This thread is to discuss what direction the current flows during inductive kickback. Or to put it another way, as the magnetic field collapses which way does the current flow?
The definition of an inductor is that an inductor opposes change in current flow. This would imply that as voltage is removed from an inductor the inductor would then try to keep the current flowing in the same direction until the magnetic field is exhausted. This can be proven with experimentation.
So let the discussion begin. Do you agree or disagree with this description of how an inductor works? Please explain why you agree or disagree. Please keep your comments to a technical discussion and not get into name calling if you disagree with someone.
Respectfully,
Carroll
Can you supply me with any links to support your theory that current and voltage can share opposite polarities?
THE CURRENT REVERSE FROM ITS ORIGINAL DIRECTION.
THE THEORY OF MAGNETIC SPIN. SO IF THE SPIN IS CW WHEN THE
CURRENT IS APPLIED THE BEMF SPIN ANTICLOCKWISE VICE VERSA.
WHEN CURRENT IS APPLIED ETHER IS FORCE OUT AND WHEN CURRENT
IS OF ETHER SPRING BACK..THERE AMEN.
HE KNOWS WHAT HE IS TALKING ABOUT EVEN IF I DO NOT UNDERSTAND IT
100% HE IS A GENIUS..https://www.youtube.com/watch?v=KooPsEE7E-Q
So sooory i have been shouting haven't i..
Quote from: citfta on November 20, 2015, 07:13:17 AM
This thread is to discuss what direction the current flows during inductive kickback. Or to put it another way, as the magnetic field collapses which way does the current flow?
The definition of an inductor is that an inductor opposes change in current flow. This would imply that as voltage is removed from an inductor the inductor would then try to keep the current flowing in the same direction until the magnetic field is exhausted. This can be proven with experimentation.
So let the discussion begin. Do you agree or disagree with this description of how an inductor works? Please explain why you agree or disagree. Please keep your comments to a technical discussion and not get into name calling if you disagree with someone.
Respectfully,
Carroll
To make things clear,it should be said that when the current source is interrupted,the current will want to keep flowing through the inductor-not if the voltage is removed.
It is said that an inductor is a current source,but that has never sat well with me,and i consider an inductor and a capacitor as a means to store energy,which can deliver power-not just current or voltage.
The flyback from an inductor can deliver power to a load,and the value/resistance of that load will determine as to what the voltage and current values will be with a set inductor. E.G-if the resistance of the load is low,then the flyback voltage will be low,and the current high. If the resistance of the load is high,then the flyback voltage will be high ,and the current low.
It should also be noted that the lower the resistive value of the load being placed on the flyback,the longer the magnetic field takes to fully collapse around the inductor. This results in a higher current due to the slower collapsing field,but a lower voltage for the same reason-the slower changing magnetic field with time. The higher the value of the resistive load on the flyback,the quicker the magnetic field around the inductor will collapse. This results in a lower current but a higher voltage due to the faster changing magnetic field with time.
2
When the current source to an inductor is interrupted, the current will continue to flow through the inductor in the same direction,but the voltage across that inductor will invert. This is how the simple circuit below is able to work,and the LED can be lit from the flyback<--im guessing you guys want to use the term !flyback! so as to keep it simple?.
Diagram 1 shows the switch closed,and current flows from the battery into the top of the inductor,and out the bottom of the inductor,and into the negative of the battery. At this point in time,i also assume that we will stick to conventional current flow throughout this thread-so as to keep it simple?.
Diagram 2 shows the current flow through the inductor,and voltage polarity across that inductor the moment the switch becomes open. This current flow,and voltage polarity will remain the same until such time as the magnetic field around the inductor has fully collapsed-all stored energy has been depleted. We know the current flowing through the inductor must be in the same direction,and the voltage must have inverted in order for the LED to light.
Hi guys
When the switch opens, inductor will try to keep its magnetic field alive, and to retain its magnetic poles at the same place as when the switch is closed. There is no other way for this to happen than the current to keep going on same direction as before. This is simple physics. What path it will take depends on the circuitry. In this example bemf just can not return to the negative of the battery because the circuit is open!!! It will go through diode because the circuit closes through the capacitor back to the base of the inductor.
Sometimes our minds get stack in simple matters. It happens to all of us. :)
Quote from: tinman on November 20, 2015, 09:05:57 AM
To make things clear,it should be said that when the current source is interrupted,the current will want to keep flowing through the inductor-not if the voltage is removed.
It is said that an inductor is a current source,but that has never sat well with me,and i consider an inductor and a capacitor as a means to store energy,which can deliver power-not just current or voltage.
The flyback from an inductor can deliver power to a load,and the value/resistance of that load will determine as to what the voltage and current values will be with a set inductor. E.G-if the resistance of the load is low,then the flyback voltage will be low,and the current high. If the resistance of the load is high,then the flyback voltage will be high ,and the current low.
It should also be noted that the lower the resistive value of the load being placed on the flyback,the longer the magnetic field takes to fully collapse around the inductor. This results in a higher current due to the slower collapsing field,but a lower voltage for the same reason-the slower changing magnetic field with time. The higher the value of the resistive load on the flyback,the quicker the magnetic field around the inductor will collapse. This results in a lower current but a higher voltage due to the faster changing magnetic field with time.
