Hi everyone,
it has been a while since I posted new information but yet again here is another interesting effect I have come upon.
As usual, the video explains most of what you would need to know if you would like to try this.
Video Link: http://www.youtube.com/watch?v=STI3koWbzE4
I personally think it is worth much more investigation to come to understand why the BEMF has so much more effect on a magnet when re-circulated back in the coil then the original energy input ???
Could this be what Ed Gray did?
If anyone does understand this and can explain it, please do not hold back!
Thanks for looking
Luc
Hi Luc.
Thanks for the video.
What you are seeing there is a conventional effect.
I am assuming that what you mean by "feeding the inductive kickback (or BEMF as most call it) back into the coil", is that you are placing the diode (flyback diode) across the coil. If so then yes you are recirculating the coil current.
Pulsing an inductive coil is a two phase process; energize, and de-energize. I think WHEN these two phases occur is obvious, but what isn't always obvious is that with a flyback diode across the coil, during the de-energize phase roughly the same amount of current keeps circulating, and in the SAME direction it was in the energize phase.
What this means is that the effective magnetic field produced by the coil is "ON" for roughly twice the amount of time it is when the diode is not present. This would have a marked effect on the neo magnet sitting on top of it, and that effect results in more lift on the magnet.
What happens when the diode is absent from the circuit is the IK results in very little current, but a large inverted voltage spike. This does not produce a magnetic force on the neo magnet because with inductors (coils), the magnetic field strength is proportional to the current through it.
Regarding the current measurement with and without the diode, I doubt either case is accurate with that meter and with the wave forms it's measuring.
Regards,
.99
Hi .99,
thanks for looking and posting your understanding of what is going on in the coil.
I'm looking for others to replicate this simple circuit experiment and hope you have time to try it! as you will find that the current does indeed drop ;).... I have not only used the meter in the video to confirm this. I used that meter in the video because it was simple. Try the experiment and post what you find, just make sure to use short duty cycles to see the effect.
Luc
Quote from: gotoluc on July 02, 2009, 06:24:29 PM
If anyone does understand this and can explain it, please do not hold back!
Thanks for looking
Luc
Hi Luc.
No offense, but there's no need for me to try the experiment :) I have a good understanding of currents in coils and my experiment efforts are already spoken for. I offered to share this knowledge to help explain what you are seeing in your experiment.
Explaining what your meter is interpreting is a little more complex, but I will get to that a little later.
In the mean time I would encourage you to place your scope probe across the meter leads in order to get an idea what the meter is reading, or trying to read ;) This might also help make sense of the difference between diode and no diode. Could you post a scope shot?
.99
Luc,
Try this.
Do your test with and without the diode feedback again, but this time hold the magnet down against the coil (so it can not vibrate or float) for both.
Observe the current reading on the meter. Does it still read lower for the case where the diode is in-circuit?
.99
Hi .99 and everyone,
@.99, the test you have suggested have been done before I started this topic. I have been testing this for the past 7 days. I do appreciate your input and have made a new video to which I dedicated a large part to demonstrate to you by using a resistive load that the current does indeed drop slightly when the Inductive collapsing field is recirculated back into the coil.
To everyone else, I am not getting much feedback from the rest of you. Is there something that I'm missing or misunderstanding here. Please be honest and share your thoughts :)
New video: http://www.youtube.com/watch?v=_WW8z36n6z8
Luc
Hi Luc
great video
I wonder how much more weight you could put on the magnet ?
may be you could run a piezoelectric
all the best
cat
Hi Cat,
thanks for posting your suggestion ;)
What I'm thinking of making is an electric motor using this principal but this can be applied to many other things.
The point that I'm trying to demonstrate here is this BEMF, Inductive Collapsing Field, Flyback Voltage or what ever you want to call it! has many more times the power to react with a magnet then the energy it takes to create it.
To me, that is amazing enough!... it is real and right in front of us :o
Lets use it!
Luc
Quote from: gotoluc on July 04, 2009, 05:42:53 PM
Hi .99 and everyone,
@.99, the test you have suggested have been done before I started this topic. I have been testing this for the past 7 days. I do appreciate your input and have made a new video to which I dedicated a large part to demonstrate to you by using a resistive load that the current does indeed drop slightly when the Inductive collapsing field is recirculated back into the coil.
To everyone else, I am not getting much feedback from the rest of you. Is there something that I'm missing or misunderstanding here. Please be honest and share your thoughts :)
New video: http://www.youtube.com/watch?v=_WW8z36n6z8
Luc
Hi Luc.
Thanks for doing the second test. You work quickly my friend :)
I never said I didn't believe that the current dropped with the diode in-circuit, but I was trying to see if the magnet itself had anything to do with it. It does not.
I'm happy to report that I was able to confirm almost all your findings, and I will post my results in the next hour or so.
Regards,
.99
Hi .99,
sounds interesting :)
Looking forward to your test results.
Luc
Luc.
In my tests I've confirmed all your findings, except the magnet elevation, but since the magnetic field in a coil is proportional to the current through that coil, I think it's safe to say that is covered as well.
What you've built is essentially a DC-DC converter. You've taken a high voltage low current source and converted it to low voltage / high current in a load. In this case the load and conversion element are one in the same, the coil. The diode completes the appropriate circuit path that allows the conversion to actually take place.
Some numbers I obtained from my tests:
Ave VS Ave IS Ave WS Ave VC Ave IC Ave WC
Without Diode: 170V 15.5mA 2.635W 157mV 15.5mA 2.43mW
With Diode: 170V 10.6mA 1.802W 2.932V 285mA 0.836W
Where:
VS = Supply Voltage
IS = Supply Current
WS = Supply Power
VC = Coil Voltage
IC = Coil Current
WC = Coil Power
(these are averaged values)
Indeed the current from the supply decreases with the diode in-circuit, but the output power from the coil never exceeds the input from the source.
Notice the huge increase in coil current with the diode in-circuit compared with it out of circuit? This explains the force it has on your neo magnet. With coils, it's all about current. A substantial amount of power has been transferred to the coil with the diode present (about 46% of the input), as opposed to nearly none without the diode.
Are there any gains in power or energy? No, in fact there is quite a substantial loss due to the DC resistance of the coil and connecting wires.
I hope this explains all that is happening with your experiment, but I'd be happy to expand on or run tests on any aspect if you want. Scope shots are also available if you wish.
Regards,
.99
Hi .99,
thank you for your test report ;) ... and to think you did all these tests without pictures or a video ;D
I do understand your point and I think you know that I was not claiming extra energy coming out of the coil compared to what is going in! that is just not possible using this simple circuit.
I am suggesting to do something else and I think many of the members at another Forum are seeing that possibility and it has lit a spark.
From my testing I also see a problem using an electronic component as a switch as it is truly not disconnecting both sides like a mechanical switch and in the end I don't think it will give the full desired effect I am thinking of, this is why I took the time to build the mechanical switch. I must say I am very please with its results other then the limit of voltage I can send through those tiny reeds.
Thank you once again for your report.
@everyone else, looking forward in seeing what your test show and ideas how to best use this effect.
Luc
@gotoluc
Yes good work as usual.
With both the first and second videos, I think you have also demonstrated in this how Thanes magnets are reacting when his coil is shorted. There is an indirect parallel here with that magnet rebound effect.
I would suspect that if and when Thane puts his scope on a shorted coil while in the acceleration mode, we will see a similar waveform.
Also, thanks for making a conscious effort in reducing you use of the term BEMF. lol
I have dome some testing using a solenoid coil taken from a cuttler-hammer breaker switch. I will try and finally work out a standard video method and make a youtube video. I am curious to know if you have tried with that big bugger of a neo magnet, when returning flyback, if you put another magnet one the coil windings, how will it affect the neo movement. I have seen greater movement when adding one and even two magnets on the coils side. But I guess approaching any other magnets to that neo is an exercise in major havoc.
You may also try this with a small neo put inside the coil and see what happens with and without magnets on the sides of the coils.
I put up two image grabs of your scope shots so people can discuss these first hand.
Quote from: gotoluc on July 05, 2009, 11:56:28 AM
Hi .99,
thank you for your test report ;) ... and to think you did all these tests without pictures or a video ;D
I do understand your point and I think you know that I was not claiming extra energy coming out of the coil compared to what is going in! that is just not possible using this simple circuit.
I am suggesting to do something else and I think many of the members at another Forum are seeing that possibility and it has lit a spark.
From my testing I also see a problem using an electronic component as a switch as it is truly not disconnecting both sides like a mechanical switch and in the end I don't think it will give the full desired effect I am thinking of, this is why I took the time to build the mechanical switch. I must say I am very please with its results other then the limit of voltage I can send through those tiny reeds.
Thank you once again for your report.
@everyone else, looking forward in seeing what your test show and ideas how to best use this effect.
Luc
Luc,
A DC-DC down converter is handy when you need to be able to vary your output voltage between the ranges of your supply and almost 0V. You do this of course by changing the pulse width. 100% will yield your full 170V, and 0% will yield 0V output. This is indeed what you've built. Try increasing the pulse width and watch your magnet rise even further. It should do this continuously until either your supply gives out due to current limitations, or your coil burns up.
As a test, connect your coil directly to your variable DC power supply and begin at 0V. Increase the voltage until the magnet levitates to the same height you had with the PWM supply you built, and I am sure you will find that the current and voltage from the DC supply is very close to the numbers I gave above, i.e. ~3VDC @ ~300mADC.
The only advantage gained from a switching power supply (i.e. the PWM supply you have built with your 555, switch and coil) and a linear variable DC power is efficiency, IF you need to vary the supply voltage. If for example you only needed a fixed 3V @ 300mA supply, it would be more efficient and practical to design a linear supply tailored to the application.
I am not sure what else you will want to do with your circuit, but if you want an efficient way of varying the strength of the magnetic field from your coil, then you have accomplished that. Linear supplies become inefficient when you turn the voltage quite low and draw high current, but at full voltage they are just as efficient as a switching power supply, or at least very close.
I've attached two scope shots matched to the shots wattsup posted, as it seems you have doubts that I did any testing. It's the best I can do with the time I have available, but I can assure you that the information is accurate. The yellow trace (coil voltage) scale is the far left, while the green trace (gate voltage) scale is to the right (-2V to 30V).
I encourage you to perform the measurements I did in addition to the DC power supply test. Measuring the coil voltage and current should be straight forward for you, and you will see that indeed you've converted 170V to about 3V or so, and 10mA to about 300mA or so.
Regards,
.99
Quote from: wattsup on July 05, 2009, 12:07:32 PM
@gotoluc
Yes good work as usual.
With both the first and second videos, I think you have also demonstrated in this how Thanes magnets are reacting when his coil is shorted. There is an indirect parallel here with that magnet rebound effect.
I would suspect that if and when Thane puts his scope on a shorted coil while in the acceleration mode, we will see a similar waveform.
Also, thanks for making a conscious effort in reducing you use of the term BEMF. lol
I have dome some testing using a solenoid coil taken from a cuttler-hammer breaker switch. I will try and finally work out a standard video method and make a youtube video. I am curious to know if you have tried with that big bugger of a neo magnet, when returning flyback, if you put another magnet one the coil windings, how will it affect the neo movement. I have seen greater movement when adding one and even two magnets on the coils side. But I guess approaching any other magnets to that neo is an exercise in major havoc.
You may also try this with a small neo put inside the coil and see what happens with and without magnets on the sides of the coils.
I put up two image grabs of your scope shots so people can discuss these first hand.
Hi wattsup,
thanks for dropping in and adding your positive comments ;)
I think you're right about it resembling the effect that Thane has found and developed over the past years.
Thanks for taking the time to capture and post the scope shots ;)
Two people came to mind when I found this effect, Richard Willis of Magnacoaster Motor and Bob Teal.
They both used mechanical switches and timing is very important. I think it maybe the best route to take from seeing how my mechanical switch is working out.
I'll play around with small magnets on the coil sides to see if something new comes from it. However I do believe that a precise mechanical timing multiple switching is needed to also get the magnet flux kickback energy into play.
Looking forward to your findings.
Thanks for sharing
Luc
Quote from: poynt99 on July 05, 2009, 12:57:06 PM
Luc,
A DC-DC down converter is handy when you need to be able to vary your output voltage between the ranges of your supply and almost 0V. You do this of course by changing the pulse width. 100% will yield your full 170V, and 0% will yield 0V output. This is indeed what you've built. Try increasing the pulse width and watch your magnet rise even further. It should do this continuously until either your supply gives out due to current limitations, or your coil burns up.
As a test, connect your coil directly to your variable DC power supply and begin at 0V. Increase the voltage until the magnet levitates to the same height you had with the PWM supply you built, and I am sure you will find that the current and voltage from the DC supply is very close to the numbers I gave above, i.e. ~3VDC @ ~300mADC.
The only advantage gained from a switching power supply (i.e. the PWM supply you have built with your 555, switch and coil) and a linear variable DC power is efficiency, IF you need to vary the supply voltage. If for example you only needed a fixed 3V @ 300mA supply, it would be more efficient and practical to design a linear supply tailored to the application.
