking about or aware of.
MileHigh
QuoteI am a little confused though because I thought that Itsu's measurements are showing more energy in the back spike than energy expended to energize the coil. Am I missing something? As a result I am not sure what the better explanation is about since things seem to have returned to normal? Sorry, I don't have the energy to read the entire thread.
Yes MH,Itsu's measurements are showing more energy captured in the back spike by the cap.
Verpies was showing what should be stored as magnetic energy in the coil,and what was actually consumed by the coil
Quote- :the energy expended by the power supply to bring L1's current from 0 to ~600mA is 17.745mJ-->however according to the standard formula for energy stored in a coil E=½Li2, the magnetic energy stored in a 37.8mH coil conducting ~600mA of current is 6.804mJ.
The recovered energy MH in C2 was- E2 = ½ * 10μF * (99V)2 = ½ * 0.000010 * 9801 = 0.049005J = 49.005mJ
So 17.745mJ in,coil should only have 6.804mJ of stored magnetic energy,but returned 49.005mJ to C2
Brad
Quote from: MileHigh on December 23, 2015, 06:44:40 AM
I am not following the thread closely but I will assume that the timing is still coming from Itsu's signal generator and the rotor is slaving to that.
Yes AFAIK, except he is using a 555 astable to avoid grounding issues. I think he is working on triggering from the rotor's angular position sensor for his next version.
Quote from: MileHigh on December 23, 2015, 06:44:40 AM
To me that suggests an important test and I hope Itsu would consider doing it (if not already done)
Already done, albeit not with the latest version.
Quote from: MileHigh on December 23, 2015, 06:44:40 AM
I am a little confused though because I thought that Itsu's measurements are showing more energy in the back spike than energy expended to energize the coil. Am I missing something?
You are not.
According to the most recent scopeshot:
A) 17.745mJ Energy expended by the power supply during one ON-pulse, to bring coil's current from 0 to ~600mA,
B) 6.804mJ Calculated energy stored in the 37.8mH coil at the end of the ON-pulse,
C) 49.005mJ Recovered energy in C2 after one ON-pulse.
The energy lost to the rotor's motion as well as the energy loss due to hysteresis are unknown.
The energy loss in the resistance could've been easily calculated from the RMS value of the current, but I prefer to have the peak current displayed on the scope and Itsu's scope can display only one statistic per channel ...I think.
I divided C by A in order to get the overall efficiency of 276%. Do you realize what would've happened if I had divided C by B ? ;)
Quote from: conradelektro on December 21, 2015, 03:17:08 PM
....
I tried that several times with different pulse motors and different drive circuits. I never could prove that input power to the motor was less if one tries to feed back power from the spikes (from the back EMF of the drive coil).
I think that the power in the spikes is forever lost.
...
Hi Conrad,
Would like to return to your above post, and I am aware of verpies's answer to it here:
http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg469191/#msg469191 with which I agree.
I have drawn a possible schematic (attached below) where you can see two coils in close magnetic coupling (simplest way to construct them is to use bifilar windings). The second coil is used to feed the voltage spike back to the positive supply rail via the flyback diode. There is a capacitor across the supply rail and the flyback energy can charge it up whenever the magnetic field collapses just after the switch is off.
So from the second ON pulse onward, the main coil in series with the switch is fed from a higher supply voltage level what the input supply would normally establishes, at least for a certain part during the ON time till the capacitor discharges to as low as the input supply voltage level if that is allowed by the timing.
The second diode put in series with the positive input supply rail automatically isolates the capacitor from the supply voltage whenever the voltage level across the capacitor is higher than the input supply voltage.
The second diode certainly causes a forward voltage drop in the input supply, low barrier Schottky or a germanium diode is to be used to minimize this loss.
Low duty cycle i.e. short ON time is needed to reduce coil losses, this needs the L/R time constant of the main coil to be considered.
Other circuit solutions are certainly possible for feeding the flyback pulse back to the input supply rail, to be able to increase overal performance, hence increasing efficiency.
Greetings
Gyula
Quote from: verpies on December 23, 2015, 07:31:31 AM
Yes AFAIK, except he is using a 555 astable to avoid grounding issues. I think he is working on triggering from the rotor's angular position sensor for the next version.
Already done but not with the latest version.
You are not.
According to the most recent scopeshot:
A) 17.745mJ Energy expended by the power supply during one ON-pulse, to bring coil's current from 0 to ~600mA,
B) 6.804mJ Calculated energy stored in the 37.8mH coil at the end of the ON-pulse,
C) 49.005mJ Recovered energy in C2 after one ON-pulse.
The energy loss in the resistance could've been easily calculated from the RMS value of the current, but I prefer to have the peak current displayed on the scope and Itsu's scope can display only one statistic per channel ...I think.
I divided C by A in order to get the overall efficiency of 276%. Do you realize what I would've gotten if I had divided C by B ? ;)
QuoteThe energy loss to the rotor as well as the energy loss due to hysteresis are unknown.
1 simple test would show if any energy is lost to the rotor. Take measurements with the rotor in play,and then take the same measurements without the rotor in play. I think you will find that there will be no energy loss as far as the rotor go's,as i still believe that the energy gain is to do with the external alternating magnetic field's-as i have show in a couple of my test already--although not as precise as you are doing here.
QuoteDo you realize what I would've gotten if I had divided C by B ?
yes = 720.238%
First-an error of 176% just to come out even,is very large. But an error of 620%+ is even larger.
I simply do not see you making this error Verpies,and i also see no other EE guy has stepped up to the plate to help you out. Now if it were me making these calculation,and providing the same outcome's,then im sure i would get run over by EE guy's trying to find the fault in my calculations. But it seems that when some one of your stature and ability comes up with these numbers,all the other EE guys are silent ::) If there is an error,then it would be an equipment reading error--not yours Verpies.
Just have Itsu run the test without the rotor,and crunch the same numbers. Then you will start to get some answers.
Brad.
Quote from: verpies on December 21, 2015, 05:20:18 PM
Quote from: partzman on December 21, 2015, 05:15:00 PM
You can use the vertical cursors to define any measurement window in time that you desire with any of the available measurement resources on your scope. It is really powerful IMO!
Show me the cursors re-positioned in such manner, that they encompass only the positive excursion of the red Math waveform and you will make a believer out of me if the Math Mean becomes all positive.
You have made a believer out of me.
Could you help Gotoluc to get his scope screenshots in normal resolution with the statistics on screen ?
Quote from: tinman on December 23, 2015, 08:38:02 AM
If there is an error,then it would be an equipment reading error--not yours Verpies.
I think Itsu has measured the C2 peak voltage with 2 scopes and verified C2 capacitance with a DMM.
Maybe the CSR is 0.01Ω and not 0.1Ω as expected and we don't know about it, but the input current waveform was also measured by a Hall probe in addition to the CSR and there was not much difference, so Itsu would have to make some systematic measuring mistake here to err.
...like his x10 switch on the probe making a bad contact, or bad probe attenuation setting in the scope ...or other silly thing like that.
I think I should ask Itsu to do a sanity check and put that scope probe, that measures the voltage from the CSR, across his power supply (or across an AA battery) and see if he gets the same voltage reading as the power supply's voltage (or AA battery).
Anyway, as soon as Itsu makes his "C2 discharger", the C2 voltage waveform will become rectangular with high duty cycle and a regular
averaging voltmeter will be able to measure its peak voltage (minus the duty cycle error), so we will have another instrument confirming or denying these measurements.
I was thinking about putting a light bulb in place of R2 because 49.005mJ being recovered 40 times per second, equals 1.96J per second and that is the same as 1.96W of average power. So a small 2W incandescent light bulb could be lit.
Who wants to calculate the ideal voltage rating of such small bulb ?
Quote from: verpies on December 23, 2015, 08:41:11 AM
Show me the cursors re-positioned in such manner, that they encompass only the positive excursion of the red Math waveform and you will make a believer out of me if the Math Mean becomes all positive.
You have made a believer out of me.
Could you help Gotoluc to get his scope screenshots in normal resolution with the statistics on screen ?
Verpies,
I can try! I haven't been following this thread closely so what type of scope does Gotoluc have?
partzman
Quote from: partzman on December 23, 2015, 09:03:13 AM
... so what type of scope does Gotoluc have?
From
these scopeshots (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg469295/#msg469295) of his, I gather that it is the Tektronix TDS 2024B
Quote from: verpies on December 23, 2015, 08:53:23 AM
I think Itsu has measured the C2 peak voltage with 2 scopes and verified C2 capacitance with a DMM.
Maybe the CSR is 0.01Ω and not 0.1Ω as expected and we don't know about it, but the input current waveform was also measured by a Hall probe in addition to the CSR and there was not much difference, so Itsu would have to make some systematic measuring mistake here to err.
...like his x10 switch on the probe making a bad contact, or bad probe attenuation setting in the scope ...or other silly thing like that.
I think I should ask Itsu to do a sanity check and put that scope probe, that measures the voltage from the CSR, across his power supply (or across an AA battery) and see if he gets the same voltage reading as the power supply's voltage (or AA battery).
Anyway, as soon as Itsu makes his "C2 discharger", the C2 voltage waveform will become rectangular with high duty cycle and a regular averaging voltmeter will be able to measure its peak voltage (minus the duty cycle error), so we will have another instrument confirming or denying these measurements.
I was thinking about putting a light bulb in place of R2 because 49.005mJ being recovered 40 times per second, equals 1.96J per second and that is the same as 1.96W of average power. So a small 2W incandescent light bulb could be lit.
QuoteWho wants to calculate the ideal voltage rating of such small bulb ?
6 volts.
Current close to 1/2 that of the input
Resistance close to 3x the voltage.
Brad
P.S--warning,light bulbs create confusion lol.
Ok, found the error, the C2 capacitor is 1uF, not 10uF as i mentioned before, i mixed up with some 10uF caps i also had on order but not yet received :-[ Sorry about that.
It measures 1002.4nF when i rechecked and screwed up on the decimal point.
Itsu
Quote from: itsu on December 23, 2015, 09:19:14 AM
Ok, found the error, the C2 capacitor is 1uF, not 10uF as i mentioned before, i mixed up with some 10uF caps i also had on order but not yet received :-[ Sorry about that.
It measures 1002.4nF when i rechecked and screwed up on the decimal point.
Itsu
Mmm
Are you sure on this Itsu,as you have now gone from 276% to 27.6% efficiency,and that sounds very low. But if that is the case,then it is good that you found the error.
But before you give this all away,can you give us the measurements with and without the rotor?
Cheers
Brad.
Hi Brad,
yes i am sure, but i think 27% is more in line with what to expect.
I will shoot a video when running and show the measurements and then stop the rotor with the same measurements taken.
Itsu
Quote from: verpies on December 23, 2015, 09:13:20 AM
From these scopeshots (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg469295/#msg469295) of his, I gather that it is the Tektronix TDS 2024B
Verpies and Gotoluc,
OK. The low resolution is probably due to the "Fast Trigger" acquisition mode being current selected. To change, under the front panel "Acquire", press the "Menu" button. This will bring up a bottom screen menu that has the current acquisition mode indicated in the 2nd from left position. The choices are in the right hand vertical screen menu so select the "Normal" (10k point) which will give you the highest horizontal resolution.
To get the Math functions working with vertical cursors, you can refer to the instructions I gave Itsu but in general, you must have "Cursor Gating" selected in the Measure menu and "VBar" selected in the Quickmenu to activate the vertical cursors. From there you should be able make any selective measurement on screen between the cursors.
partzman
Quote from: itsu on December 23, 2015, 09:41:05 AM
Hi Brad,
yes i am sure, but i think 27% is more in line with what to expect.
I will shoot a video when running and show the measurements and then stop the rotor with the same measurements taken.
Itsu
Thanks Itsu.
Quote from: itsu on December 23, 2015, 09:19:14 AM
Ok, found the error, the C2 capacitor is 1uF, not 10uF as i mentioned before, i mixed up with some 10uF caps i also had on order but not yet received :-[ Sorry about that.
Shit happens !
So all efficiency numbers have to be divided by 10.
New measurements taken both in running and stopped state.
First 2 screenshots are when running (overall and zoomed in on ontime), second 2 screenshots are when stopped (overall and zoomed in on ontime).
As mentioned earlier, the power consumption when stopped goes up
Video here: Https://www.youtube.com/watch?v=Can3bdMC7HY&feature=youtu.be
Itsu
OK, so crunching numbers for the zoomed scopeshot with the rotor stopped, now gives:
E1 = 5.382W * 4.32ms = 5.382 * 0.00432 = 0.02325024J = 23.25024mJ
E2 = ½ * 37.8mH * (824mA)2 = ½ * 0.0378 * 0.8242 = ½ * 0.0378 * 0.678976 = 0.0128326464J = 12.8326464mJ
E3 = ½ * 1μF * (99.6V)2 = ½ * 0.000001 * 9920.16 = 0.00496008J = 4.96008mJ
E3 / E2 = 4.96008mJ / 12.8326464mJ = 0.38652 = 39% (stored to recovered energy)
E3 / E1 = 4.96008mJ / 23.25024mJ = 0.21333 = 21% (expended to recovered energy)
Quote from: partzman on December 23, 2015, 09:41:41 AM
Verpies and Gotoluc,
OK. The low resolution is probably due to the "Fast Trigger" acquisition mode being current selected. To change, under the front panel "Acquire", press the "Menu" button. This will bring up a bottom screen menu that has the current acquisition mode indicated in the 2nd from left position. The choices are in the right hand vertical screen menu so select the "Normal" (10k point) which will give you the highest horizontal resolution.
To get the Math functions working with vertical cursors, you can refer to the instructions I gave Itsu but in general, you must have "Cursor Gating" selected in the Measure menu and "VBar" selected in the Quickmenu to activate the vertical cursors. From there you should be able make any selective measurement on screen between the cursors.
partzman
Thanks for the reply partzman
The below pic is the scope and what comes up when I press acquire. No "Menu"
Luc
Quote from: gotoluc on December 23, 2015, 11:46:08 AM
Thanks for the reply partzman
The below pic is the scope and what comes up when I press acquire. No "Menu"
Luc
Gotoluc,
I assume that your model was the same as the 3000 series but I was wrong. I found an online manual and it appears the best resolution would be in the "Average" acquisition mode unless you have high speed captures then you'd use "Peak Detect" but I'm sure you already know all this!
With the cursor button you can select "Time Cursors" to do any measurements that model will allow but they do state that the automatic measurements are more accurate than the cursor measurements and I'm sure you're aware of this as well.
I did not see any math capabilities in the specs but you do have mean, cycle rms, plus other measurements which should allow you to do manual energy and/or power calculations on your waveforms.
partzman
Okay so I see the crisis has been resolved.
Old data:
A) 17.745mJ Energy expended by the power supply during one ON-pulse, to bring coil's current from 0 to ~600mA,
B) 6.804mJ Calculated energy stored in the 37.8mH coil at the end of the ON-pulse,
C) 49.005mJ Recovered energy in C2 after one ON-pulse.
New data:
E1 = 5.382W * 4.32ms = 5.382 * 0.00432 = 0.02325024J = 23.25024mJ
E2 = ½ * 37.8mH * (824mA)2 = ½ * 0.0378 * 0.8242 = ½ * 0.0378 * 0.678976 = 0.0128326464J = 12.8326464mJ
E3 = ½ * 1μF * (99.6V)2 = ½ * 0.000001 * 9920.16 = 0.00496008J = 4.96008mJ
E3 / E2 = 4.96008mJ / 12.8326464mJ = 0.38652 = 39% (stored to recovered energy)
E3 / E1 = 4.96008mJ / 23.25024mJ = 0.21333 = 21% (expended to recovered energy)
So, Brad, I think it's appropriate to rub it in this one time: You jumped the gun. I am not trying to be nasty or anything like that. Let this be a lesson for you. Your mocking comments about "the books" ring hollow.
Now to get to my main theme, the differences in the current waveform between no rotor and the rotor in place. I made a quick attempt to see if the "Dia" program would allow me to make black a transparent colour on one of the two scope captures to superimpose one on the other but quickly gave up. I attached the two close-up graphics again.
Why is the current a lower value when the coil is driving the rotor? I bet that many people would believe that when the coil is driving the mechanical load of the rotor that the coil should draw more power and therefore draw more current. So why is it backwards?
The answer lies in the nature of the way the coil works and what it means when it is exporting power to the outside world to make the rotor turn.
When you apply voltage across the coil you are overcoming the electrical inertia of the coil and the current slowly rises. The voltage is "pulling the current up" in the coil and we know that the applied voltage is a constant DC value.
When the coil is driving the rotor, a portion of that voltage is not "pulling the current up" in the coil. Instead, some of the voltage is being "consumed" to drive the rotor. That means there is less voltage available to "pull the current up" in the coil and therefore there is less current flowing through the coil. The instantaneous "missing voltage" times the instantaneous current is the instantaneous power that is driving the coil. (I am simplifying and ignoring other losses.)
If you look carefully at the two current traces, and temporarily ignoring the effects of the wire resistance, the slope of the current curve is telling you the approximate amount of voltage "available" to increase the current flow. The slope of the current curve when the rotor is in place is lower than the slope when there is no rotor. Therefore less voltage is "available" to increase the current flow in the coil when the rotor is in place. So, in a manner of speaking, you are indirectly "seeing" how much voltage is being "consumed" to make the rotor spin. Voltage that is consumed to make the rotor spin is not available to increase the amount of current in the coil and therefore the slope of the rising current waveform is less compared to when there is no rotor.
That's why the DSO shows more energy put into the coil during the energizing cycle than the calculated 1/2 L i-squared value at the end of the energizing cycle. The difference is the energy "exported to the outside world" to make the rotor turn.
MileHigh
Quote from: MileHigh on December 23, 2015, 01:09:06 PM
I made a quick attempt to see if the "Dia" program would allow me to make black a transparent colour on one of the two scope captures to superimpose one on the other but quickly gave up. I attached the two close-up graphics again.
Below are the two scopeshots superimposed using
this (http://www192.lunapic.com/editor/?action=transparent) online image editor.
Quote from: verpies on December 23, 2015, 02:01:26 PM
Below are the two scopeshot superimposed using this (http://www192.lunapic.com/editor/?action=transparent) online image editor.
Thanks Santa!
I tried the "C2 discharger" using a reed relais, but the instance i remove R2 the contacts fuse together, even when the contacts have a series choke which has a diode/resistor in parallel across it.
The fusing is not permanent, but it makes it impossible to use it.
So i must follow one of verpies his "C2 discharger" designs.
The hall sensor to drive the reed relais did work ok, so i can use that to tune/trigger that setup from verpies.
Itsu
Quote from: MileHigh on December 23, 2015, 01:09:06 PM
Okay so I see the crisis has been resolved.
Old data:
A) 17.745mJ Energy expended by the power supply during one ON-pulse, to bring coil's current from 0 to ~600mA,
B) 6.804mJ Calculated energy stored in the 37.8mH coil at the end of the ON-pulse,
C) 49.005mJ Recovered energy in C2 after one ON-pulse.
New data:
E1 = 5.382W * 4.32ms = 5.382 * 0.00432 = 0.02325024J = 23.25024mJ
E2 = ½ * 37.8mH * (824mA)2 = ½ * 0.0378 * 0.8242 = ½ * 0.0378 * 0.678976 = 0.0128326464J = 12.8326464mJ
E3 = ½ * 1μF * (99.6V)2 = ½ * 0.000001 * 9920.16 = 0.00496008J = 4.96008mJ
E3 / E2 = 4.96008mJ / 12.8326464mJ = 0.38652 = 39% (stored to recovered energy)
E3 / E1 = 4.96008mJ / 23.25024mJ = 0.21333 = 21% (expended to recovered energy)
Now to get to my main theme, the differences in the current waveform between no rotor and the rotor in place. I made a quick attempt to see if the "Dia" program would allow me to make black a transparent colour on one of the two scope captures to superimpose one on the other but quickly gave up. I attached the two close-up graphics again.
Why is the current a lower value when the coil is driving the rotor? I bet that many people would believe that when the coil is driving the mechanical load of the rotor that the coil should draw more power and therefore draw more current. So why is it backwards?
The answer lies in the nature of the way the coil works and what it means when it is exporting power to the outside world to make the rotor turn.
When you apply voltage across the coil you are overcoming the electrical inertia of the coil and the current slowly rises. The voltage is "pulling the current up" in the coil and we know that the applied voltage is a constant DC value.
When the coil is driving the rotor, a portion of that voltage is not "pulling the current up" in the coil. Instead, some of the voltage is being "consumed" to drive the rotor. That means there is less voltage available to "pull the current up" in the coil and therefore there is less current flowing through the coil. The instantaneous "missing voltage" times the instantaneous current is the instantaneous power that is driving the coil. (I am simplifying and ignoring other losses.)
If you look carefully at the two current traces, and temporarily ignoring the effects of the wire resistance, the slope of the current curve is telling you the approximate amount of voltage "available" to increase the current flow. The slope of the current curve when the rotor is in place is lower than the slope when there is no rotor. Therefore less voltage is "available" to increase the current flow in the coil when the rotor is in place. So, in a manner of speaking, you are indirectly "seeing" how much voltage is being "consumed" to make the rotor spin.
That's why the DSO shows more energy put into the coil during the energizing cycle than the calculated 1/2 L i-squared value at the end of the energizing cycle. The difference is the energy "exported to the outside world" to make the rotor turn.
MileHigh
QuoteSo, Brad, I think it's appropriate to rub it in this one time: You jumped the gun. I am not trying to be nasty or anything like that. Let this be a lesson for you. Your mocking comments about "the books" ring hollow.
Lol-right on que ::)
QuoteYour mocking comments about "the books" ring hollow.
Well if your explanation below is from books MH,then i my self now have to have a laugh.
QuoteVoltage that is consumed to make the rotor spin is not available to increase the amount of current in the coil and therefore the slope of the rising current waveform is less compared to when there is no rotor.
MH
We all know that voltage alone is not power,so this part makes no sense at all. Also,as this !! voltage !! that is being consumed by the rotor is after the measuring equipment,then any power the rotor is suppose to be consuming must go through the measuring equipment first,unless you have found a way to some how go straight from the source,bypass the measuring equipment,and jump straight into the rotor. There is also the fact that even in your own words,there is more energy being put back into the system from the rotor than was used to drive it.
If you look at the two scope shot's below,and can understand what it is you are looking at,you can see where some of the energy is coming from to spin the rotor--i will also attach the schematic, with probe locations. Every thing above the red line is indicating a current flow through the charge battery !! and what !! ? ;) When the yellow line is flat during the off time,is showing you the supply battery voltage. So can you understand what these two scope shots are showing you?.
This thing about the rotor stealing battery voltage is nothing more than fairy tails MH. The rotor is generating current flow through the coil before the transistor switches on,and so this electrical inertia has already been taken care of by the rotor,and thus the supply voltage dose not have to work so hard to get this electrical inertia flowing-so to speak.
So MH-you too jumped the gun with your analogy,as it is wrong.
Brad
Quote from: tinman on December 23, 2015, 06:13:51 PM
If you look at the two scope shot's below,and can understand what it is you are looking at,you can see where some of the energy is coming from to spin the rotor--i will also attach the schematic, with probe locations. Every thing above the red line is indicating a current flow through the charge battery !! and what !! ? ;) When the yellow line is flat during the off time,is showing you the supply battery voltage. So can you understand what these two scope shots are showing you?.
Tinman,
I disagree with your "red line' comment. It looks to me as though no current can flow into the charge battery until the voltage at the coil/collector junction exceeds the voltage of both batteries in series plus one diode drop (approx. 25V). This would mean that current only flows into the charge battery at and during the mostly squared off top of the voltage waveform.
PW
Tinman,
Also, during the off portion of the cycle with the rotor installed, where we see the sinusoidal waveform produced by the passing magnets, no current flow is occurring anywhere (other than leakage) so we are only seeing an unloaded voltage waveform being generated by the rotor magnets during that time.
PW
It might be interesting if someone could edit Tinman's "'with rotor" scope capture by copying and pasting the sinusoidal waveform seen during the off time onto and aligned with the left side of that sine wave so that we can see if its phase/frequency remains constant during the on time.
In other words, if a copy of the observed sine wave is shifted to the left, does it match up, is it time continuous, with where the preceding sine wave stops at the beginning of the on time?
This may provide evidence as to whether the rotor's speed is constant or not.
Trying to do this by counting scope divisions is a bit difficult at the sweep rate used, but it does look like there may be one minor division or so missing, which would indicate rotor acceleration during the on time.
PW
Brad:
I think you had a tough day and are giving me some excess push-back.
QuoteMH
We all know that voltage alone is not power,so this part makes no sense at all. Also,as this !! voltage !! that is being consumed by the rotor is after the measuring equipment,then any power the rotor is suppose to be consuming must go through the measuring equipment first,unless you have found a way to some how go straight from the source,bypass the measuring equipment,and jump straight into the rotor. There is also the fact that even in your own words,there is more energy being put back into the system from the rotor than was used to drive it.
If you look at the two scope shot's below,and can understand what it is you are looking at,you can see where some of the energy is coming from to spin the rotor--i will also attach the schematic, with probe locations. Every thing above the red line is indicating a current flow through the charge battery !! and what !! ? (http://overunity.com/Smileys/default/wink.gif (http://overunity.com/Smileys/default/wink.gif)) When the yellow line is flat during the off time,is showing you the supply battery voltage. So can you understand what these two scope shots are showing you?.
This thing about the rotor stealing battery voltage is nothing more than fairy tails MH. The rotor is generating current flow through the coil before the transistor switches on,and so this electrical inertia has already been taken care of by the rotor,and thus the supply voltage dose not have to work so hard to get this electrical inertia flowing-so to speak.
So MH-you too jumped the gun with your analogy,as it is wrong.
I think what I wrote is pretty clear and I suspect that you are not getting it. I am not going to reword it so I suggest that you read it again and think about it. You are off the scent of the trail but I am sure you can get back on track. I am quite confident about what I am saying.
QuoteThere is also the fact that even in your own words,there is more energy being put back into the system from the rotor than was used to drive it.
No, no, no, no, no. I don't know how you see that in what I write and it looks to me like you are still projecting your wishes onto this situation as if they are real. I am on the edge of a "going bonkers" breakdown over the rotor.
This all goes back to schematics and timing diagrams. In the realm of electronics, when you are trying to truly analyze a circuit, the schematics and timing diagrams rule. Sometimes the timing diagram does not or cannot show you all the information you are looking for. You still can use your knowledge of the circuit and electronics and your knowledge of what the pulse motor itself is doing in the physical realm to extract "hidden" information in the timing diagram.
MileHigh
Quote from: picowatt on December 23, 2015, 08:56:31 PM...In other words, if a copy of the observed sine wave is shifted to the left, does it match up, is it time continuous, with where the preceding sine wave stops at the beginning of the on time?...
Don't know if this is what you mean as I have no idea where the on and off time is. His image is so small that aligning them isn't very accurate.
From the image resolution accuracy, I could tell no difference in the graph. each blip in the graph looks the same and the space between blips is the same. I would need much larger images or closer views to do a more accurate match up.
Quote from: tinman on December 24, 2015, 02:10:52 AM
The scope is across the emitter/collector junction,so the on time is when you see 0 volt's,and the off time is the rest...
Thanks for the info. Just looked at your thread and I'll check it out some more. Good info in this one and yours.
Quote from: MagnaProp on December 24, 2015, 01:45:32 AM
Don't know if this is what you mean as I have no idea where the on and off time is. His image is so small that aligning them isn't very accurate.
From the image resolution accuracy, I could tell no difference in the graph. each blip in the graph looks the same and the space between blips is the same. I would need much larger images or closer views to do a more accurate match up.
The scope is across the emitter/collector junction,so the on time is when you see 0 volt's,and the off time is the rest. I will continue on my thread,so as not to fill Lucs thread up with a different setup.
Brad
Hi everyone and Merry Christmas
I built a flyback test device that consists of two MOT face to face to continue the study of different setups like attraction vs repel (bucking) to see how it affects the flyback.
I noticed just by changing mosfet's it causes a problem, see first scope shot using a IRF840 (switch) which looks good then the next scope shot is using a IRF3710 for less on time resistance but look at what it does ???
The input is 12.98vdc through 0.1 Ohm CSR and flyback cap is 57uf
Yellow CH1 is CSR, Blue CH2 is not used, Purple CH3 is Voltage and Green CH4 is Cap charge voltage
Any ideal what causes this and why?
Thanks for your time
Luc
The slower (and curved) rise time and higher current is likely due to the much lower on resistance of the IRF3710.
The truncated flyback spike might be due to the avalanching of the body diode, but it should tolerate at least 100V, so not sure 100%. The 3710 is a 100V device, while the 840 is a 500V device. The 3710 also requires 5x the gate charge (190nC vs. 38nC).
For the voltage rating alone, I don't think the 3710 is a good choice.
BUT, I am not familiar with your circuit, so if this is with the flyback load connected, then the truncation could be caused by the flyback capacitor absorbing the spike, which is what you want anyway I think, isn't it?
Quote from: poynt99 on December 25, 2015, 09:07:20 PM
The slower (and curved) rise time and higher current is likely due to the much lower on resistance of the IRF3710.
The truncated flyback spike might be due to the avalanching of the body diode, but it should tolerate at least 100V, so not sure 100%. The 3710 is a 100V device, while the 840 is a 500V device. The 3710 also requires 5x the gate charge (190nC vs. 38nC).
For the voltage rating alone, I don't think the 3710 is a good choice.
Thanks for the reply .99
I was thinking it could be caused by the body diode but why is it happening at that low of a voltage, since like you say it should be able to deal with up to 100v.
Quote from: poynt99 on December 25, 2015, 09:07:20 PM
BUT, I am not familiar with your circuit, so if this is with the flyback load connected, then the truncation could be caused by the flyback capacitor absorbing the spike, which is what you want anyway I think, isn't it?
Yes, we want most absorption of flyback to the cap at off time but the 3710 charging the cap to only 60v when the 840 is charging it to 97v and costing less input current ???
Oh well, I'll just have to find a better mosfet then the IRF3710 and IRF840
Merry Christmas
Luc
Would you mind reposting your circuit diagram?
Merry Christmas! :)
This would be the closes to what I'm working with.
Thanks
Luc
Here are some interesting results using a test device I put together to observe the flyback and Pin of two bucking MOT's (opened) E cores.
Circuit is attached, believe it or not lol... and a picture of what the test device looks like, with E core to E core, scope shots of each test and a picture of the E cores with an I core added in between.
Both upper and lower MOT's and circuits are a mirror image. I used a 2 channel signal generator to trigger each IRF840 (Q1) mosfet's
Each MOT has their own 5Ah 12v battery (C1) which are at 12.98vdc each.
Each CSR (R1) are 0.1 Ohm
Mot primary coils (L1) are 0.35 Ohm, about 13mH with a cardboard air gap of 0.023in. (0.6mm) in between to simulate a motor rotor to stator coil gap. MOT secondaries are not used.
The capacitors (C2) are 57uf (tested) each and have a load resistor of 4.7k to drain them before the next pulse
The flyback diodes (D1) are RHRG5060
What I find interesting and Gyula may also is, if I would of built the bucking reluctance motor without a I core rotor in between the two Bucking MOT E cores, the motor would of never worked and would be a power hog with next to no flyback recovery.
Look at the scope shot difference with E core to E core (with cardboard air gap) compared to adding the I core in between the two E cores with cardboard air gap on each side.