@Tinman,
The speed that the Reed switch contacts separate determines the "Flyback Voltage".
Quote from: Jeg on November 20, 2015, 09:46:26 AM
Hi guys
When the switch opens, inductor will try to keep its magnetic field alive, and to retain its magnetic poles at the same place as when the switch is closed. There is no other way for this to happen than the current to keep going on same direction as before. This is simple physics. What path it will take depends on the circuitry. In this example bemf just can not return to the negative of the battery because the circuit is open!!! It will go through diode because the circuit closes through the capacitor back to the base of the inductor.
Sometimes our minds get stack in simple matters. It happens to all of us. :)
@Jeg,
There's no longer any current flowing after the Reed switch is open. You maintain that the magnetic field is sustained by some imaginary current!
A link to get you started: https://en.wikipedia.org/wiki/Inductor (https://en.wikipedia.org/wiki/Inductor)
I had to leave for a while but now I am back.
To answer the question about current flow and opposite polarity, answer my question about which direction does current flow inside a battery. Does it flow from positive to negative or the other way round? And which way does it flow outside the battery? When you get the answer to that you will understand why polarity can switch on an inductor but the current still flows the same way.
Also do you have a way to conduct the test with a 2 channel scope?
Carroll
Quote from: citfta on November 20, 2015, 09:56:25 AM
A link to get you started: https://en.wikipedia.org/wiki/Inductor (https://en.wikipedia.org/wiki/Inductor)
I had to leave for a while but now I am back.
To answer the question about current flow and opposite polarity, answer my question about which direction does current flow inside a battery. Does it flow from positive to negative or the other way round? And which way does it flow outside the battery? When you get the answer to that you will understand why polarity can switch on an inductor but the current still flows the same way.
Also do you have a way to conduct the test with a 2 channel scope?
Carroll
Yes i do,and will be doing that tomorrow by way of video-as it is late at night here in OZ ATM.
Quote from: synchro1 on November 20, 2015, 09:51:08 AM
@Jeg,
There's no longer any current flowing after the Reed switch is open. You maintain that the magnetic field is sustained by some imaginary current!
When the switch opens,the current continues to flow through the inductor--this is the flyback current we are talking about.
What do you think flyback is Synchro?-->it is the power produced by the collapsing magnetic field around the inductor when the switch is opened. It is how i have drawn it a few post back.
Current continues to flow through the inductor after the switch is open. The magnetic field is sustained some what due to the current loop through the inductor and LED.
Synchro,
The current that flows when the switch is open is not from the battery. Isn't an inductor an energy tank? How that energy manifests itself during a field collapsing?
Quote from: Jeg on November 20, 2015, 10:03:50 AM
Synchro,
The current that flows when the switch is open is not from the battery. Isn't an inductor an energy tank? How that energy manifests itself during a field collapsing?
@Jeg,
The energy that appears in the coil from the collapsing field is new energy and has an opposite polarity.
Quote from: synchro1 on November 20, 2015, 10:06:45 AM
@Jeg,
The energy that appears in the coil from the collapsing field is new energy and has an opposite polarity.
The voltage across the inductor invert's,but the current continues to flow in the same direction. The energy from the collapsing magnetic field is not new energy,it is stored energy from the initial input energy.
Quote from: synchro1 on November 20, 2015, 10:06:45 AM
@Jeg,
The energy that appears in the coil from the collapsing field is new energy and has an opposite polarity.
And where the stored energy goes?
Hi all
Youp seems not to be so easy this flyback spike understanding.
For me what i see on the scope is that at the end of the pulse, when the reed sharply opens, the current trace goes to almost instantly (verticaly) to zero. And only AFTER this shut down, begins the flybackspike.
So to me the current who build up the magneticfield is gone , totally dissipated, finished at the end of the pulse. He was totally used to precisely build the magnetic field. So he has no more direction at all.
Than it stays the expanded magnetic field around the coil (??) and what exactly happens at that point is ?? But on the scope , suddenly a strong narrow "high negative voltage" trace appears and also a strong very very narrow and strong current trace (not shown on the pic) also. What does create this event is still puzzling to me.
If it is the collapsing of the magneticfield please explain the process with simple words if possible.
But anyway to me that is a new current who has nothing to do directly with the one that created the magneticfield. It is what i call hes "son"
Youp, please don't shoot me down it's only an idea.
Laurent
Quote from: Jeg on November 20, 2015, 10:09:53 AM
And where the stored energy goes?
@Jeg,
The coil energy is stored in the magnetic field. The inductor can't store electricity. A capacitor can store electrical energy. Where does the new reversed polarity "Flyback" go? To the reverse pole.
Quote from: woopy on November 20, 2015, 10:33:00 AM
Hi all
Youp seems not to be so easy this flyback spike understanding.
For me what i see on the scope is that at the end of the pulse, when the reed sharply opens, the current trace goes to almost instantly (verticaly) to zero. And only AFTER this shut down, begins the flybackspike.
So to me the current who build up the magneticfield is gone , totally dissipated, finished at the end of the pulse. He was totally used to precisely build the magnetic field. So he has no more direction at all.
Than it stays the expanded magnetic field around the coil (??) and what exactly happens at that point is ?? But on the scope , suddenly a strong narrow "high negative voltage" trace appears and also a strong very very narrow and strong current trace (not shown on the pic) also. What does create this event is still puzzling to me.