I am not sure what else you will want to do with your circuit, but if you want an efficient way of varying the strength of the magnetic field from your coil, then you have accomplished that. Linear supplies become inefficient when you turn the voltage quite low and draw high current, but at full voltage they are just as efficient as a switching power supply, or at least very close.
I've attached two scope shots matched to the shots wattsup posted, as it seems you have doubts that I did any testing. It's the best I can do with the time I have available, but I can assure you that the information is accurate. The yellow trace (coil voltage) scale is the far left, while the green trace (gate voltage) scale is to the right (-2V to 30V).
I encourage you to perform the measurements I did in addition to the DC power supply test. Measuring the coil voltage and current should be straight forward for you, and you will see that indeed you've converted 170V to about 3V or so, and 10mA to about 300mA or so.
Regards,
.99
Hi .99,
sorry you feel I was doubting you actually doing the test research since that was not my intent. I was just joking around since I'm not good at making reports like you have presented. It would take me a few pictures at least. You did a great job ;)
You are also 100% correct that this circuit is not making a coil perform magnet levitation height more efficiently then using strait DC as far as current and voltage are concerned. I have tested what you mention above
before starting this topic and have found that strait DC to be slightly more efficient which demonstrates that the circuit has losses (heat at the switch). I'm not claiming either to have found the holy grail of electronic circuits as I don't believe there is one.
What I'm trying to communicate here is that an air core coil inductive kickback has more punch on a permanent magnet then the energy used to create it.That's it :D
I'm also suggesting that using an electronic component to do the switching (even though I used it in the video test 2) is not the ideal switch to use in this application even tough it shows an interesting effect.
You are a fast and smart individual as I can see and it is good to be honest and looking at this from all angles as a good show can be deceiving if we don't look at it all.
So the question is, can we use this inductive kickback energy to power a motor more efficiently then using strait DC will remain unknown I would say until we build and test it.
Thanks for sharing
Luc
Quote from: gotoluc on July 05, 2009, 04:00:23 PM
What I'm trying to communicate here is that an air core coil inductive kickback has more punch on a permanent magnet then the energy used to create it.
Luc
Luc, I've been around long enough to know when to "throw in the towel" so to speak, so I will graciously bow out of this one. I'm not here to aggravate anyone, and there comes a time when folks just need to find the facts themselves rather than hearing them from others, and that is fine :) .
I am not 100% certain of what you mean be the quote, but if it is what I think it is, it goes directly against my published numbers up at the top of this page in my findings. Either that or I just don't understand it :D
Regards,
.99
PS. Always look forward to your new videos ;) They are top notch!
Okay .99,
I understand and I'm sure you understand what I'm talking about.
I'll try to build something that can somewhat work on this principle and you can enjoy the video show ;D when I post it.
Thanks for your time and willingness to help out.
Luc
Hi Luc
Your tests etc.are interesting,doe's the mag.behave the same if it's north or south pole on the coil?
peter
Hi Peter,
well, no since one side would be in attraction mode but yes! if I change the coil polarity.
Luc
Quote from: poynt99 on July 05, 2009, 04:49:30 PM
Luc, I've been around long enough to know when to "throw in the towel" so to speak, so I will graciously bow out of this one. I'm not here to aggravate anyone, and there comes a time when folks just need to find the facts themselves rather than hearing them from others, and that is fine :) .
I am not 100% certain of what you mean be the quote, but if it is what I think it is, it goes directly against my published numbers up at the top of this page in my findings. Either that or I just don't understand it :D
Regards,
.99
PS. Always look forward to your new videos ;) They are top notch!
Hi .99,
if you can help with one more thing I would be gratefull since there is still one thing that no one has explained yet and I'm hopping you can. Why does the sens coil not pickup the rerouted energy and extend the scopes wave form more if it is the case that the energy is extended or what ever it is doing?
Thanks for your time.
Luc
Quote from: gotoluc on July 06, 2009, 11:32:21 AM
Hi .99,
if you can help with one more thing I would be gratefull since there is still one thing that no one has explained yet and I'm hopping you can. Why does the sens coil not pickup the rerouted energy and extend the scopes wave form more if it is the case that the energy is extended or what ever it is doing?
Thanks for your time.
Luc
Hi Luc.
Monitoring a coil's voltage does not always indicate what is going on there in terms of its current. The fact that the bottom portion of the coil voltage disappears when the diode is in-circuit, actually
does indicate that a conversion is taking place. The conversion is from high voltage/low current, to high current/low voltage. If you zoom in when the diode is in-circuit, you will probably see a negative swing of about -0.65 Volts (from the previous -300V or so), but the current has shot up significantly. This is the indicator that "something" different is happening in the coil that is causing the increased magnetic power.
With no load (i.e. without a flyback diode), the coil sees almost an open circuit during its inductive kickback cycle. The same amount of energy (minus losses) must be conserved, so the coil voltage extends quite high in the reverse direction, but the current is quite small.
When the coil IS loaded during its inductive kickback cycle by placing the diode across it in reverse, the diode creates nearly a short circuit across the coil during this cycle, so the current has no choice but to increase by a large amount. It is this increase of current that you can not see by looking at the coil voltage, but this increased current is what is responsible for the much larger magnetic force being applied to your neo magnet.
Remember with coils, it is current that energizes them and produces a corresponding magnetic field, not voltage. With a heavy enough wire you could produce quite a strong magnetic field with only a few volts, as long as the source can supply a large current.
Hope that helps,
.99
Quote from: poynt99 on July 06, 2009, 01:22:34 PM
Hope that helps,
.99
Hi .99,
yes indeed that helps :D ... your written explanation was so good that I could actually visualize it as I was reading.
Thank you my friend for all your expertise you have given to this and other topics.
Luc
Quote from: gotoluc on July 06, 2009, 07:04:37 PM
Hi .99,
yes indeed that helps :D ... your written explanation was so good that I could actually visualize it as I was reading.
Thank you my friend for all your expertise you have given to this and other topics.
Luc
Yep, that goes for me too.
Hi .99,
I know I said one more question but I just remembered something interesting that happened when I was first testing my large Neo magnet demo board which is in test 2 video.
I had a short happen when I touch a wire to the coil with my soldering iron but still had the power supply of the circuit plugged into the grid. It fried the mosfet and it shot the magnet out of the tube about 3 feet in the air. I was lucky it didn't fall on my head or any other part of me but here is the interesting part. My eyes were quite close to the bottom of the coil where I was going to solder when it happened and I'm quite positive I saw a flame of fire for the first 3 inches of its launch and I thought the coil was on fire but it wasn't the case since my amp meter was in series with the supply capacitor and the 250ma fuse blew before any damage could happen to the coil.
Have you ever heard or know anything about this fire from a sudden strong EMF opposing a very strong permanent magnet pole?
The coil is # 20 AWG, 800 feet long at 8.3 ohms and 63.5 mH. The voltage was 170 vdc from a 330uf cap and limited by the 250ma fuse
Thanks for your time
Luc
Error
Luc,
It would be a tough call to identify exactly what happened there without actually being there and examining the setup for burn marks etc. Also it would be prudent to use a volt meter and measure the voltage between your soldering iron tip and either of the leads or wires of the coil that may have come in contact with the iron.
But if I was to guess based on your description of what happened, I would say that you had either 120VAC alone (due to contact with the soldering iron), or a combination of 120VAC and 170VDC going through your coil for a brief instant. It all depends on earth ground paths etc. Not sure about the flash you saw, but it was more than likely a spark/arc from the transient high voltage and current.
.99
Hi .99,
thanks for the reply ;)
that's true! it could of been a mix of AC and DC. The flame left no markings what so ever. It was orange in color, much like Hydrogen burning.
In my setup half the full wave bridge rectifier positive side goes direct to the coil and the negative side goes through the mosfet. The soldering Iron goes to ground, so if I touch the coil on the mosfet side I activate the coil with half the DC voltage (80 vdc) of positive side of the bridge.
Let me know if you think this is worth reproducing to study it more.
Thanks
Luc
Luc,
Are you sure only half of the full wave bridge goes to the coil? I'm not sure I understand what you mean. I thought you had +170V at the coil?
If your negative side of the 170V supply is tied to earth ground, then by touching the iron on the MOSFET side, you are applying 170V across the coil and fully discharging your capacitor into it. Yep, this would give the neo quite a kick!
You could try this again, but in a controlled manner (i.e no MOSFET present and don't use the iron as the ground path) just to prove that this is what most likely happened.
.99
Hi .99
yes a total of 170vdc is at the capacitor but if I put my voltmeter between Earth Ground and each of the leads going to the coil (leads now disconnected from coil) I can only measure the positive side of the capacitor as the other side is blocked by the mosfet. So I can only be sure half the capacitor voltage is going through but since the mosfet fried when this originally happen then it is quite possible the full 170vdc of the cap was at the coil.
I hope this help you to understand ???
Let me know
Luc
Luc,
Is your setup something like this?:
Cap (charged to 170V, and commoned to "gnd")=>+coil lead=>-coil lead=> MOSFET Drain=> MOSFET Source (commoned to "gnd")
If this is how you have connected everything, then you have 170V across the coil when the MOSFET is ON.
If you accidentally connected the MOSFET Drain/-coil lead to ground through the iron tip, then this would have caused the full 170V to go through the coil. I am not sure why the MOSFET would have blown unless somehow the gate voltage went way up in voltage also.
.99
Hi everyone,
I have made a new video with a better quality camera I now have for all future videos.
I made this new video because something was bothering me about my previous conclusion using my good quality meter to measure the Watts Energy of the pulsing circuit compared to using straight DC. I talked to my Engineer friend and he has confirmed to me what I was thinking. Watts energy has nothing to do with Voltage since you can have thousands of volts going through a resistor and there could be no heat if there is no Watts energy. So Watts energy should be measured by heat dissipation at the Resistor regardless of the voltage.
So what I'm getting at is, since it's so difficult to measure the Energy used by these kinds of circuits lets just measure the heat dissipated in a resistor at the input of the circuit compared to the heat dissipated at the output.
I have redone a video using a visual restive load on the input with the circuit functioning with an 8% duty cycle to compare it using the same load with straight DC and the results are quite different then what my quality meter was calculating.
I know many will be saying that this is not the way to measure. However I'm not measuring! the bulb is only a visual display of Energy and nothing else regardless of the voltage going through it because it's resistive.
Also note that I will be redoing the test over but using standard resistors and measuring the heat dissipated at the input compared to the heat dissipated at the output and posting the video next week at some time. In the meanwhile this gives us something to think and talk about.
Video link: http://www.youtube.com/watch?v=HpaP__5Kd38
Luc
@gotoluc
Your last video was spot on (expect for using the BEMF word - lol). Old habits are hard to break.
What you have shown is the super efficient way of raising the magnet using pulsed DC combined with the off-pulse flyback, compared to the normal energy hog method of raising the magnet using straight DC.
Of course we cannot concluded any type of OU aspect here and this is not even important at this stage, it is just a confirmation of your efficiency level.
What would be interesting is to now run this same test but with your batteries again and add a return charging diode to the battery to see if the flyback can still maintain the magnet height while having some extra flyback energy that the coil could maybe not have taken advantage of, to then recharge the battery. A switch on the batttery charging diode would show the difference of normal flyback mode and then with added flyback battery recharge. But maybe that's taking it to a very next level that you are not near yet.
Good work.
wattsup
Hi everyone,
I made a new video using standard resistors of equal value on the input side and on the recirculated Inductive flyback side pre entry of the coil. I have measured the heat on each but have not wet come up with a better way of measurement that will satisfy all but I am working on it.
New Video: http://www.youtube.com/watch?v=WsmPyUzZtgQ
In this test it is very very difficult for me to understand how energy could slip though the input 22 Ohm resistor and leave basically no heat and end up in the collapsing field side resistor at higher temperature then the boiling point and also do work pushing up a one pound magnet over 3/8 of an inch off the coil. We cannot consider voltage as having anything to do with the effect since it has no heat Energy. So what is doing this?
Please find the attached scope shot measured across the 22 Ohm input resistor and let me know if you can calculate this.
Luc
Quote from: wattsup on July 10, 2009, 01:42:27 PM
@gotoluc
Your last video was spot on (expect for using the BEMF word - lol). Old habits are hard to break.
What you have shown is the super efficient way of raising the magnet using pulsed DC combined with the off-pulse flyback, compared to the normal energy hog method of raising the magnet using straight DC.
Of course we cannot concluded any type of OU aspect here and this is not even important at this stage, it is just a confirmation of your efficiency level.
What would be interesting is to now run this same test but with your batteries again and add a return charging diode to the battery to see if the flyback can still maintain the magnet height while having some extra flyback energy that the coil could maybe not have taken advantage of, to then recharge the battery. A switch on the batttery charging diode would show the difference of normal flyback mode and then with added flyback battery recharge. But maybe that's taking it to a very next level that you are not near yet.
Good work.
wattsup
Thanks for the post wattsup :)... ya that term keeps slipping out of my mouth if I 'm doing too many things in the same time ;D
As for charging a battery with flyback. That's something I've tried many times in the past and have not shown any gain. I think the flyback is best used to create heat and or electro mechanical power with an air core coil and powerful magnet.
I'm happy with the results so far anyways. Others can test the battery thing!... by the way, wattsup with the others ;D you can't be the only one interested in this ???