Yellow CH1 is CSR, Blue CH2 is not used, Purple CH3 is Voltage and Green CH4 is Cap charge voltage
So the big question is, where does the magnetic field go when it's E to E core?... it's not in the flyback that's for shure and notice the voltage flattens out!... why?
The other thing is, during the E to E test you would think they would push each other away (not stay in place) but nothing like that happens and and when I put my hand on the core I feel next to nothing and can barely hear a sound compared to when the I core is added in between.
Luc
Quote from: gotoluc on December 26, 2015, 12:10:12 AM
Here are some interesting results using a test device I put together to observe the flyback and Pin of two bucking MOT's (opened) E cores.
Circuit is attached, believe it or not lol... and a picture of what the test device looks like, with E core to E core, scope shots of each test and a picture of the E cores with an I core added in between.
Both upper and lower MOT's and circuits are a mirror image. I used a 2 channel signal generator to trigger each IRF840 (Q1) mosfet's
Each MOT has their own 5Ah 12v battery (C1) which are at 12.98vdc each.
Each CSR (R1) are 0.1 Ohm
Mot primary coils (L1) are 0.35 Ohm, about 13mH with a cardboard air gap of 0.023in. (0.6mm) in between to simulate a motor rotor to stator coil gap. MOT secondaries are not used.
The capacitors (C2) are 57uf (tested) each and have a load resistor of 4.7k to drain them before the next pulse
The flyback diodes (D1) are RHRG5060
What I find interesting and Gyula may also is, if I would of built the bucking reluctance motor without a I core rotor in between the two Bucking MOT E cores, the motor would of never worked and would be a power hog with next to no flyback recovery.
Look at the scope shot difference with E core to E core (with cardboard air gap) compared to adding the I core in between the two E cores with cardboard air gap on each side.
Yellow CH1 is CSR, Blue CH2 is not used, Purple CH3 is Voltage and Green CH4 is Cap charge voltage
So the big question is, where does the magnetic field go when it's E to E core?... it's not in the flyback that's for shure and notice the voltage flattens out!... why?
The other thing is, during the E to E test you would think they would push each other away (not stay in place) but nothing like that happens and and when I put my hand on the core I feel next to nothing and can barely hear a sound compared to when the I core is added in between.
Luc
Without the I core section,there is no flux path for the fields to follow,and so with the bucking field without the I core section,you have a net magnetic flux path of zero,and have made a nice induction heater. With the I core piece in place,you now have a path for the flux from each coil to follow,where the flux from each coil will share the I core piece that is now in place,but the path is complete for each coil.
Brad
Thanks for your brave ;) reply Brad
That's about how I understood it as far as the I core being there in place or not but you would think these Large MOT E cores could store up a little flux of this short pulse, no? and where does the flux go? does it just leaks in space?
Did you notice the voltage pretty much fell flat instantly and how can the current instantly go to maximum on an inductor?
What part of a bucking field is Chris suggesting to be used?
Luc
Hi Luc,
I also think what Brad wrote on the lack of real magnetic path for the cardboard gap case: the bucking fields must leave the prongs of the facing E cores i.e. leak out of the gap to the outside. This is why the the inductance of a single MOT coil in the setup is only 13 mH. And the mass of the cores is most probably way too big with respect to the repel force between the facing surfaces, i.e. resulting repel force is too low be able to influence the core masses.
The repel force can be a low value anyway due to the low 13 mH self inductance and due to the short pulse ON time (about 20 msec).
The low self inductance for the cardboard gap case also manifests in the 9 Amper peak current: the inductive reactance can only be also a low value for the same input pulse frequency.
It would be interesting to learn what the inductance value for the MOT coils were when the I core was in place instead of the cardboard gap.
These are what I could add to your questions, certainly there are some unanswered ones left.
Gyula
Quote from: gotoluc on December 25, 2015, 10:32:47 PM
This would be the closes to what I'm working with.
That probe placement could not have produces an oscillogram like that (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/154231/), because when C2 voltage is increasing there should be current flowing through the CSR ...and there isn't any.
So there is some difference between your circuit and the schematic.
Quote from: gyulasun on December 26, 2015, 11:35:09 AM
Hi Luc,
I also think what Brad wrote on the lack of real magnetic path for the cardboard gap case: the bucking fields must leave the prongs of the facing E cores i.e. leak out of the gap to the outside. This is why the the inductance of a single MOT coil in the setup is only 13 mH. And the mass of the cores is most probably way too big with respect to the repel force between the facing surfaces, i.e. resulting repel force is too low be able to influence the core masses.
The repel force can be a low value anyway due to the low 13 mH self inductance and due to the short pulse ON time (about 20 msec).
The low self inductance for the cardboard gap case also manifests in the 9 Amper peak current: the inductive reactance can only be also a low value for the same input pulse frequency.
It would be interesting to learn what the inductance value for the MOT coils were when the I core was in place instead of the cardboard gap.
These are what I could add to your questions, certainly there are some unanswered ones left.
Gyula
Thanks for your input Gyula
When the I core is in between to two E cores with the same cardboard spacer but now one on each side (so two) the inductance is 16mH. So only a 3 mH increase as appose to E to E with one spacer.
Luc
Quote from: verpies on December 26, 2015, 12:47:19 PM
That probe placement could not have produces an oscillogram like that (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/154231/), because when C2 voltage is increasing there should be current flowing through the CSR ...and there isn't any.
So there is some difference between your circuit and the schematic.
Yes, you're right!... and we went over this before. I always have problems looking at circuits and what I'm making. I guess that's why I don't posts circuits ;D
Let me go back and read what I had done to correct it as it looks to be the same problem.
Thanks for pointing it out once again.
Luc
Okay, I've corrected the same thing I did before. I guess I'm not use to the flyback going through the CSR ;D
Below are the new scope shots reflecting this change and I've also included powering one side (one MOT) only in both configurations. Obviously I can only scope one MOT at the time. So when both are powered it would be double the input.
Here are the test device details one again:
Both upper and lower MOT's and circuits are a mirror image. I used a 2 channel signal generator to trigger each IRF840 (Q1) mosfet's
Each MOT has their own 5Ah 12v battery (C1) which are at 12.98vdc each.
Each CSR (R1) are 0.1 Ohm
Mot primary coils (L1) are 0.35 Ohm, about 13mH (E to E) and 16mH (E-I-E) with a cardboard air gap of 0.023in. (0.6mm) in between (x2 for I test) to simulate a motor rotor to stator coil gap. MOT secondaries are not used.
The capacitors (C2) are 57uf (tested) each and have a load resistor of 4.7k to drain them before the next pulse
The flyback diodes (D1) are RHRG5060
Luc
Quote from: gotoluc on December 26, 2015, 01:01:56 PM
....
When the I core is in between to two E cores with the same cardboard spacer but now one on each side (so two) the inductance is 16mH. So only a 3 mH increase as appose to E to E with one spacer.
....
Hi Luc,
I must have been in "Xmas mood" and overlooked the cardboard gap for the case when the I core was used, and this is why I did not notice what verpies mentioned, sorry for these mistakes.
Gyula
Quote from: gotoluc on December 26, 2015, 09:59:42 AM
Thanks for your brave ;) reply Brad
That's about how I understood it as far as the I core being there in place or not but you would think these Large MOT E cores could store up a little flux of this short pulse, no? and where does the flux go? does it just leaks in space?
Did you notice the voltage pretty much fell flat instantly and how can the current instantly go to maximum on an inductor?
What part of a bucking field is Chris suggesting to be used?
Luc
I was thinking the same thing at first,but have a look at your last scope shot's you posted ???
Only one of the four show's near 0 volts across the coil--just a small voltage pulse at the start.
And it just so happens that the one with near 0 volts across the coil is also the only one where the flux path is incomplete for the two coils-->that being E to E with power to both coils. This one with near 0 volts across the coil will be the one that consumes the least amount of power,yet the current tells us that it will also be the one that produces the greatest !!radiant!! magnetic field(field not contained within the core). If this field is used to produce work ,will it reflects on the P/in?-->will it raise the voltage seen across the two coils?. If it dose not raise the voltage,but increases the current-->what happens to the voltage across the coils when the current is increased?-my guess is that the voltage would actually go down as the current increases ;)
I think Luc that you may be on the verge of seeing if what EMJ is saying is actually true.
I would like to see (if you still have them) what a secondary coil (a coil behind one of the primaries)shows across it in this bucking coil mode- E to E,both coils powered.
Brad
Quote from: gotoluc on December 26, 2015, 09:59:42 AM
Thanks for your brave ;) reply Brad
That's about how I understood it as far as the I core being there in place or not but you would think these Large MOT E cores could store up a little flux of this short pulse, no?
Did you notice the voltage pretty much fell flat instantly and how can the current instantly go to maximum on an inductor?
What part of a bucking field is Chris suggesting to be used?
Luc
Quoteand where does the flux go? does it just leaks in space?
In fact-yes.
But this space is the air gap between the two coils and the outer legs of the E cores that carry the opposite field,as the field must join/merge. It is this field you must tap into,although i dont think you would have the room between the two coils and outer legs of the E core's to place an air core coil.
Brad
Hi Brad
thanks for your additional posts and interest.
I'm away for a day but Monday afternoon I'll be back and can further test this.
In the meantime I've enlarged the pics and added a few notes.
Luc
Quote from: gotoluc on December 26, 2015, 11:15:45 PM
Hi Brad
thanks for your additional posts and interest.
I'm away for a day but Monday afternoon I'll be back and can further test this.
In the meantime I've enlarged the pics and added a few notes.
Luc
Mmm-so the voltage drops to near zero,but the current is still strong right until switch off.
When you get back into it Luc,could you try the simple circuit below,so as we can see both the current and voltage across both coils when switch 1 is open and closed--just so as we remove all the caps and diodes. Use a mosfet,and not the transistor i have depicted. Dont take note of coil wind direction,as i just ghosted and flipped one drawing. Just make them bucking as you did before.
Enjoy your time away ;)
Cheers
Brad
Quote from: gotoluc on December 26, 2015, 11:15:45 PM
In the meantime I've enlarged the pics and added a few notes.
These notes are correct.
I will not elaborate now...still in Xmas mode
Quote from: gotoluc on December 26, 2015, 11:15:45 PM
Hi Brad
thanks for your additional posts and interest.
I'm away for a day but Monday afternoon I'll be back and can further test this.
In the meantime I've enlarged the pics and added a few notes.
Luc
@Verpies, Tinman and Milehigh:
Luc has a scope shot above labeled "Reverse direction Flyback". How does this equate with your "Same direction" theory for BEMF?
Quote from: synchro1 on December 27, 2015, 09:00:53 AM
@Verpies, Tinman and Milehigh:
Luc has a scope shot above labeled "Reverse direction Flyback". How does this equate with your "Same direction" theory for BEMF?
Hi synchro1,
you are still under moderation but I will allow your posts as long as they conform to the rules I wrote in my first post of this topic.
It's true I've suggested a reverse flyback on the bucking E cores but with a question mark. Let's allow the EE to think about it a little and see what they make of it.
I will be performing the test TinMan has suggested to see what other results come of it.
Kind regards
Luc
Quote from: synchro1 on December 27, 2015, 09:00:53 AM
Luc has a scope shot above labeled "Reverse direction Flyback". How does this equate with your "Same direction" theory for BEMF?
It would be a violation with one coil, but with two coils one can push flux and current into the other coil.
Without seeing two gate waveforms (or d-s) on the same scopeshot, it is difficult to conclude much more.
Loosely coupled inductors ( a transformer with low k coupling coefficient ) are notoriously hard to analyze when they have capacitors across these inductors on both sides. Without the diodes in series they
behave (http://www.ee.bgu.ac.il/~intrlab/lab_number_7/Two%20inductively%20coupled%20RLC%20circuits.pdf) like a
tuned transformer (https://en.wikipedia.org/wiki/Inductance#Tuned_transformer).
I build the below verpies "C2 discharge" version (so with the MOSFET driver), and found the below screenshot signals.
As normal, blue is across CSR (point A to B)
Purple is across the Q1 MOSFET drain (point C to B)
Yellow is across the C2 cap (point D to B).
C2 cap is a 2.2uF one
2e screenshot is a zoomed in one.
Is this the kind of signal you would expect for the yellow one?
Regards Itsu
Quote from: itsu on December 27, 2015, 05:08:23 PM
Is this the kind of signal you would expect for the yellow one?
Almost.
Notice that C2 is not discharging all the way down to 0V.
To discharge C2 all the way down to 0V:
- start discharging C2 sooner
or
- discharge C2 longer. (if I knew Q3's gate waveform then I could decide whether increasing L2's inductance is necessary for this).
or
- discharge C2 faster by decreasing R2 (Warning: don't decrease R2 so much that Q3's maximum pulsed drain current is exceeded)
or
...all of the above or combination thereof
Also, put some small but good bypass caps to ground on the source of Q3 and on cathode of D2, because there is an unwanted voltage spike on the purple trace. While you are at it, decrease the inductance of the entire drain circuit of Q3 including C2 (short leads, small loop area, etc...)
Below is a revised schematic with the correct probe placement and countermeasures against the voltage spike form L2.
If C3 and C4 do not help, then adding R4 will slow down the closing of Q3 and its drain current spike that gets converted to a voltage spike by stray inductances in the drain circuit. For R4 try anything between 0Ω to 1K.
What are you using for U2 ? Do you use its inverting input or non-inverting?
Quote from: synchro1 on December 27, 2015, 09:00:53 AM
@Verpies, Tinman and Milehigh:
Luc has a scope shot above labeled "Reverse direction Flyback". How does this equate with your "Same direction" theory for BEMF?
Because you need to understand the circuit,scope probe placement,and the way the two inductors are coupled to each other.
Brad
Quote from: verpies on December 27, 2015, 03:36:24 PM
It would be a violation with one coil, but with two coils one can push flux and current into the other coil.
Without seeing two gate waveforms (or d-s) on the same scopeshot, it is difficult to conclude much more.
Loosely coupled inductors ( a transformer with low k coupling coefficient ) are notoriously hard to analyze when they have capacitors across these inductors on both sides. Without the diodes in series they behave (http://www.ee.bgu.ac.il/~intrlab/lab_number_7/Two%20inductively%20coupled%20RLC%20circuits.pdf) like a tuned transformer (https://en.wikipedia.org/wiki/Inductance#Tuned_transformer).
Good point verpies!... I can confirm even thought both E cores are very close to being identical I noticed one is using a little more current then the other and I'm quite sure it's the one I didn't provide the scope shots.
Luc
I'm a little confused about the premise here....
We are trying to improve efficiency of an electric motor,... by adding losses to the system adding a transformer???
my experience with this matter has led me to believe the "most efficient" motor that mankind has so far created
is the one we use in nearly every electric device that has a motor in it....
like a stand-alone fan, or a vaporizer, im sure you've seen these things, they are brushless A/C motors, just an inductor, with two large copper links,
and an aluminum rotor with arc-shaped grooves in it.
when these are operated at a frequency that is compatible (multiple, divisor, or octave of) with the self resonant frequency of the inductive core.
they run at 95+ % efficiency. That means if you have a supercapacitor storing an hours worth of run-energy,
you can cycle back and forth a "q-Mogen" of two of these motors for 54 minutes, (minus heat losses in your circuit)
before the energy is depleted.
We have no better motor than that.(except some magnetically suspended stuff NASA did in space...)
So it seems like a good place to start.
What I don't get, is how adding something that increases losses is supposed to raise efficiency?? like a flyback transformer?
Flybacks are notoriously sources of system losses. They are discontinuous by nature, and generally over-tax the maximum frequency of the inductor.
This is because of the high conversion ratio, I.E. # of turns on the primary compared to secondary. The voltage has to step-up faster than the induction time factor.
When the timing of the inductive core is faster than the time it takes for the signal to circle around all those turns in the secondary coil,
it has to "skip a beat" to re-synch. That sounds weired, but that's what happens inside the inductor. When the flux is saturated, and has not gone down yet, and it switches again, the inductor can't keep up. so lets say you run a fly-back for 100 cycles.
there will be a number of these cycles where the energy is unrecoverable, because it switched faster than the secondary coil could dump its energy.
it will depend of course on the ratio of turns, and the operating frequency, but lets just for example assume it's 3 out of 100.
so you're flyback runs at 97% efficiency before you attach it to the motor. that's - 3% just by adding a flyback
This effect is well known in the television industry, and is calculated into the total energy consumption of devices, as displayed on the FCC tag.
if you have a flyback of X ratio, operating at frequency Y, you will have Z losses.
Quote from: sm0ky2 on December 28, 2015, 12:22:21 AM
We are trying to improve efficiency of an electric motor,... by adding losses to the system adding a transformer???
No, I think this is an experimental tangent that investigates the simulation of a rotor by a transformer
Quote from: sm0ky2 on December 28, 2015, 12:22:21 AM
Flybacks are notoriously sources of system losses. They are discontinuous by nature, and generally over-tax the maximum frequency of the inductor.
No, "flybacks" last as long as the engineer wants. Precisely t=½π(LC)
½ In circuits like this (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/154314/) the "flyback" current is not discontinuous but quarter-sinusoidal.
What is the maximum frequency of the inductor, anyway ?
Quote from: verpies on December 27, 2015, 05:50:26 PM
Almost.
Notice that C2 is not discharging all the way down to 0V.
To discharge C2 all the way down to 0V:
- start discharging C2 sooner
or
- discharge C2 longer. (if I knew Q3's gate waveform then I could decide whether increasing L2's inductance is necessary for this).
or
- discharge C2 faster by decreasing R2 (Warning: don't decrease R2 so much that Q3's maximum pulsed drain current is exceeded)
or
...all of the above or combination thereof
Also, put some small but good bypass caps to ground on the source of Q3 and on cathode of D2, because there is an unwanted voltage spike on the purple trace. While you are at it, decrease the inductance of the entire drain circuit of Q3 including C2 (short leads, small loop area, etc...)
Below is a revised schematic with the correct probe placement and countermeasures against the voltage spike form L2.
If C3 and C4 do not help, then adding R4 will slow down the closing of Q3 and its drain current spike that gets converted to a voltage spike by stray inductances in the drain circuit. For R4 try anything between 0Ω to 1K.
What are you using for U2 ? Do you use its inverting input or non-inverting?
Thanks,
i will add the additional components and take the extra scope shots.
U2 is presently an inverting ucc37321, as a non-inverting ucc37322 did not produce the required "discharge C2" signal.
I used a hall sensor feedback system for this signal at first, but the rotor did not want to run properly then allthough i could
tune the sensor anywhere inbetween the L1 pulse train.
I now use the same 555 output for both the Q1 and U2 (inverting)/Q3 inputs.
A 10 Ohm ¼W resistor for R2 lasted only a few minutes before it opened up (invisible damage), so i use a 10W resistor there now,
not sure it was coincidence.
Itsu
Quote from: sm0ky2 on June 17, 1970, 07:20:01 PM
my experience with this matter has led me to believe the "most efficient" motor that mankind has so far created
is brushless A/C motors, just an inductor, with two large copper links,
and an aluminum rotor with arc-shaped grooves in it.
What you're describing sounds like a shaded pole motor, is this correct?
Quote from: sm0ky2 on June 17, 1970, 07:20:01 PM
When these are operated at a frequency that is compatible (multiple, divisor, or octave of) with the self resonant frequency of the inductive core.
They run at 95+ % efficiency.
Can you provide a demonstration or the details on the 95%+ efficiency test
Thanks
Luc
Quote from: itsu on December 28, 2015, 04:49:36 AM
U2 is presently an inverting ucc37321,
Note, that the maximum supply voltage of this driver chip is only 16V.
That might seem like enough when you are supplying it from 12V but remember that D2 and ZD12 allow the L2 voltage to swing up 18V higher (depending what ZD2 you have installed) and this gives the total of 30V to ground, that can appear on the output of this driver (minus the voltage drop of R3 and D3) so you are pushing the output transistors of this chip and don't be surprised they break. ...but if this driver is inexpensive, I would not go to the trouble of adding Q4 since it apparently works well alone.
Quote from: itsu on December 28, 2015, 04:49:36 AM
I now use the same 555 output for both the Q1 and U2 (inverting)/Q3 inputs.
Pretty clever but this way you charge and discharge C2 at the same time. Q3 should be open and C2 should be discharged,
before Q1 opens.
Also, this way you do not have control how long Q3 remains closed. The inductance of L2 determines that for you.
Quote from: itsu on December 28, 2015, 04:49:36 AM
A 10 Ohm ¼W resistor for R2 lasted only a few minutes before it opened up (invisible damage), so i use a 10W resistor there now,
not sure it was coincidence.
I don't think it was coincidence either.
When I was doing the power calculations for putting a small light bulb in place of R2, I calculated approximately 2W.
Quote from: sm0ky2 on December 28, 2015, 12:22:21 AM
the "most efficient" motor that mankind has so far created is the one we use in nearly every electric device that has a motor in it.... like a stand-alone fan, or a vaporizer, im sure you've seen these things, they are brushless A/C motors, just an inductor, with two large copper links, and an aluminum rotor with arc-shaped grooves in it.
A 3-phase brushless motor with a permanent magnet rotor (PMSM) is more efficient than a shaded pole motor (a type of AC induction motor) because there are no resistive eddy current losses in the PM rotor and in the shaded pole shunt winding.
The resistive conduction of electric current in a copper or aluminum winding/shunt, cannot compete with the atomic bound magnetizing current in a permanent magnet which is lossless and perpetual.
Quote from: verpies on December 28, 2015, 09:20:17 AM
Note, that the maximum supply voltage of this driver chip is only 16V.
That might seem like enough when you are supplying it from 12V but remember that D2 and ZD12 allow the L2 voltage to swing up 18V higher (depending what ZD2 you have installed) and this gives the total of 30V to ground, that can appear on the output of this driver (minus the voltage drop of R3 and D3) so you are pushing the output transistors of this chip and don't be surprised they break. ...but if this driver is inexpensive, I would not go to the trouble of adding Q4 since it apparently works well alone.
Pretty clever but this way you do not have control how long Q3 remains closed. The inductance of L2 determines that for you.
Even worse, this way you charge and discharge C2 at the same time. Q3 should be open and C2 should be discharged, before Q1 opens.
I don't think it was coincidence either.
When I was doing the power calculations for putting a small light bulb in place of R2, I calculated approximately 2W.
OK, C3 and c4 in place, but it did not make much difference, the ugly spike is still there, going for R4.
I have a 15V 1A zener for ZD2.
L2 has a core which i can move in or out, its now max in (max inductance), which sets the C2 voltage max towards zero.
EDIT, i have removed the screenshots as the setup was wrong so they seem useless.
Screenshot 1 has added point F (compared to B) in the green trace
screenshot 2 is as 1 but zoomed in
screenshot 3 has added point E (compared to B) in the green trace
screenshot 4 is as 3 but zoomed in
Itsu
Quote from: verpies on December 28, 2015, 09:20:17 AM
Note, that the maximum supply voltage of this driver chip is only 16V.
That might seem like enough when you are supplying it from 12V but remember that D2 and ZD12 allow the L2 voltage to swing up 18V higher (depending what ZD2 you have installed) and this gives the total of 30V to ground, that can appear on the output of this driver (minus the voltage drop of R3 and D3) so you are pushing the output transistors of this chip and don't be surprised they break. ...but if this driver is inexpensive, I would not go to the trouble of adding Q4 since it apparently works well alone.
Pretty clever but this way you charge and discharge C2 at the same time. Q3 should be open and C2 should be discharged, before Q1 opens.
Also, this way you do not have control how long Q3 remains closed. The inductance of L2 determines that for you.
I don't think it was coincidence either.
When I was doing the power calculations for putting a small light bulb in place of R2, I calculated approximately 2W.
Ok seen this updated bold line, so this is not going to work this way, i have to build in a delay for U2 to fire or use my hall sensor feedback system again.
I will retry that latest option first.
Itsu
Quote from: itsu on December 28, 2015, 10:40:58 AM
...i have to build in a delay for U2 to fire or use my hall sensor feedback system again.
The easiest way to do this is make an additional 555/556 monostable, that will slightly extend the length of the 555/556 astable's output. See
pulse extender (http://www.555-timer-circuits.com/monostable-555.html).
Now the trick is to use the output of the monostable to provide the "Energize L1" signal and use the output of the astable to provide the "Discharge C2" signal. This is because the astable's output pulse will end
before the monostable's output pulse ends.
I'm back and before doing Brad's suggested test I've done a few other tests on the DUT and moved the two CSR on the negative side of the batteries to be able to probe both MOT'S current and voltage at the same time since negative is common ground at the mosfet's by the 2 channel SG.
The first scope shot is E to E core with the 0.6mm air gap and next is no air gap. I see no difference :-\
The flyback diodes are still connected to the 57uf caps with 4.7k load.
I also did a single shot test to measure the voltage by DMM on each cap (no load resistor).
For channel 1 and 2 cap charged to 28v and channel 3 and 4 cap charged to 35v
Channel 1 and 2 is what was probed in the previous MOT scope shots.
The only other test I can think of doing before I make changes to test Brad's suggestion is a reducing the pulse width to where I suggested in the enlarged scope shots with added details.
If you can think of anything else please post your suggestions.
Next post will be the small pulse width.
Lux
ADDED: This may be of importance? each mot are not identical, their primaries are wound in a different direction when connected in bucking. CH 1 & 2 positive starts on inner winding and CH 3 & 4 positive starts on outer winding. Could that be why the difference at off time?
@ verpies
Got some HCF4047BE
Luc
Quote from: gotoluc on December 28, 2015, 02:55:06 PM
I'm back and before doing Brad's suggested test I've done a few other tests on the DUT and moved the two CSR on the negative side of the batteries to be able to probe both MOT'S current and voltage at the same time since negative is common ground at the mosfet's by the 2 channel SG.
The first scope shot is E to E core with the 0.6mm air gap and next is no air gap. I see no difference :-\
The flyback diodes are still connected to the 57uf caps with 4.7k load.
I also did a single shot test to measure the voltage by DMM on each cap (no load resistor).
For channel 1 and 2 cap charged to 28v and channel 3 and 4 cap charged to 35v
Channel 1 and 2 is what was probed in the previous MOT scope shots.
The only other test I can think of doing before I make changes to test Brad's suggestion is a reducing the pulse width to where I suggested in the enlarged scope shots with added details.
If you can think of anything else please post your suggestions.
Next post will be the small pulse width.
Lux
ADDED: This may be of importance? each mot are not identical, their primaries are wound in a different direction when connected in bucking. CH 1 & 2 positive starts on inner winding and CH 3 & 4 positive starts on outer winding. Could that be why the difference at off time?
I think ch2 and 4 have to be inverted-would be easier to work out then :)
brad
re: others doing similar work.
This guy was exploring along similar lines :
https://www.youtube.com/channel/UCpcT7LePUsT8nkTS614_7ew
Discussion of it here, not a well-informed discussion judging by the first couple of pages of arguing, name-calling and all the usual crap that gets in the way :
http://overunity.com/12147/this-might-blow-your-mind/#.VoKlXlLq-vg
I'm still searching for the experiments that were almost exactly along the line Luc et al are following.
Y.
Quote from: gotoluc on December 28, 2015, 06:43:02 PM
@ Verpies
Got some HCF4047BE
I was on the road when you posted this...
Below is a schematic depicting how you can use 4 of these chips to make a motor pulse sequencing circuit (warning: U3 is wired differently from the others).
If you do not need the "Discharge C" output signal, then just use only the first two chips (U1 and U2).
Calculate the capacitors (C1 - C4) according to the pulse widths that you want (or delays you want) using the formula below:
C = t / 595
where the time (t) is given in
ms..and the resulting capacitance (C) is in
μFFor Example: If you need the pulse width (or delay) to be 10ms then the relevant capacitor should be 0.017μF , ...because 10/124=0.017
Of course 0.017μF = 17nF.
If you don't have an exact capacitor that you've calculated, then put the closest one you have and adjust to the desired pulse width (or delay) with the corresponding potentiometer (multi-turn pot recommended).
Something odd Im finding with my electric bike motor setup.
At the time I had shown the waveforms I had shown that wide open throttle there was no pulse width modulation happening, where it was just switching battery power to the coils in their respective phases. Then when I backed off the throttle some, the pwm kicks in and varies a shorter turn on pulsingas you throttle down.
Well while riding recently I noticed something odd. All this time, 8 yrs, I never really seen it till now. When ridding full throttle, no pwm, the bike will get up to full speed on flat road. But if I back off the throttle, about 3/8 in, it went a bit faster. I can feel it if I gauge it while in acceleration also.
At first I thought it was the magnet in the throttle going too far past the hall sensor. But when I lift the wheel off the ground and do the scope shots, full throttle is wide open no pwm.
Could it be that while in the highest power pwm state before going full no pwm, that it provides more motive force to the wheel than wide open no pwm???
Strange but true.
So the next test will be to ride with one of my tiny scopes connected to see if things are the same as I think they are.
Mags
Thanks for the circuit verpies.
If I understand the circuit correctly, the discharge is to activate a mosfet to dump the flyback cap?... if so, why would we want to do this during Energize L on time? would it not always be after L is opened and flyback has charged the cap?
Speaking of flyback, what will activate the mosfet to capture flyback so I can get rid of the diode to make it more efficient? would pin 11 be fast enough to do this? also, keep in mind that I've got lots of MIC4451YN (attached pdf) which I think have two out that are inverted. Could this also work and should I use them to trigger all the mosfets?
Thanks for your help
Luc
Quote from: Magluvin on December 29, 2015, 05:43:24 PM
Something odd Im finding with my electric bike motor setup.
At the time I had shown the waveforms I had shown that wide open throttle there was no pulse width modulation happening, where it was just switching battery power to the coils in their respective phases. Then when I backed off the throttle some, the pwm kicks in and varies a shorter turn on pulsingas you throttle down.
Well while riding recently I noticed something odd. All this time, 8 yrs, I never really seen it till now. When ridding full throttle, no pwm, the bike will get up to full speed on flat road. But if I back off the throttle, about 3/8 in, it went a bit faster. I can feel it if I gauge it while in acceleration also.
At first I thought it was the magnet in the throttle going too far past the hall sensor. But when I lift the wheel off the ground and do the scope shots, full throttle is wide open no pwm.
Could it be that while in the highest power pwm state before going full no pwm, that it provides more motive force to the wheel than wide open no pwm???
Strange but true.
So the next test will be to ride with one of my tiny scopes connected to see if things are the same as I think they are.
Mags
That's interesting Mags
Maybe when the PWM kicks off the motor cores go above the max saturation point which would actually weakens the motors torque output.
If you can change your gear ratio so your motor can turn at a higher RPM, it may deliver max torque at full throttle.
To me it sounds like your motor is bogged down with too much of a load and that may be corrected with a gear change.
Does it make any sense to you?
Luc
Quote from: gotoluc on December 29, 2015, 11:44:26 PM
That's interesting Mags
Maybe when the PWM kicks off the motor cores go above the max saturation point which would actually weakens the motors torque output.
If you can change your gear ratio so your motor can turn at a higher RPM, it may deliver max torque at full throttle.
To me it sounds like your motor is bogged down with too much of a load and that may be corrected with a gear change.
Does it make any sense to you?
Luc
Hey Luc
Below is a pic that is very close to what im running on my bike. So there are no gears. The motor is direct drive in the wheel.