If it is the collapsing of the magneticfield please explain the process with simple words if possible.
But anyway to me that is a new current who has nothing to do directly with the one that created the magneticfield. It is what i call hes "son"
Youp, please don't shoot me down it's only an idea.
Laurent
@Woopyjump,
Thanks for the test results. They confirm everything I've been saying!
Faraday's Law (loosely) states that current is generated when a magnetic field moves across copper windings. The collapsing magnetic field is moving across the coil windings as it shrinks inward. Like a balloon: You blow air in, it expands. You release the air and it blows out the hole with a force. This force blows across the coil wraps from outward to inward. This is in the opposite direction from the initial field expansion.
Quote from: synchro1 on November 20, 2015, 10:42:05 AM
@Woopyjump,
Thanks for the test results.
Faraday's Law (loosely) states that current is generated when a magnetic field moves across copper windings. The collapsing magnetic field is moving across the coil windings as it shrinks inward. Like a balloon: You blow air in, it expands. You release the air and it blows out the hole with a force.
QuoteThey confirm everything I've been saying!
What they confirm is that you are wrong.
QuoteThis force blows across the coil wraps from outward to inward. This is in the opposite direction from the initial field expansion.
It is also opposite to that which created it. This is where you are lost. Power was supplied to the inductor to create the magnetic field.Then the magnetic field returned the energy which created it in the first place.
Let me try and explain it this way Synchro.
Lets say a motor vehicle is our inductor,and the wheels are the current. The engine is our power in when the switch is closed. When we use the engine(power input) to move the car forward,which way will the wheels rotate. Now,we switch the engine off,and with the car stationary,we move the road backwards under the car-which way are the wheels rotating?. Thats right-in the same direction.
Synchro,
Let's take things one step at a time. Which way does the current flow "inside" a battery? Now which way does it flow outside of the battery.
@Tinman,
"It is also opposite to that which created it. This is where you are lost. Power was supplied to the inductor to create the magnetic field.Then the magnetic field returned the energy which created it in the first place".
What is this some kind of attempt at burlesque?
Synchro
Please look at the diagram below,and note the scope probe placements.
Question 1- What wave form will be seen on channel 1-->an AC or DC wave form?.
Question 2- What wave form will be seen on channel 2-->an AC or DC wave form?
Synchro will you please put together the simple circuit Tinman has provided? I have reattached it here.
Now short out the cap to make sure it is fully discharged. Then momentary tap the switch closed and then leave it open. Now check the voltage on the cap. If you are correct about the current reversing then the top of the cap should be positive and the bottom negative. However you will see the bottom of the cap is positive and the top of the cap is negative. This is because the coil has become a source for the short time it takes the magnetic field to collapse. Please answer my question about which way the current flows inside of a battery.
Carroll
Quote from: synchro1 on November 20, 2015, 11:18:06 AM
@Tinman,
"It is also opposite to that which created it. This is where you are lost. Power was supplied to the inductor to create the magnetic field.Then the magnetic field returned the energy which created it in the first place".
What is this some kind of attempt at burlesque?
The collapsing magnetic field is what inverts the voltage across that inductor. Although the magnetic field is collapsing around that inductor,and this inverts the voltage,the magnetic polarity remains the same,and this maintains the current flow direction,so as it is the same as it was before the power source was interrupted. In order for the current flow to change direction,the magnetic field polarity would also have to invert-which it dose not.
I think there is a lot of confusion around the "reversing voltage" and the explanation for that is quite simple.
When you energize the inductor you have an external voltage source across the terminals of the inductor. So an external power supply defines the voltage across the inductor.
Then when the switch opens, the external voltage source is gone - it has no influence at all on the voltage across the inductor.
When the switch opens the inductor now acts as a power source. The inductor as a power source outputs current. The inductor works to keep the current flowing in the same direction. The load creates the counter-EMF under the imposed current flow from the inductor.
So, in the truest sense, the inductor is just a hapless rag doll along for the ride when it comes to the voltage across its terminals. In the first case, the external power supply is imposing its voltage across the terminals of the inductor. In the second case, the load itself is imposing its voltage across the terminals of the inductor as the inductor discharges a current flow into the load.
In both cases, the inductor is a "follower" when it comes to the voltage across its terminals - like a hapless rag doll.
An entire cottage industry is built around the "secrets" of the inductor.
Here is a fun thought demonstration:
In one hand you have a one-farad capacitor with one volt across it discharging across a one-ohm resistor.
In the other hand you have a one-henry inductor with one amp traveling through it discharging through a one-ohm resistor.
If you put a scope across the resistor in both cases you would see an identical exponentially decaying voltage waveform. Since it is a one-ohm resistor, you would also be looking at an identical decaying current waveform.
For the capacitor circuit, as you decrease the value of the load resistor, it starts to discharge faster and faster with higher and higher currents through the load resistor.
For the inductor circuit, as you increase the value of the load resistor, it starts to discharge faster and faster with higher and higher voltages across the load resistor.
If you can wrap your head around this then you will have an understanding about how capacitors and inductors work as energy storing devices and how they discharge their stored energy into a load.