Luc
@gotoluc
Don't take it personally. I think a lot of guys are on their work benches these days. I have myself lots of catching up to do after losing the last few months to this Mylow wheel thing. I am planning slowly to make a Thane Heins variation but from now on, I want to spend most of my time on TPU studies.
So don't lose heart. Whatever you are showing does help us in ways we don't know yet and every angle of effects is important to discover.
Regarding what you have shown, you know those big cranes with huge electromagnets that are used to move scrap cars in crushers. Well they must use huge amounts of electricity. I wonder if you used your pulsing method, you could probably save them loads of energy and money. May be a new business right there. Imagine the size of the pulsing circuit. lol
Added: Look at todays Google Logo.
Quote from: wattsup on July 10, 2009, 06:37:42 PM
@gotoluc
Regarding what you have shown, you know those big cranes with huge electromagnets that are used to move scrap cars in crushers. Well they must use huge amounts of electricity. I wonder if you used your pulsing method, you could probably save them loads of energy and money. May be a new business right there. Imagine the size of the pulsing circuit. lol
;D ;D ;D ya! that would take a large mosfet.
Thanks for taking the time to post :)
Luc
Quote from: gotoluc on July 10, 2009, 03:45:05 PM
Hi everyone,
I made a new video using standard resistors of equal value on the input side and on the recirculated Inductive flyback side pre entry of the coil. I have measured the heat on each but have not wet come up with a better way of measurement that will satisfy all but I am working on it.
There is no need to prove any further that the output resistor is much hotter than the input resistor. This is what would be expected.
Quote
In this test it is very very difficult for me to understand how energy could slip though the input 22 Ohm resistor and leave basically no heat and end up in the collapsing field side resistor at higher temperature then the boiling point and also do work pushing up a one pound magnet over 3/8 of an inch off the coil. We cannot consider voltage as having anything to do with the effect since it has no heat Energy. So what is doing this?
Voltage and current have to be considered together in a load of any kind when dealing with power. If you were to scope across your input resistor, you would see that the voltage on average is much higher than that across the input 22 Ohm resistor, and since the two are of equal value, you can surmize which one will be hotter. See the attached sim scope shot.
However, the power in the input resistor is not an indication of how much power is being taken from your power supply. You have more power dissipated in the output resistor than the input one, but the total power from the power supply has to be taken into account. If done so, it will reveal that indeed a lot of power is being used.
To do this you need a small current sense resistor of about 10 milli-Ohms so you can measure the current from the power supply. Then you multiply that by the supply voltage and you have instantaneous power. Now integrate it and you have a direct readout of energy. See below Joules plot of the 170V supply (red) vs. that of the 22 Ohm output (bemf) resistor (green).
Are you surprised?
Below is the schematic, voltage plot across both 22 Ohm resistors (illustrating why the output R (green) is much hotter than the input R (red), and finally a plot of the energy in Joules dissipated by the output resistor vs. the energy taken from the 170V power supply. This is measured directly, even though it indicates mV it is mJ.
Think about a car stereo aftermarket amplifier that puts out a true 100W rms per channel. How do you suppose that can be achieved with only a 12V battery source? With Ohm's law it becomes obvious that this is not possible unless the battery voltage is boosted up somehow ;)
Sorry for jumping in again. I'll back off and let some others explain it if they wish.
.99
Hi .99,
I do appreciate your help but I'm sorry I don't understand conventional EE
Instead can you please answer the questions below. This will help me much more.
Let us take the fine filament 12 volt bulb I used in test 3
We know that if we place that bulb in series in a circuit we can pass several hundreds of volts through it and it won't necessarily light up, correct?
If this is correct then voltage is not Energy!... correct?... however I do understand that voltage and current can be present together in a circuit if the voltage source has the ability to provide Energy.
Current is a heat Energy, correct?... and if so then voltage is not a heat energy, correct?
So if voltage cannot create heat at whatever voltage it is at then why would one need to consider it to calculate Energy used in a circuit.
Lets use the bulb in this example:
For Energy to partake in a circuit I believe we need two conditions, one is, it has to be available from the voltage supply and two, the circuit would need to create a wast (resistance) of some kind. Now if our bulb which is in series in the circuit starts to glow it is now dissipating heat because of resistance somewhere in the circuit. The bulb is a visual display or indicator of Energy now present in the circuit. This Energy is dissipating in the bulbs filament as heat, correct?... so if we have 10 volts in a circuit under load and the filament starts to glow and we reduce the duty cycle since the circuit is a pulse circuit and we then raise the voltage to 100 volts and the bulb starts to glow to the same level as when 10 volts was running through the circuit, do you really believe that we now have more Energy at the bulb because the voltage was raised 10 time higher then before???
If you believe this then can you please explain to me how voltage can contain Energy since something is not making any sense to me anymore ::)
Thanks for your time and sharing
Luc
Quote from: gotoluc on July 10, 2009, 11:06:12 PM
Hi .99,
I do appreciate your help but I'm sorry I don't understand conventional EE
Instead can you please answer the questions below. This will help me much more.
Let us take the fine filament 12 volt bulb I used in test 3
We know that if we place that bulb in series in a circuit we can pass several hundreds of volts through it and it won't necessarily light up, correct?
If this is correct then voltage is not Energy!... correct?... however I do understand that voltage and current can be present together in a circuit if the voltage source has the ability to provide Energy.
Current is a heat Energy, correct?... and if so then voltage is not a heat energy, correct?
the heat depends also of the profile of your wire, if it is big enough then there is less or nearly no heat.
think of current as the quantity of electrons moving in a wire, the more elctron the more heat because of friction and voltage is just the pressur. Hope this helps.
Norbert
Quote from: gotoluc on July 10, 2009, 11:06:12 PM
Hi .99,
I do appreciate your help but I'm sorry I don't understand conventional EE
Instead can you please answer the questions below. This will help me much more.
Ok Luc, I'll try.
Quote
Let us take the fine filament 12 volt bulb I used in test 3
We know that if we place that bulb in series in a circuit we can pass several hundreds of volts through it and it won't necessarily light up, correct?
This depends on what the total circuit resistance (or impedance) is and what the RMS value of the wave form is.
Quote
If this is correct then voltage is not Energy!... correct?... however I do understand that voltage and current can be present together in a circuit if the voltage source has the ability to provide Energy.
Current is a heat Energy, correct?... and if so then voltage is not a heat energy, correct?
Voltage alone is not necessarily energy, that is correct. A 12V battery sitting with open leads is not supplying much if any energy external to itself. So it depends on the circuit that the "voltage source" is connected to.
Current is a better indicator that energy is being supplied by a source, but voltage and current are closely tied in a relationship that together amounts to power output or input. If either the voltage or the current are zero, there will be no power, no matter how large the other parameter is. You have to have at least some voltage AND some current simultaneously in a circuit in order to "make power".
Quote
So if voltage cannot create heat at whatever voltage it is at then why would one need to consider it to calculate Energy used in a circuit.
I'm getting to this as it is related to the previous paragraph. Power in Watts does equal heat if that power is dissipated in a dissipative element such as a resistor.
Quote
Lets use the bulb in this example:
For Energy to partake in a circuit I believe we need two conditions, one is, it has to be available from the voltage supply and two, the circuit would need to create a wast (resistance) of some kind.
More or less, yes. The circuit is such that it draws power from the source because of its loading effect on it.
Quote
Now if our bulb which is in series in the circuit starts to glow it is now dissipating heat because of resistance somewhere in the circuit. The bulb is a visual display or indicator of Energy now present in the circuit. This Energy is dissipating in the bulbs filament as heat, correct?...
This is not necessarily correct. As you know you can connect the bulb by itself across the supply and it will glow. No resistance is required in the circuit to allow the bulb to glow. If you used superconductor wires (i.e. zero resistance) in the circuit, the bulb would glow, and glow brighter. The bulb is it's own resistance, and yes it is this resistance which gets hot (dissipates power) and glows as a result.
Quote
so if we have 10 volts in a circuit under load and the filament starts to glow and we reduce the duty cycle since the circuit is a pulse circuit and we then raise the voltage to 100 volts and the bulb starts to glow to the same level as when 10 volts was running through the circuit, do you really believe that we now have more Energy at the bulb because the voltage was raised 10 time higher then before???
No. Assuming that you were able to set the bulb intensity roughly the same as before, then the
average current and voltage in the circuit would also be measured as roughly the same as before.
Quote
If you believe this then can you please explain to me how voltage can contain Energy since something is not making any sense to me anymore ::)
Thanks for your time and sharing
Luc
Getting back to the relationship between voltage and current in producing power in a circuit element, it is necessary to be aware of and understand a few simple equations.
Power (
P) in Watts =
V x
I (voltage times current)
From this you can see that if either the voltage-across OR the current-through a circuit resistor is
0, the resulting power dissipated in that resistor will also be zero. So for power to manifest in the circuit and be dissipated in any circuit resistance, there must be
some voltage-across the resistor AND
some current-through it.
Hope that helps.
.99
Hi .99,
Thank you for taking the time to answer the questions.
This question was the most important to me and you answered it as I understand it.
My Question:
so if we have 10 volts in a circuit under load and the filament starts to glow and we reduce the duty cycle since the circuit is a pulse circuit and we then raise the voltage to 100 volts and the bulb starts to glow to the same level as when 10 volts was running through the circuit, do you really believe that we now have more Energy at the bulb because the voltage was raised 10 time higher then before???
Your Answer:
No. Assuming that you were able to set the bulb intensity roughly the same as before, then the average current and voltage in the circuit would also be measured as roughly the same as before.
Your answer is worth Gold to me and I will make a new video to show you why.
Thanks again for sharing.
Luc
QuoteNo. Assuming that you were able to set the bulb intensity roughly the same as before, then the average current and voltage in the circuit would also be measured as roughly the same as before.
Luc,
Since it was important to you, I'd like to expand a bit.
A pulsed-DC supply can be replaced by a linear (straight DC) supply of the same average value.
So a supply that is pulsed at 10V peak with a pulse width of 10ms and a period of 100ms (10% duty cycle) will have the same
average voltage output as a pulsed-DC supply with 100V peak pulses and 1ms pulse width and the same 100ms period (1% duty cycle).
The
average voltage output in both pulsed-DC cases is 1V.
So using a plain DC supply set to 1V output is the same as either case using the pulsed-DC supplies above.
The difference in applying that 1V source to a load is evident by how much power can be drawn from it.
Remember the car amplifier I mentioned? Are you familiar with power inverters? You can buy them at many automotive stores etc. They convert automotive 12VDC to 110VAC for obvious uses. Some of these units can supply 1000's of Watts to a load.
Let's use a household incandescent bulb as an example. A 100W bulb will have a hot resistance of about 120 Ohms. If we were to connect this bulb up to a automotive 12V battery, the most power we could obtain from this bulb would be (from P = V
2 /R) about 1.2 Watts. This is also the maximum we can draw from this 12V source, unless we begin to parallel many of these bulbs together.
But if we now use one of these 12VDC to 120VAC inverters, we can draw 100W from the battery with a single bulb connected. The inverter is able to step up the RMS voltage (by a factor of about 10) from the battery which allows more power to be drawn from it.
When doing the type of tests that you are, it is very important to always know how much power is being taken from your source. The best way to do this is to use a low resistance shunt in series with the supply (either + or - side) and measure the voltage across this shunt. Your good meter should be able to handle this (use "AC voltage" setting, but try DC as well). That will be a direct current readout in RMS current, which you will then multiply by the supply voltage (170VDC?) to get total power taken from the supply.
I don't recommend using resistors in the legs of your MOSFET.
.99
@poynt99
Thanks also for your explanation which is the basic volts x amps that we know very well.
The difference Luc is showing is what you can do more with the pulsed DC in that your are now creating a steady flyback condition that he capitalizes to make the magnet rise with very low overall wattage, when compared to the standard straight DC supply method that does not favor any flyback production.
In the pulsed method the bulb which is the overall wattage consumption visual indicator was barely lit, while in the straight DC method, to lift the magnet to the same height, the overall wattage consumed and shown by the very brightly lit bulb indicates that more energy is consumed.
Again, this is not intended to show any OU effect but to simply show that the same work (raising and holding the magnet levitated) can be accomplished with drastically lower power consumption. This in my view is a perfect differentiation that could give a certain direction for further experiments leading to OU. But that's my opinion only, for what it's worth.
wattsup
Hi wattsup,
your explanation is exactly correct ;)
I have just completed 2 videos to demonstrate what has been interesting me so much in this simple circuit and will start uploading them now.
Stay tuned ;D
Luc
can t wait to see them goto :-)
Hi everyone,
a testing base line needs to be established to better study this circuit effect. That is why I first needed to confirm with .99 what I understood of using a fine filament bulb in series at the input as visual indicator of energy entering the circuit.
I have found that using meters no mater how expensive they are have been giving me false results. Once you makes a change to the circuit, like frequency and or voltage you want to re-tune the duty cycle to send back exactly the same amount of energy to compare the previous test. I know many will say your eye cannot be that accurate. That is right! ... if the intensity of the bulb is high... but if you keep the bulb intensity to the point where it just starts glowing and not go above that you will see that it has an amazing accuracy to show any change in the circuit. Just stay with this technique I just explained and it should work well enough. Also, know that you are not measuring anything with the bulb, it is just a visual indicator of change or a way to re-tune duty cycle after you make a change in order to send back exactly the same amount of energy in the circuit to be able to observe if the change you made is giving you more or less work for the same amount of energy. That is all you want to do at this time. Forget Over Unity proof.