Im also reposting the previous scope shots. The one with no lines of pwm is when the throttle is pulled all the way on. I would think that not having any off time would be more motive force in the motor or to say the most torque. But when riding and I pulled the throttle back a bit, i go faster. More go. So Ill have to put the tiny scope on the bike and ride to see if loading the motor with me on it will still show no pwm at full on throttle and see if it kicks into pwm when I back it off for more boost. If thats still the case when riding then I think we have something to aim for in all this.
The pwm on and off time periods would be mostly on and very short off when just backing off of full throttle. Seems to be around 15 to 16 kz on the scope. Varies a bit for some reason. So if it is what I think then I would want to experiment with different freq also to see if there is more to get by varying that parameter.
All this time im pulling the throttle all the way back when I wanna go, burning the most current from the batteries when I could have been saving one energy use by limiting the throttle and also going faster. Seems like a good deal to me. ;) So Im looking further into it.
Mags
Quote from: Magluvin on December 30, 2015, 12:40:20 AM
Hey Luc
Below is a pic that is very close to what im running on my bike. So there are no gears. The motor is direct drive in the wheel.
Im also reposting the previous scope shots. The one with no lines of pwm is when the throttle is pulled all the way on. I would think that not having any off time would be more motive force in the motor or to say the most torque. But when riding and I pulled the throttle back a bit, i go faster. More go. So Ill have to put the tiny scope on the bike and ride to see if loading the motor with me on it will still show no pwm at full on throttle and see if it kicks into pwm when I back it off for more boost. If thats still the case when riding then I think we have something to aim for in all this.
The pwm on and off time periods would be mostly on and very short off when just backing off of full throttle. Seems to be around 15 to 16 kz on the scope. Varies a bit for some reason. So if it is what I think then I would want to experiment with different freq also to see if there is more to get by varying that parameter.
All this time im pulling the throttle all the way back when I wanna go, burning the most current from the batteries when I could have been saving one energy use by limiting the throttle and also going faster. Seems like a good deal to me. ;) So Im looking further into it.
Mags
Yes, that won't work on a hub motor!... unless you changed the rim to a smaller size ;D
I guess you don't have amp metering capabilities while you drive right?... or else you would of noticed the extra current wasted?
Is your Hub motor one of those that say you can use it with 24 to 36 volt input?... what is your input voltage?
I don't believe motors can be that flexible in voltages as transportation weight changes, added wind resistance, tire pressure and all kinds of things can bog down the motor to a point the cores are over max flux capability. I think that's what you're experiencing.
So you're still thinking of taping into flyback capture?
Luc
I have a WattsUp meter that will work good for those measurements. My system is 48v and the WattsUp can do 50 amp, 100 amp peak.
You might have something with the saturation bit. More power, less output. But maybe not.
Mags
Quote from: gotoluc on December 29, 2015, 11:33:23 PM
If I understand the circuit correctly, the discharge is to activate a MOSFET to dump the flyback cap?
Yes, the "Discharge C" is signal is for dumping the energy accumulated in the "flyback cap" into a resistor heater or a 2
nd coil (a.k.a. "assistance coil")
Quote from: gotoluc on December 29, 2015, 11:33:23 PM
... if so, why would we want to do this during Energize L on time? Would it not always be after L is opened and flyback has charged the cap?
That's exactly what is happening with this pulse sequencer: The "flyback cap" is dumped after "L is opened and the flyback has charged the cap".
What you don't realize in this scheme, is that the flyback has already charged the cap
in the previous cycle.
This way, the energy stays in the "flyback cap" for almost the entire cycle and is dumped right before the next recharge, which facilitates good measurement of the peak voltage, when an averaging voltmeter is connected across this "flyback cap".
Of course, if you do not care about the accuracy of the voltmeter readings, you can change the time when the "flyback cap" is dumped, to any other time you want, just by rewiring this pulse sequencer a little ...as long as the "dumping" is not happening
while the "flyback cap" is recharging (...like Itsu has done accidentally in the previous version).
Quote from: gotoluc on December 29, 2015, 11:33:23 PM
Speaking of flyback, what will activate the MOSFET to capture flyback so I can get rid of the diode to make it more efficient?
would pin 11 be fast enough to do this?
Yes, pin 11 is fast enough. Actually it might be too fast and not give the main coil's MOSFET enough time to turn off.
The problem with substituting MOSFET for the "flyback diode" is elsewhere. Namely for that purpose, you'd have to use 2 MOSFETs connected in series back-to-back, because of their internal body diodes.
Providing suitable gate drive signals to such MOSFETs with non-isolated drivers and a single power supply is a challenge.
Quote from: gotoluc on December 29, 2015, 11:33:23 PM
also, keep in mind that I've got lots of MIC4451YN (attached pdf) which I think have two out that are inverted. Could this also work and should I use them to trigger all the mosfets?
Yes, big MOSFETs should not be driven directly from the 4047 chips. They should be driven thru some stronger driver chips.
The MIC4451 drivers are suitable for this purpose but be warned that they are inverting chips, which means that all the output signals from the Pulse Sequencer should be inverted too, e.g.: as in the schematic below.
@Itsu
In the circuit you are using now, the "Discharge C" signal should be inverted (compared to Gotoluc's diagram (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/154379/)), because the L2 requires a precharge period in order to build up energy in L2 which is used later to generate a gate pulse for Q3. This gate pulse begins when the precharge period ends and your MOSFET driver interrupts the grounding current to L2 (when point F goes high).
Additionally in your circuit, the width of Q3's gate pulse is not determined by the width of the "Discharge C" signal but by the inductance of L2.
Note, that in the Pulse Sequencer circuit, that I had posted for Gotoluc, the width of the "Discharge C" signal is determined by U4, which is suitable for discharging the "flyback cap" by a relay.
So don't be tempted to use Gotoluc's Pulse Sequencer with your motor circuit as is (without minor changes). I will post these changes if you ever build Gotoluc's Pulse Sequencer (...or Gotoluc builds your motor circuit)
Thanks verpies for all the updates.
Luc
Quote from: verpies on December 30, 2015, 07:42:34 AM
@Itsu
In the circuit you are using now, the "Discharge C" signal should be inverted (compared to Gotoluc's diagram (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/154379/)), because the L2 requires a precharge period in order to build up energy in L2 which is used later to generate a gate pulse for Q3. This gate pulse begins when the precharge period ends and your MOSFET driver interrupts the grounding current to L2 (when point F goes high).
Additionally in your circuit, the width of Q3's gate pulse is not determined by the width of the "Discharge C" signal but by the inductance of L2.
Note, that in the Pulse Sequencer circuit, that I had posted for Gotoluc, the width of the "Discharge C" signal is determined by U4, which is suitable for discharging the "flyback cap" by a relay.
So don't be tempted to use Gotoluc's Pulse Sequencer with your motor circuit as is (without minor changes). I will post these changes if you ever build Gotoluc's Pulse Sequencer (...or Gotoluc builds your motor circuit)
Ok, i did not succeed in getting the signals right by using 2 555 timers, as the wrong side of the pulse was moveable.
Inverting the signal did not invert the moveable side, just the signal :o
I went back to using one 555 timer for the Q1 drive (40Hz @ 20% duty cycle) and a Hall sensor amplified/inverted by some transistors as the input (feedback) for U2/Q3.
I think i now realized the signals as meant to be, see below screenshots (1st is overall, 2th is zoomed in).
Blue is the signal across the CSR (A) compared to point B
purple is the signal of the Q1 drain (C) compared to point B
yellow is the signal across the C2 (2.2uF) cap (D) compared tp point B
green is the signal at point F compared to point B
Due to the placements of the magnets, there is some jitter on the green/yellow signals).
Itsu
Quote from: itsu on December 30, 2015, 03:08:48 PM
I think i now realized the signals as meant to be, see below screenshots (1st is overall, 2th is zoomed in).
Yes the signals are correct now.
Where is the zero volts for the yellow channel?
How narrow can you get the 0V period in the yellow channel?
What does a regular voltmeter connected across C2 show now ?
Ok, the yellow zero is underneath the green zero marker
The screenshot below shows about the most possible narrow period.
A DDM across C2 shows the same as the P2P value of the scope (± 80V), see the video (last part).
https://www.youtube.com/watch?v=qG7Xs8MnRWI&feature=youtu.be
Itsu
Which MOSFET drivers are inverting ?
None, i have only 1 driver in, U2 which now is a non-inverting one (ucc37322).
Q1 is directly being driven by the 555 via a 50 ohm resistor.
Itsu
Quote from: itsu on December 30, 2015, 03:08:48 PM
OK, i did not succeed in getting the signals right by using 2 555 timers, as the wrong side of the pulse was moveable.
See
this sim (http://falstad.com/circuit/circuitjs.html?cct=$+1+0.000005+3.5993318835628396+64+5+50%0A165+336+144+448+144+0+0%0Aw+336+176+336+272+0%0Aw+336+176+304+176+0%0Aw+304+176+304+272+0%0Ac+304+272+304+320+0+2.2e-7+0.005711565920990004%0Ag+304+320+304+336+0%0Ar+304+176+304+112+0+21000%0Aw+304+112+400+112+0%0Aw+224+144+224+208+0%0Aw+160+144+224+144+0%0A165+96+176+112+176+2+12%0Aw+64+144+160+144+0%0Ar+64+208+64+144+0+100000%0Ag+64+368+64+384+0%0Ac+64+304+64+368+0+2.2e-7+7.7006369083001545%0Aw+64+304+96+304+0%0Aw+64+272+64+304+0%0Aw+64+272+96+272+0%0Ar+64+208+64+272+0+22000%0Aw+96+208+64+208+0%0AR+400+112+400+64+0+0+40+12+0+0+0.5%0AR+160+144+160+96+0+0+40+12+0+0+0.5%0Af+528+208+576+208+0+1.5%0Ar+464+208+528+208+0+50%0Ag+576+224+576+288+0%0Al+576+192+576+112+0+0.001+-1.4095121503548535e-7%0AR+576+112+576+64+0+0+40+12+0+0+0.5%0Ad+640+192+640+112+1+0.805904783%0Aw+576+192+640+192+0%0Aw+576+112+640+112+0%0Aw+256+384+624+384+0%0Ad+752+288+752+336+1+0.805904783%0Ar+752+208+752+288+0+47%0Al+752+208+752+128+0+0.001+1.961050679483156e-17%0AR+752+128+752+96+0+0+40+12+0+0+0.5%0Az+832+160+832+208+1+0.805904783+15%0Ad+832+160+832+128+1+0.805904783%0Aw+752+208+832+208+0%0Aw+832+128+752+128+0%0Aw+224+240+256+240+0%0Aw+256+240+336+240+0%0Aw+256+240+256+384+0%0Af+896+208+960+208+0+1.5%0Aw+832+208+896+208+0%0Aw+960+224+1008+224+0%0Aw+1008+224+1008+128+0%0Aw+832+128+1008+128+0%0Ar+960+192+960+144+0+100%0AR+960+144+960+96+0+0+40+80+0+0+0.5%0Aa+624+368+704+368+0+12+0+1000000%0Aw+704+368+704+320+0%0Aw+624+352+624+320+0%0Aw+624+320+704+320+0%0Aw+704+368+752+368+0%0Aw+752+368+752+336+0%0AO+256+384+256+416+0%0AO+464+208+464+336+0%0Ax+594+221+626+224+0+24+Q1%0Ax+916+256+948+259+0+24+Q3%0Ao+55+16+0+550+20+0.00009765625+0+-1%0Ao+56+16+0+550+20+0.00009765625+0+-1%0A) of two 555 working without inverting drivers.
I did not complete the full drain circuits of Q1 and Q3 - I just completed their gate circuits.
The adjustment of the astable pulse width is done by varying the 22K resistor and the monostable pulse width is done by adjusting the 21K resistor.
( Yes, I know that changing that 22K resistor will change not only the pulse width/duty cycle, but it will also change the frequency of the entire astable waveform - that's why I hate working with these 555s and why people come up with such monstrosities as in
Fig. 4.4.8 (http://www.learnabout-electronics.org/Oscillators/osc44.php) )
BTW: The "out" labels what is shown on the sim scope.
Thanks, the signal coming out of my first 555 is inverted from yours, so i have a 40Hz @20% positive duty cycle while you have a 80% or so positive duty cycle.
I need to see if my pots can be adjusted enough to match your 80%, i thought i tried that already, but am not sure what happened.
But don't spend to much time on this as i have ordered some 4047's.
Regards Itsu
Quote from: itsu on December 31, 2015, 05:27:38 AM
Thanks, the signal coming out of my first 555 is inverted from yours, so i have a 40Hz @20% positive duty cycle, while you have a 80% or so positive duty cycle.
Inverted without an inverter?
According to datasheets, the 555 is incapable of outputting
less than 50% duty cycle in its classical astable configuration. That's why people come up with things like in Fig. 4.4.8 (http://www.learnabout-electronics.org/Oscillators/osc44.php)
Anyway, the high duty cycle of the astable (>80%), suits well the requirements of the Q3 drive.
Quote from: verpies on December 31, 2015, 05:33:55 AM
Inverted without an inverter?
According to datasheets, the 555 is incapable of outputting less than 50% duty cycle in its classical astable configuration. That why people come up with things like in Fig. 4.4.8 (http://www.learnabout-electronics.org/Oscillators/osc44.php)
Anyway, the high duty cycle of the astable (>80%), suits well the requirements of the Q3 drive.
Yes, no inverter after the first 555 timer, see this video at the 2:20 mark where i have the raw output of the first 555 timer on the blue probe/trace (connection goes at the back side of the pcb also to the yellow wire)
https://www.youtube.com/watch?v=9klpMi8zeLU&feature=youtu.be
I use this diagram for the first 555:
http://www.electroschematics.com/5834/pulse-generator-with-555/
Itsu
Quote from: itsu on December 31, 2015, 05:47:59 AM
I use this diagram for the first 555:
http://www.electroschematics.com/5834/pulse-generator-with-555/ (http://www.electroschematics.com/5834/pulse-generator-with-555/)
This is not a classical 555 astable configuration, so it is not surprising to me anymore that you are obtaining less than 50% duty cycle at its output.
Quote from: itsu on December 30, 2015, 04:21:58 PM
the yellow zero is underneath the green zero marker
So C2 is not discharged all the way down to 0V.
Would deleting R4 or increasing L2 (or decreasing R2) help to get it all the way down to 0V ?
I do not have R4 in (just C3/C4), and L2 is adjustable by a ferrite core which is now in max (max inductance, guestimate at 2.5mH) which indeed
lowered C2 towards zero as it is now, so probably increasing L2 even more could do it, so need to look for a higher inductance l2.
In the mean time i could try decreasing R2 to see what effect that has.
Itsu
decreasing R2 from 10 to 3.3 Ohm made the C2 cap pratically go to zero, see below screenshot where i have ch1 (yellow) set at max visible amplitude.
Itsu
Looks good itsu ;)
Thanks for sharing all your tests
Luc
Quote from: itsu on December 31, 2015, 10:27:15 AM
decreasing R2 from 10 to 3.3 Ohm made the C2 cap pratically go to zero, see below screenshot where i have ch1 (yellow) set at max visible amplitude.
That poor Q3 must be carrying a Helluva current pulse now and I hope that it does not break, but the waveforms look really nice.
Next year, I shall calculate the "flyback" energy recovery efficiencies and losses, in the circuit's resistances and diode voltage drops.
P.S.
I'd like to notice that it is the designer's fault (my) that Q3 closes only for such a short time (proportionally to L2).
If I had added more diodes to Q3's gate circuit, then Q3 would stay closed until the Pulse Sequencer had explicitly turned it off (not the decaying current in L2). I did not want to do that in order not to complicate the circuit too much, but maybe I'll do it next year.
Happy New Year!
Thanks,
I measured this Helluva current pulse in Q3 (IRF840) with my new current probe, see screenshot below.
The yellow trace is the voltage across the cap as before, purple the Q1 Drain signal as before and green
the current through Q3 with the current controller set to 5A/div., so we see about 15A pulse current.
The IRF840 should be able to withstand 32A pulse current, so should be fine.
Looking forward to next year
Happy New Year!
Quote from: gotoluc on December 28, 2015, 08:14:52 AM
What you're describing sounds like a shaded pole motor, is this correct?
Thanks
Luc
something that looks similar to this picture at the bottom
they are in almost every household appliance that has a fan in it.
not this one in particular, they come in different shapes and sizes, some are double-motors attached tandem.
I think this one is a Braun, which kind of sucks because of the windings.
Westinghouse makes some good ones that can be scoped out to find a resonant node.
speed increase with simultaneous current drop is the prime indicator.
the other is a "cleaner" waveform at the frequency node.
if you have a good scope you can expand the scope image you can see variances in the signal, tiny spikes, that reduce in a resonance mode
when you are dealing with a pre-manufactured device, you don't have the options of engineering everything to a specific set of resonant frequencies.
the best you can do, with a complex system of multiple inductors and coils, is to scope the signals across a broad frequency spectrum.
and try to find a resonant node.
pay attention to voltage, current, and signal coherency.
resonant nodes will be noticeable when you are looking for them
the inductor or housing of the device may physically vibrate at that freq.
if your system has multiple frequencies, you can do a little math, and/or scope out both frequencies while you change freqs and record results.
Hi sm0ky2,
please make a video of your shaded pole motor demonstrating the advantages.
Thanks
Luc
Quote from: verpies on December 30, 2015, 05:19:57 PM
See this sim (http://falstad.com/circuit/circuitjs.html?cct=$+1+0.000005+3.5993318835628396+64+5+50%0A165+336+144+448+144+0+0%0Aw+336+176+336+272+0%0Aw+336+176+304+176+0%0Aw+304+176+304+272+0%0Ac+304+272+304+320+0+2.2e-7+0.005711565920990004%0Ag+304+320+304+336+0%0Ar+304+176+304+112+0+21000%0Aw+304+112+400+112+0%0Aw+224+144+224+208+0%0Aw+160+144+224+144+0%0A165+96+176+112+176+2+12%0Aw+64+144+160+144+0%0Ar+64+208+64+144+0+100000%0Ag+64+368+64+384+0%0Ac+64+304+64+368+0+2.2e-7+7.7006369083001545%0Aw+64+304+96+304+0%0Aw+64+272+64+304+0%0Aw+64+272+96+272+0%0Ar+64+208+64+272+0+22000%0Aw+96+208+64+208+0%0AR+400+112+400+64+0+0+40+12+0+0+0.5%0AR+160+144+160+96+0+0+40+12+0+0+0.5%0Af+528+208+576+208+0+1.5%0Ar+464+208+528+208+0+50%0Ag+576+224+576+288+0%0Al+576+192+576+112+0+0.001+-1.4095121503548535e-7%0AR+576+112+576+64+0+0+40+12+0+0+0.5%0Ad+640+192+640+112+1+0.805904783%0Aw+576+192+640+192+0%0Aw+576+112+640+112+0%0Aw+256+384+624+384+0%0Ad+752+288+752+336+1+0.805904783%0Ar+752+208+752+288+0+47%0Al+752+208+752+128+0+0.001+1.961050679483156e-17%0AR+752+128+752+96+0+0+40+12+0+0+0.5%0Az+832+160+832+208+1+0.805904783+15%0Ad+832+160+832+128+1+0.805904783%0Aw+752+208+832+208+0%0Aw+832+128+752+128+0%0Aw+224+240+256+240+0%0Aw+256+240+336+240+0%0Aw+256+240+256+384+0%0Af+896+208+960+208+0+1.5%0Aw+832+208+896+208+0%0Aw+960+224+1008+224+0%0Aw+1008+224+1008+128+0%0Aw+832+128+1008+128+0%0Ar+960+192+960+144+0+100%0AR+960+144+960+96+0+0+40+80+0+0+0.5%0Aa+624+368+704+368+0+12+0+1000000%0Aw+704+368+704+320+0%0Aw+624+352+624+320+0%0Aw+624+320+704+320+0%0Aw+704+368+752+368+0%0Aw+752+368+752+336+0%0AO+256+384+256+416+0%0AO+464+208+464+336+0%0Ax+594+221+626+224+0+24+Q1%0Ax+916+256+948+259+0+24+Q3%0Ao+55+16+0+550+20+0.00009765625+0+-1%0Ao+56+16+0+550+20+0.00009765625+0+-1%0A) of two 555 working without inverting drivers.
I did not complete the full drain circuits of Q1 and Q3 - I just completed their gate circuits.
The adjustment of the astable pulse width is done by varying the 22K resistor and the monostable pulse width is done by adjusting the 21K resistor.
( Yes, I know that changing that 22K resistor will change not only the pulse width/duty cycle, but it will also change the frequency of the entire astable waveform - that's why I hate working with these 555s and why people come up with such monstrosities as in Fig. 4.4.8 (http://www.learnabout-electronics.org/Oscillators/osc44.php) )
BTW: The "out" labels what is shown on the sim scope.
I managed to adjust my duty cycle pot of the first 555 so i got a similar out signal as your sim (40Hz @ 80% positive duty cycle).
This signal is fet to U2 (non-inverting)/Q3.
It is also fet to the input of the 2e 555 timer (pulse extender) and its output adjusted by its 100K pot to give a 18% positive duty cycle signal
which is used to drive (no driver in between) Q1.
So the feedback hall sensor is not in use anymore.
Below screenshot 1 are the signals again using this setup (C2 is 2.2uF cap).
Blue voltage across the csr point A to B
yellow the voltage across the C2 cap point D to B
purple the voltage across the drain of Q1 point C to B
green the signal at point F to B
A DDM across the C2 cap reads 78V DC
Using a 10uF / 400V cap makes the yellow signal fail to go to zero again, so probably need more L2 inductance and/or less R2 resistance.
Screenshot 2 is the current spike (green trace) through Q1 measured with my current probe at R2 (3.3 Ohm), the current controller still set to 5A/Div.
Yellow is the C2 voltage when shorted by Q1
Itsu
Here the overall picture with the 10uF C2 capacitor.
Green trace is the current through Q3 which peaks 10A (5A/Div.)
Notice the not return to zero for the yellow trace.
The DDM across C2 shows 41V DC
Itsu
Here the same picture as above, but now zoomed in on the current.
Itsu
I was trying to see where the power after stopping the rotor (if any) was going to, and therefor i made some input power measurements and temperature measurements
of specific components both when running and when stopping the rotor.
The below table shows that the input (on time only!) indeed increases after the rotor was stopped (6.17W versus 5.15W).
The temperature measurements show where the power went to, see also graph below.
Running (input 5.15W) | Stopped (input 6.17W)
-------------------------------------- | ------------------------
time 20:00 20:30 21:00 | time 21:30 22:00
|
ambient 19.2°C 18.8°C 19.4°C | 19.4°C 19,6°C
|
L1 coil 21.1°C 21.6°C 21.8°C | 24.5°C 24.7°C
R2 27.1°C 26.5°C 25.8°C | 30.1°C 30.2°C
L2 20.9°C 20.3°C 20.5°C | 21.5°C 21.1°C
Csr 20.3°C 19.1°C 19.3°C | 20.1°C 20.1°C
Q3 24.4°C 24.7°C 25.9°C | 28.5°C 27.7°C
Q1 21°C 19.7°C 20.5°C | 21.6°C 21.6°C
So we see 20% more input when stopped mostly going to R2 and Q3, (so current through the shortening MOSFET), L1 and L2.
The 2 555 timers and the MOSFET driver U2 are being fed by a different battery, so not measured by the input.
Screenshots show input power when running and when stopped, the current controller was set to 500mA/Div, so the displayed values for current (green) and power (red) needs to be taken times 50!
Diagram is the present circuit as i have it running with some modifications, like:
R2 = 3.3 Ohm
ZD2 = 15V / 1A
D2 = 1n4002
L2 = 2.5mH
Q3 = irf840
C2 = 10uF (yes really)
Q1 = irf540
csr = 0.1 Ohm / 1%
U1 = not existing (direct from 555 timer via a 50 Ohm resistor)
Video here: https://www.youtube.com/watch?v=RWihYSKtPuk&feature=youtu.be
Regards Itsu
Quote from: itsu on January 02, 2016, 04:46:35 PM
I was trying to see where the power after stopping the rotor (if any) was going to, and therefor i made some input power measurements and temperature measurements
of specific components both when running and when stopping the rotor.
The below table shows that the input (on time only!) indeed increases after the rotor was stopped (6.17W versus 5.15W).
The temperature measurements show where the power went to, see also graph below.
Running (input 5.15W) | Stopped (input 6.17W)
-------------------------------------- | ------------------------
time 20:00 20:30 21:00 | time 21:30 22:00
|
ambient 19.2°C 18.8°C 19.4°C | 19.4°C 19,6°C
|
L1 coil 21.1°C 21.6°C 21.8°C | 24.5°C 24.7°C
R2 27.1°C 26.5°C 25.8°C | 30.1°C 30.2°C
L2 20.9°C 20.3°C 20.5°C | 21.5°C 21.1°C
Csr 20.3°C 19.1°C 19.3°C | 20.1°C 20.1°C
Q3 24.4°C 24.7°C 25.9°C | 28.5°C 27.7°C
Q1 21°C 19.7°C 20.5°C | 21.6°C 21.6°C
So we see 20% more input when stopped mostly going to R2 and Q3, (so current through the shortening MOSFET), L1 and L2.
The 2 555 timers and the MOSFET driver U2 are being fed by a different battery, so not measured by the input.
Screenshots show input power when running and when stopped, the current controller was set to 500mA/Div, so the displayed values for current (green) and power (red) needs to be taken times 50!
Diagram is the present circuit as i have it running with some modifications, like:
R2 = 3.3 Ohm
ZD2 = 15V / 1A
D2 = 1n4002
L2 = 2.5mH
Q3 = irf840
C2 = 10uF (yes really)
Q1 = irf540
csr = 0.1 Ohm / 1%
U1 = not existing (direct from 555 timer via a 50 Ohm resistor)
Video here: https://www.youtube.com/watch?v=RWihYSKtPuk&feature=youtu.be
Regards Itsu
Great test Itsu.
But remember one thing-->in order for the waste heat to increase,there first has to be an increase in current. So the rotor is reducing current flow,even though that current flow is what is driving the rotor. The extra heat is just a bi-product of the increased current flow.
Brad
WOW Itsu!... that's a lot of data 8)
Thanks for going all the way and sharing it
Most excellent work
Luc
Nice, but why not just eliminate the heat?
artv
Quote from: gotoluc on December 31, 2015, 05:44:53 PM
Hi sm0ky2,
please make a video of your shaded pole motor demonstrating the advantages.
Thanks
Luc
im not saying these have any particular advantage, other than the ones we already take advantage of...
im just saying, if you want to make a "more efficient" motor, you have to beat those ones first.....
if you can do that, then you have a point of sale in every appliance in every home in America......
Quote from: tinman on January 02, 2016, 06:21:13 PM
Great test Itsu.
But remember one thing-->in order for the waste heat to increase,there first has to be an increase in current. So the rotor is reducing current flow,even though that current flow is what is driving the rotor. The extra heat is just a bi-product of the increased current flow.
Brad
Off course, for the components to get hotter, there must flow a greater current through them, but its not easy to individually measure the current through each component.
So somehow the running rotor cause the current in the whole system (not only through L1) to go down.
To me this means that the "entry" (being the battery) is being influenced.
So this could be a "negative-induced voltage in series with the battery voltage" as Poynt99 mentioned, or "EMF superposition" as verpies did call it.
(http://overunity.com/16261/rotating-magnetic-fields-and-inductors/msg469735/#msg469735)
This "negative-induced voltage in series with the battery voltage", or "EMF superposition" seems stronger then the driving power for the rotor.
Another thing i noticed and needs further investigation is the fact that it "looks like" that the increase in current in the whole system starts immediately
when the rotor is "out of sync" with the SG, so we see already an increased current while the rotor is still running.
This to me supports the "negative-induced voltage in series with the battery voltage", or "EMF superposition" as it needs to be in sync with driving the rotor to have that effect.
Regards Itsu
Quote from: itsu on January 03, 2016, 06:01:34 AM
(http://overunity.com/16261/rotating-magnetic-fields-and-inductors/msg469735/#msg469735)
Regards Itsu
QuoteOff course, for the components to get hotter, there must flow a greater current through them, but its not easy to individually measure the current through each component.
This is true,and a very well equipped lab would be required,or maybe put the whole device in a box,and calculate all heat dissipated from each component as one device?.
QuoteSo somehow the running rotor cause the current in the whole system (not only through L1) to go down.
To me this means that the "entry" (being the battery) is being influenced.
So this could be a "negative-induced voltage in series with the battery voltage" as Poynt99 mentioned, or "EMF superposition" as verpies did call it.
I am not sure that verpies was agreeing with Poynt's theory,or the fact that verpies like to call this negative-induced voltage the EMF superposition voltage?. At any rate,if you look at Poynts own test he carried out,you can see that his theory on the negative-induced voltage being the cause of the effect is not correct. Below are his scope shot's,and you can see that while the P/in may have gone down with his simmed rotor in play,the P/out also went down(i have drawn a white line across the inductive kickback across the 200 ohm resistor). So what poynt achieved was to reduce the P/in,and at the same time reduce the P/out. So the overall efficiency remained much the same.
In my(and yours) with the rotor in play,we see a decrease in P/in,but an !increase! in P/out--as seen by the width of the inductive kickback spike in my below scope shot's,and also in the video's when using both the scope and DMM's to measure P/in and P/out.
So no-Poynts test did not show the results we are seeing.
QuoteThis "negative-induced voltage in series with the battery voltage", or "EMF superposition" seems stronger then the driving power for the rotor.
Yes-exactly
Some of the EE guys are saying that the rotor is an energy storage device,and it is just returning it's energy back into the system.
I have asked the question many times now-->how can the rotor return more energy than it took to spin it in the first place?,not to mention the fact that there will be losses in the rotor due to windage and bearing friction. But some how we can put X amount of energy into the rotor,and get Y amount out :o.
QuoteAnother thing i noticed and needs further investigation is the fact that it "looks like" that the increase in current in the whole system starts immediately
when the rotor is "out of sync" with the SG, so we see already an increased current while the rotor is still running.
This to me supports the "negative-induced voltage in series with the battery voltage", or "EMF superposition" as it needs to be in sync with driving the rotor to have that effect.
Do as i did Itsu,and you will see if it is the negative induced voltage that is the cause for the current decrease. All you have to do is measure(scope) the voltage across the coil during the on time,with and without the rotor in play.
I to have seen a massive increase in current when the rotor is not in sync,and this is because the coil is now trying to suck the rotor back toward it,or pushing against the apposing magnet on the rotor. Once it becomes out of sync,it will stop very quick-or quicker than a free wheeling spin down.
At a 13% duty cycle,i cn get my rotor to !hunt!,meaning that the rotor is falling in and out of sync--you can hear it hunting. At the point where you can hear it out of sync,you also see that the input current go's up,and when it go's quiet(back in sync),the input current go's down again.
Brad
Quote from: itsu on January 03, 2016, 06:01:34 AM
Another thing i noticed and needs further investigation is the fact that it "looks like" that the increase in current in the whole system starts immediately
when the rotor is "out of sync" with the SG, so we see already an increased current while the rotor is still running.
This to me supports the "negative-induced voltage in series with the battery voltage", or "EMF superposition" as it needs to be in sync with driving the rotor to have that effect.
Regards Itsu
Yes, it comes down to phasing, or timing of the induced current. In my simulation I can change the phasing by varying either the inductance or capacitance of the resonant tank, and if adjusted so that the phase is 180° to what was shown, the input power increases.
Good observations itsu! ;) I admire your objectivity.