Hi Tinman :)
There is really no sense for voltage to change polarity while the field is collapsing. Voltage and current have a strict relation between the two. Current always flow from higher voltage potential to a lower one and never the opposite even momentarily. As current keeps going the same direction after collapsing, the same is with voltage. Not only it doesn't alter its polarity, but also becomes magnitudes higher also momentarily like current does.
Quote from: Jeg on November 20, 2015, 12:26:14 PM
Hi Tinman :)
There is really no sense for voltage to change polarity while the field is collapsing. Voltage and current have a strict relation between the two. Current always flow from higher voltage potential to a lower one and never the opposite even momentarily.
That is not true. Suppose that you have two batteries that have a common ground. One battery is +1 volt and the other battery is +2 volts.
What will happen if you connect a 1-ohm resistor between the two batteries?
The answer is that one amp of current will flow into the +1 volt battery. So in this example, current is flowing in the "opposite" direction through the +1 volt battery.
Quote from: MileHigh on November 20, 2015, 12:34:01 PM
That is not true. Suppose that you have two batteries that have a common ground. One battery is +1 volt and the other battery is +2 volts.
What will happen if you connect a 1-ohm resistor between the two batteries?
The answer is that one amp of current will flow into the +1 volt battery. So in this example, current is flowing in the "opposite" direction through the +1 volt battery.
Hi MH.
I said the same. Current flows from higher to lower potential always.
Ps. Ok MH I got it. Thanks.
Inductor is now a power source itself, so yes Tinman is right. It changes voltage polarity but current flows the same direction inside the inductor. This is true.
Quote from: tinman on November 20, 2015, 10:09:03 AM
The voltage across the inductor invert's,but the current continues to flow in the same direction. The energy from the collapsing magnetic field is not new energy,it is stored energy from the initial input energy.
@Tinman,
Everything is everything!
From what i have found in the past is that the output of the coil when input is taken away doesnt necessarily have to be a spike. It depends on the load it is sent to.
Like putting a diode across a relay coil to recirculate the bemf, the the field collapse is slower than with a higher ohm load. Most relays are high ohm already so it does drop off very fast. But a very low ohm winding with high induction should slope while diminishing through the diode rather than a quick spike reaction of a higher ohm coil.
I have coils that I had shown in Lucs thread that are .5ohm 2mh. The bemf from those coils will heat up a 5w 5 ohm resistor that you cant touch when hot. And its not due to quick spikes.
Mags
Quote from: tinman on November 20, 2015, 09:25:40 AM
When the current source to an inductor is interrupted, the current will continue to flow through the inductor in the same direction,but the voltage across that inductor will invert.
I agree.
I would add though, that
the cause of current flow through an inductor can be:
- an external current source,
- an external voltage source,
- an external magnetic flux source,
- the inductor itself, since a current flowing through an ideal shorted inductor will flow forever.
Quote from: verpies on November 20, 2015, 01:41:53 PM
I agree.
I would add though, that the cause of current flow through an inductor can be:
- an external current source,
- an external voltage source,
- an external magnetic flux source,
- the inductor itself, since a current flowing through an ideal shorted inductor will flow forever.
From Woopyjump:
"For me what I see on the scope is that at the end of the pulse, when the reed sharply opens, the current trace goes to almost instantly (verticaly) to zero. And only AFTER this shut down, begins the flybackspike.
So to me the current who build up the magnetic field is gone , totally dissipated, finished at the end of the pulse. He was totally used to precisely build the magnetic field. So he has no more direction at all.
Than it stays the expanded magnetic field around the coil (??) and what exactly happens at that point is ?? But on the scope , suddenly a strong narrow "high negative voltage" trace appears and also a strong very very narrow and strong current trace (not shown on the pic) also. What does create this event is still puzzling to me.
If it is the collapsing of the magneticfield please explain the process with simple words if possible.
But anyway to me that is a new current who has nothing to do directly with the one that created the magnetic field. It is what I call his "son".
The current passes into the inductor; The inductor stores the current in a magnetic field; The current is cut off; The magnetic field collapses and a new current and voltage are generated; The new current and voltage share the same polarity! Stop trying to falsely maintain that the new voltage reverses polarity while the new current does something else!
Go back and look at Woopyjump's scope shots!
Simple test for Woopy.
Put one channel of your scope across the coil and the other channel across the 10 ohm resistor. Now pulse the coil and watch the two waveforms. Please post your results.
Quote from: Jeg on November 20, 2015, 12:26:14 PM
Hi Tinman :)
There is really no sense for voltage to change polarity while the field is collapsing. Voltage and current have a strict relation between the two. Current always flow from higher voltage potential to a lower one and never the opposite even momentarily. As current keeps going the same direction after collapsing, the same is with voltage. Not only it doesn't alter its polarity, but also becomes magnitudes higher also momentarily like current does.
@Jeg,
Go back and have another look at Woopy's scope shots!
Quote from: citfta on November 20, 2015, 02:21:15 PM
Simple test for Woopy.
Put one channel of your scope across the coil and the other channel across the 10 ohm resistor. Now pulse the coil and watch the two waveforms. Please post your results.
@Citfta,
What's the voltage of the power source? What's the inductance of the coil?
Quote from: synchro1 on November 20, 2015, 02:13:09 PM
The current passes into the inductor; The inductor stores the current in a magnetic field; The current is cut off; The magnetic field collapses and a new current and voltage are generated; The new current and voltage share the same polarity! Stop trying to falsely maintain that the new voltage reverses polarity while the new current does something else!