You will understand more after seeing the video's below how I have been using this technique.
Link to test 5: http://www.youtube.com/watch?v=JwE5WlHL48k
Link to test 6: http://www.youtube.com/watch?v=_OsN_K3wNLA
Thanks
Luc
Here's a question: Does the inductance of the magnet-lifting coil change, as the magnet lifts? I really don't know, and I think it would be interesting to try to find out. I can sort of argue both ways, at the present state of my knowledge.
I do know that bringing a powerful magnet close to the toroid of an operating Joule Thief causes interesting effects...it can quench or substantially brighten the LED...
It would seem on casual inspection that straight DC would lift the magnet higher than the reversing field from AC, at the same power levels. I think.
Got to build me one. Where did I put that mile of magnet wire....
Hi
Sorry for jumping in on this thread, i havnt really followed the story so far, but wanted to clear up a question i have.
Is the bulb in series with the supply voltage cap and the circuit under test.?
Peter
Ok just read some more of the thread and it does look like the bulb is in series with the supply cap and the device under test.
If this is so then the bulb is only telling half the story for the consumpion of power drained from the supply cap, you must also account for the voltage supplied by the power supply itself.
The bulb is lighting because of a voltage drop across itself, which is related to the current drawn through it and it's own resistance, but you then would need to multiply this current by the power supply voltage to get a true value of power consumed.
So keeping the bulb at a certain brightness but altering the supply voltage is indeed increasing the power consumed by the device under test.
Hope that helps
Peter
Quote from: TinselKoala on July 11, 2009, 03:57:04 PM
Here's a question: Does the inductance of the magnet-lifting coil change, as the magnet lifts?
Humm, I don't think a magnet has inductance but I think you maybe referring to the coil. Like I said in the video, the magnet does not affect the effect. It is just used as a visual aid as to be able to calculate change when changes have been made. Anyways, I've tried it also without the magnet and it shows actually a very small efficiency.
Quote from: TinselKoala on July 11, 2009, 03:57:04 PM
It would seem on casual inspection that straight DC would lift the magnet higher than the reversing field from AC, at the same power levels. I think.
You may think that! but not according to my tests. I even have that test on video! you may have missed it. Here it is again: http://www.youtube.com/watch?v=HpaP__5Kd38
Quote from: TinselKoala on July 11, 2009, 03:57:04 PM
Got to build me one. Where did I put that mile of magnet wire....
Just make sure your 555 timer circuit is the one that gives you 10% or less duty cycles ;D
Luc
Quote from: TinselKoala on July 11, 2009, 03:57:04 PM
Here's a question: Does the inductance of the magnet-lifting coil change, as the magnet lifts? I really don't know, and I think it would be interesting to try to find out. I can sort of argue both ways, at the present state of my knowledge.
I do know that bringing a powerful magnet close to the toroid of an operating Joule Thief causes interesting effects...it can quench or substantially brighten the LED...
It would seem on casual inspection that straight DC would lift the magnet higher than the reversing field from AC, at the same power levels. I think.
Got to build me one. Where did I put that mile of magnet wire....
@TinselKoala
On your first question: the inductance of an air core coil does not change but very little when moving a permanent magnet in it because the permeability of permanent magnets approach pretty close to unity i.e. to that of the air.
If a coil has got a ferromagnetic core and you approach a permanent magnet to it, then the flux changes inside the core (thus the permeability of the core) so the result is you move the magnetic working point of the core on the B-H curve, depending on the poles, and even you can saturate it.
rgds, Gyula
EDIT Permeability of Neo magnets ranges from 1.03 to 1.1 or so, ceramic magnets has 1.1-1.2 or so.
Quote from: Peterae on July 11, 2009, 04:47:26 PM
Ok just read some more of the thread and it does look like the bulb is in series with the supply cap and the device under test.
If this is so then the bulb is only telling half the story for the consumpion of power drained from the supply cap, you must also account for the voltage supplied by the power supply itself.
The bulb is lighting because of a voltage drop across itself, which is related to the current drawn through it and it's own resistance, but you then would need to multiply this current by the power supply voltage to get a true value of power consumed.
So keeping the bulb at a certain brightness but altering the supply voltage is indeed increasing the power consumed by the device under test.
Hope that helps
Peter
Hi Peter,
I'm sorry but that is not correct! if you read this posts: http://www.overunity.com/index.php?topic=7713.msg190619#msg190619 you will find that user .99 has addressed that since I specifically asked the question before doing my last two videos.
Luc
Luc,
Could I get a bit of info?
I'd like to know what that 22 Ohm resistor and the auto bulb each read on your inductance meter. Also, could you measure what the hot resistance is of the bulb? The "hot" resistance would be soon after disconnecting it from the circuit and measuring with your Ohm-meter.
One last thing, is the bulb the first thing in series with the + side of your storage capacitor for all your tests so far?
[Edit] I also assume that the 22 Ohm in series with the MOSFET Source lead is no longer in-circuit and not being used?
Thanks,
.99
Quote from: gotoluc on July 11, 2009, 04:51:54 PM
Humm, I don't think a magnet has inductance but I think you maybe referring to the coil. Like I said in the video, the magnet does not affect the effect. It is just used as a visual aid as to be able to calculate change when changes have been made. Anyways, I've tried it also without the magnet and it shows actually a very small efficiency.
You may think that! but not according to my tests. I even have that test on video! you may have missed it. Here it is again: http://www.youtube.com/watch?v=HpaP__5Kd38
Just make sure your 555 timer circuit is the one that gives you 10% or less duty cycles ;D
Luc
Thanks, Luc and Gyula.
Hmm...I see.
But there is a secondary lifting effect that occurs when AC is applied to a coil and a (non-magnetic) conductive surface is placed above it--eddy current levitation, which can be quite strong. I wonder if the eddy current levitation due to the AC is adding to the normal electromagnet-permanent magnet repulsion.
As you turn the frequency down and down, approaching DC, does the height of the magnet vary, at the same input power? (Hmm--lock-in amplifiers, positive feedback loop, phase-locked loop, optical sensing, HPIB data logging and experimental control....might be interesting to explore....hmmm.......)
And I see that the low permeability of the magnet shouldn't affect the coil's inductance...but surely the magnetic field of the PM must be interacting with the coil's field in some manner. Besides simple repulsion or attraction I mean. I mean, an increase in permeability means basically that more field lines are in there; sticking a ferromagnetic core increases the flux density (increases inductance) within the coil by providing a high permeability path for the field... the opposing fields of the coil and PM surely change the flux density around and perhaps in the coil...?
I think it's an interesting feature, regardless.
And I agree with an earlier point of Luc's : If the bulb brightness is kept quite low, eyeballing its relative brightness will be more accurate than it is at brighter light levels. I'm still thinking about instrumenting this measurement, though.
In fact it should be possible to construct a simple feedback loop that would vary input peak current to the bulb to keep the bulb at a constant repeatable brightness; then the bias current would be the dependent measure.
Hmm--looks like I've had too much coffee.
(adjourns for medication adjustment.)
Quote from: poynt99 on July 11, 2009, 06:35:08 PM
Luc,
Could I get a bit of info?
I'd like to know what that 22 Ohm resistor and the auto bulb each read on your inductance meter. Also, could you measure what the hot resistance is of the bulb? The "hot" resistance would be soon after disconnecting it from the circuit and measuring with your Ohm-meter.
One last thing, is the bulb the first thing in series with the + side of your storage capacitor for all your tests so far?
[Edit] I also assume that the 22 Ohm in series with the MOSFET Source lead is no longer in-circuit and not being used?
Thanks,
.99
Hi .99
the bulb seems to be unmeasurable on my inductance meter. The lowest setting on my meter is 2mH and it works its way down to .000 then the -.000 comes up. The 22 Ohm resistor is .015mH
The bulbs cold DC resistance is 4.5 Ohm and if I have it at the same intensity with 200vdc switching around 1.8KHz and disconnect it the fastest I can (about 3 seconds) it measured 5.5 Ohm and worked its way down.
No! the bulb was first thing in series with the - side of your storage capacitor to the MOSFET Source for test 5 and 6. In test 3 it was as you say above.
Yes the 22 Ohm in series resistor is no longer in-circuit on the input side.
Thanks
Luc
Quote from: TinselKoala on July 11, 2009, 07:59:12 PM
In fact it should be possible to construct a simple feedback loop that would vary input peak current to the bulb to keep the bulb at a constant repeatable brightness; then the bias current would be the dependent measure.
A good idea TinselKoala! can you design one. Even better, I wounder if it would be possible to make one that would be giving an Energy measurement also.
Luc
Hi everyone,
a YouTube user posted a comment which explains very well and my reply may also help some.
User: Can101276
I like your unconventional testing, it is crude, however it will alow you to see work being done and see how your changes effect the work performed.
Correct me if im wrong, but what I got out of your experiment was that you can get more work performed with high voltage, high frequency pulses, than with lower frequency, higher duty cycle pulses at the same voltage.
So in dummy speak, more work for same watts!!!
You are on the right track.
My reply:
Yes Can101276, you are understanding this correctly. However there maybe limits as my higher voltage tests over 250vdc the efficiency starts going down but that maybe an electronic component issue causing this. I also see efficiency starting to go down when frequencies over 5KHz are used. This could have something to do with the coil characteristics. More testing will be required to understand this better.
Luc
Quote from: gotoluc on July 11, 2009, 09:33:18 PM
A good idea TinselKoala! can you design one. Even better, I wounder if it would be possible to make one that would be giving an Energy measurement also.
Luc
I think the circuit would be very simple. A 2n2222 controlling the voltage to the bulb, the photoresistor in the gate circuit of the transistor, the ammeter in the same circuit, some appropriate trimming resistors and a cap or two just for grins, and Bob's yer uncle as they say around here. Bulb gets dimmer, photoresistor increases, gate current decreases, transistor lets more current thru to bulb, bulb gets brighter, photoresistor decreases, gate current increases, transistor lets less current thru to bulb.... trimmers set brightness point, microammeter in gate circuit can be calibrated to a known power setpoint. (So many microamps on meter, not directly caused by power circuit, but can be accurately correlated with power, because as the power circuit increases voltage to try to make the bulb brighter, the photocell circuit uses gate current to make the transistor cut the power to the bulb by the corresponding amount.) This is off the top of my head, it's probably backwards or something. (NPN, PNP, always confuses the heck out of me)
I'd breadboard one up in an instant (and a half) if I only had the photoresistor on hand. I've got some in another location and I'll pick them up tomorrow and try it, let you know.
But even better would be to just record the bulb's brightness level using the photocell and the software chart recorder function that might have come with your scope software.
Funny that you should mention the energy measurement. In the old days people would sometimes look at the oscilloscope screen using a photoresistor. The brightness of the trace picked up by the photoresistor could be correlated with the energy in the signal giving the scope trace if the scope was set up right. The photoresistor acted like an optical electronic integrator. If your scope has a "Z" input this is one thing it can be used for.
Quote from: gotoluc on July 11, 2009, 09:33:18 PM
A good idea TinselKoala! can you design one. Even better, I wounder if it would be possible to make one that would be giving an Energy measurement also.
Luc
Luc,
You've already got an instrument for measuring the "brightness" of your input bulb.
Replace the bulb with a 1W 10 Ohm resistor (or experiment with Wattage and resistance values between 3 and 15 Ohm), and use your laser temp gauge/meter to measure and adjust the pulse width to obtain the same baseline temperature in the resistor each time a change is made. This is probably more accurate than eyeballing the bulb intensity anyway.
Thanks for the component specs btw ;)
.99
@TK
Sound good TK ;) let me know how the circuit works out.
Thanks for taking the time to help :)
Luc
@.99
That also sounds good .99 the only thing that is a problem is the IR temperature meter laser is not aligned with the sensor. You can see me struggling with it in the videos to find the right point and that is with those large ceramic resistors.
I hope the data will help
Luc
This is my last post for today.
a user at Energetic posted this interesting test idea and I would like to know your opinion on this .99 or TK since I don't like cutting up my coils if not necessary ;D
Luc
Suppose your coil had several TAPS ... one at 50 turns, one at 75 turns, one at 100 turns, one at 125 turns, etc. etc. etc.
Now from each of these TAPs, bring a diode out and route it back into the
coil.
That will really piss off the magnet. :-)
@gotoluc
I also agree with @poynt99. If someone grabbed an image of the bulb each time you made a change and blew up the image and put them side by side, I am sure there will be some differences. At that low brightness level, a small change could easily mean 50% more or less and it would still be hard to make out the difference visually.
Also, during both video tests was the return diode on or off. This was not clear to me. It may have been good to switch it in and out to see the difference at each setting. But very good work.