Quote from: tinman on January 03, 2016, 07:39:34 AM
Do as i did Itsu,and you will see if it is the negative induced voltage that is the cause for the current decrease. All you have to do is measure(scope) the voltage across the coil during the on time,with and without the rotor in play.
Seems you missed what I said in my post.
I clearly stated that
you will not be able to measure any change in induced voltage across the coil because you can not isolate the coil's inductance from its resistance in your measurement. You are measuring across a relatively low impedance during the ON time, i.e. that of the battery and the ON resistance of the transistor or MOSFET, so that is why you can not see a change in the battery voltage. However, you can measure the current, and clearly it has decreased. Why? Because the voltage induced by the rotor during that portion of the cycle is in direct opposition to the battery voltage, and this reduces the current. Reduced current through the coil's DC resistance means less waste energy burned by the coil.
Starting at ground potential, and through the battery to the coil load on down to the collector, the polarities of the voltages are as follows:
- +(bat)-->+ -(coil)
Indeed, the measurable voltage across the coil is effectively clamped to the battery voltage when the MOSFET is ON. However, the coil still can generate a "hidden" internal EMF that ultimately affects the rate of rise of current through the coil. By observing the rate of rise of current through the coil you can make an indirect measurement of the EMF generated inside the coil that is in opposition to the battery voltage.
And like I stated a while back some of that "hidden" opposing EMF can be directly attributed to the pushing on the rotor magnets to do mechanical work, i.e.; the export of energy to the outside world with each battery voltage pulse and associated current pulse. There are many ways to look at it and describe it - I like to use the metaphor of some of the battery voltage during the pulse being "eaten" by the export of energy to the outside word - leaving less remaining battery voltage to go towards increasing the current in the coil. That is effectively the motor action.
Quote from: itsu on January 03, 2016, 06:01:34 AM
Off course, for the components to get hotter, there must flow a greater current through them, but its not easy to individually measure the current through each component.
.....
Another thing i noticed and needs further investigation is the fact that it "looks like" that the increase in current in the whole system starts immediately when the rotor is "out of sync" with the SG, so we see already an increased current while the rotor is still running.
This to me supports the "negative-induced voltage in series with the battery voltage", or "EMF superposition" as it needs to be in sync with driving the rotor to have that effect.
Regards Itsu
yes exactly
its like when the brake pads on your car are rubbing against the rotor. it adds "friction" that heats up the system.
when the phase shift drifts in and out of synch, there are associated changes in instantaneous current, frequency, and duty cycle.
each of them can be plotted on a graph to "see" whats happening at each moment in time.
current goes up when the EM fields are working against each other, then lowers as they synch up and momentum works with it.
each "wobble" has its own forces involved, which shows up as spikes, or partial wave-forms, over each instant.
load on the rotor will create heat and increase current for the same reason.
Superposition analysis makes this process easier to deal with, and enables us to see the end result or "cumulative effects"
without all the tedious calculations in between.
kind of like a "black-box" approach
In either case we see that Both frequencies must be "synchronized" for maximum efficiency.
the "ideal" situation, would be to try to stabilize the rpm near the max. at freqs. when current (with respect to duty cycle) are at a min.
Trial & Error is not the right way to do this sort of thing, and while it might seem like the answer to just turn knobs and make adjustments until you find something that works....
This approach has, and always will only lead to more questions, and more misunderstanding.
It is better to calculate the frequencies that work together, and know that all others Don't.
and set the parameters of your system, based on That.
Only then can we establish the appropriate mathematical baseline.
Quote from: tinman on January 03, 2016, 07:39:34 AM
This is true,and a very well equipped lab would be required,or maybe put the whole device in a box,and calculate all heat dissipated from each component as one device?.
I am not sure that verpies was agreeing with Poynt's theory,or the fact that verpies like to call this negative-induced voltage the EMF superposition voltage?. At any rate,if you look at Poynts own test he carried out,you can see that his theory on the negative-induced voltage being the cause of the effect is not correct. Below are his scope shot's,and you can see that while the P/in may have gone down with his simmed rotor in play,the P/out also went down(i have drawn a white line across the inductive kickback across the 200 ohm resistor). So what poynt achieved was to reduce the P/in,and at the same time reduce the P/out. So the overall efficiency remained much the same.
In my(and yours) with the rotor in play,we see a decrease in P/in,but an !increase! in P/out--as seen by the width of the inductive kickback spike in my below scope shot's,and also in the video's when using both the scope and DMM's to measure P/in and P/out.
So no-Poynts test did not show the results we are seeing.
Yes-exactly
Some of the EE guys are saying that the rotor is an energy storage device,and it is just returning it's energy back into the system.
I have asked the question many times now-->how can the rotor return more energy than it took to spin it in the first place?,not to mention the fact that there will be losses in the rotor due to windage and bearing friction. But some how we can put X amount of energy into the rotor,and get Y amount out :o.
Do as i did Itsu,and you will see if it is the negative induced voltage that is the cause for the current decrease. All you have to do is measure(scope) the voltage across the coil during the on time,with and without the rotor in play.
I to have seen a massive increase in current when the rotor is not in sync,and this is because the coil is now trying to suck the rotor back toward it,or pushing against the apposing magnet on the rotor. Once it becomes out of sync,it will stop very quick-or quicker than a free wheeling spin down.
At a 13% duty cycle,i cn get my rotor to !hunt!,meaning that the rotor is falling in and out of sync--you can hear it hunting. At the point where you can hear it out of sync,you also see that the input current go's up,and when it go's quiet(back in sync),the input current go's down again.
Brad
Brad, here i have the "on time" shown when running (white trace) and when stopped (purple trace) on top of each other.
It does not show any difference, only the spike (outside the "on time") is higher.
Its point C compared to point B (ground lead) in the earlier shown diagram.
Itsu
Quote from: poynt99 on January 03, 2016, 09:51:15 AM
Seems you missed what I said in my post.
I clearly stated that you will not be able to measure any change in induced voltage across the coil because you can not isolate the coil's inductance from its resistance in your measurement. You are measuring across a relatively low impedance during the ON time, i.e. that of the battery and the ON resistance of the transistor or MOSFET, so that is why you can not see a change in the battery voltage. However, you can measure the current, and clearly it has decreased. Why? Because the voltage induced by the rotor during that portion of the cycle is in direct opposition to the battery voltage, and this reduces the current. Reduced current through the coil's DC resistance means less waste energy burned by the coil.
Starting at ground potential, and through the battery to the coil load on down to the collector, the polarities of the voltages are as follows:
- +(bat)-->+ -(coil)
If we look at your scope shots,and the way you have tried to show the effect,then yes,youe explanation may fit. But it dose not fit in with my DUT. If we look at your scope shot's,then we can clearly see that the reduction in current flow on the P/in side has also resulted in a reduction of P/out on the output side. So if what you say is true,then i should also see a reduction on the P/out side of my DUT--which i do not. In fact,i see an increase in current flow from the output,and this can be clearly seen in the two scope shots below. You can see that the inductive kickback current flow time duration is longer with the rotor in play,but the current flow on the input side is lower.
You show less P/in and less P/out with your simulated rotor
I show less P/in,but more P/out with my real rotor.
You say to me on the other thread that-Quote:As I said Brad, the simulation could use some fine tuning to not only increase the effect, but maintain or increase the flyback power as well.
And yet my DUT is just a throw together job with no fine tuning,and i get the results you cannot--> i wounder how far i could go with some fine tuning.
Like Itsu said-the rotor seems to be putting more energy back into the system than it takes to run it. The induced reverse voltage has nothing to do wit it as far as i can see,as when the transistor switches on,the coil is shorted across the battery,and that induced reverse voltage is no longer there. Lets say that what you say is true,and as you can see in my scope shot,the voltage across the coil battery combo is at close to 10 volts when the transistor switches on. So now we dont see a reduction in voltage across the coil,as when the transistor switches on,the battery voltage is across the coil,but in stead,we see a reduction in current flow. In your sim,this results in a reduction of current flow on the output side as well--as it should,but in my DUT,we see not a reduction,but an increase in current flow from the output-all while the coil is putting energy into the rotor,so as the rotor can return this energy into the coil.
Brad
Quote from: itsu on January 03, 2016, 10:47:57 AM
Brad, here i have the "on time" shown when running (white trace) and when stopped (purple trace) on top of each other.
It does not show any difference, only the spike (outside the "on time") is higher.
Its point C compared to point B (ground lead) in the earlier shown diagram.
Itsu
And that is what we would expect to see.
You can see that the induced reversed voltage dose not change the voltage across the coil during the on time,as the coil is basically shorted across the battery,and would be seeing the battery voltage only. The induced reverse voltage is an open voltage only,and voltage alone is not power. So at this point in time,the rotor is not returning any of it's stored energy back to the system,as the circuit is open. But what has happened is this-
Lets say your coil is producing a north field at the rotor end of the coil when the mosfet switches on. I will also have a guess that this is when the north out magnet on the rotor has just passed the coils core--as mine dose. So the coil now pushes the north magnet away,and draws the south magnet toward it. But also,as the north magnet has just passed the coils core,the core will already have the correct induced magnetic flux from that passing north magnet,and so,less power from your supply is required to align the magnetic domains within the core,as they have already been mostly aligned by the passing magnet.
Brad
Quote from: tinman on January 03, 2016, 10:56:58 AM
Like Itsu said-the rotor seems to be putting more energy back into the system than it takes to run it.
I did not see itsu say that anywhere.
Quote from: poynt99 on January 03, 2016, 11:35:57 AM
I did not see itsu say that anywhere.
Post 656: http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg470141/#msg470141 (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg470141/#msg470141)
Quote:
This "negative-induced voltage in series with the battery voltage", or "EMF superposition" seems stronger then the driving power for the rotor.
OK, thanks.
I'm not sure how one would come to that hypothesis. Obviously it can not be true. Only a portion of the stored energy is returned each cycle.
Well, thats the problem with words, they can be explained differently as one had originally intended, if not selected carefully, like in this case.
What i was trying to say was that it looks like the magnets (fields) are inducing this "negative-induced voltage in series with the battery voltage"
against the power needed to drive the coil.
So at that instance, this "negative-induced voltage" seems stronger then the driving power for the rotor (otherwise we would not "see" the effects
of this "negative-induced voltage"), but it does not mean that "the rotor seems to be putting more energy back into the system than it takes to run it"
Itsu
Blast from the past from JLN and Steorn:
https://www.youtube.com/watch?v=nqAF_c5ThoI (https://www.youtube.com/watch?v=nqAF_c5ThoI)
When you hear the big buzzing sound the coil is doing electrical-to-mechanical work to make that happen - it is exporting power to the outside world.
Note how the rise time of the current waveform increases significantly (the slope decreases) when the coil is doing real world work. AKA, "negative-induced voltage in series with the battery voltage."
Quote from: itsu on January 03, 2016, 01:02:31 PM
Well, thats the problem with words, they can be explained differently as one had originally intended, if not selected carefully, like in this case.
What i was trying to say was that it looks like the magnets (fields) are inducing this "negative-induced voltage in series with the battery voltage"
against the power needed to drive the coil.
So at that instance, this "negative-induced voltage" seems stronger then the driving power for the rotor (otherwise we would not "see" the effects
of this "negative-induced voltage"), but it does not mean that "the rotor seems to be putting more energy back into the system than it takes to run it"
Itsu
Thanks for clearing that up Itsu. ;)
That makes sense.
Quote from: MileHigh on January 03, 2016, 01:37:21 PM
Blast from the past from JLN and Steorn:
https://www.youtube.com/watch?v=nqAF_c5ThoI (https://www.youtube.com/watch?v=nqAF_c5ThoI)
When you hear the big buzzing sound the coil is doing electrical-to-mechanical work to make that happen - it is exporting power to the outside world.
Note how the rise time of the current waveform increases significantly (the slope decreases) when the coil is doing real world work. AKA, "negative-induced voltage in series with the battery voltage."
Hi MH,
this is indeed an interesting demo and effect. However, I don't think it's relevant to what is being tested here as I'm quite sure the effect demonstrated in JLN video you posted is due to a rise in inductance which occurs when a magnet approaches the Finemet Nanocrystalline toroid cores which were used in the Orbo experiment.
So I don't think you have the answer.
Luc
No, if anything, the presence of an external magnetic field should interfere with the core material's normal domain flipping and effectively reduce the inductance and therefore make the current waveform rise more quickly.
For some strange reason JLN states, "When the lag of the current is max, here is the angle where the coil must be energized by the controller." He is looking for the best performance of that strange pulse motor demoed by Steorn, and to get that all that you would really want would be to energize the coil a smidgen before TDC so that it "disappears" for the second half of the rotor magnet fly-by. So he seems to be off in the clouds, something that has happened before with him.
He is simply not realizing that the current is rising more slowly because of the motoring effect where the magnet is seriously vibrating due to the pulsing coil. It takes power to make the magnet vibrate, and we can hear it in the clip. That power is coming from the coil itself.
The moral of the story is this: For any pulse motor where you are using a fixed pulse timing with a constant supply voltage, if the coil is going to push on a rotor to make it spin or do some kind of action that exports power to the outside world, then when that happens the average power consumption of the coil will go down, not up. With no power being exported to the outside world, and ignoring the back spike, the coil does nothing except produce waste heat. When the coil starts to do real work and export power to the outside world, and ignoring the back spike, then the average power consumption of the coil goes down, and that power is now split into useful work and waste heat.
Quote from: MileHigh on January 03, 2016, 04:44:21 PM
No, if anything, the presence of an external magnetic field should interfere with the core material's normal domain flipping and effectively reduce the inductance and therefore make the current waveform rise more quickly.
Are you absolutely sure there is no possibilities of an increase in inductance when approaching a magnet to these kind of cores?
Luc
Quote from: gotoluc on January 03, 2016, 05:14:12 PM
Are you absolutely sure there is no possibilities of an increase in inductance when approaching a magnet to these kind of cores?
Luc
The whole point of that Steorn toroidal coil + core configuration was to be impervious to the effects of external magnetic fields and have a near-net-zero production of EMF from changing external magnetic fields. I am not absolutely sure in the sense that you may be able to intentionally set up something that increases the inductance. But then that is just loading the dice, right? In a generic sense for a magnet approaching that toroidal coil + core configuration I can't see the inductance increasing.
What I am absolutely sure of is that that clip demonstrates a big honking vibratory motor powered by the pulsing toroidal coil + core. When the current is off, there is attraction and the magnet gets pulled in. When the current is switched on there is no attraction and the magnet goes back. The tape/tire acts as a lossy elastic material that pulls on the magnet to keep it in place. The bottom line is that the magnet does not vibrate for free, the power to make it vibrate comes from the coil and in the clip there is a direct correlation between the vibrating and the slower rise time of the current. I am willing to bet you if the experiment was repeated but this time when the magnet was brought closer to the coil it did not vibrate at all, then the observed current waveform would not change or it would rise slightly faster. It was a big honking motor effect and JLN was not aware of it.
Quote from: MileHigh on January 03, 2016, 05:44:01 PM
In a generic sense for a magnet approaching that toroidal coil + core configuration I can't see the inductance increasing.
Okay, here is a generic test done by me demonstrating an increase of Inductance when approaching a magnet to a coil wound on a Finmet toroid core.
If you pulse a coil on a Finemet toroid and approaches a magnet the result will be a decease of current just as JLN and myself have demonstrated.
Link to my demo done back in 2011: https://www.youtube.com/watch?v=_cCYKChCFqk (https://www.youtube.com/watch?v=_cCYKChCFqk)
Regards
Luc
Quote from: itsu on January 03, 2016, 01:02:31 PM
Well, thats the problem with words, they can be explained differently as one had originally intended, if not selected carefully, like in this case.
What i was trying to say was that it looks like the magnets (fields) are inducing this "negative-induced voltage in series with the battery voltage"
against the power needed to drive the coil.
Itsu
QuoteSo at that instance, this "negative-induced voltage" seems stronger then the driving power for the rotor (otherwise we would not "see" the effects
of this "negative-induced voltage"), but it does not mean that "the rotor seems to be putting more energy back into the system than it takes to run it"
As i said before,there is no work being done by the rotor during this !negative voltage! stage,as the circuit is open,and voltage alone is not power--as we are told often,and no power means no work being done. As the transistor or mosfet closes,then the voltage across the coil instantly becomes the same as the voltage across the supply battery--as your own test showed Itsu. With or without the rotor,the voltage across the coil remains the same during the on time.
What needs to be looked at is what can the induced magnetic field do to or change in that inductor as a whole that would lower the current required to maintain that voltage across the coil during the on time. Well one thing would be that some how(and i dont see how),the PM's field is raising the resistance of the coil,and so less current would be required to obtain the voltage across the coil-or the fact that the magnetic domains have already become some what aligned by the PM before current starts to flow through the coil. It take energy to align the domains within the core,and at the same time eddy currents are produced within the core. Without the rotor,this energy comes from your supply source,and with the rotor in place,some of this energy comes from the rotor in stead,and so less is required from the source.
Some times in life you have to make hard choices,and this is one of those times.
What i mean by this is,you now have to either believe in what you see on your bench,and work it out for your self--or you choose to believe in those that you deem apt in the art of EEing.
I guess it's time to weigh up all the info and results presented on both sides of the fence,and start putting them into perspective,and start asking yourself questions-like
1-how can the rotor in some way reduce the P/in by more than is required to drive it in the first place?
2-How can the open circuit voltage across the battery/coil combo reduce the P/in,when the voltage across the coil remains the same with and without the rotor in play?
3- Why dose Poynts sim not show the same effect as we see with our DUT's-in that while he can show a reduction in P/in,he also gets a reduction in P/out,where as we get a reduction in P/in while maintaining or increasing the P/out when the rotor is in play. What is the one thing his sim dose not have that we do have?.
4- Why has no EE to date been able to recreate the effect without the use of PM's ?.
5- Is the inductance of the coil increasing or decreasing as the PM approaches it. This needs to be thought about very carfully,and what is taking place as the coil is switched on. The coil creates one pole at the end of the rotor,and the approaching magnet on the rotor is of the opposite pole. So is the core going to see an increase of magnetic flux at the rotor end,or a decrease-->remember,the coil has produced a magnetic field of one polarity,and the PM approaching it is of the opposite magnetic polarity. Also-will a coil with less inductance draw more current during that pulse,or less current during that pulse to that of the same coil that has a higher inductance value?.
Added
Here is something that MH wrote a few post back
Quote: No, if anything, the presence of an external magnetic field should interfere with the core material's normal domain flipping and effectively reduce the inductance and therefore make the current waveform rise more quickly.
It is interesting that we see the opposite to that,in where the current waveform rises slower when the rotor is in play.
Things to think about.
Brad
Quote from: gotoluc on January 03, 2016, 05:58:57 PM
Okay, here is a generic test done by me demonstrating an increase of Inductance when approaching a magnet to a coil wound on a Finmet toroid core.
If you pulse a coil on a Finemet toroid and approaches a magnet the result will be a decease of current just as JLN and myself have demonstrated.
Link to my demo done back in 2011: https://www.youtube.com/watch?v=_cCYKChCFqk (https://www.youtube.com/watch?v=_cCYKChCFqk)
Regards
Luc
Great demo Luc,and once again seems to show the opposite to what the EE guys say should happen. I will have to give this a go myself today,but use a ferrite toroid to see if the effect is the same.
Brad
Quote from: gotoluc on January 03, 2016, 05:58:57 PM
Okay, here is a generic test done by me demonstrating an increase of Inductance when approaching a magnet to a coil wound on a Finmet toroid core.
If you pulse a coil on a Finemet toroid and approaches a magnet the result will be a decease of current just as JLN and myself have demonstrated.
Link to my demo done back in 2011: https://www.youtube.com/watch?v=_cCYKChCFqk (https://www.youtube.com/watch?v=_cCYKChCFqk)
Regards
Luc
I looked at your clip and it does appear that the inductance is increasing with the addition of the magnet. However, it's highly unlikely that what you did in that test is directly comparable to the JLN test. I don't get a sense that there is any motoring action in your clip, but it is a possibility.
So, motoring action will slow down the increasing current in a coil, which on the surface appears to look like increasing electrical inductance, but it's not really. If anything, you can say that through the motoring action you have coupled to a mechanical inductor (the mass of the magnet) to the electrical inductor. A mass in motion has the exact properties of inductance.
You apparently demonstrated increasing inductance wile adding a cylindrical magnet directly to the center axis of the toroid. This is not directly comparable to the JLN clip. The lesson is that you can't do one experiment and then apply the results of that experiment with a wide brush to all other situations.
In your clip, you need a schematic to allow people to make sense of what your scope display is showing. I even made that comment on your clip in 2011. What you are calling the back spike looks more like a damped LC resonance. When you add the magnet, the frequency of the LC resonance decreases, which also supports the increasing inductance claim. I hope that you learned from Verpies that going forward you need a schematic 100% of the time, no matter how simple the circuit is.
MileHigh
Quote from: tinman on January 03, 2016, 07:09:59 PM
Great demo Luc,and once again seems to show the opposite to what the EE guys say should happen. I will have to give this a go myself today,but use a ferrite toroid to see if the effect is the same.
Brad
Nope, you have the same trait that Luc has where you do one experiment and believe that you have created a "rule" that applies to all other experiments and situations. It doesn't work like that.
One more time, you think something might be amiss and the EE guys are stumped again but you apparently haven't put any serious thought into the whole thing before you made that statement.
For example, the magnet when placed in the tube could have been vibrating giving you motoring action just like in the JLN clip. Is that increased inductance? In a way it is, but not in the conventional sense that we are talking about.
You notice that there was a big piece of magnetic material attached to the back of the cylindrical magnet. That may not have been saturated, just like the magnet itself may not have been saturated. So it's possible that what was really going on was that the extra ferrite material was adding inductance through the "secondary magnetic field" path of the toroid. That path is along the central axis of the toroid. So you went from an "air core" to a "ferrite core" along the secondary magnetic path. However, I think that unlikely because of the dramatic increase in measured inductance. Nonetheless, it's still worth mentioning and putting under consideration.
For sure, the magnet may have increased the inductance because of the magnetic field influencing the domains in the crystalline core material. However, from what I understand, "occupying" the domains in the core material normally
decreases the effective inductance of the coil. I am pretty sure that that subject has been discussed and covered three or four times over the past few years. So, if what I am saying is indeed true, that seems to present a real problem and the mechanism would have to be properly understood and explained.
MileHigh
Quote from: tinman on January 03, 2016, 06:44:31 PM
As i said before,there is no work being done by the rotor during this !negative voltage! stage,as the circuit is open,and voltage alone is not power
Someone provided an explanation about this a little while back. Guess you missed it?
The "work" being done by the rotor is during the ON portion of the cycle.
QuoteWith or without the rotor,the voltage across the coil remains the same during the on time.
Someone already provided an explanation (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg470149/#msg470149) as to the reason the measured voltage is the same in both cases. Did you miss that one also? Just in case, I provided a link. ;)
Quote
What needs to be looked at is what can the induced magnetic field do to or change in that inductor as a whole that would lower the current required to maintain that voltage across the coil during the on time.
Someone provided an explanation as to how and why the current is being reduced during the ON time. Surely you didn't miss that one too?
Quote
What i mean by this is,you now have to either believe in what you see on your bench,and work it out for your self--or you choose to believe in those that you deem apt in the art of EEing.
Believe absolutely what you see on your bench.
BUT! Don't just look at numbers_in_boxes and assume everything is as it seems, or try to make your observations or (mis)-interpretations fit into some esoteric pet theory of yours. Unfortunately, I see this happening more often than not here. Ask questions and figure things out. The only assumptions that you should be making about your results are that what you are observing is
normal, and not extraordinary. This of course applies to observations that seem out of the ordinary. But keep your common sense and knowledge about you. For example, if the total measured capacitance of two parallel capacitors is not greater than the value of either one, then don't assume that is "normal".
Quote
1-how can the rotor in some way reduce the P/in by more than is required to drive it in the first place?
That is a false unfounded assumption. Where is your proof or evidence of this?
Quote
2-How can the open circuit voltage across the battery/coil combo reduce the P/in,when the voltage across the coil remains the same with and without the rotor in play?
Again, a false assumption or faulty analysis, particularly disappointing when it has already been explained.
Quote
3- Why dose Poynts sim not show the same effect as we see with our DUT's-in that while he can show a reduction in P/in,he also gets a reduction in P/out,where as we get a reduction in P/in while maintaining or increasing the P/out when the rotor is in play. What is the one thing his sim dose not have that we do have?.
Already explained. Do you read my posts? The sim requires tweaking to match the effect, which is often the case when dealing with non-linear cores. My sim has no magnets and no rotor, but provides a good emulation of both.
Quote
4- Why has no EE to date been able to recreate the effect without the use of PM's ?.
Why have you not attempted to recreate the effect using a non-rotor method already provided to you? Would that not be one way to "check" your theory about the magic being in the magnets?
Quote
Things to think about.
Indeed
Quote from: tinman on January 03, 2016, 07:09:59 PM
Great demo Luc,and once again seems to show the opposite to what the EE guys say should happen. I will have to give this a go myself today,but use a ferrite toroid to see if the effect is the same.
Brad
Dear Brad,
to date Finemet tape cores are the only cores I've ever seen that inductance of a coil wound on them increases when a magnet approaches the core.
Ferrites are the worse, complete opposite. Transformer steel laminations are not affected very much by a magnet.
If we can prove there's a benefit to a coils increasing in Inductance between the on and off time, then Finemet cores are the way to go.
Luc
Quote from: poynt99 on January 03, 2016, 08:09:30 PM
Indeed
QuoteSomeone provided an explanation about this a little while back. Guess you missed it? The "work" being done by the rotor is during the ON portion of the cycle.
The explanation provided has not been proven to be correct--and that is a fact.
QuoteSomeone already provided an explanation[/url] as to the reason the measured voltage is the same in both cases. Did you miss that one also?
Once again,that explanation is not backed up with proof.
QuoteSomeone provided an explanation as to how and why the current is being reduced during the ON time. Surely you didn't miss that one too?
No-once again i did not miss it,and no proof has been provided to back up the explanation.
QuoteBelieve absolutely what you see on your bench. BUT! Don't just look at numbers_in_boxes and assume everything is as it seems, or try to make your observations or (mis)-interpretations fit into some esoteric pet theory of yours. Unfortunately, I see this happening more often than not here. Ask questions and figure things out. The only assumptions that you should be making about your results are that what you are observing is normal, and not extraordinary. Then figure out why.
I am yet to see this !normal! reproduced,and every explanation given has not been backed up with proof.
Quote1-how can the rotor in some way reduce the P/in by more than is required to drive it in the first place?
That is a false unfounded assumption. Where is your proof or evidence of this?
Quote Poynt:
QuoteThe "work" being done by the rotor is during the ON portion of the cycle.
So i ask once again--how can the rotor do more work,or return stored energy back to the coil during the ON portion of the cycle,when it is during the ON portion of the cycle that the rotor receives it's energy from the coil?. So my assumptions are not unfounded at all !thank you very much!. And this is the reason your explanations make no sense. You have clearly stated that the rotor is doing work,or returning energy back to the system during the ON part of the cycle. But it is during the ON part of the cycle that the rotor receives it's energy from the coil. So when i ask you to explain as to how (at the same time) the rotor returns more energy to the coil than it received from it,i mean exactly that. If it did not return more energy than it received,then we would see an increase in P/in,or a reduction in P/out.
QuoteAgain, a false assumption or faulty analysis, particularly disappointing when it has already been explained.
There has been no explanation that fits what is happening,and when it is happening. Like your statement that the stored energy,or work being done by the rotor is during the ON time-->no it is not,as during the ON time is when the rotor receives it's energy to store,if it was not,then the rotor would not spin. When the rotor syncs up with the coil,the magnet on the rotor is at TDC when the coil switches off,so there is no point in time during the ON time that the rotor can return any of it's energy to the system,as at that ON time period is when the rotor is receiving energy--not giving it back.
QuoteAlready explained. Do you read my posts? The sim requires tweaking to match the effect, which is often the case when dealing with non-linear cores. My sim has no magnets and no rotor, but provides a good emulation of both.
Once again-yes i read your post,but i guess you missed my reply to your post?. If you didnt miss my reply,then you would not be asking me if i read your post.
QuoteWhy have you not attempted to recreate the effect using a non-rotor method already provided to you? Would that not be one way to "check" your theory about the magic being in the magnets?
Why would i waste my time doing that,when you your self have been unable to recreate the effect without the use of PM's. So the method provided dose not resemble or show the same effect as seen with the rotor and PM's.
If you are going to post !! theories !! about what is happening and why,then have the evidence to back it up. At this point in time,my theory of the magnets doing the work to align the magnetic domains within the core is more credible than what you have provided so far.
Brad
Quote from: tinman on January 03, 2016, 08:52:13 PM
There has been no explanation that fits what is happening,and when it is happening. Like your statement that the stored energy,or work being done by the rotor is during the ON time-->no it is not,as during the ON time is when the rotor receives it's energy to store,if it was not,then the rotor would not spin. When the rotor syncs up with the coil,the magnet on the rotor is at TDC when the coil switches off,so there is no point in time during the ON time that the rotor can return any of it's energy to the system,as at that ON time period is when the rotor is receiving energy--not giving it back.
For a fellow that supposedly has an open mind to possibilities, you appear to be having problems thinking outside the box.
Has it occurred to you that the rotor and coil are exchanging energy at the same time? Apparently not.
Wish there was, but I see no reason to continue my discussion here.
Quote from: MileHigh on January 03, 2016, 07:18:44 PM
I looked at your clip and it does appear that the inductance is increasing with the addition of the magnet. However, it's highly unlikely that what you did in that test is directly comparable to the JLN test. I don't get a sense that there is any motoring action in your clip, but it is a possibility.
So, motoring action will slow down the increasing current in a coil, which on the surface appears to look like increasing electrical inductance, but it's not really. If anything, you can say that through the motoring action you have coupled to a mechanical inductor (the mass of the magnet) to the electrical inductor. A mass in motion has the exact properties of inductance.
You apparently demonstrated increasing inductance wile adding a cylindrical magnet directly to the center axis of the toroid. This is not directly comparable to the JLN clip. The lesson is that you can't do one experiment and then apply the results of that experiment with a wide brush to all other situations.
In your clip, you need a schematic to allow people to make sense of what your scope display is showing. I even made that comment on your clip in 2011. What you are calling the back spike looks more like a damped LC resonance. When you add the magnet, the frequency of the LC resonance decreases, which also supports the increasing inductance claim. I hope that you learned from Verpies that going forward you need a schematic 100% of the time, no matter how simple the circuit is.
MileHigh
Quite the contrary, my test is very comparable to JLN. We are just using different sizes of magnets, mounted on different surfaces and placing them in different locations.
His large magnet will make his wheel vibrate which makes the sound you think is motor action. Mine is at most 10% of the size of his, so obviously won't be affected as much and it is also being held by a small steel lamination dampened by my fingers. You're also not comparing how much power he is putting in his coil compared to mine.
So all this is giving you a false impression that these test are not the same. Seeing a schematic will also not conclude anything. Both are coils being turned on and off.
How different and complicated is that ::)
You were wrong about a magnet not being able to increase a cored coils inductance and your wrong about these tests not being the same.
What are the chances, since I got the exact core that were recommended for the Orbo build and don't you think JLN did the same?
Here you are arrogantly making a mockery of JLN test based on your ignorance that a magnet cannot increase inductance:
Quote from: MileHigh on January 03, 2016, 04:44:21 PM
For some strange reason JLN states, "When the lag of the current is max, here is the angle where the coil must be energized by the controller." So he seems to be off in the clouds, something that has happened before with him.