Go back and look at Woopyjump's scope shots!
Synchro, instead of just repeating what you believe please take the time to think about what we are saying. You have never answered the question about which way the current flows inside the battery. Jeg now understands what is going on because he took the time to consider that idea.
Quote from: synchro1 on November 20, 2015, 02:28:56 PM
@Citfta,
What's the voltage of the power source? What's the inductance of the coil?
It doesn't really matter. The results will be the same. A voltage of 12 volts or so is usually convenient. And the higher the inductance the easier it will be to see the scope traces and follow what is happening.
Quote from: citfta on November 20, 2015, 02:30:56 PM
Synchro, instead of just repeating what you believe please take the time to think about what we are saying. You have never answered the question about which way the current flows inside the battery. Jeg now understands what is going on because he took the time to consider that idea.
@Citfta,
I don't know anything about batteries.
Quote from: citfta on November 20, 2015, 02:33:13 PM
It doesn't really matter. The results will be the same. A voltage of 12 volts or so it usually convenient. And the higher the inductance the easier it will be to see the scope traces and follow what is happening.
@Citfta,
I can adjust the values to produce a spark across the Reed contacts.
Quote from: synchro1 on November 20, 2015, 02:34:11 PM
@Citfta,
I don't know anything about batteries.
OK.
If the conventional idea of current flow is for current to flow from the positive pole to the negative pole through the circuit then which way would the current have to flow inside the battery to complete the path of current flow?
Quote from: synchro1 on November 20, 2015, 02:35:52 PM
@Citfta,
I can adjust the values to produce a spark across the Reed contacts.
I doubt you can get the reed to arc with a 100 load resistor. Not unless you are using a much higher voltage than 12 volts.
Quote from: citfta on November 20, 2015, 02:40:05 PM
I doubt you can get the reed to arc with a 100 load resistor. Not unless you are using a much higher voltage than 12 volts.
@Citfta,
We're not mind readers. Upload a video. Can you explain why the "Flyback Spike" would arc the Reed contacts and travel back to the positive electrode of the power source to begin with?
Quote from: synchro1 on November 20, 2015, 02:42:27 PM
@Citfta,
We're not mind readers. Upload a video. Can you explain why the "Flyback Spike" would arc the Reed contacts and travel back to the positive electrode of the power source to begin with?
It is NOT traveling back. It is trying to continue on. That is the basic definition of an inductor. It opposes change in current flow.
Instead of me posting a video why don't you do the simple test I proposed using Tinman's circuit and then measure the voltage on the cap so that you can see for yourself that the current does not reverse. Did you think about how current has to flow inside a battery to complete the current path? HINT It has to be opposite of the voltage polarity shown by a meter.
Quote from: citfta on November 20, 2015, 02:54:38 PM
It is NOT traveling back. It is trying to continue on. That is the basic definition of an inductor. It opposes change in current flow.
Instead of me posting a video why don't you do the simple test I proposed using Tinman's circuit and then measure the voltage on the cap so that you can see for yourself that the current does not reverse. Did you think about how current has to flow inside a battery to complete the current path? HINT It has to be opposite of the voltage polarity shown by a meter.
@Citfta,
I'm in political exile in Costa Rica. What do you mean it's not traveling back? You pretend it's really trying to follow the path you outlined in your schematic? Come on, put your thinking cap on.
You don't think you might have your DMM electrodes or scope probes reversed?
@Citfta,
This is my all time favorite pulse motor build by Igor Moroz; I've watched it at least a hundred times. It's as close to a self runner as anything I've ever seen and it's simpler then anyone can imagine. I love it. Please look at it more then once. It's only two minutes long. Study it:
https://www.youtube.com/watch?v=vWvI7T7h3tk
Two people believing the wrong thing does not make it true. I have had this discussion several times before in the past. Some have learned and some have not.
Quote from: tinman on November 20, 2015, 09:05:57 AM
The flyback from an inductor can deliver power to a load,and the value/resistance of that load will determine as to what the voltage and current values will be with a set inductor. E.G-if the resistance of the load is low,then the flyback voltage will be low,and the current high. If the resistance of the load is high,then the flyback voltage will be high ,and the current low.
It should also be noted that the lower the resistive value of the load being placed on the flyback,the longer the magnetic field takes to fully collapse around the inductor. This results in a higher current due to the slower collapsing field,but a lower voltage for the same reason-the slower changing magnetic field with time. The higher the value of the resistive load on the flyback,the quicker the magnetic field around the inductor will collapse. This results in a lower current but a higher voltage due to the faster changing magnetic field with time.
You are right about the current not changing direction but you are not right about the magnitude of the current.
When you disconnect the power source from the coil, whatever the magnitude of the current flowing through the coil is, that's the initial amount of current that will flow through the load.
If you have one amp flowing through the coil and you disconnect your power source and the load is a 100-ohm resistor, then initially the current flow through the load will be one amp and the voltage across the load will be 100 volts. If the load is 5000 ohms, then initially the current flow through the load will still be one amp and the voltage across the load will be 5000 volts.
Hi Cifta and Synchro and all
If i posted above it is not to get what i can get from all text books about physic.
My questions are very clear :
1- at the end of the pulse, on the scope shot, the input current from the power source has totally vanished yes or not ?