Quote from: gotoluc on July 11, 2009, 11:14:08 PM
This is my last post for today.
a user at Energetic posted this interesting test idea and I would like to know your opinion on this .99 or TK since I don't like cutting up my coils if not necessary ;D
Luc
Suppose your coil had several TAPS ... one at 50 turns, one at 75 turns, one at 100 turns, one at 125 turns, etc. etc. etc.
Now from each of these TAPs, bring a diode out and route it back into the
coil.
That will really piss off the magnet. :-)
Luc,
I wouldn't go hacking up my coil just yet. At least until after my next post if you're still tempted.
To be honest though, I think it would amount to the same degree of boost as the single feedback point as you have it right now anyway.
.99
When I wind mine I'll put taps in.
Quote from: wattsup on July 11, 2009, 11:32:30 PM
@gotoluc
I also agree with @poynt99. If someone grabbed an image of the bulb each time you made a change and blew up the image and put them side by side, I am sure there will be some differences. At that low brightness level, a small change could easily mean 50% more or less and it would still be hard to make out the difference visually.
Also, during both video tests was the return diode on or off. This was not clear to me. It may have been good to switch it in and out to see the difference at each setting. But very good work.
Hi wattsup,
go ahead and grab a frame of each test since I'm quite positive the levels are all the same. I have been using this technique for over 6 months and I know how to keep the level the same. If you look at the bulbs barley glowing filament you will see a slightly darker spot always at the same spot. That is the right side filament support and I use it for fine tuning, one micro change and it starts to glow. I keep it just before that, so I'm sorry but I disagree with you on this one. It cannot be off by that amount. For beginners using this technique I would recommend to bring the bulb to the point you can just barely see a glow, since this is more accurate but one needs to find the right bulb to do this.
In both test 5 and 6 the diode is recirculating. I did not switch off because I have demonstrated that in the previous videos, it also takes up time and it is the recirculating diode effect that I'm studying.
@.99
Thanks for the advice.
@TK
Looking forward to see that demo ;D
Luc
@gotoluc
Thanks for your reply.
I put in some overtime and grabbed each bulb image, saved them at 1200 dpi resolution, blew em up 3fold, then grabbed the whole thing to show. I also summarized your Video6?
This is just a small technicality and would never show any sign of major disparity between what you have shown and the results, etc. But when you say the bulb is the same intensity, you then have to show how the same they are. I could see it by eye but wanted to put it here. We can make our own judgments. The bulb was never as bright as when you put it on straight DC so that is already a winner. But academia will demand more precision even though the end result is obvious.
Hope this helps.
wattsup
Added:
I forgot to mention that these photos would not be possible if your handy cam work was not so precise at each test giving the same angle shot. I must say, the whole video was very well done. An example to all.
Hi everyone,
Rosemary was asking me about the circuit used in my test 4 video and I drew it by hand for her and though to also post it here if anyone is interested.
I also just saw her circuit today for the first time! not much difference ... the only difference is in her circuit the flyback is created by the small inductance of a heating element and in mine it is created by a much higher inductance coil.
Luc
WOW wattsup :o
that is a massive amount of work :P... I never thought you would do it ;D
Thank you so much for putting this together. I'm sure it will help for those how want to look and compare the results.
I'm also surprised to see the intensity is quite similar since none of the shots were done exactly at the same distance and would produce some differences. I can use a tripod if ever I do this again or better yet as TK has suggested a photo resistive cell and monitor the voltage output.
Anyways, thanks for doing such an amazing job and to think you used only the Youtube resolution.
Luc
Hi all,
I just found out that in fact there is a very large difference with my circuit and Rosemary's:
Originally Posted by SkyWatcher
Hi folks, Hi gotoluc, I did not realize that is the way you had the output resistor within circuit gotoluc. I was using Rosemary's circuit which does give more heat because I tried both ways although not sure if the coil field would reduce. Here is a cad pic of the circuit I've been using, which I believe is the same as Rosemary's circuit.
Hi SkyWatcher,
humm... you're right!... I see now there is a very big difference. In Rosemary's circuit the resistor causes resistance during the on pulse, which means it will be heated by the supply energy and then heated by the flyback when it switches off. In mine, I only wanted the flyback to go though the resistor.
Wow SkyWatcher, thanks for bringing my attention back to this since I thought that both circuits were the same.
If I were you I would think about this as you maybe getting more heat in Rosemary's circuit but if you think about it you are paying for it since it comes directly from the power supply during the on cycle.
I chose not to do that since flyback effects is what I've been studying.
Anyways, do what you think is best.
Luc
For your entertainment ;D and or correcting my understanding.
The below was posted by a YouTube user that I had been believing to be an engineer.
Posted by Drevtoobe (1 day ago)
Hey Luc, just a few pointers on measurement techniques for your consideration. For your temperature measurements, everything is relative to the ambient temperature in the room. How hot the resistors are is not about their absolute temperature, it is about their temperature difference from the ambient temperature. All of the information is in the difference temperature. That gives you a sense of how much energy is being dissipated.
One of your big resistors is elevated in the air, and the other one is on your table top. The average experimenter might noght realise that this makes a huge difference in your temperature comparisons between the two resistors. To make the comparison fairer the resistor on the table should also have been elevated in the air.
When you elevate the second resistor, then both electrical resistors with then have a comparable "thermal resistance" to the external environment.
Posted by gotoluc (1 hour ago)
I agree with what you are saying but that is not what is causing the huge differences.
Do you want me to do a video to prove to you how insignificant of a different what you are saying will do?... it will be a pleasure! others will see what a joke this HUGE difference you are imagining this will be.
Luc
Posted by Drevtoobe (6 hours ago)
Luc: Yes, the second resistor is hotter than the first resistor because the first resistor is not showing you all of the energy being supplied by the battery. Look at your schematic, when Q1 switches on, at the beggining, most of the battery energy is being stored in L1, and a smaller amount is being dissipated in R1. When Q1 switches off R1 does nothing and L1 pumps all of it's stored energy into R2, it discharges. You absolutely have to understand this to advance.
Posted by gotoluc (1 hour ago)
Drevtoobe, your arguments are becoming weak! Here are your explanations with my Yes or No replies at the end.
#1, Q1 switches on, Yes!
#2, energy is being stored in L1, Yes!
#3, a smaller amount is being dissipated in R1, No!
#4, When Q1 switches off R1 does nothing, Yes!
#5, L1 pumps all of it's stored energy into R2, Yes and more also.
#3 if all the energy that is stored in L1 is first going through R1 then that amount of energy will be affected by the Resistance of R1 and will heat the resistance of R1. You cannot bypass the energy that is FIRST going though R1 which IS THE SAME Resistance as R2 and dissipate it in R2 and say R2 is heating more because some of that same amount of energy that FIRST went through R1 was able to bypass the resistance of R1. Do you realize how absurd of an argument this is!
#5, are you also considering that when Q1 is switched off and L1 is sending back all its stored energy though R2 which is outputting more heat then R1, the first resistor that SAME AMOUNT OF ENERGY it had to go though (which you believe to be normal) but you are forgetting that some of the energy makes it through R2 and goes BACK in L1 and keeps a one pound magnet levitating! So in fact I'm not even close to dissipating all the energy in R2 that was able to mysteriously slip through R1.
Luc
To everyone: I am open to corrections if I'm not understanding correctly.
Okay, here is the new video with all the measurements hopefully :P
The only thing I thought of after the video is I did not show the temperature of an non heated item in the room. So I measured one and it is at 24 degrees Celsius.
Video Link: YouTube - Effect of Recirculating BEMF to Coil test 7 (http://www.youtube.com/watch?v=9UH_v5EO7-A)
Luc
If his next reply is..."oops, my bad..I thot R1 was in parallel with the line", then he be no doctor.
Great work as usual Luc...I like where this is going.
Regards...
Okay, here is my last video for the day for your enjoyment purpose only ;)
Let's not say the words please, since we may get visitors ;D
Video Link: http://www.youtube.com/watch?v=xvE7IGCra14
Love, Peace and Light to all
Luc
Luc,
Finally got this thing finished.
My response to your video #3 and #6 was getting a little long for a post so I put it in a pdf document.
I hope I have covered all that will explain what you are seeing in your circuits and why. I hope also that it will give you a better understanding of these circuits and how in general to test them and design with them.
I'm not perfect, and likely neither is the document. I hope it helps out though, that was the goal.
Feel free to ask for clarification on any part of it all. If I've confused more than enlightened, well sorry about that ;) .
Cheers,
.99
Wow .99,
that looks like an amazing :o amount of work you have put into this document.
Blow me away :P I don't know how much time it will take me to understand this information if ever, since integrating by reading is my weakest side.
With that said, I would suggest to you to post it at the Energetic Forum for their viewing and evaluation.
Once again, thanks for doing this.
Luc
Quote from: gotoluc on July 13, 2009, 08:51:38 PM
Wow .99,
that looks like an amazing :o amount of work you have put into this document.
Blow me away :P I don't know how much time it will take me to understand this information if ever, since integrating by reading is my weakest side.
With that said, I would suggest to you to post it at the Energetic Forum for their viewing and evaluation.
Once again, thanks for doing this.
Luc
You're under-selling yourself Luc ;)
You've obviously learned a heck of a lot in a short period of time on your own, and I think that's great. From your posts I can tell you'll be able to handle what I wrote there, otherwise I wouldn't have spent so much damn time working on it :D
Take one paragraph at a time and remember this stuff isn't difficult. I know it's a big chunk of stuff all at once, but If it makes it seems any less :-\ think of it as an accumulation of about 10 or 15 posts :)
.99
Dear Poynt99
Could you please send us the complete formula of this system so that we can analyticly play with formula to increase ringing.
I mean can we derive a complete formula which includes all L,C,R values, freq, duty cycle, Voltage parameters and it draws resulting ringing currents?
Also does Lenz law work in space? Vacuum?
What is the cause of this Lenz reply? Is space is elastic?
If space is elastic to magnetic triggering, then we should find its formula or constant?
Gotoluc's space elasticity constant= ?
Great work gotoluc.. Congradulations..
Could you please speak with Thane about his BTT?
Love and brothership for all humans..
NT,TR
Quote from: samedsoft on July 14, 2009, 12:36:03 PM
Great work gotoluc.. Congradulations..
Could you please speak with Thane about his BTT?
Love and brothership for all humans..
NT,TR
Hi samedsoft,
thank you for your positive comments :)... the questions you are asking .99 are dificult to answer I think.
Thane is aware of the circuit but I have not done a live demo for him yet since it kept developing each day. I don't want to crate false hopes so I've kind of also been waiting for those skilled in the art, like .99 to evaluate it.
Love and Light
Luc
Hi gotoluc,
what is your claim in your video comparing the temperature of the two 22 Ohm resitors ( accoring to your circuit in reply 67 ) ?
in order to get real input-power data you have to put a 0.5 Ohm resistor in the plus-line between plus-terminal of the elko and the upper feed-in-point of the coil and measuring with your scope the periodic voltage-drop across this 0.5 ohm-resistor. Just measuring the temperature of the resistor in the source-line gives a total wrong impression regarding in- and output of energy-flow in this circuit
The total work of the levitating magnet consists of the power-in during the on-cycle and the recycled energy via the flyback. The energy flowing into the circuit during the on-cycle is split between the E_Mag in the coil and the caloric energy in the source-resistor.
Most of the input energy is consumed in the coil during the forward-cycle
There is no ou in this setup
Regards
Kator01
Hi Kator01,
I make no claim as I do not feel qualified ::)
Your conclusion seems that you are.
Luc
Hi All,
I have just uploaded a new video for your evaluation and opinions.
Please post your comments.
Link to Video: http://www.youtube.com/watch?v=gIbFIXJC6IA
Luc
Hi Luc.
Stellar video as usual.
One thing I recommend be differentiated is the difference between "energy" and "current". Do you remember the following relationships?:
P = V x I
E = P x t
I think what you meant to say was that the series string of 9V batteries has the same "current" capability as a single 9V battery, but the voltage is raised to a higher value.
Take a look at that first equation again. You'll notice that since your voltage has increased, you will have a corresponding increase in available power, and hence energy.
So what you have shown is similar to your other recent experiments with the same coil. In this case the load power (bulb) is evenly distributed among all the 9V batteries there, so indeed you should see a much more gradual decline of any one battery's terminal voltage. ;)
Are you beginning to get the idea that it's pretty handy to be aware of and be able to utilize a few simple equations such as the first one there? That one equation explains what you are seeing.
.99
Dear Poynt99,
In P=V.I.Cos(Phi) equation one has to measure current and voltage lag and put Phi on Cosine function to measure active power.
If you connect Gotoluc system to grid, you may draw reactive power from grid. It is forbidden in Turkey. Thats why we have compensation panels in industrial facilities.
But as in Thane's bifilar transformer, you can convert reactive power to active power. I asked Thane to use DC battery or Capacitor to measure and calculate how much power is transferred.
In short, Gotoluc's system may be drawing reactive power from grid. So DC source like battery is much more preferred or you have to declare the lag between Current and Voltage drawn from grid!
Love & Brothership
NT,TR
Hi Luc and All,
there has been a lot of discussion here about measuring efficiencies and possible OU? I can kind of follow along. I am not an electronics expert. But I found a post by user named bolt that was elegant in its simplicity here:
http://www.overunity.com/index.php?topic=455.0
"
Another way is find a value resistor on the input that just gets hot then take that out and put the same value in the output. If the output one goes up in smoke you have OU "
Makes sense to me
tishatang
@gotoluc
Again nice video.