He is simply not realizing that the current is rising more slowly because of the motoring effect where the magnet is seriously vibrating due to the pulsing coil. It takes power to make the magnet vibrate, and we can hear it in the clip. That power is coming from the coil itself.
and you come up with your own BS motoring hypothesis.
The one who is off the clouds is you! and you could of saved your face a little but you've openly admitted (above) that you've already seen my video back in 2011 and failed to remember what should of been obvious at the time.
What are you doing here??? are you not at least capable of learning something form someone who is supposed to know less than you?
May this be a lesson to you and all your followers that you don't have all the correct answers based on what you have learned from the past. There are new products like Finemet that you obviously know nothing about.
So it's fine if you don't want to do experiments but don't think you know all the answers as things are changing fast and in time you'll be an old school dinosaur.
So better stop your BS now before you really sink your ship... or should I say shit?
Luc
Quote from: poynt99 on January 03, 2016, 09:10:09 PM
QuoteFor a fellow that supposedly has an open mind to possibilities, you appear to be having problems thinking outside the box.
Really :o
It looks to me that you are the one that insist on sticking to the !!known!! laws--not me.
In stead of trying to think outside the box,just remove the box altogether.
QuoteHas it occurred to you that the rotor and coil are exchanging energy at the same time? Apparently not.
So we have 1 guy putting two cup's of water into a bucket at the same time another guy is taking one cup of water out. So we would be loosing water from the supply to the bucket. But in this case we seem to be taking 2 cups of water out of the bucket,and only putting in 1 cup-but the bucket remains full-the rotor continues to spin,and the P/in go's down.
So once again--how is it that we can transfer energy from the coil to the rotor,and at the same time have the rotor return energy back into the coil,and have the P/in go down,while the P/out remains the same or greater?.
We now have Luc just show us that the inductance increases in his toroid core when an external magnetic field is induced into the core. We also had JLN show the very same thing,and i am also saying that this is the reason for the reduced current flow into the coil--the inductance of the coil rises during the on time,as there is a magnetic field of the opposite polarity approaching the core of the coil. The closer the rotor magnet gets to the core of the coil,the higher the inductance will rise within that coil,as the field/flux induced by the current flowing through the coil is being reduced/neutralized by the approaching magnet of opposite polarity. Of course MH is trying to find ways or reasons ,so as the books remain correct,and dismiss the test results shown by others as some sort of motoring action.
QuoteWish there was, but I see no reason to continue my discussion here.
Well i was hoping that you would be able to recreate the effect with your sim setup,but as you wish.
Quote from: gotoluc on January 03, 2016, 09:16:01 PM
Quite the contrary, my test is very comparable to JLN. We are just using different sizes of magnets, mounted on different surfaces and placing them in different locations.
His large magnet will make his wheel vibrate which makes the sound you think is motor action. Mine is at most 10% of the size of his, so obviously won't be affected as much and it is also being held by a small steel lamination dampened by my fingers. You're also not comparing how much power he is putting in his coil compared to mine.
So all this is giving you a false impression that these test are not the same. Seeing a schematic will also not conclude anything. Both are coils being turned on and off.
How different and complicated is that ::)
You were wrong about a magnet not being able to increase a cored coils inductance and your wrong about these tests not being the same.
What are the chances, since I got the exact core that were recommended for the Orbo build and don't you think JLN did the same?
Here you are arrogantly making a mockery of JLN test based on your ignorance that a magnet cannot increase inductance:
and you come up with your own BS motoring hypothesis.
The one who is off the clouds is you! and you could of saved your face a little but you've openly admitted (above) that you've already seen my video back in 2011 and failed to remember what should of been obvious at the time.
What are you doing here??? are you not at least capable of learning something form someone who is supposed to know less than you?
May this be a lesson to you and all your followers that you don't have all the correct answers based on what you have learned from the past. There are new products like Finemet that you obviously know nothing about.
So it's fine if you don't want to do experiments but don't think you know all the answers as things are changing fast and in time you'll be an old school dinosaur.
So better stop your BS now before you really sink your ship... or should I say shit?
Luc
So very correct Luc.
Once again we see that MH (and others) insist that power is required to cause the vibrations(MH's motoring) in JLN's demo,but once again,we see a reduction of P/in by way of the reduced current.
Some how the EE guys here are trying to tell us that the PMs receive energy from the coil,and then that energy is returned back to the system. If we take into account losses involved in this back and forth energy swapping,we should see an increase in P/in. But no-every time we see a decrease in P/in,while the P/out either remains the same or is greater. How can you put energy into a rotor,and have that rotor return that energy--all while there is a reduction in P/in?
JLN shows the reason--,you have shown very clearly the reason,and i have stated that the reason is because of an increase in inductance of the core when the PM is approaching the core. All this we have seen and shown right in front of us,and yet we have the EE guys trying to tell us the opposite is true--even though they have provided no evidence to back up there claim.
The increase in inductance you showed with your toroid is far to great to be a mistake Luc-->great job indeed.
Brad
Quote from: tinman on January 04, 2016, 12:03:34 AM
The increase in inductance you showed with your toroid is far to great to be a mistake Luc-->great job indeed.
Brad
Just keep in mind that I have only observed an increase in Inductance by magnet to core with Finmet toroids.
To date, no other core material I have tested with magnet has shown an increase.
Ferrite is the worst, it drops Inductance like a bad leak at the presence of any magnet.
Luc
Quote from: gotoluc on January 04, 2016, 12:14:50 AM
Just keep in mind that I have only observed an increase in Inductance by magnet to core with Finmet toroids.
To date, no other core material I have tested with magnet has shown an increase.
Ferrite is the worst, it drops Inductance like a bad leak at the presence of any magnet.
Luc
Yes,i just found that out with my ferrite toroid. How ever,i also tried a metglass tape wound toroid core from an old toroid transformer,and it showed an increase in inductance when a magnet was bought close to it.
Seems ferrite is not all it's cracked up to be in this situation.
Brad
From what I have just 'reread' in the book "Regulated Power Supplies" 4th edition Irving M Gottlieb in the chapter Devices and Components, the magnetically biased inductor is meant for dc pulsing operations. So say we have a rod core with a coil and we apply an ac current to it we should be able to measure the cores magnetization curve from one magnetic polarity to the other, NS to SN etc.
So when we pulse a coil on the rod core with dc, we are only able to use half of the magnetization curve of the core. But when we bias the core with a magnet, with biasing in the opposite polarity of what the coil produces when pulsed, we now have access to the full magnetization curve of the core. By doing so we get almost twice as much ampere turns capability from the same coil/core with magnet bias vs without the magnet.
And it also increases inductance.. Its in the book. ;) ;D
Mags
Also, there is nothing said about core material affecting the outcome of core biasing. Most power supplies use ferrite cores, and that is what the book is all about.
Mags
Quote from: tinman on January 04, 2016, 12:44:53 AM
Yes,i just found that out with my ferrite toroid. How ever,i also tried a metglass tape wound toroid core from an old toroid transformer,and it showed an increase in inductance when a magnet was bought close to it.
Seems ferrite is not all it's cracked up to be in this situation.
Brad
Are you sure on the metglas? because where I am at this time they have large metglas cores and I tested them with a magnet before posting.
The inductance goes up when approaching the magnet but once the magnet comes to rest the inductance goes back down. So to me that's not the same and more of an effect of induction taking place and affecting the Inductance meter rather then a permanent boost of inductance.
With the Finemet the boost in inductance did not come down once the magnet stopped moving or came to rest.
If you don't mind rechecking this it could be helpful as I'm hoping verpies can find a use for this effect.
Thanks mate
Luc
Quote from: Magluvin on January 04, 2016, 02:10:08 AM
From what I have just 'reread' in the book "Regulated Power Supplies" 4th edition Irving M Gottlieb in the chapter Devices and Components, the magnetically biased inductor is meant for dc pulsing operations. So say we have a rod core with a coil and we apply an ac current to it we should be able to measure the cores magnetization curve from one magnetic polarity to the other, NS to SN etc.
So when we pulse a coil on the rod core with dc, we are only able to use half of the magnetization curve of the core. But when we bias the core with a magnet, with biasing in the opposite polarity of what the coil produces when pulsed, we now have access to the full magnetization curve of the core. By doing so we get almost twice as much ampere turns capability from the same coil/core with magnet bias vs without the magnet.
And it also increases inductance.. Its in the book. ;) ;D
Mags
This is good information Mags!
If I remember correctly a few years back MH was going on and on about how magnets have zero effect or practical use on cored coils.
Thanks for sharing and get your gears in motion on how we can use this effect to our advantage.
Luc
This is still a similar pattern. Take the example of screw threads. When you turn a screw or bolt clockwise it tightens. But you can shout, "On my bicycle there is one screw thread that tightens when you turn it counter-clockwise so what you are saying is not true! Screw threads turn counter-clockwise to tighten!" It's not a valid argument.
Yes, Magluvin has a book on designing power supplies and what he quoted from the book is 100% true. Is anybody around here designing regulated switching power supplies? Finemet is used for common mode chokes for EMI filters, beads, high-frequency power transformers, pulsed power cores, and presumably many other applications in magnetics. Is anybody around here designing high-frequency power transformers? Do I hear the sound of one hand clapping?
I thought that we were trying to understand the energy dynamics of pulsing a coil to do some work like driving a pulse motor. I think several people have made some good comments about that issue but I am not sure the message is getting through. Or should we talk about the counter-clockwise screw thread industry instead?
In the past year on this forum I have seen perhaps 20 magnetic circuits in diagrams that include odd arrangements with biasing magnets in various places. My gut feel is telling me that none of them used biasing magnets with a definite useful purpose like Magluvin quoted from his power supply design book. They were mostly put in the circuits out of blind belief and if you asked the people to explain why they added the biasing magnets you would most likely get a useless answer that doesn't make sense and can't be justified. That's the state of the art. So yes, in general, using "biasing magnets" in AC or pulse magnetic circuits will have no affect because magnets produce a DC bias and magnetic circuits are AC-based. All of you should reject electronics quackery if it is introduced into something you are following.
Getting back to the dynamics of a pulsing coil - the power consumption will go down and the current will rise more slowly when it pushes on a rotor or exports power to the outside world. This is something that pulse motor designers need to understand and appreciate.
MileHigh
Is F*D "force x distance?" I am having difficulty understanding what you are saying there.
I am talking about what is typically observed in a pulse motor type of situation.
For how it's happening, a simple test that anybody can do:
Suppose you pulse a fixed coil and it produces a north field that is facing the north end of a movable magnet:
[S-coil-N] [N-magnet-S]
So naturally the magnet will get pushed away when you pulse the coil.
Now, what happens if you have the coil open circuited and on your scope, and you pull the magnet away?
My expectation is that you will see an EMF generated by the coil that is opposite the applied voltage of the battery that you used in the first part of the test.
The conclusion: When the battery pulses the coil, and the magnet gets pushed away, then the moving magnet will induce EMF in the coil that effectively reduces the voltage applied across the coil. Even though you can't see it on your scope, it's still happening "inside" the coil. That internal EMF opposing the battery voltage will reduce the rate of current rise when you energize the coil. WOW - two things are happening simultaneously inside the coil.
All of this should be in accord with Lenz's Law.
MileHigh
I am back,
As promised here is the schematic for Itsu utilizing four 4047 CMOS chips as a pulse sequencer.
The major changes are the values of ZD1 and ZD2.
The rotor shaft sensor signal can be anything in the CMOS input range: an Optoreflector, a digital Hall sensor, 555 Timer, 4047 Astable or even your Function Generator isolated by an Optocoupler.
P.S.
If L2 or R3 overheats due to high duty cycle of the "Discharge C2" signal, then this can be solved with two additional 4017 chips but I did not draw it now, since I do not know if it will be a real problem.
Quote from: verpies on January 04, 2016, 11:42:11 AM
I am back,
As promised here is the schematic for Itsu utilizing four 4047 CMOS chips as a pulse sequencer.
The major changes are the values of ZD1 and ZD2.
The rotor shaft sensor signal can be anything in the CMOS input range: an Optoreflector, a digital Hall sensor, 555 Timer, 4047 Astable or even your Function Generator isolated by an Optocoupler.
P.S.
If L2 or R3 overheats due to high duty cycle of the "Discharge C2" signal, then this can be solved with two additional 4017 chips but I did not draw it now, since I do not know if will be a real problem.
Welcome back verpies
Thanks for the new circuit.
Luc
Yes, thanks verpies,
i will use one of the 555 timers which are already on board to drive the 4047's.
For C2 i will use a 10uF/400V cap, R2 will be back to 10 Ohm (now 3.3 Ohm).
Itsu
Quote from: MileHigh on January 04, 2016, 10:10:07 AM
Is F*D "force x distance?" I am having difficulty understanding what you are saying there.
I am talking about what is typically observed in a pulse motor type of situation.
For how it's happening, a simple test that anybody can do:
Suppose you pulse a fixed coil and it produces a north field that is facing the north end of a movable magnet:
[S-coil-N] [N-magnet-S]
So naturally the magnet will get pushed away when you pulse the coil.
Now, what happens if you have the coil open circuited and on your scope, and you pull the magnet away?
My expectation is that you will see an EMF generated by the coil that is opposite the applied voltage of the battery that you used in the first part of the test.
All of this should be in accord with Lenz's Law.
MileHigh
QuoteThe conclusion: When the battery pulses the coil, and the magnet gets pushed away, then the moving magnet will induce EMF in the coil that effectively reduces the voltage applied across the coil. Even though you can't see it on your scope, it's still happening "inside" the coil. That internal EMF opposing the battery voltage will reduce the rate of current rise when you energize the coil. WOW - two things are happening simultaneously inside the coil.
What is WOW is how you guys leave things out to suit your needs.
As you and Poynt have clearly stated,when an external magnetic field is induced into the inductors core,the inductance value will go down,and that will result in a current rise each pulse-easily tested and seen. But i see you left that part out in your above quote--how convenient ::) So any kind of small gain that might be had by the magnet moving away(using your example) would be more than offset by the fact that the core !at the same time! now has that PM's field induced into it,which will result in a higher current draw.
So your own laws MH are arguing against your reasoning.
Brad
I am basing my comments on what I actually see when a pulsing coil does a motor action and makes something move by exporting power to the outside world. Are you "sure" those are the conditions that Poynt was talking about? Perhaps he was referencing more of a static condition where the magnet does not move? Perhaps it also depends on the relative directions of the magnet's field lines and the coil's field lines since magnetic fields are a vector quantity with magnitude and direction?
You need to think more than one or two steps deep for all this stuff.
Magluvin referenced a book where a a coil with a ferrite core can be "preloaded" under the influence of an external magnetic field. Considering in my example the two magnetic fields are 180 degrees diametrically opposed I think that will apply in this case. It looks like when the coil first energizes it has to do the work to "clear out" the biasing of the core which is in the opposite direction that the coil wants to fire. That sounds to me like it will increase the effective inductance - in this case.
On the other hand, if a static and unmoving magnet was "behind" the coil and biasing the core in the "right" direction, then when the coil was energized then in short order the core would get saturated and go "air core." That sounds like it would reduce the effective inductance to me. See? You actually have to think these things through.
If we want to be more complete in our description it looks like there are several effects:
1) The coil when energized has to "clear out" the biasing of the core and then start biasing the core the "right" way.
2.1) When the magnet moves away, an EMF will be induced in the coil that is opposite that of the battery.
2.2) It's likely that the summation (integration) of the "negative EMF" times the instantaneous current flow represents the energy that is put into the moving mass of the magnet - the motor action.
3) Points 1) and 2) above will work to slow the increasing of the current flow when the coil is energized, which effectively makes it look like the coil is a higher inductance - and reduce the average power consumption of the coil.
4) The distributed resistance of the wire of the coil is always there in the background dissipating energy and working to slow down the current flow.
You note that the simple test that pushes a magnet away gives you a more controlled environment to see the effects of the EMF from the magnet moving away from the coil. In a real pulse motor you actually have the magnet approaching and then leaving the coil, the associated acceleration and deceleration of the rotor, and the interaction with the timing of the firing of the pulse.
That's about all that I can think of and I intentionally ignored discussing the back spike. I have never done any of this stuff in real life. I am forced to try to visualize it in my mind.
You can investigate what is taking place to any level of detail that you want. What you can't do is cherry pick one thing and then blindly assume that it applies to all cases without thinking things through. That seems to happen way too often around here.
MileHigh
Quote from: MileHigh on January 04, 2016, 07:50:52 PM
MileHigh
QuoteI am basing my comments on what I actually see when a pulsing coil does a motor action and makes something move by exporting power to the outside world. Are you "sure" those are the conditions that Poynt was talking about? Perhaps he was referencing more of a static condition where the magnet does not move? Perhaps it also depends on the relative directions of the magnet's field lines and the coil's field lines since magnetic fields are a vector quantity with magnitude and direction?
Well as we have all be discussing magnets moving in relation to stationary core's,then one would hope that he was talking about the same thing,if he was trying to explain as to what is happening.\
QuoteYou need to think more than one or two steps deep for all this stuff.
It would appear as though i have thought deeper about this than what you and Poynt have.
I will talk about this some more toward the end of my reply to your post.
QuoteMagluvin referenced a book where a a coil with a ferrite core can be "preloaded" under the influence of an external magnetic field. Considering in my example the two magnetic fields are 180 degrees diametrically opposed I think that will apply in this case. It looks like when the coil first energizes it has to do the work to "clear out" the biasing of the core which is in the opposite direction that the coil wants to fire. That sounds to me like it will increase the effective inductance - in this case.
Lol-amazing . Here i have been,trying to tell you this on two thread's !!for how long now!!?,and all i get from you and Poynt ,is that is incorrect.
QuoteOn the other hand, if a static and unmoving magnet was "behind" the coil and biasing the core in the "right" direction, then when the coil was energized then in short order the core would get saturated and go "air core." That sounds like it would reduce the effective inductance to me. See? You actually have to think these things through.
!!We!! have to thinks things through ::)-->man,where have you been for the last two week's-->and what have myself and Luc been trying to tell you??.
We have to think things through Lol--is this some sort of joke MH?.
If we want to be more complete in our description it looks like there are several effects:
Quote1) The coil when energized has to "clear out" the biasing of the core and then start biasing the core the "right" way.
As i have been trying to tell you and Poynt for weeks now.
Quote2.1) When the magnet moves away, an EMF will be induced in the coil that is opposite that of the battery.
Not with my DUT,as it is the south field that is approaching the core of the coil that produces a north field at the end of the coil where the rotor is. But this has nothing to do with the reduction in current on the P/in side.--Detailed reason and evidence at bottom of this post--that you seem'd to have overlooked.
Quote2.2) It's likely that the summation (integration) of the "negative EMF" times the instantaneous current flow represents the energy that is put into the moving mass of the magnet - the motor action.
So once again we are being told that energy from the coil is transferred to the rotor,and at the same time the rotor returns this stored energy,!!BUT!! the I/ in drop's resulting in less P/in. Once again-some how,we have put energy into the rotor,and taken that energy back out(equal and opposite-minus losses due to windage and bearing friction of the rotor),but seem to have reduced the P/in ::).
Quote3) Points 1) and 2) above will work to slow the increasing of the current flow when the coil is energized, which effectively makes it look like the coil is a higher inductance - and reduce the average power consumption of the coil.
Thats because the coils inductance has risen.
Quote4) The distributed resistance of the wire of the coil is always there in the background dissipating energy and working to slow down the current flow.
Really?-->we shall see.
QuoteThat's about all that I can think of and I intentionally ignored discussing the back spike. I have never done any of this stuff in real life. I am forced to try to visualize it in my mind.
And that is where you fail. The vital information to understanding what is happening,is right there in the back spike. And you say we need to look further than just two step's,while you have totally ignored the very thing that is showing you that the induced reverse voltage across the coil from the moving magnet is not what is reducing the I/in-P/in. This !is! the case with Poynt's sim(that was suppose to simulate the results of my DUT,but did not),but not the case with my DUT.
QuoteYou can investigate what is taking place to any level of detail that you want.What you can't do is cherry pick one thing and then blindly assume that it applies to all cases without thinking things through. That seems to happen way too often around here.
I really cannot believe what i am reading in this whole post of your MH--it is truly unbelievable :o
Quote: What you can't do is cherry pick one thing and then blindly assume that it applies to all cases without thinking things through.
And this is exactly what you have done throughout the ages you have been here on this forum. You blindly stick to what the books tell you,and everything must obey the !known! laws-->which are based only around current observations-->thats right MH-observations,and observations are not laws. With people like you trying to sway others that see different,is it any wonder that any new observations have never been see.
You tell us that !we! need to look at more than two thing;s,and yet here you admit to leaving out the one thing that provides all the answers--the inductive kickback current.
You insist that we cant just cherry pick one thing ,and blindly assume that it applies to all cases--and yet here you are saying,or trying to preach that everything must abide by these know laws--everything.
You clearly have the inability to stand back,and have an unbiased go at working this out. You must stick to your known laws that are based only on observations so far. You preach books that are filled with information that is 100's of years old-->and we are suppose to be looking for the energy of the future.
Poynts quick little attempt at trying to replicate my DUT's test results failed. What he showed was !your! normal outcome,from your known !!laws!!. He succeeded in reducing the I/in-P/in,and as a result,he also reduced the P/out. Why did his results show this?. That is easy-the inductance of the coil remained the same when he switched on his tank circuit,that was suppose to represent a magnet moving toward and away from the core of the coil. As the inductance remained the same,then when the current was reduced,then so was the magnitude of the magnetic field built up around the coil/inductor. We know this is true,because the I/out-P/out also dropped<--this is the bit you chose to ignore,and your undoing to understanding that what i have been saying is true,and in the case of my DUT,it has nothing to do with the induced reverse voltage across the coil.
Look closely at the scope shots below MH-->do you see a reduction in the I/out P/out in my DUT/
No,in fact you see an increase,and this increase was measured by both the scope across the CVR,and also by the DMM's across the CVR-->both methods recommended by both you and Poynt.
So now all you have to do,is work out as to why or how we can have a reduction of current flowing into the coil,and yet have an increase of current flowing out of the coil--how is it that my results are opposite to that of what Poynt showed?. How can you decrease the current flowing into an inductor,and yet increase the current flowing out of the inductor during the kickback.
The only way to increase the current flowing out of the inductor(when the inductors wire and turn ratio remain the same),is to increase the magnitude of the magnetic field that is built up around the inductor. Now,how can that magnetic field be increased if we have just decreased the I/in-P/in of that inductor MH?. Well the answer is simple--the inductance of that inductor had to have increased. You wrote this your self MH--Quote: It looks like when the coil first energizes it has to do the work to "clear out" the biasing of the core which is in the opposite direction that the coil wants to fire. And this is exactly what happens with my DUT,and is exactly what i have been saying,and trying to tell you on two different thread for the past 3 weeks.
Again-->As the south field of the magnet on my rotor approaches the core of the coil,it induces that field into the core. When the coil fires,it fires a !!north!! field at the end of the coil that is closest to the rotor. You just said exactly what i have been saying for over three week's,and during that time,you !and Poynt! are trying to tell us all here that it is this !!cannot be seen!! reverse voltage across the coil that is the reason for the reduction in current draw during the ON time. Well in Poynt's Sym case,that is correct,as the I/out also went down<-- Your lenz's law.
But in my case,we have a situation where the I/in went down,but the I/out went up. If what you and Poynt are saying were true with my DUT,then we would have seen the same results Poynt showed,in that when the I/in was reduced,then the I/out would have also been reduced--but it was not-it increased.
So i hope you take some notice of your own words,and go and have a closer look at what is happening with my DUT. Try and be unbiased for once in your life MH.
Re-posted on my thread.
Brad
Quote from: gotoluc on January 03, 2016, 05:58:57 PM
Okay, here is a generic test done by me demonstrating an increase of Inductance when approaching a magnet to a coil wound on a Finmet toroid core.
If you pulse a coil on a Finemet toroid and approaches a magnet the result will be a decease of current just as JLN and myself have demonstrated.
Link to my demo done back in 2011: https://www.youtube.com/watch?v=_cCYKChCFqk (https://www.youtube.com/watch?v=_cCYKChCFqk)
That is a very interesting phenomenon that does not occur very often.
When the PM aligns the domains in
the same direction as the coil's current, then the apparent inductance
decreases.
However, when the PM aligns the domains in
the opposite direction as the coil's current, then the apparent inductance
increases.
In other words, the inductance is measured relative to the equilibrium level of the domains.
The equilibrium level is the level of domain polarization to which the domains return spontaneously, when the current in the coil falls back to zero.
When the PM polarizes the domains one direction and the coil's current - in the opposite direction, then more headroom is created to accept the energy delivered by the coil before saturation occurs.
Without the PM, the coil can polarize the domains only from 0 to some maximum positive domain polarization.
With the PM, the coil can polarize the domains from some negative domain polarization to the maximum positive domain polarization - that increases the headroom a lot.
While analyzing such problems, it is useful to be mindful of the distinction between the "differential permeability" and the "absolute permeability" of the core and the entire magnetic circuit.
Actually it is about the non-linearity of the BH curve and the directional opposition of polarization vector
components, but let's not split hairs.
Verpies:
I don't disagree with you but I believe that the explanation is somewhat different in this case. Apparently for a Finemet-type nanocrystaline magnetic material in tape form, a magnetic field passing normal to the tape can increase the energy storage properties of the material. I did a search and found the attached document.
If you look at Luc's clip, the geometry between the magnet placement and the Finement tape toroid does the trick. I got beat up for my ignorance, I hope that doesn't happen to you.
MileHigh
Quote from: MileHigh on January 05, 2016, 11:27:43 AM
Apparently for a Finemet-type nanocrystaline magnetic material in tape form, a magnetic field passing normal to the tape can increase the energy storage properties of the material. I did a search and found the attached document.
So Finemet is transversely anisotropic. That exacerbates the directional domain polarization effects.
Thanks for the paper.
Quote from: verpies on January 05, 2016, 10:57:09 AM
That is a very interesting phenomenon that does not occur very often.
When the PM aligns the domains in the same direction as the coil's current, then the apparent inductance decreases.
However, when the PM aligns the domains in the opposite direction as the coil's current, then the apparent inductance increases.
In other words, the inductance is measured relative to the equilibrium level of the domains.
The equilibrium level is the level of domain polarization to which the domains return spontaneously, when the current in the coil falls back to zero.
When the PM polarizes the domains one direction and the coil's current - in the opposite direction, then more headroom is created to accept the energy delivered by the coil before saturation occurs.
Without the PM, the coil can polarize the domains only from 0 to some maximum positive domain polarization.
With the PM, the coil can polarize the domains from some negative domain polarization to the maximum positive domain polarization - that increases the headroom a lot.
While analyzing such problems, it is useful to be mindful of the distinction between the "differential permeability" and the "absolute permeability" of the core and the entire magnetic circuit.
Actually it is about the non-linearity of the BH curve and the directional opposition of polarization vector components, but let's not split hairs.
Sounds like you read my post.
http://overunity.com/16261/rotating-magnetic-fields-and-inductors/msg470331/#msg470331
Quote from: poynt99 on January 05, 2016, 12:20:40 PM
Sounds like you read my post.
http://overunity.com/16261/rotating-magnetic-fields-and-inductors/msg470331/#msg470331 (http://overunity.com/16261/rotating-magnetic-fields-and-inductors/msg470331/#msg470331)
No, I didn't but evidently you were first with this line of thinking.
Magluvin also:
Quote from: Magluvin on January 04, 2016, 02:10:08 AM
So when we pulse a coil on the rod core with dc, we are only able to use half of the magnetization curve of the core. But when we bias the core with a magnet, with biasing in the opposite polarity of what the coil produces when pulsed, we now have access to the full magnetization curve of the core. By doing so we get almost twice as much ampere turns capability from the same coil/core with magnet bias vs without the magnet.
And it also increases inductance.. Its in the book. ;) ;D
Note in essence it's a zero sum game. If you reverse the current flow through the coil, the inductance is decreased.
Quote from: verpies on January 05, 2016, 10:57:09 AM
That is a very interesting phenomenon that does not occur very often.
When the PM aligns the domains in the same direction as the coil's current, then the apparent inductance decreases.
However, when the PM aligns the domains in the opposite direction as the coil's current, then the apparent inductance increases.
I'm glad you find it interesting and so did I when I saw it back in 2011.
Now, is there a way we can use this in our favor before they remove consumer access to Finmet cores?
BTW, I have your circuit on a project board ready to test.
Do I need any caps on the MIC4451YN mosfet drivers chip?
Will the 5vdc input be enough for the MIC4451YN mosfet drivers to switch the mosfet's on?
Thanks for your help
Luc
Let's have some fun and put this to bed. You up for a challenge Brad?
I challenge you to an efficiency battle using only flyback from a coil. You stick with your iron core coil and rotor of magnets, and I'll use whatever coil I choose, but with no rotor, and no resonant tank. What do you say?
Let me know if you're up for it, then we can agree on some goals for the challenge. ;D
Quote from: poynt99 on January 05, 2016, 02:24:26 PM
Let's have some fun and put this to bed. You up for a challenge Brad?
I challenge you to an efficiency battle using only flyback from a coil. You stick with your iron core coil and rotor of magnets, and I'll use whatever coil I choose, but with no rotor, and no resonant tank. What do you say?
Let me know if you're up for it, then we can agree on some goals for the challenge. ;D
oh, oh, oh, can I play too?
Quote from: gotoluc on January 05, 2016, 01:40:03 PM
BTW, I have your circuit on a project board ready to test.
Looks well.
It seems that you run out of the yellow jumpers.
I have to ask Grumage to make me a machine to strip and bend those jumpers with an adjustable length.
Quote from: gotoluc on January 05, 2016, 01:40:03 PM
Now, is there a way we can use this in our favor before they remove consumer access to Finmet cores?
I would have to think about it how they would behave in the classical Magnetic Amplifier application and go from there.
Quote from: gotoluc on January 05, 2016, 01:40:03 PM
Do I need any caps on the MIC4451YN mosfet drivers chip?
As usual, it is a good practice to have additional ~0.1μF power supply decoupling caps between power supply pins (pins 7 & 14) of these 4047 chips and the power supply pins of MOSFET drivers, but I almost never draw these decoupling caps on schematics unless their absence is known to cause problems.
Quote from: gotoluc on January 05, 2016, 01:40:03 PM
Will the 5vdc input be enough for the MIC4451YN MOSFET drivers to switch the MOSFETs on?
Only with the so called "logic level" MOSFETs and I don't think yours are one of those.
If I were you, I'd go for the standard 12VDC power supply to drive the MOSFETs hard and give the MIC4451YN drivers separate power supply lines (see the Starpoint below) so they don't spike-up the 4047 chips.
Take example from Itsu - he must be doing something right if his waveforms are so clean...
Quote from: verpies on January 05, 2016, 12:57:04 PM
No, I didn't but evidently you were first with this line of thinking.
Magluvin also:
Really?
Myself and Luc have been saying this for weeks,and Poynt finally thinks there may be a way that the inductance of the coil can be increased by way of a PM's field,and he is the first in this line of thinking???.
I have shown my DUT,and how it is set up right from the start of my thread. I posted a quick test carried out showing where the south magnetic field of the rotor magnet was in relation to the coil which created a north field at the rotor end when switched on. Is it not clear that the core of the coil would have that south magnetic field induced into it before the current started flowing through the coil,and that the field in the core would flip from south to north when the transistor was switched on.