2- The current has built a magnetic field in or around the inductor and this magnetic field has been " used " to propel the rotor magnet yes or not ?
3- so the current is not stored in the magneticfield but has been used the give the rotor magnet a kinetic energy yes or not ?
4-So if the magneticfield collapses and by shrinking it crosses the wires of the coil and redo a new current does it mean that the magneticfield can be used twice yes or not ?
And just for the dessert
A small video showing that the flybackspike can be really powerfull and destroying as every body knows very well
https://youtu.be/TAx7Y0UIyHA
good night at all
Laurent
Quote from: MileHigh on November 20, 2015, 03:31:50 PM
You are right about the current not changing direction but you are not right about the magnitude of the current.
When you disconnect the power source from the coil, whatever the magnitude of the current flowing through the coil is, that's the initial amount of current that will flow through the load.
If you have one amp flowing through the coil and you disconnect your power source and the load is a 100-ohm resistor, then initially the current flow through the load will be one amp and the voltage across the load will be 100 volts. If the load is 5000 ohms, then initially the current flow through the load will still be one amp and the voltage across the load will be 5000 volts.
Don't you want to rethink that last paragraph? You have just given (if true) the formula for OU. In your first example you are putting 100 watts into the load. In your second example you are putting 5000 watts into the load. Where did that extra 4900 watts come from?
Quote from: MileHigh on November 20, 2015, 03:31:50 PM
If you have one amp flowing through the coil and you disconnect your power source and the load is a 100-ohm resistor, then initially the current flow through the load will be one amp and the voltage across the load will be 100 volts. If the load is 5000 ohms, then initially the current flow through the load will still be one amp and the voltage across the load will be 5000 volts.
That's why they call it a current source - constant current regardless of the load resistance.
P.S.
That Igor guy and Synchro are wrong about the current through the inductor reversing when the inductor is opened.
How do I know? - I did the lab work and I can distinguish when my scope is measuring current vs. when it is measuring voltage.
Quote from: citfta on November 20, 2015, 03:43:56 PM
In your second example you are putting 5000 watts into the load. Where did that extra 4900 watts come from?
No, only the initial instantaneous power dissipated into the load (the 100Ω resistor) is I
2R.
The energy stored in the inductor is ½LI
2 in both cases
Quote from: synchro1 on November 20, 2015, 03:22:49 PM
@Citfta,
This is my all time favorite pulse motor build by Igor Moroz; I've watched it at least a hundred times. It's as close to a self runner as anything I've ever seen and it's simpler then anyone can imagine. I love it. Please look at it more then once. It's only two minutes long. Study it:
https://www.youtube.com/watch?v=vWvI7T7h3tk (https://www.youtube.com/watch?v=vWvI7T7h3tk)
Yes I looked at the clip and Igor is
dead wrong. A few years ago he was making clips where he was always saying that his diodes were redirecting current back to the source battery. I pointed out to him that that was wrong and he acknowledged it and then just shrugged it off. He probably had at least a dozen clips with that mistake in them.
It's safe to assume that Igor is just an ordinary guy, an amateur, so don't make the mistake of taking what he says in his clips as always being true.
Here is a comment on Igor's clip from three years ago:
<<< @Mopozco Great, thanks for sharing results. The collapsing field of a coil wants to keep current flowing in same direction, coil behaves like battery in series with your 1.5V battery ~(-)batt(+)~~~(-)coil(+)~. Please confirm direction of LED on your drawing. Is LED cathode connected to coil so that when current from collapsing field at higher voltage (bemf) exceeds LED Vf it flashes it? This would keep the current flowing in the same direction not in opposite direction as shown on your drawing. >>>
The guy is probably right and Igor had the LED in the opposite direction that he thought it was. Alternatively, I don't know if an LED will light up if you slam current through it in the wrong direction but you never know.
It all goes back to the messed up culture on the forums of being too polite to correct your peers when they make a mistake. It creates a regimen of enforced foolishness and stupidity. I can't tell you how many times I have seen inane ridiculous postings by people and people on the forum say things like, "Thanks for your interesting ideas!" It's ridiculous and totally counterproductive.
I just posted that Tinman was wrong about the current flow in an inductor. Chances are if I didn't post the correction, then nobody would have corrected him. That would turn what is supposed to be an informative educational thread into more uncorrected disinformation, messing up people's understanding of basic basic electronics, yet again.
Quote from: woopy on November 20, 2015, 03:42:07 PM
Hi Cifta and Synchro and all
If i posted above it is not to get what i can get from all text books about physic.
My questions are very clear :
1- at the end of the pulse, on the scope shot, the input current from the power source has totally vanished yes or not ?
2- The current has built a magnetic field in or around the inductor and this magnetic field has been " used " to propel the rotor magnet yes or not ?
3- so the current is not stored in the magneticfield but has been used the give the rotor magnet a kinetic energy yes or not ?
4-So if the magneticfield collapses and by shrinking it crosses the wires of the coil and redo a new current does it mean that the magneticfield can be used twice yes or not ?
And just for the dessert
A small video showing that the flybackspike can be really powerfull and destroying as every body knows very well
https://youtu.be/TAx7Y0UIyHA
good night at all
Laurent
@Laurent,
Wear your welding glasses!