I would be curious to see what would happen if you put the output bulb in series with a small transformer (110vac primary and 12vac secondary) secondary and then put the primary back to the battery terminals via a diode. This way you could light the bulb and try and send some juice back to the batteries to see if the voltage drop will be less or even increased voltage.
Or put a diode on the bulb positive pointing to the positive of battery and put the bulb negative to the battery negative without a transformer.
wattsup
Quote from: poynt99 on July 15, 2009, 01:37:35 AM
Hi Luc.
Stellar video as usual.
One thing I recommend be differentiated is the difference between "energy" and "current". Do you remember the following relationships?:
P = V x I
E = P x t
I think what you meant to say was that the series string of 9V batteries has the same "current" capability as a single 9V battery, but the voltage is raised to a higher value.
Take a look at that first equation again. You'll notice that since your voltage has increased, you will have a corresponding increase in available power, and hence energy.
So what you have shown is similar to your other recent experiments with the same coil. In this case the load power (bulb) is evenly distributed among all the 9V batteries there, so indeed you should see a much more gradual decline of any one battery's terminal voltage. ;)
Are you beginning to get the idea that it's pretty handy to be aware of and be able to utilize a few simple equations such as the first one there? That one equation explains what you are seeing.
.99
Thanks for the reply .99 :)
Should the 9 batteries in series last 9 time longer then using the same load on a single battery for it to be unity?
Luc
Luc,
Battery chemistry is a little odd and they don't work in perfect linearity.
However, your statement is fair enough.
Definitely, the 9 batteries will last quite a bit longer than "one" if used to drive the same power into the bulb for both cases.
.99
Hi Luc,
QuoteHi Kator01,
I make no claim as I do not feel qualified ::)
Your conclusion seems that you are.
Luc
Now this is an interesting statement. We do not talk about each other´s qualification, right ? We do talk here about efficient and correct means to exactly measuring input-power.
I have done hundrets of experiments with a similar circuit, the only difference was
1) I did not use a magnet as the magnet changes the inductance in a periodic way because the magnet is phyically vibrating like the coil in a speaker. In this way there are no stable parameters in the coil
2) I did not use a load in the recycling-circuit driven by the diode.
instead I loaded a second capacitor and thus could exactly compare the
input versus the output. by comparing the energy stored in the second
elko.
3) I was using very high quality Krummer-capacitors ( 200 V, 33 000 myko Farad) which can take a few Kiloampre current when fully shorted.
http://www.krummer-kondensatoren.de/ (http://www.krummer-kondensatoren.de/)
In order for all of us to become more qualified in this specific topic ( I include myself here ) it would be a good idea to know the exact power which enters this circuit in order to get the clue why you present this here. This can be easily done by ( a slight modification of what I have proposed in my last post ) by feeding your capacitor at the left via a 0.5 Ohm resistor and measure the voltage-drop Ud across the resistor and calculate the current according to I = Ud / R. Then multiplicate this current which enters your capacitor-set with the voltage directly at the plus-terminal of the cap feeding the main switching-circuit.
Another good idea would be to use a 10 000 myko-Farad Elko for your supply-capacitor in order to eliminate any voltage-ripples so this measurement and the calculation can be done according to the above simple formula without any major error.
It is just pretty easy to do this. This will give you the exact amount of power flowing into the circuit.
Also when I watched your video I could see that the laser-light-point of your infrared-thermometer was not on the resistor but somwhere above it. Usually you have to measure at a distance of about 20 centimeter in order to get the two laserbeams overlap each other in one point. Only then will you get the correct reading.
Now using a infrared-thermo-device ( which is digitally controlled ) in such a vinicity of a strong oscillator is another big questionmark regarding correct
readings.
I hate to say this luc, but I have gone through many measurement-nightmares myself but I had 2 very good professional teachers in my life which gradually upgraded my knowledge which is far from beeing considered an expert-knowledge. But I try my best.
Best Regards
Kator01
Kator,
I have suggested (based on a recommendation from aethertech) something similar for Luc to obtain an easy Pin measurement. My post and diagram is at EF:
http://www.energeticforum.com/61265-post140.html
Here is a recent post from Aaron again espousing his nonsense:
QuoteLuc,
I don't know the status of your testing at this very moment but heard about the recommended test with the meter. It gobbles the gain. Your circuit and demonstrations are easily OU. The battery is needed to show it.
Would you care to explain WHY "it gobbles the gain" Aaron?
Good grief!
.99
Hi poynt99,
yes.. what was the name of this band : foolish garden ?
it is very simple : they keep gullible people busy and attack those sceptics with a good technical background.
I have never seen such a foolish remark... actually too foolish to assume that it was
done unintentionally.
In the middle ages the inquisition-priests tested a suspicious person in an open court-session by touching the arm of the suspect with a metal bar. In the front of this bar embedded was a sharp thorn which was ejected a few millimeter upon touching the skin.
If the suspect person was screaming he was proven to be a witch.
Boy, our proposed methods of measurement gives proof : we are the wicked priests...
Poynt99, this thing is sort of developing into some sort of religion.
This sort of discussion-activities is a waste of time.
Silence ? !
Regards
Kator01
Hi everyone,
a quick video update of the last test I'm doing before my trip out of the country this Thursday.
This test is using three 12vdc batteries in Series at source and 3 identical batteries in Parallel on the flyback recirculated side.
I find the test to be already demonstrating a good result compared to my previous test using only one 12vdc battery on each side.
It has now been running for 3 hours and the voltage at the series source is at 39.5525 and when I started the test they were at 39.7534. The interesting part is I adjusted the duty cycle to maintain the charge level voltage of the 3 parallel batteries with the attached load (bulb) and it has maintained 12.99vdc for the 3 hour period.
Once you see the video you may understand more: http://www.youtube.com/watch?v=g-XxgCtPtzI
Luc
@gotoluc
Nice video as usual.
I am curious if you can try this.........
Do the same set-up but put only one battery on the output and add one wire going from negative of first series battery to the negative of the output battery, then put another wire with a diode from the positive of the output battery to the positive of the series battery that also has the negative. Try it, to see what happens, then move the negative and positive on the first series battery to the second series battery and see what happens, then move them again to the third series battery.
What I am thinking is if you have only one output battery to stabilize the output end for your bulb, then any excess can shoot back to the feed source via that diode.
Maybe one other note on the slow voltage drop on the feed side. Guys will want to make a joule calculation, etc., since those three series batteries do have a good level of energy available, but maybe with the above trial, the voltage will go down even less (or maybe go up) hic hic.
Just an idea. Always good work.
Hi wattsup,
thanks for the good suggestion :) I like it ;D
However it will have to wait, since I'm going out of the country on Thursday.
I'll be back August 9th.
Thanks
Luc
Hi everyone,
last night I made a new video but could not upload it since my internet connection is down. I found a few free wireless spots so I was able to upload it today.
New Video Link: YouTube - Effect of Recirculating BEMF to Coil test 11 (http://www.youtube.com/watch?v=qYIK40Yp-68)
After I stopped the circuit and allow the batteries to rest for 2 hour the source series battery voltage recovered to 39.3958
and here are the individual battery voltage
Battery F 13.1249
Battery E 13.0817
Battery G 13.1894
and the voltage of the 3 Batteries in Parallel recovered to 12.99
This is all I can do until I return from my trip after August 8th
Luc
Hi everyone,
I made a quick change to the circuit this morning. I removed the 3 batteries in parallel on the flyback side and replaced it with a 30,000uf capacitor to give a large storage so we can tune the duty cycle and get our 12.95vdc with load (bulb) attached to have a comparison to the previous test 11.
It was so fast to see the results that even after 5 minutes it was obvious to me that there is a real humongous efficiency bonus using the batteries.
Video link: YouTube - Effect of Recirculating BEMF to Coil test 12 (http://www.youtube.com/watch?v=StLdP6aWaec)
I had to stop the test after 1 hour and 40 minutes since the voltage on the source batteries started making large drops to 35vdc and climb back to 38vdc and back down again. I don't know what is up with that but anyways it's clear that there is an astronomical efficiency boost using the batteries in parallel on the inductive flyback side.
I hope others can replicate this so we can advance to an even more efficient way to use this effect. I'm now realizing that if I would of tested this method in my resonance experiments I may of had some better results. So many things to test now.
I'm out of the country for a week. Talk to you all soon.
Luc
Hi everyone,
I tried to replicate my test 12 video using a capacitor instead of the battery to try to understand why there was so much difference between the two tests (11 and 12) and I found out that one of my 3 batteries is defective and that was the main cause of the huge difference. So I deleted test 12 video so not to mislead anyone. Anyways, I'm replacing it with the one below.
This video demonstrates that a Resistor can attain the same amount of heat connected directly (no Coil) to the pulse circuit then connected to the inductive kickback side of a coil. Both tests were tuned to use the same amount of energy from the Series battery bank.
What I find interesting is, you would think that by adding a coil it would make it less efficient but it seems to be as efficient as without the coil.
So where am I going with this!... well, if we add a coil in the circuit and it cost nothing, and we can still produce the same amount of heat that's good! since the the coil could be doing work, like turning a motor and then the inductive kickback will keep the hot water tank hot or something of that kind.
Just sharing ;) ... let me know what you think.
Video Link: YouTube - Effect of Recirculating BEMF to Coil test 12 (http://www.youtube.com/watch?v=FYKf8zXBzEk)
Luc
@gotoluc
Welcome back and good video.
I think one question many will have is does the pulse driving circuit get its power from the same battery bank. That was kind of unclear.
I have a side question. How did you find the other wire end in that big spool of wire. I have tried finding the other wire so many times but could not on three of my 10 lbs spools but with much finer wire - 28-29-30 awg.
wattsup
Quote from: wattsup on August 16, 2009, 08:47:55 AM
@gotoluc
Welcome back and good video.
I think one question many will have is does the pulse driving circuit get its power from the same battery bank. That was kind of unclear.
I have a side question. How did you find the other wire end in that big spool of wire. I have tried finding the other wire so many times but could not on three of my 10 lbs spools but with much finer wire - 28-29-30 awg.
wattsup
Hi wattsup,
thanks for looking and the welcome back :)
In all my tests the pulse circuit uses its own 12vdc battery which is still at 12.80vdc even with about 100 hours of use.
Some spools ship with both ends exposed and some not. One way or the other I should be rewinding this wire since in this spool configuration I'm fare from obtaining Maximum Inductance.
The performance should be better once redone. Here is a great program I found for calculating best spool or coil geometry for an inductance value. If you or anyone else know of another program that would work from a fixed wire length to give max inductance please let me know.
Thanks
Luc
Luc,
A battery's voltage is probably not a reliable indicator of the load it sees. I don't believe a battery's voltage drops linearly with load, especially if it is fully charged.
Get yourself a "current shunt" or "current sensing resistor" (both non-inductive) and use it inline with your battery supply. Use your scope to probe across this current sensing resistor to obtain the voltage wave form. Measure the duty cycle and voltage peak and calculate the power being drawn from the battery. That's the sure way and it "enables" you to perform measurements at the "next level".
I and others here can help you if you're willing to take this next step in your measurement protocol.
.99
Thanks .99 for your post and offer to help as always ;)
My batteries are not fully charged as I have not re-charged them since I started the series of battery tests.
I do have a shunt but I think it's too large for my delicate tests, it's stamped 60A 75mv and FL - 2 (unreadable symbol) 0.5
I tried it with my probe across it on the negative feed side and see no peaks during the on time. However I do have a one watt carbon 1 Ohm (brown, black, gold, gold) resistor that I tried and it shows the on peaks very well.
The duty cycle is much lower when connected straight to the resistor then to the coil and the wave form on the scope is very different in each case. Strait to resistor is much like the on pulses of the 555 but with slightly rounded corners and connected to the coil it goes up on a 45 degree angle and drops straight down when switched off.
I don't trust the RMS reading my scope calculates since it changes quite drastically when increase the times division. So I'm lost as how to calculate it.
Luc
Luc.
You can try the 1 Ohm.
The wave forms you saw are correct.
You have a pretty good meter there and it would be nice to use it. So this is what I would like you to do, if you are so inclined: We need to see how your meter interprets certain wave forms.
Does your signal generator have an "offset" function? What frequency are you using from your 555?
What I want to know about your meter is if we give it the following, what will it read out?
Generator (1kHz) | Meter DC-RMS | Meter AC-RMS
________________________________________________________
50% duty, 0V-1V (pulsed-DC)
25% duty, ""
50% duty, -0.5V-+0.5V
25% duty, ""
Hopefully one of the two (AC or DC) readings will be correct (I'll crunch the values), and then we'll know which to use across the 1 Ohm current sensing resistor.
.99
Luc and .99:
That's a really great multimeter. I was doing some hunting and you might find this interesting:
Quote
The Schlumberger / Solartron 7150, 6.5 Digit Benchtop Multimeter features AC & DC voltage ranges 0.2 V to 1000 V (5 ranges, true RMS on AC). Resistance ranges 2 kohm to 20 Mohm (5 ranges). Linearity < 0.001% of range. DC current to 2 A, ac current to 2 A. Temperature measurement mode (-200 C to +600 C, requires additional platinum resistance probes).
http://www.teknetelectronics.com/DataSheet/SOLARTRON/SOLAR_7150103803.pdf
MileHigh
P.S.:
I found the Mother Load, but it is a really slow connection, best to "save as."