For weeks now we have been trying to tell you this is why the P/in is reduced,and the P/out can remain the same,and for the same amount of time MH and Poynt have been saying it is because of some fantasy revers voltage that you cant see is the reason for the effect. Only now dose Poynt work out how this might be the case,and he is the first with this line of thinking?--!!really!!.
So often we see the little guys discovering effects like this,and the so called big guns taking credit for it.
Well Luc showed it in 2011,so you EE guys are only 4 years behind.
Brad.
Quote from: poynt99 on January 05, 2016, 02:24:26 PM
Let's have some fun and put this to bed. You up for a challenge Brad?
I challenge you to an efficiency battle using only flyback from a coil. You stick with your iron core coil and rotor of magnets, and I'll use whatever coil I choose, but with no rotor, and no resonant tank. What do you say?
Let me know if you're up for it, then we can agree on some goals for the challenge. ;D
My reply to Poynt from my thread.
Sure-you know me-I never back down from a challenge.
But first you must finish the challenge you have already voluntarily half completed--> and that is to get your simulated setup to replicate my results from my DUT. That is-to get the output current-power to increase when the input current-power is decreased.
Once you have answered and completed that challenge you put upon your self, then we can set the ! Apples for Apples! parameters for the next challenge.
You will of course be required to build an actual device for the next challenge-as I do. And you will be required to post a video here on this thread of your device under test, and the results obtained from that test during the video--> as I do.
So yes, im up for the challenge as long as we are on equal ground.
Quote from: tinman on January 06, 2016, 04:42:04 AM
Really?
Myself and Luc have been saying this for weeks,and Poynt finally thinks there may be a way that the inductance of the coil can be increased by way of a PM's field,and he is the first in this line of thinking???.
I don't keep good track who stated what first because I concentrate more on the knowledge than on the chronology, so I apologize if I stepped on some toes.
I came back from a trip and jumped in by replying to one post, without analyzing the previous ones.
Quote from: tinman on January 06, 2016, 04:42:04 AM
So often we see the little guys discovering effects like this,and the so called big guns taking credit for it.
Well Luc showed it in 2011,so you EE guys are only 4 years behind.
The aforementioned "line of thinking" did not refer to the existence of this phenomenon or its application to your DUT but to its explanation along
those lines (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg470341/#msg470341).
Cheers,
P.S.
The increase of inductance is a property of the core - not the coil itself. Precisely, the increase of differential permeability of the core in one dimension...
Just read a post from Smoky in an unlikely thread and it coincides with my electric bike anomaly with backing off the throttle and getting more go.
"using pulsed-power signals to run those motors has had a lot of success in lowering power requirements per lift force.
increasing the runtime for the copters and gyros
as it turns out, pulsing the motor doesn't change the momentum a whole lot, so you still maintain a stable lift/wind force.
but run the motor at the same rpm, using a lower/shorter duty cycle."
http://overunity.com/16308/electret-antenna-out-of-an-coaxial-cable/msg470452/#msg470452
Mags ;) ;D
Quote from: verpies on January 06, 2016, 06:18:03 AM
P.S.
The increase of inductance is a property of the core - not the coil itself. Precisely, the increase of differential permeability of the core in one dimension...
Bringing a secondary field (PM or EM) near the coil/core will increase its' inductance, AND reluctance.
in a rather symmetrical manner.
what you gain in increased induction, you lose in increased current draw to make the effect possible.
regardless of the orientation. one half works with the field, the other half works against it, and they cancel each other out.
Quote from: sm0ky2 on January 07, 2016, 12:25:50 AM
Bringing a secondary field (PM or EM) near the coil/core will increase its' inductance, AND reluctance.
Shouldn't that be ?:
Bringing a secondary field (PM or EM) near some cored coils will increase their inductance AND
permeance (https://en.wikipedia.org/wiki/Permeance#Electromagnetism).
Hi everyone,
I would need help to find a simple software that I could print a circle of up to 6 inches (15cm) in diameter on white paper. It needs to be pie shape in black for my optical reflector switch off times. It's very simple, if my motor rotor has 8 poles (on times) then the circle would have 8 black pie and 8 white pies equally spaced in the circle. That's all I need but I can't seem to find anything.
So please help my out if you know of something or have some time to look.
Thank you for your help
Luc
Quote from: gotoluc on January 07, 2016, 03:04:25 PM
Hi everyone,
I would need help to find a simple software that I could print a circle of up to 6 inches (15cm) in diameter on white paper. It needs to be pie shape in black for my optical reflector switch off times. It's very simple, if my motor rotor has 8 poles then the circle would have 8 black pie shapes equally spaced in the circle. That's all I need but I can't seem to find anything.
So please help my out if you know of something or have some time to look.
Thank you for your help
Luc
Hi, you could try blender.
Although a steep learning curve, its powerfull free software.
www.blender.org (http://www.blender.org/)
Edit. Another option; Not sure how accurate it is.
But in Exel or open office you could use a circle diagram and print it out. Very easy.
Quote from: verpies on January 07, 2016, 04:05:00 AM
Shouldn't that be ?:
Bringing a secondary field (PM or EM) near some cored coils will increase their inductance AND permeance (https://en.wikipedia.org/wiki/Permeance#Electromagnetism).
That of course is an exception, not the rule.
Quote from: Cherryman on January 07, 2016, 03:07:37 PM
Hi, you could try blender.
Although a steep learning curve, its powerfull free software.
www.blender.org (http://www.blender.org/)
Thanks Cherryman,
I'll give it a try but if I can't figure it out in 20 minutes I'll need something else.
I'll post if it works.
Luc
Quote from: gotoluc on January 07, 2016, 03:14:16 PM
Thanks Cherryman,
I'll give it a try but if I can't figure it out in 20 minutes I'll need something else.
I'll post if it works.
Luc
Hi Luc, I edited my post.
More simple is just an circle diagram in exel or open office. You can easaly put in any value or color
But im not sure how accurate the size after printing will be.
Quote from: Cherryman on January 07, 2016, 03:18:05 PM
Hi Luc, I edited my post.
More simple is just an circle diagram in exel or open office. You can easaly put in any value or color
But im not sure how accurate the size after printing will be.
I just tried Blender and it's not for me.
I'll look at that but it made me think of a pie chart program, maybe there's an online version.
I'll look into it.
Thanks
Luc
ADDEDI found this online Pie Chart Maker: https://nces.ed.gov/nceskids/graphing/classic/bar_pie_data.asp?ChartType=pie (https://nces.ed.gov/nceskids/graphing/classic/bar_pie_data.asp?ChartType=pie)
and made the below. It's all I need to make but I need better resolution.
Inkscape (http://inkscape.org/en/) is a free vector drawing software, Luc.
Quote from: gotoluc on January 07, 2016, 03:04:25 PM
Hi everyone,
I would need help to find a simple software that I could print a circle of up to 6 inches (15cm) in diameter on white paper. It needs to be pie shape in black for my optical reflector switch off times. It's very simple, if my motor rotor has 8 poles (on times) then the circle would have 8 black pie and 8 white pies equally spaced in the circle. That's all I need but I can't seem to find anything.
So please help my out if you know of something or have some time to look.
Thank you for your help
Luc
try this:
https://imgflip.com/piemaker
Quote from: gotoluc on January 07, 2016, 03:04:25 PM
Hi everyone,
I would need help to find a simple software that I could print a circle of up to 6 inches (15cm) in diameter on white paper. It needs to be pie shape in black for my optical reflector switch off times. It's very simple, if my motor rotor has 8 poles (on times) then the circle would have 8 black pie and 8 white pies equally spaced in the circle. That's all I need but I can't seem to find anything.
So please help my out if you know of something or have some time to look.
Thank you for your help
Luc
Hello Luc
I may do it in a graphic vectorial program I use professionally an send it to you as a JPG with high resolution, in various sizes or number of divisions you´ll like. Just PM and I´ll send it to you asap
Cheers
Alvaro
PS: as well as any other graphic design you may need of course
Ok here you have it in attached pdf
this will work as a 50% duty cycle, if you need other % just say so
cheers
Alvaro
Quote from: kEhYo77 on January 07, 2016, 04:08:23 PM
Inkscape is a free vector drawing software, Luc.
Thanks, I'll have a look but if it's not obvious how to do it within the first minute I'll pass on it.
Quote from: Over Goat on January 07, 2016, 09:43:18 PM
try this:
https://imgflip.com/piemaker
I've already tried that one and this one is easier: https://www.meta-chart.com/pie
Quote from: ALVARO_CS on January 08, 2016, 10:10:35 AM
Ok here you have it in attached pdf
this will work as a 50% duty cycle, if you need other % just say so
cheers
Alvaro
Thanks Alvaro, that is very kind of you and the resolution is excellent.
I will let you know if I need any other patterns by PM.
Thanks for your help guys
Luc
Quote from: verpies on January 07, 2016, 04:05:00 AM
Shouldn't that be ?:
Bringing a secondary field (PM or EM) near some cored coils will increase their inductance AND permeance (https://en.wikipedia.org/wiki/Permeance#Electromagnetism).
hmm,.. I may be wrong, if so feel free to correct me, but
I believe, (below the curie temp), the presence of a permanent magnet will lower permeability of the core.
as the PM field wants to resist change, and already saturates the core.
a momentary measurement of permeance, or perpendicular flux density will show a higher value than without the magnet
however, the change in flux is lower and opposed by the PM field
Quote from: sm0ky2 on January 08, 2016, 12:10:36 PM
I believe, (below the curie temp), the presence of a permanent magnet will lower permeability of the core.
as the PM field wants to resist change, and already saturates the core.
That is true most of the time and for isotropic cores.
However when the permanent magnet magnetizes certain anisotropic cores in a direction that is different than the direction of the coil's field, then the permeability, permeance increases.
Quote from: gotoluc on January 07, 2016, 03:04:25 PM
I would need help to find a simple software that I could print a circle of up to 6 inches (15cm) in diameter on white paper. It needs to be pie shape in black for my optical reflector switch off times.
Wouldn't this be less work and far more accurate?
http://tinyurl.com/hpw5jfk
Quote from: verpies on January 08, 2016, 01:08:57 PM
Wouldn't this be less work and far more accurate?
http://tinyurl.com/hpw5jfk (http://tinyurl.com/hpw5jfk)
Yes, far more accurate that's for sure but time spent to find or make couplers to adapt to the shaft of what ever you build or test. Then if I decide to go to 3 phase I've got to get another model.
For now I'll stay with the cheap option and flexability of optical reflective switches and will keep the rotary encoder switch for the final design.
Thanks for the suggestion.
Luc
Quote from: gotoluc on January 08, 2016, 02:23:25 PM
Yes, far more accurate that's for sure but time spent to find or make couplers to adapt to the shaft of what ever you build or test.
That might be true if the shaft is mismatched to the hole
Quote from: gotoluc on January 08, 2016, 02:23:25 PM
Then if I decide to go to 3 phase I've got to get another model.
Oh no. This encoder generates 360pulses per revolution, so you can use it to trigger at any angle and as many times per revolution as you want.
Quote from: gotoluc on January 08, 2016, 02:23:25 PM
For now I'll stay with the cheap option and flexibility of optical reflective switches...
Do you realize, that it is also possible to have a fixed optical reflective switch with
only 1 pulse per revolution anywhere, and electronically divide the time between these single pulses into 360 equal periods using the 4046 chip and some counters, in order to trigger at any shaft angle degree and as many times per revolution as you want ...without a shaft encoder ?
Quote from: verpies on January 08, 2016, 08:00:05 PM
Oh no. This encoder generates 360pulses per revolution, so you can use it to trigger at any angle and as many times per revolution as you want.
I didn't know that! so that same model can do 3 phases? how do you program it to do what you want?
Quote from: verpies on January 08, 2016, 08:00:05 PM
Do you realize, that it is also possible to have a fixed optical reflective switch with only 1 pulse per revolution anywhere, and electronically divide the time between these single pulses into 360 equal periods using the 4046 chip and some counters, in order to trigger at any shaft angle degree and as many times per revolution as you want ...without a shaft encoder ?
No, I did not know that!... I'm relying on people like you who know about electronics to inform me of these things when you see me struggling with making timing reflector wheels on the devices I built.
It's a good thing you're telling me now because I'm presently modifying an off the shelf 12 pole 3 phase Switch reluctance motor and spent the day building a 3 optical switch (see pick) to control each phase.
How many of these 4046 chips and any other components do I need to make a 3 phase work? so I can look into it to see if it's worth going that direction at this point.
Thanks for your time.
Luc
Quote from: gotoluc on January 08, 2016, 11:02:15 PM
I didn't know that! so that same model can do 3 phases?
Yes.
Quote from: gotoluc on January 08, 2016, 11:02:15 PM
how do you program it to do what you want?
The rotary shaft encoder is commonly used by counting the pulses, that it outputs. BTW: There are also rotary encoders that output 3600p/rev. ...and more.
The counting is done digitally by several cheap CMOS chips and BCD switches (without programming) ...or alternatively by one microcontroller (with programming).
For example if you set it to output the 1
st, 121
st, 241
st pulse then it produces a 3 phase output 120º apart and if you want to advance this timing by 5º then you can set it to output the 6
th, 126
th, 246
th pulse.
Quote from: gotoluc on January 08, 2016, 11:02:15 PM
How many of these 4046 chips and any other components do I need to make a 3 phase work? so I can look into it to see if it's worth going that direction at this point.
Just one 4046 chip and four counter chips like 4059 (or 40102), which are cheap, but the BCD switches (http://www.ebay.com/sch/sis.html?_nkw=10pcs+22mm+x+8mm+0+9+Digits+8421BCD+Code+Pushwheel+Thumbwheel+Switches+KM1&_trksid=p2047675.m4100) to program them can be expensive.
The alternative, one chip solution with a microcontroller costs ~$50 but it requires a connection with a computer to program and work.
Be aware that dividing 1 revolution into e.g. 360 smaller time periods with the 4046 chip, gives you an accurate shaft angle down to 1º only when the rotor's speed is constant. If the rotor is accelerating or decelerating, then this method of computing the shaft angle makes an error that increases with the acceleration.
The rotary encoder method does not make this error, regardless of acceleration.
The digital timing diagram below illustrates how that appears on a scope with a x18 pulse multiplication.
If a rotary shaft encoder is used instead of the 4046 pulse multiplication, then the purple waveform comes directly from the encoder.
P.S.
Your 3 optoreflectors and the pie-wheel, constitute a home-made optical rotary encoder of low pulse count: 12pulses / rev.
Just another way to look at it ;)
Hi luc,
I am not sure if this will be any help or not. How much control do you want to have over the pulses going to the motor? If you are willing to let a circuit make those adjustments automatically there are some 3 phase motor controllers already built that are pretty cheap. They are used in Radio Control vehicles. I fly RC airplanes and the little motors on them are very powerful and controlled by what are called ESCs or electronic speed controls. They come in all sizes from ones for 15 amp maximum output up to 150 amp output and probably even higher. The only problem with them is they are usually for 12 volts except for the larger and more expensive ones that are for higher voltages like 30 volts or so. You also need some way to control the esc if you don't have a RC transmitter and receiver. But I think a simple circuit could be built to control the esc. If you think you could use something like this let me know and I'll do some more research to see what would be needed to control the esc. The better escs let you make adjustments to the program for fine tuning to different motors. The controllers automatically adjust the pulses going to the 3 phases to control the motor from 0 rpm up to 20,000 rpm or more depending on the design of the motor.
Carroll
Hi Luc
citfta is right about using an RC controller for model airplanes. If you go that route, use a controller for a model car, it has a fan for cooling. Airplanes have the air stream for cooling.
I went this route to drive my alternator setup. I converted a car alternator to run as a 3 phase motor. It works fine. However you are limited to around 12 volts or so and the electronics are frail.
If I were to do this rig again I would go for a DC motor controller for an Ebike or scooter. They are cheap and rugged. And motors sometimes for free as many scooters are in the recycle yards when they break, also with controllers in them.
Then I would use this DC motor to drive a car alternator without the regulator and diode bridge in it. The alternator now becomes a 3 phase generator to drive your test motor with pure 3 phase power. All this very rugged and cheap.
If you are interested I could post photos and videos how to convert the car altenator and hookup the RC ESC using a simple $5 device. No need to have a radio transmitter to run the ESC.
Hope this helps,
Chris
I just realized that the in-hub motors in Ebikes are brushless 3 phase motors, the same as RC models. One of these controllers might just drive your motor straight away without the use of the alternator?
On the other hand, it might limit your experiments. For example, maybe you could put a diode on each leg to the test motor. Maybe it will run in attraction mode? Then, when you collect the back spike, you can easily compare output to input. You can't do this experiment with the ESC controller because they depend on feedback to sense the position of the rotor. Here the DC motor/alternator makes more sense.
Just tossing out some ideas.
Quote from: citfta on January 09, 2016, 07:31:37 AM
How much control do you want to have over the pulses going to the motor?
That is a very good question because many home-made rotors often are not made symmetrically, and e.g. a 8-magnet rotor might not have the magnets spaced exactly 45º apart or the magnets are not magnetized evenly, requiring the stator pulses to be precisely tailored to the magnet positions.
Quote from: citfta on January 09, 2016, 07:31:37 AM
There are some 3 phase motor controllers already built that are pretty cheap. They are used in Radio Control vehicles.
What shaft position sensor do they use?
Do they allow the energizing pulse to be positioned at specific shaft angles ? How about the "flyback" pulse ?
Quote from: verpies on January 09, 2016, 08:56:56 AM
That is a very good question because many home-made rotors often are not made symmetrically, and e.g. a 8-magnet rotor might not have the magnets spaced exactly 45º apart or the magnets are not magnetized evenly, requiring the stator pulses to be precisely tailored to the magnet positions.
What shaft position sensor do they use?
Do they allow the energizing pulse to be positioned at specific shaft angles ? How about the "flyback" pulse ?
It looks like tishatang has done more research on these controllers than I have so he could probably answer your questions better than I. I was under the impression the controller sensed the signal coming back just as tishatang has said. There are no other position sensors on the little motors. They only have the 3 phase wires going to them. I do know there is a lot of information available on the RC forums about the controllers and rewinding the motors and so forth. If luc thinks these can be used I would be glad to do more research to see how they could be used.
Tishatang I would like to see the simple $5 control circuit you mentioned. I might have a good use for that sometime and not have to connect up the transmitter and receiver just to control a motor.
Carroll
Quote by Verpies
"That is a very good question because many home-made rotors often are not made symmetrically, and e.g. a 8-magnet rotor might not have the magnets spaced exactly 45º apart or the magnets are not magnetized evenly, requiring the stator pulses to be precisely tailored to the magnet positions."
I just realized, the test motor has 8 poles and the stator has 12 poles. The idea of an ESC will not work because of the timing issue based on non-sensored motors. They measure the back emf of the non-energized pole. This gives close enough timing to work. Maybe an ESC will work if a position sensor is added? Some ESC can handle both types of sensing. In this case make sure the ESC is for motor with sensor.
I am fairly certain the DC motor/alternator system will work because the rotor, regardless the number of poles, will try to follow the rotating magnetic field within the stator. I would guess there will be speeds that will lock on at specific rpm's that are in sync with multiples of 8 and 12? Harmonic steps that will be more efficient and stable?
Chris
@Carroll
Here is a link to different kinds. They are called servo testers. They are used on the bench to check the flaps and speed of motor without the use of radio. First, I used the cheap light blue one. Then I went to the big red knob one. Has finer control, better pot.
http://www.hobbyking.com/hobbyking/store/__470__189__Servos_Parts-Servo_Tester.html (http://www.hobbyking.com/hobbyking/store/__470__189__Servos_Parts-Servo_Tester.html)
Chris
I feel like a total dunce! I have a servo tester. I just never thought about using it for controlling the esc!!
Thanks Chris!!
Carroll
Hi everyone,
thanks for all your suggestions.
@verpies, the encoder sounds like the better way to go but then the problem is me not knowing how to program them becomes the next issue.
As for the ESC's recommendations. I am aware and have some that I tested. The problem with them is they output multiple on times operating at high frequencies between 8kHz to 20kHz and also reverse phases (AC). How do you capture flyback with such switching conditions?
For now I'll stay with my simple plan of using the 3 optical switches. However, I do have a question for verpies about my TCRT5000 optical sensors.
Can the photo-transistor of the TCRT5000 be good enough to trigger the 4047 chip directly, or would the photo-transistor need to trigger a transistor which will then trigger the 4047?
Thanks for everyone participation.
Luc
Quote from: gotoluc on January 09, 2016, 11:10:42 AM
@verpies, the encoder sounds like the better way to go but then the problem is me not knowing how to program them becomes the next issue.
Yes, that is a problem if you are using a microcontroller (MCU) ...unless someone sends a programmed one to you.
The alternative to MCU is using several of these CMOS chips and programming them in hardware with jumpers and BCD switches (http://www.ebay.com/sch/sis.html?_nkw=10pcs+22mm+x+8mm+0+9+Digits+8421BCD+Code+Pushwheel+Thumbwheel+Switches+KM1&_trksid=p2047675.m4100) ...without software. I already know you can do it that way, but I don't know if it is any fun for you.
Quote from: gotoluc on January 09, 2016, 11:10:42 AM
I do have a question for verpies about my TCRT5000 optical sensors. Can the photo-transistor of the TCRT5000 be good enough to trigger the 4047 chip directly?
Maybe.
The amplitude of the photo-transistor is strong enough to drive the the 4047 with a 1k pullup resistor, but the rise/fall time of its pulse might not be fast enough. Doing it "by the book" requires a Schmitt trigger such as the 4106 (http://www.onsemi.com/pub_link/Collateral/MC14106B-D.PDF), that "squares up" a slowly changing signal.
Quote from: verpies on January 09, 2016, 01:59:38 PM
Yes, that is a problem if you are using a microcontroller (MCU) ...unless someone sends a programmed one to you.
The alternative to MCU is using several of these CMOS chips and programming them in hardware with jumpers and BCD switches (http://www.ebay.com/sch/sis.html?_nkw=10pcs+22mm+x+8mm+0+9+Digits+8421BCD+Code+Pushwheel+Thumbwheel+Switches+KM1&_trksid=p2047675.m4100) ...without software. I already know you can do it that way, but I don't know if it is any fun for you.
I don't think that's my kind of fun!... but I appreciate you bringing it up and who knows, maybe for a larger version if this test device proves to have an advantage.
Quote from: verpies on January 09, 2016, 01:59:38 PM
Maybe.
The amplitude of the photo-transistor is strong enough to drive the the 4047 with a 1k pullup resistor, but the rise/fall time of its pulse might not be fast enough. Doing it "by the book" requires a Schmitt trigger such as the 4106 (http://www.onsemi.com/pub_link/Collateral/MC14106B-D.PDF), that "squares up" a slowly changing signal.
Will these work the same? http://www.ebay.com/itm/221769763079 (http://www.ebay.com/itm/221769763079)
If so, could you please draw me a schematic starting with the TCRT5000 to CD40106 to HCF4047BE with all components needed in between so I can order it all one time.
Thanks for your help
Luc
Quote from: gotoluc on January 09, 2016, 10:33:29 PM
I don't think that's my kind of fun!
You don't know what you're missing ;(
Quote from: gotoluc on January 09, 2016, 10:33:29 PM
Will these work the same? http://www.ebay.com/itm/221769763079 (http://www.ebay.com/itm/221769763079)
Yes, but I'd have to draw a different schematic diagram for the 40106, since 40106 are not the same as 4106 (the former are inverting and the latter are not).
That is of course if you choose to build a circuit that uses one of them.
Quote from: gotoluc on January 09, 2016, 10:33:29 PM
If so, could you please draw me a schematic...
Below are two ways of conditioning the signals from your optos - one is adjustable and the other is not.
Both circuits are using only one chip (4106 or AD8032).
Quote from: verpies on January 10, 2016, 06:01:53 PM
You don't know what you're missing ;(
Yes, but I'd have to draw a different schematic diagram for the 40106, since 40106 are not the same as 4106 (the former are inverting and the latter are not).
That is of course if you choose to build a circuit that uses one of them.
Below are two ways of conditioning the signals from your optos - one is adjustable and the other is not.
Both circuits are using only one chip.
Thanks verpies for the new circuits.
In the first (non adjustable) circuit I see you have 6 opto's... why do I need that if I'm only switching 3 phases?
The second circuit, unfortunately I don't know what threshold and hysteresis adjustments are for or what they would affect when adjusted?
Care to explain in a simple lawmen way.
EDIT I think I got it, see below post.
Thanks for all your help
Luc
Quote from: gotoluc on January 10, 2016, 06:41:26 PM
unfortunately I don't know what threshold and hysteresis adjustments are for
I gave it some thought and threshold could be to adjust at what point the opto's rising voltage you want it to switch on and hysteresis could be the amount of time you want the on time to last.
Do I pass?
Luc
Quote from: gotoluc on January 10, 2016, 06:41:26 PM
In the first (non adjustable) circuit I see you have 6 opto's... why do I need that if I'm only switching 3 phases?
You don't but the the 4106 chip has 6 channels so I had to draw them.
You can use only 3 channels and leave the remaining channel inputs grounded.
Quote from: gotoluc on January 10, 2016, 06:51:49 PM
I gave it some thought and threshold could be to adjust at what point the opto's rising voltage you want it to switch on
Yes
Quote from: gotoluc on January 10, 2016, 06:51:49 PM
Do I pass?
1 of 2
Quote from: gotoluc on January 10, 2016, 06:51:49 PM
... and hysteresis could be the amount of time you want the on time to last.
No, although hysteresis affects the on-time a little.
On Diag.17, Hysteresis is the vertical distance between Threshold 1 and Threshold 2.
BTW: If you connect the conditioned signal from an opto to the 4047 chain, then the on-time of the conditioned signal will not matter anyway, because the 4047 chain pays attention only to the
rising edge of this signal and stretches its final pulse width to whatever you set it to with the
Width1 (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/dlattach/attach/154384/) pot.
Thanks verpies for making everything clear ;)
You're a good teacher!... I've learned more in the past few weeks with your component suggestions and explinations then I have in years.
I appreciate your help
Luc
I finally completed the "pulse sequencing circuit" as designed by verpies here:
http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg470273/#msg470273
The resulting diagram is slightly different due to some inferior components (zeners), so we had to add an extra MOSFET (Q5), see below diagram.
The goal was to have a circuit which smoothly controls a pulse motor and to manipulate/calculate the flyback spike from the rotor coil (L1) to be able to put it to good use lateron.
For now we settled for each cycle to:
# short out the capture capacitor C2,
# capture the spike in this emptied capacitor C2
# measure the recovered voltage in this capacitor.
Verpies also designed a spreadsheet with a pie chart which, after plugging in the measured data, visually shows the measured data
and calculate the energy distribution in this pulse motor including the recovered energy in the C2 capacitor.
Be aware that its an European based spreadsheet, so the decimal point is a comma!
I made a video which in 4 pieces shows/explains:
# the diagram and physical setup
# the controls/signals of the 4047 chips
# the signals (voltage/current/timing/capturing) on a running system
# the input power measurements for inputting into the spreadsheet for a running system.
video here: https://www.youtube.com/watch?v=UlEl1V44pPE&feature=youtu.be
Below screenshot 1 is the partial (on time only) input power measurement on a running system
Below screenshot 2 is the partial (on time only) input power measurement on a stopped system
The first picture is a snapshot of the spreadsheet made with the above inputted data and shows the energy distribution in a running system.
Note that the energy which is unaccounted for is probably mostly the kinetic energy stored in the running rotor, but this would be very hard to measure (and thus account for).
The second picture is a snapshot from a stopped system where there is much less unaccounted for energy.
A big thanks to verpies for designing the circuit, the spreadsheet and to assist in ironing out the encountered problems.
Hopefully we are able to further zoom in on certain energy streams and to put the captured/recovered flyback pulse energy to good use.
Regards Itsu
Thanks Itsu for taking the time to complete the circuit, making a video demo and share your results.
Top notch!
Thanks also goes to verpies for his helping us to get this done.
On my side, just minutes ago I received the last components to complete the opto circuits.
I hope to have something to share next week.
Luc
Thanks Luc,
glad you liked it.
Looking forward to your build/results.
Itsu
Hi Itsu,
Great work, thanks for showing the results. Thanks also goes to Verpies of course.
The spread sheet data clearly show that in coil L1, about 45-50% of the input power is lost. This indicates again the importance of using thick wire whenever there is room for it to minimize copper loss in coils.
Gyula
Thanks Gyula,
so i have to build myself a 38mH coil which instead of 10 Ohm has a much lower resistance, like 1 Ohm.
Any suggestions on how or what type of coil that might be (which core)?
Regards Itsu
Quote from: itsu on January 21, 2016, 04:57:38 AM
Thanks Gyula,
so i have to build myself a 38mH coil which instead of 10 Ohm has a much lower resistance, like 1 Ohm.
Any suggestions on how or what type of coil that might be (which core)?
Regards Itsu
http://peakoil.com/forums/new-electric-motor-smaller-cheaper-more-powerful-t68369.html#p1152117
Quote from: itsu on January 21, 2016, 04:57:38 AM
....
Any suggestions on how or what type of coil that might be (which core)?
....
Hi itsu,
I think we need to think of coils which are mainly used in audio crossover circuits. However, the usual inductance involved is not higher than 5-10 mH and you need 38 mH. Here is a link to an air core inductor calculator designed mainly for crossover coils: http://www.colomar.com/Shavano/inductor_info.html
And my suggestion to reduce a little the received mechanical sizes for a 38 mH coil is as follows: measure your present L1 coil without the core you have in it now and enter the air core mH value into the above link. From the results you can choose between some wire gauge and/or mechanical sizes.
Of course, there are other multilayer air cored coil calculators on the web like http://coil32.net/multi-layer-coil.html and use your thick wire size you may have in the 0.9 to 1.3 mm gauge range to get the air core inductance referred to above, then you insert a core to increase it to 38 mH or so.
All in all, your new coil will still have a little higher size (volume) than the present L1 coil has, preferably with a DC resistance of indeed around 1 Ohm. For a core you could use the one from the 10 Ohm L1 coil or similar like there is in the choke coil, or lamination core of I shape.
Gyula
Gyula,
i have some audio crossover air coils, but they are specified as 0.39mH, so i will see what happens if i put a core inside, else i have to try on somewhat lower inductance.
Itsu
Yes, there are as low value crossover coils as you wrote, and no amount of a rod type core would increase it up the 38 mH I am afraid.
Even such coils like this would probably be too small initial values to insert a core:
http://www.ebay.com/itm/Air-Core-Cross-over-Inductor-1-5-mH-1-5-ohms-/381517136716?hash=item58d431834c:g:FOIAAOSwJkJWhrDU (http://www.ebay.com/itm/Air-Core-Cross-over-Inductor-1-5-mH-1-5-ohms-/381517136716?hash=item58d431834c:g:FOIAAOSwJkJWhrDU)
Well, perhaps with such bobbins http://tinyurl.com/gp7hhjg (http://tinyurl.com/gp7hhjg) you could make a suitable coil, depending on what wire size you happen to have and insert a core.
I know a low DC resistance high mH coil needs thick copper wire hence large size.
Gyula
Quote from: gyulasun on January 21, 2016, 09:14:09 AM
I think we need to think of coils which are mainly used in audio crossover circuits. However, the usual inductance involved is not higher than 5-10 mH and you need 38 mH. Here is a link to an air core inductor calculator designed mainly for crossover coils: http://www.colomar.com/Shavano/inductor_info.html (http://www.colomar.com/Shavano/inductor_info.html)
I like air coils, too, because they are devoid of hysteresis loss and are linear.