Quote from: woopy on November 20, 2015, 03:42:07 PM
Hi Cifta and Synchro and all
If i posted above it is not to get what i can get from all text books about physic.
My questions are very clear :
1- at the end of the pulse, on the scope shot, the input current from the power source has totally vanished yes or not ?
Yes
2- The current has built a magnetic field in or around the inductor and this magnetic field has been " used " to propel the rotor magnet yes or not ?
Yes
3- so the current is not stored in the magneticfield but has been used the give the rotor magnet a kinetic energy yes or not ?
The current builds the magnetic field and the magnetic field gives the rotor magnet the kinetic energy
4-So if the magneticfield collapses and by shrinking it crosses the wires of the coil and redo a new current does it mean that the magneticfield can be used twice yes or not ?
Yes that is why we can send current to the cap after the reed switch has opened.
And just for the dessert
A small video showing that the flybackspike can be really powerfull and destroying as every body knows very well
https://youtu.be/TAx7Y0UIyHA (https://youtu.be/TAx7Y0UIyHA)
good night at all
Laurent
Quote from: MileHigh on November 20, 2015, 03:55:56 PM
Yes I looked at the clip and Igor is dead wrong. A few years ago he was making clips where he was always saying that his diodes were redirecting current back to the source battery. I pointed out to him that that was wrong and he acknowledged it and then just shrugged it off. He probably had at least a dozen clips with that mistake in them.
It's safe to assume that Igor is just an ordinary guy, an amateur, so don't make the mistake of taking what he says in his clips as always being true.
Here is a comment on Igor's clip from three years ago:
<<< @Mopozco Great, thanks for sharing results. The collapsing field of a coil wants to keep current flowing in same direction, coil behaves like battery in series with your 1.5V battery ~(-)batt(+)~~~(-)coil(+)~. Please confirm direction of LED on your drawing. Is LED cathode connected to coil so that when current from collapsing field at higher voltage (bemf) exceeds LED Vf it flashes it? This would keep the current flowing in the same direction not in opposite direction as shown on your drawing. >>>
The guy is probably right and Igor had the LED in the opposite direction that he thought it was. Alternatively, I don't know if an LED will light up if you slam current through it in the wrong direction but you never know.
It all goes back to the messed up culture on the forums of being too polite to correct your peers when they make a mistake. It creates a regimen of enforced foolishness and stupidity. I can't tell you how many times I have seen inane ridiculous postings by people and people on the forum say things like, "Thanks for your interesting ideas!" It's ridiculous and totally counterproductive.
I just posted that Tinman was wrong about the current flow in an inductor. Chances are if I didn't post the correction, then nobody would have corrected him. That would turn what is supposed to be an informative educational thread into more uncorrected disinformation, messing up people's understanding of basic basic electronics, yet again.
@Milehigh,
The LED is reverse biased in Igor's video. This is the trick of the build. You have too much to say about this intrepid inventor for someone who never uploaded even so much as one test video of any kind.
You pasted some copied comments from Igor's video, but you neglected to copy this one by Lidmotor:
"I have been doing this same thing and it seems to work really well. It saves the reed switch contacts also".
Quote from: verpies on November 20, 2015, 03:52:28 PM
No, the instantaneous power dissipated into the load (the 100Ω resistor) is I2R in both cases ...or 100W.
That would be the case in any constant current source.
The formula P=V*I or P=E2/R could be applied if V represented the magnitude of a voltage source.
In Milehigh's example, there are no such sources.
Ha! Ha! Verpies you tripped up! ;D
The initial instantaneous power for the 100-ohm load is 100 watts. The initial instantaneous power for the 5000-ohm load is 5000 watts.
But of course the total energy in the current pulses for both cases will be the same - no OU.
Quote from: MileHigh on November 20, 2015, 04:03:54 PM
Ha! Ha! Verpies you tripped up! ;D
Yes, I did but I will make an example of myself how quickly such mistakes should be admitted.
Quote from: MileHigh on November 20, 2015, 03:55:56 PM
I just posted that Tinman was wrong about the current flow in an inductor. Chances are if I didn't post the correction, then nobody would have corrected him.
You'll get no argument from me.
I did not even read that post of his, but if I did I would've objected to his statement about the load resistance affecting the current of a constant current source (an inductor in this case).
I'll second that idea of admitting mistakes quickly. Milehigh I wasn't meaning to say you were wrong with what you said it was the way you said it that could lead some (or many) to believe that you believed in OU from the discharge of a coil. I am pretty sure you didn't mean that.
Thank's a lot Citfta
Finally somebody answering directly my questions.
So let's go further if you don't mind.
in my question 3
you seems to agree that the current builds up effectively the magnetic field which propels and gives the kinetic energy to the rotor.
But is the current or his energy still stored in the magnetic field after the magnetic field has propelled the rotor ?? Any entropy ?
So in my question 4
you seems to say that the input energy could be used 2 time ?
Thank's very much for your input and for this thread , because i have so much to understand
Laurent
Laurent,
I my opinion, if you can understand how an inductor and a mechanical flywheel are directly related, that will make all the difference. So the challenge is to just understand that first, and then once you understand the concept, then you can apply it to understanding circuits.
You saw the "Woopy's Linear Track Experiment" drawing that I made?
Look at the attached image that I found on Google. This is just a teaser, does the image look familiar?