[EDIT: The link below gives you access to the 7150 Plus user manual. Click on "7150" at the botttom of the page.]
http://bama.sbc.edu/solartron.htm
Hey MH.
Yeah,
I've had the manual downloaded some time ago shortly after Luc first showed it on his videos. ;)
.99
Smart ass. :)
P.S.: Did you ever notice how you have to use "mind power" to keep a download with a slow and flaky connection going? I swear, if you take your eyes off of it for too long it stops. Then a good "mind power" stare-down session gets it going again. Do you think we can tap into that? lol
P.P.S.: It's not me it's them. I have a vanilla DSL connection. In my case very occasionally I come across a site that has a good server and nobody around. I downloaded a game demo once from an Australian web site at a nice blazing 600K. That distinctive modem negotiation sound is now a distant memory.
Hey MH.
I don't have to worry about slow connections, not since about 12 years ago ;) But yeah I remember playing that game too. ::)
Glad those days are gone :P
.99
Quote from: poynt99 on August 16, 2009, 12:26:02 PM
Luc.
You can try the 1 Ohm.
The wave forms you saw are correct.
You have a pretty good meter there and it would be nice to use it. So this is what I would like you to do, if you are so inclined: We need to see how your meter interprets certain wave forms.
Does your signal generator have an "offset" function? What frequency are you using from your 555?
What I want to know about your meter is if we give it the following, what will it read out?
Generator (1kHz) | Meter DC-RMS | Meter AC-RMS
________________________________________________________
50% duty, 0V-1V (pulsed-DC)
25% duty, ""
50% duty, -0.5V-+0.5V
25% duty, ""
Hopefully one of the two (AC or DC) readings will be correct (I'll crunch the values), and then we'll know which to use across the 1 Ohm current sensing resistor.
.99
Okay .99,
since you have the manual to my meter you know more then I about it.
My SG is a Wavetek 134 and I don't think it has an offset but what do I know :P
The frequency I used on my 555 for the last test was 145Hz
That is about all I have.
Luc
Hi Luc.
I found the manual for your 134 Generator here:
http://www.el.bqto.unexpo.edu.ve/tperez/LabME/MWavetek.pdf
It looks like it does have a DC offset control (+/-5V), but it is adjusted from the back of the unit ;)
This is a pretty good generator too. You did well if you got it for $25 ;)
145 Hz? OK, I thought it might be quite a bit higher.
.99
Luc,
There is also this if you recall I posted at EF a while back. This is simple and still valid for your recent testing. No special meters required if you do it this way.
QuoteLuc,
Here's a simple circuit suggested by aethertech and similar concept by gyula to ease input power measurements.
I had never tried this before so I PSpiced it up to check against the more direct method.
I'm happy to report that it seems to work quite well. All you need is a couple fairly large electrolytic capacitors and a sufficient wattage 10 Ohm resistor (or close). Connect as shown in the diagram here. If you use a wire-wound 10 Ohm it will filter even a little more.
Take two cheap DC voltage meters (both set on "DC Voltage"), one across R3 to measure the voltage across/current through it (I=V/R), and one across the Vbat point on the diagram to measure output voltage V.
Input power to the circuit is now Pi = V x I
Hope that helps.
.99
Hi .99,
I now see the DC offset ;)... what to do with it is another story :P
Hey, you even know I paid $25. for the Wavetek ;D
Is there anything you don't know ::)
Yes, you're right! I didn't try the capacitor resistor arrangement yet. I'll put it together and test it.
I'll post what I find or a video if it gets good score.
Luc
Hi everyone,
I just uploaded a new video as a re-test of the previous test 12 to re-confirm the results that we can create a magnetic field while creating heat at next to no cost. The difference in this video is I'm using the capacitor resistor circuit that .99 suggested to be able to easily measure current draw.
The results are not quite as good as it was in test 12, however we need to keep in mind that if some of the energy is going back to the batteries I would believe this circuit would not work with that effect. You be the judge!... take note that the voltage drop on the batteries stays in the exact same range from test to test as this was the calibration reference I used in test 12 to re-adjust duty cycle between tests.
Video Link: http://www.youtube.com/watch?v=9VHL2lMkepI
Luc
Here is another video that continues the test of video 13 and is also testing an effect that user: Michael John Nunnerley of the Energetic Forum believes to occur.
Here is Michael's topic: http://www.energeticforum.com/renewable-energy/4604-two-element-bemf.html
Link To video: http://www.youtube.com/watch?v=QqoXEqZ7t_0
Luc
Hi Luc.
Was wondering, are you measuring battery voltage across the batteries or across C2?
Any inductive kickback occurring will go back into C2 just as well as it would go back into the batteries, but any recharging effect that may add to the battery charge due to ion flow would be lost, if in fact this OU effect did occur.
Thanks,
.99
Quote from: poynt99 on August 19, 2009, 08:01:45 PM
Hi Luc.
Was wondering, are you measuring battery voltage across the batteries or across C2?
Any inductive kickback occurring will go back into C2 just as well as it would go back into the batteries, but any recharging effect that may add to the battery charge due to ion flow would be lost, if in fact this OU effect did occur.
Thanks,
.99
Hi .99,
battery voltage is measured across C2.
I agree that the switch flyback would go back to C2 but like you say, it would have no ability to recharge the capacitors chemistry ;D
So, is the test doing what you thought it would do! or are you surprised it's doing what it is.
I think the way I was calibrating the duty cycle in test 12 using the battery bank voltage drop between each circuit was accurate enough and a fair comparative since we can clearly see that the voltage drop is exactly the same between each test using the cap resistor method. Would you not agree?... and if we can get the resistor to heat to a higher temperature with the identical voltage drop as with the cap resistor circuit that can only support that some of the energy is making it way back to the batteries and giving them a recharge.
Hey, here's a new test. I start the inductor resitor circuit with the cap resistor circuit adjusted to draw the same 25ma till the voltage drop is very stable and pull the feed connections from the caps and connect them directly to the batteries. If the battery voltage starts to rise then the recharge effect is real and valid. Agree
Luc
Quote from: poynt99 on August 19, 2009, 08:01:45 PM
Hi Luc.
Was wondering, are you measuring battery voltage across the batteries or across C2?
Any inductive kickback occurring will go back into C2 just as well as it would go back into the batteries, but any recharging effect that may add to the battery charge due to ion flow would be lost, if in fact this OU effect did occur.
Thanks,
.99
@99
I am glad to see this comment coming from you as this shows you are really checking @gotolucs back during these trials. This then adds to the complexity of trying to measure things in a system that relies on clear passage for flyback return. I imagine that the undoubtable end will be as we do now, to try and run a load and notice little or no battery discharge.
Or is there a more non-intrusive way of checking this like passing the wire through a small ferrite coil and taking measurements off the coil? A clamp on DC ammeter would also be good.
Updated
Hi everyone,
I just deleted test 15 an 16 video as I have found an error in my calculations and only just noticed it now when testing my new meter. It appears that there is a .25 Volts drop at Capacitor No. 2 then what is actually at the batteries and if we do the calculations we will quickly see that it come to the same difference I measured yesterday at 75.877 with cap circuit to 76.12 direct from batteries.
So I'm sorry to say but I was wrong and need to withdraw my conclusions.
I will let you know if I find anything more.
Luc
Hello,
please read carefully how to measure rms-values
Regards
Kator01
Quote from: gotoluc on August 20, 2009, 05:51:19 PM
Updated
Hi everyone,
I just deleted test 15 an 16 video as I have found an error in my calculations and only just noticed it now when testing my new meter. It appears that there is a .25 Volts drop at Capacitor No. 2 then what is actually at the batteries and if we do the calculations we will quickly see that it come to the same difference I measured yesterday at 75.877 with cap circuit to 76.12 direct from batteries.
So I'm sorry to say but I was wrong and need to withdraw my conclusions.
I will let you know if I find anything more.
Luc
Hi Luc.
This is precisely what I was getting around to telling you yesterday with my post at EF (first I wanted to see if you noticed the discharge rates for each). I'm glad you figured it out on your own.
I am sure that the measured voltage would slowly decrease at about the same rate in both cases, but with a 0.25V offset due to the drop across the 10 Ohm. Sorry for not mentioning it then, but I had doubts you would have been too receptive. ;)
.99
Hi all,
here is the status at this time in a video update.
Link to Video:http://www.youtube.com/watch?v=LaCGfnrkkXU
Luc
Hi everyone,
it's been a while since I've posted. Been busy making some $ and stocking up while I can for the long Canadian winter :P
Tyson, user Skywatcher at Energetic Forum asked me if I had any ideas to use this effect and I said no but honestly I always had a motor in mind ::)
So I decided to go back to my basic idea and re-tested with a simple yet effective way to see if it has any advantages in a motor.
I made a new video and I think it's simple enough that most anyone to understand. The result are quite interesting comparing a straight DC capacitor discharge to the coil and magnet with no flyback circulation vs flyback re-circulation for the same magnet lift, measured height in the video below.
The results are as followed:
No flyback circulation using a 58uf discharge cap charged at 200V = 1.16 Joules.
With flyback re-circulation using a 58uf discharge cap charged at 84V = 0.205 Joules.
That is close to 1 Joule less for the same height :-\ ... can someone please explain what I maybe missing here ::) ... would this not be of benefit to a DC magnet motor???
Link to video: http://www.youtube.com/watch?v=7QUYkilgkzU
Thanks for your time.
Luc
Luc:
Quote from: poynt99 on July 05, 2009, 12:36:34 AM
Luc.
In my tests I've confirmed all your findings, except the magnet elevation, but since the magnetic field in a coil is proportional to the current through that coil, I think it's safe to say that is covered as well.
What you've built is essentially a DC-DC converter. You've taken a high voltage low current source and converted it to low voltage / high current in a load. In this case the load and conversion element are one in the same, the coil. The diode completes the appropriate circuit path that allows the conversion to actually take place.
Some numbers I obtained from my tests:
Ave VS Ave IS Ave WS Ave VC Ave IC Ave WC
Without Diode: 170V 15.5mA 2.635W 157mV 15.5mA 2.43mW
With Diode: 170V 10.6mA 1.802W 2.932V 285mA 0.836W
Where:
VS = Supply Voltage
IS = Supply Current
WS = Supply Power
VC = Coil Voltage
IC = Coil Current
WC = Coil Power
(these are averaged values)
Indeed the current from the supply decreases with the diode in-circuit, but the output power from the coil never exceeds the input from the source.
Notice the huge increase in coil current with the diode in-circuit compared with it out of circuit? This explains the force it has on your neo magnet. With coils, it's all about current. A substantial amount of power has been transferred to the coil with the diode present (about 46% of the input), as opposed to nearly none without the diode.
Are there any gains in power or energy? No, in fact there is quite a substantial loss due to the DC resistance of the coil and connecting wires.
I hope this explains all that is happening with your experiment, but I'd be happy to expand on or run tests on any aspect if you want. Scope shots are also available if you wish.
Regards,
.99
Quote from: poynt99 on July 06, 2009, 01:22:34 PM
Hi Luc.
Monitoring a coil's voltage does not always indicate what is going on there in terms of its current. The fact that the bottom portion of the coil voltage disappears when the diode is in-circuit, actually does indicate that a conversion is taking place. The conversion is from high voltage/low current, to high current/low voltage. If you zoom in when the diode is in-circuit, you will probably see a negative swing of about -0.65 Volts (from the previous -300V or so), but the current has shot up significantly. This is the indicator that "something" different is happening in the coil that is causing the increased magnetic power.
With no load (i.e. without a flyback diode), the coil sees almost an open circuit during its inductive kickback cycle. The same amount of energy (minus losses) must be conserved, so the coil voltage extends quite high in the reverse direction, but the current is quite small.
When the coil IS loaded during its inductive kickback cycle by placing the diode across it in reverse, the diode creates nearly a short circuit across the coil during this cycle, so the current has no choice but to increase by a large amount. It is this increase of current that you can not see by looking at the coil voltage, but this increased current is what is responsible for the much larger magnetic force being applied to your neo magnet.
Remember with coils, it is current that energizes them and produces a corresponding magnetic field, not voltage. With a heavy enough wire you could produce quite a strong magnetic field with only a few volts, as long as the source can supply a large current.
Hope that helps,
.99
and with the document:
http://www.overunity.com/index.php?topic=7713.msg191135#msg191135
Peter L. is the only one that is somewhat close so far, but not quite. It's simple: Adding the flyback diode causes the inductor current to be "rectified", similar to how voltage is that's used to charge a capacitor. Because of this rectified current, the average, or mean current in the inductor increases substantially, and as such, so does the magnetic force from the coil which kicks the magnet. Remember also that the flyback diode helps convert voltage to current, which is what is needed for a stronger mag field. See the comparison of the coil current wave forms at the end of the document.
.99
Just for fun. let me put my two bits in here.