However, their small inductance would force pulse widths to be very short and consequently, the RPM of the rotor - very high.
I have a 840mH air coil with less than 0.001Ω resistance, but it requires LH & LN :(
To get this kind of of L/R ratio in a home-made coil, a thick wire and a high-perm core must be used, e.g. Nanoperm or Metglas.
Also, keeping the air gap to the minimum, helps a lot. Using a silver wire allows to lower the DC resistance by 4% and a square wire allows to lower the DC resistance by additional 27%, but at higher frequencies, the Skin Effect and Proximity Effect do much much more damage to the effective resistance....so the Litz wire comes to the rescue.
It is important to remember, that when all things are being equal, doubling the number of turns, doubles the resistance but quadruples the inductance.
Quote from: verpies on January 21, 2016, 12:16:24 PM
I have a 840mH air coil with less than 0.001Ω resistance, but it requires LH & LN :(
840mH on a air core of 0.001Ω :o how is that acheived? and what is LH & LN?
Thanks
Luc
LH is probably Liquid Hidrogen , it could be Helium too but Hidrogen needs about 20 Kelvin "only" (while Helium needs to be cooled to about 4 Kelvin) to become liquid.
LN is Liquid Nitrogen.
Quote from: gyulasun on January 21, 2016, 01:44:28 PM
LH is probably Liquid Hidrogen , it could be Helium too but Hidrogen needs about 20 Kelvin "only" (while Helium needs to be cooled to about 4 Kelvin) to become liquid.
LN is Liquid Nitrogen.
Thanks Gyula, I now understand ;)
Luc
Quote from: gyulasun on January 21, 2016, 01:44:28 PM
LH is probably Liquid Hydrogen , it could be Helium too but Hydrogen
It is the former. Expensive stuff and volatile
Quote from: itsu on January 21, 2016, 04:57:38 AM
Thanks Gyula,
so i have to build myself a 38mH coil which instead of 10 Ohm has a much lower resistance, like 1 Ohm.
Any suggestions on how or what type of coil that might be (which core)?
Regards Itsu
Hi Itsu
Not an easy result to achieve!... unless coil size and copper mass does not matter?
try the attached coil calculator program.
Results are air core.
I've used it for years, with accurate results.
Let me know if you have any questions.
Luc
Hi everyone,
as promised, here is a test of verpies circuit which I've connected to my pulse motor to which I made a new drive coil for on Metglas core.
I also did a true motor efficient (electrical to mechanical) load test using a generator coil as load on the pulse motors magnet rotor.
@ MileHigh, finally a pulse motor that can do more the just turn a rotor!
The results are quite good I would say.
Link to video: https://www.youtube.com/watch?v=FxSccG5DNFM (https://www.youtube.com/watch?v=FxSccG5DNFM)
Enjoy
Luc
Quote from: gotoluc on January 26, 2016, 10:23:45 PM
Hi everyone,
as promised, here is a test of verpies circuit which I've connected to my pulse motor to which I made a new drive coil for on Metglas core.
I also did a true motor efficient (electrical to mechanical) load test using a generator coil as load on the pulse motors magnet rotor.
@ MileHigh, finally a pulse motor that can do more the just turn a rotor!
The results are quite good I would say.
Link to video: https://www.youtube.com/watch?v=FxSccG5DNFM (https://www.youtube.com/watch?v=FxSccG5DNFM)
Enjoy
Luc
very interesting results, what are the potential applications for this?
Quote from: gotoluc on January 26, 2016, 10:23:45 PM
Hi everyone,
as promised, here is a test of verpies circuit which I've connected to my pulse motor to which I made a new drive coil for on Metglas core.
I also did a true motor efficient (electrical to mechanical) load test using a generator coil as load on the pulse motors magnet rotor.
@ MileHigh, finally a pulse motor that can do more the just turn a rotor!
The results are quite good I would say.
Link to video: https://www.youtube.com/watch?v=FxSccG5DNFM (https://www.youtube.com/watch?v=FxSccG5DNFM)
Enjoy
Luc
@Gotoluc,
What's the COP?
Quote from: Over Goat on January 27, 2016, 02:48:21 PM
very interesting results, what are the potential applications for this?
More efficient electrical power generation.
Luc
Quote from: synchro1 on January 27, 2016, 03:27:33 PM
@Gotoluc,
What's the COP?
I'll let the experts decide!... but I'm sure you must have a general idea of it?
Luc
Quote from: gotoluc on January 26, 2016, 10:23:45 PM
as promised, here is a test of verpies circuit which I've connected to my pulse motor
I appreciate delivering on your promise.
I think only the opto conditioning and timing circuits were designed by me, the power switching circuit was designed by you.
The small cap on the 40106's input is adding a little hysteresis and low pass filtering to the opto's signal.
BTW: How do you like positioning the driving pulse electronically, without having to physically move the opto?
Quote from: gotoluc on January 26, 2016, 10:23:45 PM
to which I made a new drive coil for on Metglas core.
These are nice looking coils. Did you glue the windings together?
What are their inductances and resistances ?
Quote from: gotoluc on January 26, 2016, 10:23:45 PM
I also did a true motor efficient (electrical to mechanical) load test using a generator coil as load on the pulse motors magnet rotor.
@ MileHigh, finally a pulse motor that can do more the just turn a rotor!
This motor has three outputs:
1) The 100Ω resistor across the "generating coil"
2) "flyback" recovery cap and "draining pot"
3) Mechanical
The average power dissipated #1 is the easiest to calculate as P=V
RMS2/R which calculates to: P = (7.12V)
2/100Ω = 0.507W or 507mW.
The average "flyback" energy #2 captured in 1 cycle is the peak voltage across the recovery capacitor squared, times capacitance and divided by two or E=½CV
2 , which calculates to: E=½*34.2μF*(10V)
2 = ½ * 0.0000342F *100V
2 = 0.00171J = 1.71mJ
But the motor runs at about 65Hz under load (pls verify), so we have 65 * 1.71mJ = 111.15mJ per second or 111.15mW.
The average "flyback" power is actually more than 111mW, because that "draining pot" keeps draining the 34.2μF cap even while this cap is being charged. (no periodic discharge circuit like Itsu's).
...and if the "draining pot" is wirewound and inductive, that it forms an LCR circuit with the 34.2μF cap creating an interesting problem...
The mechanical output #3 remains a mystery unless you are willing to calculate the bearing and windage losses ...or play with prony brakes.
Can we have the schematic of the power pulsing circuit ? with the power transistor, flyback diode and all...
Also, it would be nice if you'd care to scope the signals going to these multimeters by using AC channel coupling on your scope. Watch for a groundloop!
If you keep mentioning incremental input power in comparison to absolute output power, then a cerberus will descend from a high mountain and chew you out. From what height? ...about a mile, I guess.
P.S.
The current waveforms are not easily explainable on your scopeshots. I can smell a "back EMF" discussion brewing.
Is this the same?
I'm collecting the back spike in caps, I'm using the banks of caps to drive the rotor,but when one bank is driving the other is outputting, but the output adds to the drive.
Sorry I haven't watched the video's ,can't afford it.
Lots to do.
artv
Quote from: verpies on January 27, 2016, 05:48:58 PM
BTW: How do you like positioning the driving pulse electronically, without having to physically move the opto?
I love it!... verrry accurate and saves so much time. No one knows how much time it takes to change the width of the reflective strips to adjust the on time. On the fly pulse width is a must option. Timing is also nice to have and again much more accurate.
Quote from: verpies on January 27, 2016, 05:48:58 PM
These are nice looking coils. Did you glue the windings together?
Yes, the new drive coil was super glued after it was wound and its side plates (to support the wire while winding) carefully removed, then more super glue added on the sides to keep the wire from falling out the side. Kind of a trick build.
Quote from: verpies on January 27, 2016, 05:48:58 PM
What are their inductances and resistances ?
The drive coils is 0.55 Ohm with 3.76mH
The generator coil is 8.1 Ohms with 150mH
Quote from: verpies on January 27, 2016, 05:48:58 PM
This motor has three outputs:
1) The 100Ω resistor across the "generating coil"
2) "flyback" recovery cap and "draining pot"
3) Mechanical
The average power dissipated #1 is the easiest to calculate as P=VRMS2/R which calculates to: P = (7.12V)2/100Ω = 0.507W or 507mW.
The average "flyback" energy #2 captured in 1 cycle is the peak voltage across the recovery capacitor squared, times capacitance and divided by two or E=½CV2 , which calculates to: E=½*34.2μF*(10V)2 = ½ * 0.0000342F *100V2 = 0.00171J = 1.71mJ
But the motor runs at about 65Hz under load (pls verify), so we have 65 * 1.71mJ = 111.15mJ per second or 111.15mW.
The average "flyback" power is actually more than 111mW, because that "draining pot" keeps draining the 34.2μF cap even while this cap is being charged. (no periodic discharge circuit like Itsu's).
...and if the "draining pot" is wirewound and inductive, that it forms an LCR circuit with the 34.2μF cap creating an interesting problem...
Agreed, the generator coil is delivering 507mW plus its internal resistance which you can determine with the now known wire resistance of 8.1 Ohms
I checked the frequency and it's 88Hz (9 samples in window @10ms), so we have 88 * 1.71mJ =150.48mJ per second or 150mW on flyback recovery.
I agree, we can't calculate all the Joules from the cap the way I have it set up. I can live with that ;)
Quote from: verpies on January 27, 2016, 05:48:58 PM
The mechanical output #3 remains a mystery unless you are willing to calculate the bearing and windage losses ...or play with prony brakes.
I wasn't interested of the mechanical output of the rotor. However, an idling motors rotor has zero torque.
What I was interested in knowing, is the motor electrical to mechanical Watts efficiency.
So in my test video I used a known load (a generator coil with known resistor)
I applied the generator coil to load the motors magnet rotor to give me an idea of the motors efficiency.
As we know a generator coil is never 100% efficient because of wire resistance losses (in my case 8 Ohms) and core losses.
So a small part of the power is lost in wire resistance and part is delivered to the load which we calculated to be 507mW across the 100 Ohm load resistor.
I'll let you calculate the power dissipated in the wire resistance.
Here is the problem with my test results. The generator coil is delivering 507mW plus some extra mW caused by its wire resistance.
How can we explain the motors power input only increasing by 350mW? ... like I said and I've even read MH say the same thing, an idling motor has zero torque. So if there is no continues power storage in an idling motors rotor, then when we apply a load to the rotor, it should consume more power from the input and normally a little more power then the power the load is delivering.
However, it's not the case, we have a shortage of over 150mW!... why?
Quote from: verpies on January 27, 2016, 05:48:58 PM
Can we have the schematic of the power pulsing circuit ? with the power transistor, flyback diode and all...
Also, it would be nice if you'd care to scope the signals going to these multimeters by using AC channel coupling on your scope. Watch for a groundloop!
I'll see what I can do to provide a schematic. You know how much I hate doing schematics :(
I'm not sure why or how to scope the signals going to my multimeters using AC channel coupling?
I'm aware of the groundloop problems ;)
Quote from: verpies on January 27, 2016, 05:48:58 PM
If you keep mentioning incremental input power in comparison to absolute output power, then a cerberus will descend from a high mountain and chew you out. From what height? ...about a mile, I guess.
Yes LOL
Quote from: verpies on January 27, 2016, 05:48:58 PM
P.S.
The current waveforms are not easily explainable on your scopeshots. I can smell a "back EMF" discussion brewing.
Let me know if you want me to post some zoomed in scope shots.
As long as discussion are kept civil, respectful and productive they can be entertained in this topic.
Kind regards
Luc
Quote from: shylo on January 27, 2016, 07:23:38 PM
Is this the same?
I'm collecting the back spike in caps, I'm using the banks of caps to drive the rotor,but when one bank is driving the other is outputting, but the output adds to the drive.
Sorry I haven't watched the video's ,can't afford it.
Lots to do.
artv
Sounds similar.
Do you have a laptop?
Luc
Quote from: gotoluc on January 23, 2016, 02:08:34 PM
Hi Itsu
Not an easy result to achieve!... unless coil size and copper mass does not matter?
try the attached coil calculator program.
Results are air core.
I've used it for years, with accurate results.
Let me know if you have any questions.
Luc
Thanks Luc,
i had some stuff laying around, so i used that for building my new coil.
I see you are going on very well to (very nice coils), and it seems we are using almost the same things to build the coil.
I build my core from metglas tape also, but did not chop it up, just rolled about 8m of this tape onto a fiberglass rod
to create a metglas core of 4 cm long, 18mm thick to fit a former, see video: https://www.youtube.com/watch?v=YSB-fj1PH9w&feature=youtu.be
The new coil measures 0.8 Ohm @ 4.5mH and is using 1mm diameter bonded magnet wire, bifilar wound.
Due to this low resistance i had to lower the ontime (duty cycle) of the pulse, but with verpies his sequencer that was no problem.
It now runs at 40Hz (won't go any higher) @ 4% duty cycle (was 20%, but won't go any lower then 4%).
Its amazing what can be done with a few cents worth of chips/components and some time as this "verpies sequencer" gives you full control over the pulses used.
I filled in the spreadsheet below with this new data and we see a big improvement on the "input recovered in C2"
At the same time the "unaccounted Energy" part is reduced strongly compared with the earlier coil/setup.
The screenshot show some data collection points to fill in the spreadsheet which is added to.
Regards Itsu
Excellent drive coil Itsu!... which gives excellent test results as well.
Are you going to take it further by trying 2mm wire? as I see your spool has room for twice the amount of wire ;)
Thanks for making a video of your coil build and posting all the test data results.
Kind regards
Luc
I would if i had laying it around, but i don't :(
I think i will do some load tests like you did, using my old coil, but somehow i think i am on the threshold of useability of this rotor.
Will see, regards Itsu
Quote from: itsu on January 28, 2016, 01:19:27 PM
I would if i had laying it around, but i don't :(
I think i will do some load tests like you did, using my old coil, but somehow i think i am on the threshold of useability of this rotor.
Will see, regards Itsu
Okay, just checking!
As for the load tests. Is your magnet rotor still synchronizing to the drive coil? if so, I can't imagine you can load it and expect it to stay in sync. You may need to lock the rotor and drive coil with a trigger.
BTW, I found a high impedance coil works best as gen coil and load it to around 50% of its open coil voltage.
Just my 2 cents
Luc
Quote from: gotoluc on January 28, 2016, 01:30:51 PM
BTW, I found a high impedance coil works best as gen coil and load it to around 50% of its open coil voltage.
Just my 2 cents
Luc
So the gen coil is putting out more voltage than used in the drive circuit? At half of gen open voltage, is the voltage higher than the drive stage voltage?
There was a point in your vid where it looked like 23.0 something but you stated 2 hundred something. Or was the meter set up some how differently
Mags
Quote from: Magluvin on January 28, 2016, 05:58:00 PM
So the gen coil is putting out more voltage than used in the drive circuit? At half of gen open voltage, is the voltage higher than the drive stage voltage?
There was a point in your vid where it looked like 23.0 something but you stated 2 hundred something. Or was the meter set up some how differently
Mags
Yes, open gen coil would be around 20v and the drive coil is driven by 10vdc.
Not sure on the 2 hundred something. Can you give me the vid time line of what you're referring to.
Luc
Quote from: gotoluc on January 28, 2016, 06:21:25 PM
Yes, open gen coil would be around 20v and the drive coil is driven by 10vdc.
Not sure on the 2 hundred something. Can you give me the vid time line of what you're referring to.
Luc
So try to loop it. ;D
Ill look for the time in the vid. may have just been a misquote at the time.
Mags
Luc. Look at 8min on video.
Mags
oh I think I see. measuring voltage of a csr?
Mags
Quote from: Magluvin on January 28, 2016, 07:03:16 PM
oh I think I see. measuring voltage of a csr?
Mags
Yes Mags, that's millivolts across the low pass filter cap bank 0.1 CSR which = DC mA
Been using this low pass filter cap bank meters combo for just about every test that involves switching of DC for over 6 years now!
Luc
Thanks for the follow up Luc, been too busy at work to participate. Excellent latest build and the whole series. As always its been very educational for me and I'm sure others. As soon as I can get to the bench this will be my build of choice :)
Thanks for letting me know mate!
Cheers
Luc
Here is a video demo to update my recent build which should be fired up today.
https://www.youtube.com/watch?v=O9wh23H5lhE (http://www.energeticforum.com/redirect-to/?redirect=https%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3DO9wh23H5lhE)
Luc
Got the reluctance motor turning but only on 2 phases as it looks like the timing of the optos needs to be physically changed to get it in the right range.
I think I'll have to have the optos on the outside of the motor till I have it all figured out.
Anyways, thought of giving you a quick first glance: https://www.youtube.com/watch?v=TfhIv3ihOKY
Anyone interested or reading this topic?
Luc
Quote from: gotoluc on February 05, 2016, 02:12:00 PM
Got the reluctance motor turning but only on 2 phases as it looks like the timing of the optos needs to be physically changed to get it in the right range.
I think I'll have to have the optos on the outside of the motor till I have it all figured out.
Anyways, thought of giving you a quick first glance: https://www.youtube.com/watch?v=TfhIv3ihOKY
Anyone interested or reading this topic?
Luc
I certainly am interested in this. this is very exciting research, and very nice workmanship in the videos, I appreciate all y0ur effort. I won't pretend to understand the exact ramifications of this, on the face of i the results look amazing , 50% power recovery already on the first try with this new build, with not even all the coils operable yet, I can't say enough, encouragement and gratitude here, heres' hoping for the best , hope it is replicated and verified of course , immensely grateful for you and your work
Chasing 50 thousand reads ??
I assume there is interest !
Thanks for sharing Your hard work and amazing build skills Here and elsewhere .
Chet K
Thanks guys,
Just checking to see as it's been kind of quiet around here lately.
Luc
Hi Luc
Very best work as usual
And of course following your progress which is a real stimulation to me
Thank's very much for sharing
Laurent
Quote from: gotoluc on February 05, 2016, 03:47:52 PM
Thanks guys,
Just checking to see as it's been kind of quiet around here lately.
Luc
I keep checking your youtube and this thread to see if there are any updates on your progress with the last build, you were sorting out the timing of the light sensors .
Quote from: Over Goat on February 09, 2016, 06:01:45 AM
I keep checking your youtube and this thread to see if there are any updates on your progress with the last build, you were sorting out the timing of the light sensors .
Thanks for your interest.
The next day I had the issue of the optical switches timing sorted out and all 3 Phases working well now. Nothing needed to be modified other then finer tuning and timing adjustments.
Yesterday I made a mechanism so I can manually move (on the fly) the optical switches assembly forward or back which helps tune to ideal timing.
Her is a video of that modification and explaining the next needed modification.
Link to video: https://www.youtube.com/watch?v=cYwAFqBzmdc
Luc
what an ingenious fix and with hand tools yet.i would have needed a rotary table to make that piece.great video's and research.
Luc has gone dark on this, I wonder if the final results are so good he doesn't know what to do now :)
what if he's just blown away with his numbers, he's been advised to keep a lid on this for now
Quote from: billbailey on February 09, 2016, 08:01:01 PM
what an ingenious fix and with hand tools yet.i would have needed a rotary table to make that piece.great video's and research.
Hi Bill,
Nice to see you have joined the forum.
Thanks for your comment
Luc
Quote from: Over Goat on February 18, 2016, 09:49:24 AM
Luc has gone dark on this, I wonder if the final results are so good he doesn't know what to do now :)
what if he's just blown away with his numbers, he's been advised to keep a lid on this for now
I'm still around, just been busy with a few things.
May have an update tomorrow.
Luc
Here is a video update.
Motor is now working perfectly. So now the load tests can begin so we can establish the (modified) motors efficiency. Then I will add the assist motor which will only be powered by the primary motors flyback and see if it helps boost the efficiency which is the main goal of this research.
Link to new video demo: https://www.youtube.com/watch?v=aXtkUfdlvRI
Thanks for your interest
Luc
Hi Luc,
Very good progress indeed, thanks for showing it. I would suggest one thing if you agree in connection with the load tests. You showed the 128 mA current draw from the 24.7 V input voltage for one of the 3 phases and showed also this current increased a little (to 132 mA or so) when you mechanically loaded the shaft.
What I would suggest is to make an adjustable and dependable mechanical load for the shaft, better than your fingers and set a given load on the shaft and logging the loaded rpm and current draw and THEN compare these data to the ones obtained with the assist motor also working. I guess your fingers and ears are 'calibrated' to a certain degree by your excellent experience but perhaps a better brake on the shaft with repeatable "grip" force would be better. ;)
Thanks, Gyula
Quote from: gyulasun on February 23, 2016, 03:46:42 PM
Hi Luc,
Very good progress indeed, thanks for showing it. I would suggest one thing if you agree in connection with the load tests. You showed the 128 mA current draw from the 24.7 V input voltage for one of the 3 phases and showed also this current increased a little (to 132 mA or so) when you mechanically loaded the shaft.
What I would suggest is to make an adjustable and dependable mechanical load for the shaft, better than your fingers and set a given load on the shaft and logging the loaded rpm and current draw and THEN compare these data to the ones obtained with the assist motor also working. I guess your fingers and ears are 'calibrated' to a certain degree by your excellent experience but perhaps a better brake on the shaft with repeatable "grip" force would be better. ;)
Thanks, Gyula
Thanks for your post Gyula
You're right, the finger load test does not prove much but now that the motor is operating to my liking the next step is real load tests.
Stay tuned
Luc
Here is a quick video of the first load test using a RC motor as an AC alternator which is directly attached to the motor shaft.
I'm a bit at a lost :-\ as to how I can calculate the motor efficiency if there is no change to the input power when the alternator is put under load.
Link to video: https://www.youtube.com/watch?v=Xsgk6CRfMus
If someone can suggest how this can be done that would be helpful.
Thanks
Luc
Hi Luc
amazing work
If i understand well.
1- the motor is idling and spins the generator under no load and the input power is around 9 watts.
2- When you load the generator, the input power to the motor is always around 9 watts, but now the generator outputs 5 watts
So it seems that 5 watts are created at the expense of only a decrease in RPM but not due to an increase in input power???
3- The flyback energy is not used in this test, i mean that you only redirect the flyback energy to the charge battery but not to the génerator or the motor ?
This is a "scratching head" and very interesting test
Thank's for sharing
Laurent
Hi Luc,
I think that regardless of the unchanging input power we could calculate efficiency in a 'normal way'. That is, the total input power is 3 x 9.4 W = 28.2 W (assuming the other two phases consume similarly 383 mA current from 24.3 V battery voltages which is also 9.4W for each).
Then the output power consists of the 5 W AC output from the alternator but we do not know the self efficiency of this alternator so I assume it has say 60%, this means the shaft load on your motor may have been equivalent to 5 W times 1.4= 7W output instead of the 5W dissipated in the 5 Ohm.
And the flyback recovery output which presently charges the other batteries is also an output, and to be more precise here, the charging batteries could be replaced also with real power resistors that would represent equivalent load for the flyback output the batteries do now. So summing up these V2 /Rload outputs with the roughly 7W load on the shaft would give total output. If this sum is equal or higher than the total input power, then you surely have a very decent achievent. 8)
Of course I do not mean to lessen the significance of the unchanging input power under the load of the alternator but this would be really significant if the total ouput would be at least comparable to the total input.
Hi Laurent,
the input power to the motor is 3 times the 9 Watts because the other two phases also consume the same 9 W from the other two battery banks (there are 3 separate battery banks for the input).
Gyula
Quote from: woopy on February 24, 2016, 09:06:27 AM
Hi Luc
amazing work
If i understand well.
1- the motor is idling and spins the generator under no load and the input power is around 9 watts.
2- When you load the generator, the input power to the motor is always around 9 watts, but now the generator outputs 5 watts
So it seems that 5 watts are created at the expense of only a decrease in RPM but not due to an increase in input power???
3- The flyback energy is not used in this test, i mean that you only redirect the flyback energy to the charge battery but not to the génerator or the motor ?
This is a "scratching head" and very interesting test
Thank's for sharing
Laurent
Bonjour Laurent,
I wish it was only 9 Watts total input but it is 9 Watts per phase. So 9w x 3 = 27 Watts total input.
I checked the Flyback Recovery and it is 4 Watts per Phase. So 4w x 3 = 12 Watts total Recovery.
So the motor uses 15 Watts to turn the Alternator under
no load.
When the alternator is
on load, the motor rpm decreases which lowers the frequency of the Flyback so now the Recovery drops to 3 Watts per phase. So 3w x 3 = 9 Watts total Recovery, so we lost 3 Watts with the frequency drop, however, we are delivering 5 Watts on our Alternator load.
Summary: Input is 27 Watts in all conditions
Recovery is 12 Watts (no load)
With Alternator on load delivers 5 Watts
Recovery is 9 Watts (on load)
So the motor with NO load is 27w - 12w recovery = 15 Watts
and with motor ON load it is 27w - 14w recovery = 13 Watts
Hope this helps better explain the results?
This is just a first basic test and I will get some more precise measurement data in the next several days.
Luc
Quote from: gotoluc on February 24, 2016, 10:30:03 AM
So the motor with NO load is 27w - 12w recovery = 15 Watts
and with motor ON load it is 27w - 14w recovery = 13 Watts
Luc
:) so it is looking to be 50% more efficient at least , at least 50% recovery?
and over 50% recovery in the second instance.
are there tweaks which you could to get this even better, not that this is not amazing enough,
great work Luc
Quote from: gotoluc on February 24, 2016, 10:30:03 AM
Bonjour Laurent,
I wish it was only 9 Watts total input but it is 9 Watts per phase. So 9w x 3 = 27 Watts total input.
I checked the Flyback Recovery and it is 4 Watts per Phase. So 4w x 3 = 12 Watts total Recovery.
So the motor uses 15 Watts to turn the Alternator under no load.
When the alternator is on load, the motor rpm decreases which lowers the frequency of the Flyback so now the Recovery drops to 3 Watts per phase. So 3w x 3 = 9 Watts total Recovery, so we lost 3 Watts with the frequency drop, however, we are delivering 5 Watts on our Alternator load.
Summary:
Input is 27 Watts in all conditions
Recovery is 12 Watts (no load)
With Alternator on load delivers 5 Watts
Recovery is 9 Watts (on load)
So the motor with NO load is 27w - 12w recovery = 15 Watts
and with motor ON load it is 27w - 14w recovery = 13 Watts
Hope this helps better explain the results?
This is just a first basic test and I will get some more precise measurement data in the next several days.
Luc
Hey Luc
Seems odd that recovery from the motor, with the other motor added, is less, because the switching on times should be longer if the motor is slowed down by the other loaded motor. Not sure. Maybe Im not fully understanding. Just thinking longer on time for the coils shouldnt end with less bemf than free running short pulses. Possibly the number of pulses per second being less as the rpm is lower is the culprit as in total bemf output. Like if you end up with the same out per pulse, whether it is high or low rpm, the high will have more pulses equaling more out over time.
But neat stuff to say the least. ;)
Mags
Quote from: Over Goat on February 24, 2016, 11:33:43 AM
:) so it is looking to be 50% more efficient at least , at least 50% recovery?
and over 50% recovery in the second instance.
are there tweaks which you could to get this even better, not that this is not amazing enough,
great work Luc
I'm sure Electrical Engineers would not calculate it quite that way ;)
The final concept is to use the recovery power which now seems to be around 50% and put it back as torque in the Motor shaft by using a 2nd assist motor.
Only when that is done will we know if my idea has any value. We are not there yet and before doing that we must get some consistent and accurate load test results with the motor on its own so we have something to compare to once we hook up the assist motor.
Thanks for your supportive post.
Luc
Quote from: Magluvin on February 24, 2016, 12:23:03 PM
Hey Luc
Seems odd that recovery from the motor. with the other motor added, is less, because the switching on times should be longer if the motor is slowed down by the other loaded motor. Not sure. Maybe Im not fully understanding. But neat stuff to say the least. ;)
Mags
Hey Mags,
it's only odd because of the conventional way we are use to seeing motor commutation.
The way this switch (commutating) circuit works is, the pulse width stays consistent no matter the rpm, so if we reduce the motors rpm by applying a load we are reducing the frequency of the coils on times, which also reduces the flyback frequency.
In a standard motor commutation, as you reduce the motors rpm the coil on time increases because the motor coils pulse width (on time) get wider as the rotor is turning slower, so the flyback in this case would increase.
Let me know if you have any other questions or need additional details.
Luc
Howdy Luc
that was a very nice piece of work removing those 4 areas on your motor shaft.Is there a reason why channel 1 seems to hold the 40 volts thru out the latest 2 video's,while channel 2 and 3 vary quite a lot? also if you used a different alternator and still used the same 5 ohm resistor would the numbers change?new to this and easily confused :-\
thanks bill
Quote from: billbailey on February 24, 2016, 04:16:22 PM
Howdy Luc
that was a very nice piece of work removing those 4 areas on your motor shaft.Is there a reason why channel 1 seems to hold the 40 volts thru out the latest 2 video's,while channel 2 and 3 vary quite a lot? also if you used a different alternator and still used the same 5 ohm resistor would the numbers change?new to this and easily confused :-\
thanks bill
Hi Bill,
The Scopes channel 1 is measuring current (across a 0.01 Ohm resistor) of all 3 phase.
The other 3 scope channels are measuring the voltage (input and flyback) across each motor coil.
A different kind of alternator would change the results. Each have their own characteristics as far as voltage and current output at a certain rpm.
If what you are asking is if I hooked up an alternator that produced more output, would the motor sustain it and the answer is no, there would be a point where the load would stop the motor. My best guess is, we may get another 1 or 2 Watts more before the motor stops.
Hope this answers your questions
Luc
Okay, here is the official load tests comparing the stock motor & circuit to the modified motor & new circuit.
I have done my level best to keep the tests as fair as possible.
Link to test video: https://www.youtube.com/watch?v=H55BVbgDW0E (https://www.youtube.com/watch?v=H55BVbgDW0E)
Test results:
Stock Motor
At 730 rpm it consumes 21.85 Watts (no load)
At 725 rpm it consumes 29.29 Watts (with load)
so 29.29W - 21.85W = 7.44W used from input to to deliver 1.75 Volts on 1 Ohm resistor = 3 Watts
Mod Motor
At 730 rpm it consumes 6 Watts (no load)
with 1.66 Watts recovered from Flyback
so 6W - 1.66W = 4.34 Watts used from input
At 725 rpm it consumes 16.64 Watts (with load)
with 4.49 Watts recovered from Flyback
so 16.64W - 4.49W = 12.15 Watts used from input
so 12.15W - 4.34W = 7.81W used from input to deliver 1.75 Volts on 1 Ohm resistor = 3 Watts
It is comforting to see the Mod Motor has not lost its electrical to mechanical efficiency with the modifications.
Now the next step is to add the assist motor to see if the flyback is best used to assist the motor or charge a battery.
Any bets?
Luc
Looking good Luc. I really enjoyed the video. It was very informative. I suspect you are not getting a lot of comments on this thread because most of us like myself feel anything we could add would not be any better than what you are doing so we just watch and read. I am looking forward to your next steps. My own projects have usually shown it is more beneficial to use the flyback to charge a battery. But your builds are much better than mine so you may find it more beneficial to use them for the assist motor. So no bets either way from me.
Carroll
Quote from: citfta on February 25, 2016, 09:15:42 PM
Looking good Luc. I really enjoyed the video. It was very informative. I suspect you are not getting a lot of comments on this thread because most of us like myself feel anything we could add would not be any better than what you are doing so we just watch and read. I am looking forward to your next steps. My own projects have usually shown it is more beneficial to use the flyback to charge a battery. But your builds are much better than mine so you may find it more beneficial to use them for the assist motor. So no bets either way from me.
Carroll
Thanks Carroll for the appreciation and taking the time to write a few lines.
Well, we'll soon find out what's best to do with flyback. I don't know the answer yet! so even I wouldn't know what side to bet on.
I'll just give it my best shot and we'll see what happens.
Luc
Hi Luc,
I just noticed your comment under the youtube video you did a quick test with the stock motor run with one phase and got about the same power differences. This is good to know too.
Just an observation: it is interesting that the modified motor did not maintain the same input power draw with the one phase drive like it did with the three phase drive. I mean the test you reported in Reply #808 where there was no change in input power when the alternator was put under load. Or the much higher RPM (at about 1800) in that 3 phase test can explain the no change I wonder (coils inductive reactance are higher at higher RPM).
Thanks
Gyula
Quote from: gyulasun on February 26, 2016, 07:06:52 AM
Just an observation: it is interesting that the modified motor did not maintain the same input power draw with the one phase drive like it did with the three phase drive. I mean the test you reported in Reply #808 where there was no change in input power when the alternator was put under load. Or the much higher RPM (at about 1800) in that 3 phase test can explain the no change I wonder (coils inductive reactance are higher at higher RPM).
Thanks
Gyula
Hi Gyula,
the reason for that is I wanted to keep RPM the same between the stock motor and mod motor.
The stock motor monitors the rpm, so when a load hits the shaft it instantly dumps more PWM pulses (more current) to maintain the rpm.
The mod motor has a fixed pulse width (current) that I manually adjust and why I had to stop the camera recording to add the load and readjust the pulse width and voltage so it could maintain the same rpm under load as the stock motor so we have a comparable test.
Hope this clears it up? let me know if there is anything that does not make sense.
Luc
Hi Luc,
That has been okay what you wrote as the reason, I just wanted to draw attention to the lack of similar behavior for the modified motor i.e. the power draw increased when you loaded the alternator while the power draw did not increase in the earlier test of the modified motor when you loaded its shaft with the alternator. And the difference between the two tests was only in the RPM and in the number driving phases. This is all I wished to point out.
Gyula
@gotoluc
I am posting this so this thread does not disappear. I had made a post but again, thinking it may not be what you want to read but will post it anyways.
@gotoluc
Few questions please. Did you pay attention to which wires are going to the 1st layer for each of those 9 motor coils?
Also did you jot down those same orientations when you opened the original model before you modified them so you know how your stock model is wired? Without that, it's hard to follow or compare. I understand the experiments but cannot place them in context without knowing how the coils are precisely connected. Seems to me that is the most important part of these tests is to understand those precise relations.
Maybe look at a few of my Half Coil Syndrome videos since they also apply to motors. Motors suck but you have a good chance to test a theory. Just add a transformer primary in series to your 3 pulsed motor coils (1 phase). In series on the non-pulsed side meaning the side of your motor coils that is opposite the mosfet. Then apply the same juice you applied to test 1 and see the rpm. Just use the primary of a real intact MOT (hahaha). Do your measurements on the motor and on the MOT secondary and seeeeeeeee if there is better overall performance. The MOT secondary could be dumped into a good sized 250v capped diode. Also, see the difference of the flyback level. HCS sucks in all our single pulsed coils but there are ways to beat it like instead of beating it or neutralizing it you let it happen outside of the "working primary" environment, leaving the working primary to react or change polarity throughout its coiling thus imparting more to the core and more to the effect. I call it using a Slave Coil to lessen the primary burden, could be only a coil or another primary, depending on if you need to recoupe some of the input and put it back to work like you are looking to do. The MOT is just a fast idea but you may have better transformer coils laying around so the idea is it's open to a wide ranging effect.
Orrrrrrrrrr. Try using the other 6 motor coils as the slave coil. But you need to answer the first question first since you absolutely need to have full control of how the coils are wired up and 1st layer versus outer layer is important to know at all times. I am sure when they made the stock motor they applied full attention to their winding method. Maybe draw 2 rows of 9 dots makes your 18 conductors and say which is going to 1st and which is going to last of each coil.
Anyone who plays with coils on a core needs to always know which lead is which. If you are not conscious of that point, you will not play those variables and you will not learn their effects.
Always great work man.
wattsup
PS: Mind exercise: With a good sized DC motor and a torque measurement device (pony break or other), you apply DC voltage, positive to the red motor lead and negative to the black motor lead. You measure the torque. You then apply the reverse voltage to the motor, positive to black and negative to red wires of the DC motor. Now the motor turns in reverse. Question: Will the torque be the greater, the same or less then when it is driven the first way. WHY????????? Can this reasoning be applied to @gotolucs' modified motor? Hmmmmmmmm.
Quote from: wattsup on March 02, 2016, 10:16:00 AM
Did you pay attention to which wires are going to the 1st layer for each of those 9 motor coils?
Correction, the Stator is 12 coils (4 coils per Phase)
Quote from: wattsup on March 02, 2016, 10:16:00 AM
Also did you jot down those same orientations when you opened the original model before you modified them so you know how your stock model is wired? Without that, it's hard to follow or compare. I understand the experiments but cannot place them in context without knowing how the coils are precisely connected. Seems to me that is the most important part of these tests is to understand those precise relations.
Yes, for each phase the stock motor has 2 coils connected in series then the two are connected in parallel.
Quote from: wattsup on March 02, 2016, 10:16:00 AM
Maybe look at a few of my Half Coil Syndrome videos since they also apply to motors. Motors suck but you have a good chance to test a theory. Just add a transformer primary in series to your 3 pulsed motor coils (1 phase). In series on the non-pulsed side meaning the side of your motor coils that is opposite the mosfet. Then apply the same juice you applied to test 1 and see the rpm. Just use the primary of a real intact MOT (hahaha). Do your measurements on the motor and on the MOT secondary and seeeeeeeee if there is better overall performance. The MOT secondary could be dumped into a good sized 250v capped diode. Also, see the difference of the flyback level. HCS sucks in all our single pulsed coils but there are ways to beat it like instead of beating it or neutralizing it you let it happen outside of the "working primary" environment, leaving the working primary to react or change polarity throughout its coiling thus imparting more to the core and more to the effect. I call it using a Slave Coil to lessen the primary burden, could be only a coil or another primary, depending on if you need to recoupe some of the input and put it back to work like you are looking to do. The MOT is just a fast idea but you may have better transformer coils laying around so the idea is it's open to a wide ranging effect.
I'll let you know if I find anything interesting.
Quote from: wattsup on March 02, 2016, 10:16:00 AM
Orrrrrrrrrr. Try using the other 6 motor coils as the slave coil. But you need to answer the first question first since you absolutely need to have full control of how the coils are wired up and 1st layer versus outer layer is important to know at all times. I am sure when they made the stock motor they applied full attention to their winding method. Maybe draw 2 rows of 9 dots makes your 18 conductors and say which is going to 1st and which is going to last of each coil.
Anyone who plays with coils on a core needs to always know which lead is which. If you are not conscious of that point, you will not play those variables and you will not learn their effects.
Every single wire is labeled with a number 1 to 12 corresponding to the numbers I physically labeled the stator with. You should be able to see the labels and numbers in my demo videos.
Quote from: wattsup on March 02, 2016, 10:16:00 AM
PS: Mind exercise: With a good sized DC motor and a torque measurement device (pony break or other), you apply DC voltage, positive to the red motor lead and negative to the black motor lead. You measure the torque. You then apply the reverse voltage to the motor, positive to black and negative to red wires of the DC motor. Now the motor turns in reverse. Question: Will the torque be the greater, the same or less then when it is driven the first way. WHY??? Can this reasoning be applied to @gotolucs' modified motor? Hmmmmmmmm.
No, you cannot reverse the direction of my motor by swapping the leads.
Regards
Luc
Hi everyone,
a small update on the testing.
For those who have been following the topic you would know I've recently modified the 8 segment rotor to a 4 segment.
After many days of testing each rotor version under load, I can now confirm the 8 segment is 10 to 20% more efficient.
It was important to confirm this before moving on to the flyback assist motor.
Hopefully next week I should have the assist motor attached for its first tests.
Luc
Here are the revised test data.
First are original tests then the revised (8 segment rotor) test results.
Original Test results
Stock Motor
no load the motor consumes 21.85 Watts at 730 rpm
with load the motor consumes 29.29 Watts at 725 rpm
the difference between 29.29W (on load) - 21.85W (off load) = 7.44W used to deliver the 3W load (1.75 Volts on 1 Ohm = 3 Watts)
Mod Motor
no load the motor consumes 6 Watts at 730 rpm
and 1.66 Watts is recovered from Flyback
so 6W - 1.66W = 4 .34 Watts used to turn the motor with no load at 730 rpm
with load the motor consumes 16.64 Watts at 725 rpm
and 4.49 Watts is recovered from Flyback
so 16.64W - 4.49W = 12.15 Watts used from input
the difference of 12.15W (on load)- 4.34W (off load) = 7.81W used to deliver the 3W load (1.75 Volts on 1 Ohm = 3 Watts)
New Mod Motor test results with 8 segment rotor
Mod Motor
no load the motor consumes 6 Watts at 730 rpm
and 2 Watts is recovered from Flyback
so 6W - 2W = 4 Watts used to turn the motor with no load at 730 rpm
with load the motor consumes 15 Watts at 725 rpm
and 5 Watts is recovered from Flyback
so 15W - 5W = 10 Watts used from input
the difference of 10W (on load)- 4W (off load) = 6W used to deliver the 3W load (1.75 Volts on 1 Ohm = 3 Watts)
These revised results beat the stock motor by 1.44 Watts
Now we can attache the assist motor which will use the 5 Watts flyback recovery. So with the combined motors we need to achieve 10 Watts (or less) in consumption while delivering the 3 Watts to the load to prove if this concept is worthy.
Luc
Here are the final and best results I was able to obtain.
What I decided to do is recirculate the 5 watts of flyback back in the same motor by using one of the two extra phases. So one phase is the input and its 5 watts flyback is collected and re-switched back in the second phase which also has a flyback collected and loaded across a 100 Ohm resistor to calculate the final output power.
So with the second phase assisting the input the first phase dropped from our previous best score of 15 Watts down to 10.69 Watts and we also collect the flyback from the second phase which is 13 volts DC across a 100 Ohm load resistor = 1.69 Watts.
So the Mod Motors with flyback assist is down to 9 Watts while under the same load of turning the alternator which delivers 3 watts across a load resistor.
This represents a additional 10% efficiency improvement over the previous mod motor test of just collecting flyback.
I must admit I thought a 30% improvement would be possible but still 10% is probably better then some may of thought possible.
For those who do not quite understand the results, here's the bottom line.
The stock motor uses 2.44 Watts more to do the same work (turn alternator with 3 watts load) as the final modified motor.
This makes the mod motor 21% more efficient then the stock motor.
The 10% saving I mentioned above is only the second stage modification in efficiency boost.
Hope the results are clear and understandable
It's been an educational experience and hope it's been the same for others who have been following the topic.
Regards
Luc
Hi Luc,
I think it is okay you did not choose to use an assist motor but use another stator coil on the same stator to further drive the rotor by the collapsed energy of the first stator coil (if this is correct).
Even if you expected a bit higher improvement I think the 10% efficiency increase is still nice in your present series of tests and this could surely be improved by running the setup at a higher RPM (the lower than 1000 RPM was a compromise due to the comparison possibility to the unmodified motor which lacks its RPM control circuit).
So if one of the goals was to prove whether the collapsed energy of a switched stator coil when collected could aid the overall performance of this motor, then you proved it surely could, right?
Thanks for showing these experiments.
Gyula
Quote from: gyulasun on March 09, 2016, 09:04:10 AM
I think it is okay you did not choose to use an assist motor but use another stator coil on the same stator to further drive the rotor by the collapsed energy of the first stator coil (if this is correct).
Yes, you have it correct.
The reason for deciding to simply use another phase for the flyback was I had 2 phases available and the other main reason is, if I attached a separate motor to assist, we are only dealing with 5 watts of flyback power, so I'm quite sure the 5 watts would of been consumed by the motors baseline losses and in the end not demonstrate any advantage.
One way or the other if I was to design a motor using this concept, I think it would be more efficient to have the assist coils within the same motor.
So what mostly needs to be done is design a stable switching circuit to accomplish this.
I think it's doable with all the component improvements we have had these recent years.
Quote from: gyulasun on March 09, 2016, 09:04:10 AM
Even if you expected a bit higher improvement I think the 10% efficiency increase is still nice in your present series of tests and this could surely be improved by running the setup at a higher RPM (the lower than 1000 RPM was a compromise due to the comparison possibility to the unmodified motor which lacks its RPM control circuit).
Yes, I agree! it's unfortunate I don't have the communication protocol to increase to rpm of the stock motor. The results could be better at higher rpm and under much larger loads?
Quote from: gyulasun on March 09, 2016, 09:04:10 AM
So if one of the goals was to prove whether the collapsed energy of a switched stator coil when collected could aid the overall performance of this motor, then you proved it surely could, right?
Yes, I am definitely convinced that a
switched coil motor performance can be improved by at least 10% (or more) by collecting and re-using its flyback to assist.
If anyone is ready to assist with a $200. donation, I could buy a used Maytag Neptune front loader washing machine which is what these motor are from. This would give me all the circuits to operate the motor at higher rpm which would produce much more test data then what I presently have.
Luc
Hi everyone,
I'm building a larger version of one of the concepts I demonstrated in late January. Link to the January video: https://www.youtube.com/watch?v=FxSccG5DNFM
But it takes time to get parts when you're on a small budget and need to order from China but I may have an update in a week or so.
However, today I received some other parts to test a concept that Robert Murray Smith shared in early February. His video demo: https://www.youtube.com/watch?v=n4YD8Nvyfa4
Here is my version of it: https://www.youtube.com/watch?v=0VlhMI5tv5Y
Luc
Hi Luc, That is very interesting , Tesla showed something similar, I'll have to go back and look for it.
I'm curious as to what pole gets projected out the ends of the I's without putting the keeper on top?
Also instead of draining that cap through the resistor use it to fire your coil and by doing so collect the flyback again and keep using it , it will diminish over time but thats when you hit it with a battery pulse to bring it back to the start.
Can just regular ceramic magnets be used ,I don't have any others except one big neo ring magnet?
Thanks for sharing.
artv
Nice work once again gotoluc!
On the Robert Murray device, I wonder how much shorter you can cut the Alnico magnet? I wonder if soft iron spacers will let you get away with a shorter Alnico and smaller coil so you can use less power to switch it?
Quote from: shylo on March 25, 2016, 05:27:52 AM
I'm curious as to what pole gets projected out the ends of the I's without putting the keeper on top?
When the flux projects out it's the North and South of the Neo that goes through the I's. At this point the Alnico is in opposite poles and why the Neo's flux is projected out.
Quote from: shylo on March 25, 2016, 05:27:52 AM
Also instead of draining that cap through the resistor use it to fire your coil and by doing so collect the flyback again and keep using it , it will diminish over time but thats when you hit it with a battery pulse to bring it back to the start.
Yes, flyback could be used to assist but it needs about 2.3 Joules of energy to magnetize the Alnico and each flyback recoveries are only about 0.1 Joule.
Quote from: shylo on March 25, 2016, 05:27:52 AM
Can just regular ceramic magnets be used ,I don't have any others except one big neo ring magnet?
I don't know and haven't tried it with ceramic magnets. Give it a try and see.
Luc
Quote from: MagnaProp on March 25, 2016, 06:14:50 AM
Nice work once again gotoluc!
On the Robert Murray device, I wonder how much shorter you can cut the Alnico magnet? I wonder if soft iron spacers will let you get away with a shorter Alnico and smaller coil so you can use less power to switch it?
I think it's all relative, less magnet mass = less flux strength. As it is, if I shorten the pulse to the Alnico I get less magnetization force since some of the Neo's flux conducts through the Alnico if it's less magnetized. A longer pulse creates the opposite but obviously to a certain point, as once you reach the Alnico's maximum magnetization any more input is a waste of power.
Luc
Does it take more energy to reverse the Alnico when the neo is uses as apposed to using no neo? I'm assuming it does.
Quote from: gotoluc on March 25, 2016, 09:56:54 AM
I think it's all relative, less magnet mass = less flux strength...
Good point. If we cut the Alnico to short then it won't be strong enough to stop the neo magnetic flow. Hmmm
When the flux projects out it's the North and South of the Neo that goes through the I's. At this point the Alnico is in opposite poles and why the Neo's flux is projected out.
So on 1 side of the coil 1 I is n , and the other I is s?
It projects one pole on one , and the opposite pole on the other?
I'm using stored flyback to fire my coils to cause rotation, this might give better acceleration.
But like you say the amount of stored flyback is small ,compared to what you need to input for drive.
Every added coil produces alot more flyback, it's a doubling effect.
Looking forward to your further test's.
Thanks your a great contributor.
artv
Hello Luc, thanks for your experiment sharing :)
Don't we have to know the mass of the magnetic shunt and check at which height the U shape will lift it completely?
Let's say the barre is 100 g and you lift it up to stick it to the U shape, right when you approches it at 1 cm from the top while place on the table:
Weight Potential Energy in Joules =
{Mass in Kg} * {Gravity in m.s^-2} * {Distance, or Length, in meters}
= M [kg] * g [m.s^-2] * L [m] = W [J]
Let's say:
M = 200 g = 0.2 kg
g = 10 m.s^-2 (we only seek about orders of magnitude, so don't care for the "cents" )
L = 1 cm = 0.01 m
W [J] = 0.2 kg * 10 m.s^-2 * 0.01 m = 0.02 [J]
2 [J] of consumption of an impulse minus the flyback recovery (0.1 J), divided by 0.02 [J] of lifting energy = 100.
Thus the efficiency would be the inverted ratio: 1/100 = 1 %.
Thus the limit distance for over unity efficiency would be around, for 200 g of magnetic shunt:
L [m] = W [J] / (M [kg] * g [m.s^-2])
= 2 [J] / (0.2 [kg] * 10 [m.s^-2])
= 1 [m]
Am I missing something? Am I wrong your your numbers or my calculations?
Regards,
Didier
Hi Didier, what if we consider magnetization time and and make the rotor and stator segment very long.
In most electric motors the rotor and stator segments are at most 30 degrees or less in length. So if you build it that way I don't see an advantage. But what if the segments were 90 degrees long, since once the magnetization has occurred there is no extra cost to attract a very long rotor segment, no?
If you wish I can give you some measurements on how much weigh the magnetization can lift including a standard motor air gap.
Let me know
Luc
Quote from: gotoluc on March 26, 2016, 09:24:28 PM
Hi Didier, what if we consider magnetization time and and make the rotor and stator segment very long.
In most electric motors the rotor and stator segments are at most 30 degrees or less in length. So if you build it that way I don't see an advantage. But what if the segments were 90 degrees long, since once the magnetization has occurred there is no extra cost to attract a very long rotor segment, no?
Hi Luc, as said elsewhere, I am not skill enough in this area to answer you with any certainty.
But what about just two rotating barres (for the balance) with 4 reluctant switched coils magnets as stator at 90°, that you can change with ordinary electromagnets, wouldn't be easier to test?
Quote
If you wish I can give you some measurements on how much weigh the magnetization can lift including a standard motor air gap.
Let me know
Luc
In my humble opinion, it would be the minimum do to in any case.
Thanks for your Interest, Luc, for your experiments,
Regards,
Didier
Yes Didier, your test suggestion is good and I may do something like that if it is needed.
However, as you may know Robert Murray-Smith (https://www.youtube.com/user/RobertMurraySmith) is presently looking into the matter and I'll wait till he posts his new video demo with his calculations.
Thanks for your interest
Luc
Hi Luc, I did a test where I put 4 I strips on my big neo (2 on each side) to match the length of a secondary coil from a MOT (on it's edge),and placed ceramic's between the I's in the hole of the secondary. Then I put another I across the ends of the 2 on each side. It still has 2 poles ,but when the coil is pulsed , it makes the poles change.
The way you have it , it's either magnetic or it's not?
A very weak field but it is there.
artv
Quote from: shylo on March 28, 2016, 07:07:09 PM... It still has 2 poles ,but when the coil is pulsed , it makes the poles change.
The way you have it , it's either magnetic or it's not?
...
Can you draw a simple image of what you have done? I can't tell from the written description. Thanks in advance for any help. I'm looking for the switching of poles as you describe. The best I can come up with so far is basically two of gotoluc's devices but that would be twice as heavy with only half of the magnets being used at any given time. Would also take more power to switch two Alnico's.
@shylo:
I too would like to see a drawing or picture of how you
QuoteIt still has 2 poles ,but when the coil is pulsed , it makes the poles change.
truesearch
Here's a different one seems to work better.(see attached)
I'm going to try and see if I can maybe use it to assist rotation of my rotor.
I was just powering it with some of the stored flyback.
artv
Quote from: shylo on March 30, 2016, 04:26:37 AM
Here's a different one seems to work better...
Thanks for the image.
You're using all Alnico magnets and no Neo's? So you switch the poles by switching all magnets?
Quote from: MagnaProp on March 30, 2016, 06:13:58 AM
You're using all Alnico magnets and no Neo's? So you switch the poles by switching all magnets?
No, he wrote he is using all ceramic magnets.
Luc
The poles at the end of the I's are opposite each other, N & S, when pulsed they switch polarity.
The big ring is a neo,
The little fillers in the center of the coil are ceramic's.
It makes for a weak field but the closer the better.
artv
Quote from: shylo on March 30, 2016, 03:47:09 PM
...It makes for a weak field but the closer the better...
Still not sure how it works. When you switch the ceramic magnets, instead of closing the loop with the neo, the ceramic magnets over power the round neo to the point that more ceramic field exits the I instead of the neo field?
If I still have it wrong, then I'll leave it at that and thank you for posting the info. @gotluc- Thanks for the clarification.
There is no way the little chunks of ceramic (cut up speaker magnets) could over power that neo ,it is very strong.
I put one I-core on one side of the neo, lay the coil on, place the ceramics in the hole of the coil , then put the other I on.
The ends of the I's each have their own pole a N & S, power the coil and the poles flip at the ends of the I's.
Without powering the coil , if you bring a magnet of the same pole close to the I, you can feel the repulsion , power the coil and the magnet gets sucked in.
Pretty simple to test, I'm gonna try it without even using the ceramics' , maybe even placing more I's all around.
Hope that's clear enough.
artv
The ideal way to get the strongest flux out of this design would be to use only neo magnets. However, to re-magnetize (flip the pole) of a neo magnet would require a very large amount of power. So there would be no point to even try this.
Alnico magnets are a step down from neo's. So if you compare an equal size alnico to a neo, the alnico would be about half the strength of the neo. So this is part of what makes the alnico easier to re-magnetize compared to the neo.
Now if you compare equal size ceramic to a alnico, the ceramic would be about half the strength of the alnico.
So syhlo, I hope you understand that this is the reason you're not having much success with your test device, as you are using magnets of two different extremes. Not only that but the neo looks to be larger in size then the ceramic magnets.
You may want to consider using all ceramics and keep the size ratio the same. Or try 2 to 3 times the size (mass) of ceramic to size ratio of neo.
Luc
He's right:
The coercivity of AlNiCo magnets is much lower than NdFeB magnets.
Also, NdFe magnets have a higher remanent flux density than AlNiCo magnets and ceramic magnets.
However what is not commonly known, is that AlNiCo magnets have lower coercivity than ceramic magnets, despite having higher remanent flux density.
Last, but not least, AlNiCo magnets and NdFeB magnets both conduct electricity and that makes them suffer from Eddy current losses whenever flux changes.
Thanks verpies for posting the information.
This link http://www.duramag.com/alnico-magnets/available-alnico-magnet-grades/ (http://www.duramag.com/alnico-magnets/available-alnico-magnet-grades/) was posted by a youtube member. It contains details on the different Grades of Alnico manufacturing process.
He also wrote:
"Just a note, it appears that Alnico magnets require a length to diameter ratios of at least 4:1. Apparently if they are shorter relative to diameter they don't function as well. I noticed that you cut down your magnets which may be affecting the performance.
I would like a second opinion on the above.
Also, if you can think of any other kind of material which could be re-magnetized easier then Alnico please let us know.
Luc
@shylo:
Maybe I'm not understanding your magnet set up completely, but it reminds me of Flynn's switchable / parallel fux path device.
Isn't it simular to this?
trueseach
Quote from: truesearch on April 04, 2016, 11:53:14 AM
@shylo:
Maybe I'm not understanding your magnet set up completely, but it reminds me of Flynn's switchable / parallel fux path device.
Isn't it simular to this?
trueseach
shylo has most likely based his design after I shared this video from the previous page: https://www.youtube.com/watch?v=0VlhMI5tv5Y
The operating principle is explained and demonstrated in my video.
Shylo has tried to replicate it with ceramic magnet instead of an Alnico. However, as I have explained above it won't give the best results.
Luc
Hi truesearch, Luc is correct ,I simply tried his test but,I don't have any Alnico's.
What are these Alnico commonly used in, I would like to salvage some.
Verpies, thanks for that info good to know.
Luc in the setup I had the ends of the I's have a constant field when the coil is powered the field reverses,but it doesn't shut off.
Your's shuts off the field or should I say blocks it?
Just wondering why yours shuts off whereas my just flips?
Think I'll try it without the neo but then basically it's just an electro-magnet right?
Thanks artv
Quote from: shylo on April 05, 2016, 03:57:41 AM
Hi truesearch, Luc is correct ,I simply tried his test but,I don't have any Alnico's.
What are these Alnico commonly used in, I would like to salvage some.
Verpies, thanks for that info good to know.
Luc in the setup I had the ends of the I's have a constant field when the coil is powered the field reverses,but it doesn't shut off.
Your's shuts off the field or should I say blocks it?
Just wondering why yours shuts off whereas my just flips?
Think I'll try it without the neo but then basically it's just an electro-magnet right?
Thanks artv
I thought I explained why your results are not so good in this post: http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg479140/#msg479140
Basically your neo is too large and your ceramics are too weak.
Alnico magnets are not commonly used. You will have to buy them. Here is the one I purchased: http://www.ebay.com/itm/172050785362
Luc
Quote from: gotoluc on April 05, 2016, 01:10:15 PM
...Basically your neo is too large and your ceramics are too weak...
We don't understand why this would cause his poles to switch as shylo claims.
If the neo is too large and the ceramics too weak, then we would expect the poles to stay in the same place and not switch sides. The poles from the neo might be a little weaker as some neo flux is channeled through the ceramics, but the dominate poles from the neo shouldn't be able to change positions.
The only way I see the poles switching are if the ceramics were able to over power the neo or the neo its self is being switched. Both of which shouldn't be possible in shylo's set up. Perhaps shylo's neo is an alnico and he doesn't know it? He could test this easily enough by removing the ceramics and see if his coil switches his neo.
////Edit
Quote from: gotoluc on April 05, 2016, 03:57:26 PM
...So shylo's device the I's are always magnetized because the Neo is overpowering and what I wrote previously...
I agree with all you wrote. However, not only are the I's always on but the strange part is that the polarity of the I's are switching.
He says it's a weak effect so I'll just chalk it up to a mystery of life.
Quote from: MagnaProp on April 05, 2016, 02:05:22 PM
We don't understand why this would cause his poles to switch as shylo claims.
If the neo is too large and the ceramics too weak, then we would expect the poles to stay in the same place and not switch sides. The poles from the neo might be a little weaker as some neo flux is channeled through the ceramics, but the dominate poles from the neo shouldn't be able to change positions.
The only way I see the poles switching are if the ceramics were able to over power the neo or the neo its self is being switched. Both of which shouldn't be possible in shylo's set up. Perhaps shylo's neo is an alnico and he doesn't know it? He could test this easily enough by removing the ceramics and see if his coil switches his neo.
Is it me that doesn't understand?
In shylo's device the neo is overpowering the ceramic magnets in both ways. They cannot store the neo's flux, because if they did this is the time the
I's are NOT magnetized. When he tries to re-magnetize the ceramics, again they cannot repel the neo's flux as they are too weak as this would be the time the
I's ARE magnetized.
So shylo's device the I's are always magnetized because the Neo is overpowering and what I wrote previously.
This is how I understand it to work, sylo is free to correct me if I don't understand.
I hopes to clear the confusion.
Luc
I think the ends of the I's switch polarity, is due to the distance.
The field at the ends of the I's is considerably weaker than at the neo itself.
By powering the coil in the opposite polarity the neo's flux flips the ceramics poles and leaves a very weak opposite pole at the ends.
Not sure about that just what I think.
Luc is your magnet in the center of your coil (the Alnico) repelling or attracting to the I that is attached to your neo? My ceramics are attracted , if I try to place them in the repulsion mode the neo is too strong and they won't even stick to the I they just flip back to attraction.
artv
Quote from: shylo on April 05, 2016, 04:29:27 PM
Luc is your magnet in the center of your coil (the Alnico) repelling or attracting to the I that is attached to your neo? My ceramics are attracted , if I try to place them in the repulsion mode the neo is too strong and they won't even stick to the I they just flip back to attraction.
artv
Yes, as explained in the video the Alnico is the magnet in the center of the coil.
The operating principle is the same as a permanent magnet electromagnet flux gate. However, the difference here is, instead of the coils core to be Iron, it's an Alnico which get re-magnetizes by a single pulse of the coil and holds the magnetization without additional power.
You can consider it a latching permanent magnet flux gate.
Hope this makes it clears now?
Luc
Dear Luc,
I was thinking about your system and I am afraid that the energy and power needed to demagnetise the ordinary magnet is of to overcome its own magnetisation PLUS the flux of the neo which is passing through (which would explain why so less efficiency).
Thus, I wouldn't be surprised if you would find similar result without an ordinary magnet but just a steel bar instead; at least you wouldn't have to overcome the flux of the ordinary magnet.
Regards,
Didier
I think what you say may be the case Didier.
the concept is something Robert Murray Smith shared. After I posted the video demo he said he would do the power calculations to see if it's worth pursuing. Waiting for his results to further tests.
Luc
Hello Dear Luc, thanks for your reply.
yeah, I am afraid indeed to be right :/
Do You have news now from Jim?
Note:
I know You're much about HONEST Free Energy Quest; do You know the work of my Big Friend Kurt? "kdkinen" in YouTube.
Did You see his "wireless system"?
Would You be interested to replicate it and then completely explain it, up to now, he was not able, from my own point of view, to really explain what is going on?
When I say "to explain", I mean mathematically, been able to predict the results thanks to mathematical formulas which would demonstrate a True Accuracy in the understanding.
Quote from: Khwartz on May 07, 2016, 01:12:34 PM
Do You have news now from Jim?
I guess you mean Robert Murray Smith?... if so, no, he has not gotten back to me and it does not look like he has anything new about it.
Luc
All we need to try is a solid state setup like this (in the attachment).
Quote from: kEhYo77 on December 09, 2016, 12:17:07 PM
All we need to try is a solid state setup like this (in the attachment).
Very nice design kEhYo77
Are you going to build it?
Thanks for sharing
Luc
Hi Luc! I got the alnicos and "I" ferrite bars and the output coil so who knows? ;)
That's great!... please keep us updated on your build and tests
Wishing you the very best
Luc