Without the fly-back diode in place, you have an LC resonator that also pushes the magnet up when it first engages. Keep that in mind.
For both setups you can split the sequence of events into Step 1, from initial relay closure to the point in time where the cap has discharged down to zero volts, and Step 2, whatever happens after that.
Case A: Without the diode in the circuit.
Step 1: The cap discharges, some of the energy pushes the magnet up, some of the energy gets stored because the coil gets energized with current.
Step 2: The current reverses and the cap and the coil and wire resistance form an RLC oscillator which starts a ring down at a high frequency. The ring down alternately pushes and pulls on the magnet for a net zero upwards push. Therefore none of the energy stored in the coil at the beginning of Step 2 helps push the coil upwards and simply becomes resistive heat in the wires.
Case B: With the diode in the circuit
Step 1: The cap discharges, some of the energy pushes the magnet up, some of the energy gets stored because the coil gets energized with current. (The same as Step 1 above)
Step 2: The "LC oscillator" is now charged with current flowing through the inductor. The capacitor voltage starts to reverse but the moment it hits 0.6 volts, the diode starts to conduct and the coil starts to recirculate the current flow through itself and the diode. This sustains the magnetic field longer and you assume that this keeps a steady upward push on the magnet for a certain amount of time. This is in contrast to a net zero upward push on the magnet in Case A.
The net result: In Case A all of the available energy at the start of Step 2 becomes heat in the wires. In Case B some of the available energy at the start of Step 2 pushes up on the magnet, and some of the available energy becomes heat in the wires.
I think that's the way the cookie crumbles.
MileHigh
@gotoluc
Always good video and very well explained.
Also, @MileHigh came up with a pretty good explanation, but maybe about the LC resonator part is not right if there is only one pulse (one pulse, plus lots of time charging, then another pulse, etc.).
I have a question about the relay. Is the relay switching both sides of the caps or just one side? What I mean by that is does the first cap, second cap and the coil share any common connection or are they all isolated by the relay switching two sides?
Wattsup,
MH and I are mostly saying the same thing, and he is correct about the oscillation. However, I will correct him on one point. Without the diode, the ringdown is indeed there, but it has no ability to do work in this phase. It does not push and pull the magnet in oscillation because the current and voltage are not in the right alignment to make real power. The real push only comes right at the start as he said.
When the flyback diode is introduced, the initial current pulse is recirculated within the coil and not only is it present for a longer duration, but it is higher in amplitude as well. It's that rectified current I spoke of.
See the two plots here of average coil current with a single discharge pulse, one without, and one with the flyback diode present.
.99
There are some great follow-up questions to this test. What is the true efficiency or COP for each of the two configurations? That leads to the following question, how do you define the efficiency?
The best place to start answering these two questions is to define what the goal is for this experiment. I'll answer that one to put my line of reasoning into a proper context. The goal is to lift the magnet as high as possible.
MileHigh
.99:
QuoteMH and I are mostly saying the same thing, and he is correct about the oscillation. However, I will correct him on one point. Without the diode, the ringdown is indeed there, but it has no ability to do work in this phase. It does not push and pull the magnet in oscillation because the current and voltage are not in the right alignment to make real power.
I disagree with you here. The magnet will be influenced (have force act upon it) by whatever type of magnetic field is created by the current running through the coil. The magnet is only influenced by the current flow in the coil and not by any IV timing relationship. If you had a much slower resonant frequency and held the magnet in your hand you would be able to feel the alternating push and pull force on the magnet as the ringdown takes place - like putting your hand on a speaker cone and feeling a 60 Hz vibration.
To get a bit esoteric, the physical mass of the magnet is a mechanical low pass filter, i.e.; a mechanical capacitor. The mechanical low pass filter in the actual experimental setup filters out the high frequency oscillations to a great extent so that the up and down wobble of the magnet is imperceptible to the naked eye.
I'll also add that "low pass filter" can be restated as "high frequency cut-off filter" - they both mean the same thing and the second definition makes it clearer for this example. The mass of the magnet is filtering out, a.k.a. "cutting off" any high frequency vibrations of the magnet so that they are imperceptable to the naked eye. Nor would you likely be able to feel them if you were holding the magent in your hand during the ring-down - they would be way too small in amplitude and way too fast for your nerve cells in your fingers to register the sensation.
QuoteWhen the flyback diode is introduced, the initial current pulse is recirculated within the coil and not only is it present for a longer duration, but it is higher in amplitude as well. It's that rectified current I spoke of.
I am mystified by the plots you posted, can't wrap my brain around them. Can you discuss them a but more? The "average current" is throwing me off. If it is possible, can you superimpose the regular coil current vs. time for both cases? I think that would help.
MileHigh
Quote from: wattsup on September 26, 2009, 10:56:37 AM
@gotoluc
I have a question about the relay. Is the relay switching both sides of the caps or just one side? What I mean by that is does the first cap, second cap and the coil share any common connection or are they all isolated by the relay switching two sides?
Hi wattsup,
the caps share a common connection, I'm connecting and disconnecting the ground side only.
Luc
@99
Thanks for your clarification. i see on your wave that the slope is gradually going down at that would in fact show the resonance after one pulse.
@gotoluc
Quote from: gotoluc on September 26, 2009, 12:49:13 PM
Hi wattsup,
the caps share a common connection, I'm connecting and disconnecting the ground side only.
Luc
Of course my next question would be, what do you think the difference would be if both the positive and negative sides were switched by the relay. I think that relay is a DPDT so you may have enough to try it.
One other thing @gotoluc. You have everything on the table now to try The Tesla Project on the thread here; http://www.overunity.com/index.php?topic=3972.0
If you can master the Tesla battery shorting principle and current reversal and flyback that happens, along with your highly disciplined method of testing, observing and showing, I am convinced this will give you a whole new outlook on the potential and teach many here what maybe we could not teach during that thread. The Tesla Ozone Patent is a basis for what I think is in many of the OU device attempts we see today, including some aspects of which are used in the SM TPU.
Anyways, just thought I'd bring it up in case you are looking for something else.
Quote from: wattsup on September 26, 2009, 01:49:31 PM
@gotoluc
Of course my next question would be, what do you think the difference would be if both the positive and negative sides were switched by the relay. I think that relay is a DPDT so you may have enough to try it.
One other thing @gotoluc. You have everything on the table now to try The Tesla Project on the thread here; http://www.overunity.com/index.php?topic=3972.0
If you can master the Tesla battery shorting principle and current reversal and flyback that happens, along with your highly disciplined method of testing, observing and showing, I am convinced this will give you a whole new outlook on the potential and teach many here what maybe we could not teach during that thread. The Tesla Ozone Patent is a basis for what I think is in many of the OU device attempts we see today, including some aspects of which are used in the SM TPU.
Anyways, just thought I'd bring it up in case you are looking for something else.
Ya, I could see that question coming ;D
I'm 99% sure that switching both poles off and on won't make any difference in the effect I'm demonstration. The effect has nothing to do with using capacitors in the first place. I only used them so to be able to calculate power in for each test.
I real busy working on a new setup which will demonstrate the use of this effect. So I may not reply to all post until I have it done.
Luc
Quote from: MileHigh on September 26, 2009, 12:32:10 PM
.99:
I disagree with you here. The magnet will be influenced (have force act upon it) by whatever type of magnetic field is created by the current running through the coil. The magnet is only influenced by the current flow in the coil and not by any IV timing relationship. If you had a much slower resonant frequency and held the magnet in your hand you would be able to feel the alternating push and pull force on the magnet as the ringdown takes place - like putting your hand on a speaker cone and feeling a 60 Hz vibration.
MileHigh
MH, I think you would agree that if no
energy is being "supplied by" or "depleted in" the coil that it can not possibly be doing any work?
Take a look at the plot. I think it clearly illustrates that no energy is available (except for the static charge in the cap) to do any work during the ringdown phase. Therefore, it is not possible for the coil to act with any force on any magnet in it's proximity.
I'll address your second point shortly ;)
.99
Hey Poynt,
Thanks a lot for doing that plot. I am more confused now! lol
For starters, I am under the impression that the relay connects the charged cap to the coil, you get the magnet jump and the ringdown. The relay is on for 1/2 second or thereabouts.
Are you modeling the actual magnet-coil interaction in PSpice? I am assuming that you can't do that, but I know next to nothing about the actual program.
I can't make heads or tails of the coil current taking that big triangular waveform with the turn-around at 120 us. Nor do I get the parasitic capacitance retaining some energy. Isn't the capacitance in parallel with the coil and the coil has it's own innate series resistance? Are you adding resistors for the wire resistance?
Since I am so confused, just a few thoughts, only discussing the "no diode" circuit.
The end of the Step 1 discharge gives you current in the coil and delta-h and delta-v energy in the moving mass of the magnet. This means that there will be less current flowing in the coil as compared to a no-magnet case.
At the beginning of Step2 you have a "launched" magnet and a ringdown event. Yes, the NET work done on the launched magnet is more or less zero (the ringdown is not a constant amplitude sine wave), because the push and pull cancel each other out.
If you ignore the gravitational acceleration, the magnet velocity is wobbling 90 degrees out of phase with the AC magnetic field.
The modelling is a little bit tricky here. The mass of the magnet is experiencing an sinusoidal AC force. This is equivalent to an sinusoidal AC current source connected to a capacitor. The higher the frequency of the AC current source the lower the resultant AC voltage across the capacitor will be. That translates into a lower AC velocity of the wobbling magnet. So indeed there is no net energy expended here, the mass of the capacitor is a purely reactive load. By the same token, there is a tangible force that the mass experiences.
If you are still with me, it means that the rigdown event is being split up. When the coil starts its discharge cycle, the energy goes into the electrical capacitor and the mechanical capacitor. When the coil has depleted all of it's energy in terms of a cycle in the ringdown event, let's assume that most of the energy is stored in the electrical capacitor, and some of the energy is stored in the moving magnet (a.k.a. mechanical capacitor). Not forgetting all the time that we are factoring out the acceleration of the moving magnet due to gravity.
I am not going to the other half of the cycle, the electrical capacitor part is obvious, and the moving magnet will induce current flow in the coil.
Ultimately, at the start of Step 2, the magnet continues with it's "launch phase" with this energy-neutral magnetic-field-induced wobble as it continues skyward. When it reaches the peak (you assume the ringdown event is long gone), a certain proportion of the energy that originally came from the charged cap is now happily sitting in the mass of the magnet in the form of delta-h.
The magnet then drops back down and induces a final little rush of current through the coil and has a hard landing. That's your heat creation event. Of course when the magnet was going up during the launch phase that in itself was inducing current in the coil in the opposite direction as compared to the drop down. That was superimposed on top of the cap discharge current also.
Anyway, either you think this makes sense or I am nuts! lol
MileHigh
Here are your two coil current shots. I eliminated some superfluous circuitry that Luc had before, so the two peak currents come out the same now. The circuit is back to "basic" now. There is no "load" on the coil, but I will try to add one.
If I can manage, then we should be able to see if the "load" is geting any energy during the ringdown.
I don't think it is necessary to fully simulate all that Luc is doing in this test with the discharging cap etc, but if you are so inclined, I can do that. Right now I am just pulsing the coil with 170V supply and using 120us pulse width. Just one single pulse.
.99
Hey Poynt,
Now I get it. It's funny because now going back to your previous posts I get it, and it seems so obvious that you were just pulsing the coil to get some juice going through it.
There is a lesson for all here: Document! Document! Document! lol
Can you see how I made a disconnect when going from Luc's clip to your plots? I was expecting something different and it threw me off. I think Glen also went back and annotated some plots and stated what his probes were connected across for his scope shots. It really helps. When you do an experiment you have to step outside of your fishbowl and try to see what it looks like from an observer's perspective.
As far as the plots go, yes, very cool. Notice everybody that the current does not follow a standard exponential decay curve for the case with the fly-back diode? That's because the diode is a non-linear device where it's voltage drop is not proportional to the current flowing through it. The net result is that the current flows for a longer time and takes a more or less linear drop with respect to time.
I am not a fan of your red average current plot for these two cases though. It looks like they are the "average current over all time" and I don't think they are conveying too much information.
Anyway I hope somebody appreciated the detailed energy analysis I did in the previous posting. I think that explains a phenomenon that I think we have all seen in a YouTube clip or two. When someone has a self-resonating oscillator based on a transistor and coil + capacitor and are listening to the sound it makes on a speaker, sometimes you hear the frequency change when they bring a magnet close to it. By bringing a magnet close to the coil, you are effectively adding to the capacitance of the LC oscillator via magnetic coupling with the external magnet, which should lower the resonant frequency. Is that far out or what? lol
Going back to the plots, you can easily see how you have much more upwards pushing power on the launched magnet when the diode is in the circuit.
100% percent efficiency would be defined by having all of the energy in the capacitor being used to lift the magnet up to a certain height. In other words, 1/2 C v-squared = Mgh
Therefore the maximum height the magnet could be raised is as follows:
h = (1/2 C v-squared)/Mg
If you measure how high the magnet jumps and compare it to "h" then you can calculate the efficiency for each of the two variations of the circuit.
MileHigh
Sorry MH.
You're right I should have stated the difference in the setup.
The end result is pretty much the same though for both.
.99