MileHigh
Quote from: woopy on November 20, 2015, 04:37:30 PM
Thank's a lot Citfta
Finally somebody answering directly my questions.
So let's go further if you don't mind.
in my question 3
you seems to agree that the current builds up effectively the magnetic field which propels and gives the kinetic energy to the rotor.
But is the current or his energy still stored in the magnetic field after the magnetic field has propelled the rotor ?? Any entropy ?
The magnetic field is maintained as long as there is power applied to the coil. When power is no longer applied the magnetic field begins to weaken as it tries to maintain the current through the coil. If you continue to apply power after the current has reached the maximum value which is also when the field has gotten the strongest then any power after that is wasted as heat because of the resistance of the wire in the coil. So you want the pulse to be just long enough to cause the current to reach its peak and that will also mean the field has reached its peak. Then when you turn off power to the coil the kickback will be the strongest and can thus give the most power to the cap.
So in my question 4
you seems to say that the input energy could be used 2 time ?
Yes by capturing that kickback spike and reusing it you can make your system more efficient. I think Luc just posted some more videos about that today.
If you have time try the test I proposed for you earlier with a 2 channel scope. You can see the polarity switch with scope channel that is across the coil and you can see the current continue flowing through the resistor with the scope channel across the resistor. If you can set this up with a 555 timer or some other means to fire the coil regularly it will be easier to get the scope picture you want.
Thank's very much for your input and for this thread , because i have so much to understand
Laurent
For an inductor, you can measure the current flowing through it, and the voltage across it.
Look at the picture of the laboratory flywheel. You can see there is a smaller disk so you can use your hands to spin the flywheel.
When you apply voltage across an inductor, the current does not flow instantly. Instead, it slowly rises.
If you have a stationary flywheel, and then you want to make it spin by applying torque to it with your hands, it does not spin at full speed right away. Instead, the rotational speed slowly rises.
This is the first thing you have to understand:
Applying voltage to a coil makes the current start to flow, starting from zero. Voltage is like "pressure" on the coil.
Applying torque to the flywheel makes the flywheel spin, starting from zero. The torque is a pressure on the flywheel.
Therefore voltage is like torque.
If you can understand that, it's a good start.
Hi all,
If you take a single coil , and pass a magnet over it , as the magnet approaches it induces a potential (voltage) correct?
There is no current if the ends of coil aren't connected (open) correct?
If not open but connected to a load current will flow. correct?
But when the magnet is approaching the coil it builds , hits it's peak then builds in the opposite direction, short the coil right after the peak (disconnect load and short coil) then reconnect load.
That gives twice the work out of the coil.
Now have recovery diodes and caps to collect the short, for both pulses.
Passing a magnet over a coil produces two pulses.
When the magnet is at the center of the coil that is where you short the coil leads together removed from the load , reconnect and do it again, before the passing magnet leaves(or stops influencing the coil)
Then short again.
artv
What's wrong with this picture? The arrow in (c) looks llike it's moving in the same direction as (b) through the inductor. This is an illusion caused by confusing path with current direction.
YES!! Now you have it! Of course with no where to go except through the switch you will get arcing at the switch that way. That is why we are looking at using the energy from the collapsing field to charge another cap. We don't waste the energy and the reed switch will last longer too.
Great!! I am so glad you got it.
Respectfully,
Carroll
Quote from: citfta on November 20, 2015, 08:36:14 PM
YES!! Now you have it! Of course with no where to go except through the switch you will get arcing at the switch that way. That is why we are looking at using the energy from the collapsing field to charge another cap. We don't waste the energy and the reed switch will last longer too.
Great!! I am so glad you got it.
Respectfully,
Carroll
@Citfta,
You just turned that smart!
Quote from: MileHigh on November 20, 2015, 03:31:50 PM
When you disconnect the power source from the coil, whatever the magnitude of the current flowing through the coil is, that's the initial amount of current that will flow through the load.
If you have one amp flowing through the coil and you disconnect your power source and the load is a 100-ohm resistor, then initially the current flow through the load will be one amp and the voltage across the load will be 100 volts. If the load is 5000 ohms, then initially the current flow through the load will still be one amp and the voltage across the load will be 5000 volts.
QuoteYou are right about the current not changing direction but you are not right about the magnitude of the current.
You do understand that i am referring to the average current over one cycle ?.
Quote from: verpies on November 20, 2015, 04:11:58 PM
You'll get no argument from me.
I did not even read that post of his, but if I did I would've objected to his statement about the load resistance affecting the current of a constant current source (an inductor in this case).
Once again,i am referring to average current over one cycle.
This should help clear everything up:
Quote from: synchro1 on November 20, 2015, 08:47:28 PM
This should help clear everything up:
I do like that picture! Every time I see it it makes me feel like my brain is tilted a little. LOL
Here you go Synchro-test one complete.
What do you see?
1- a 30% on time from the FG(switch closed)
2-a 70% off time(switch open)
During the 30% on time(switch closed) we can see that the voltage(yellow trace) across the inductor(supply voltage) is on the top side of the zero volt line.The current is flowing in a forward direction,as the current trace(blue trace) is also above the zero volt line.
During the 70% off time(switch open) we can now see that the voltage is inverted across the inductor/resistor series circuit-!BUT! the current is still flowing in the same direction as it was during the 30% on time,as the current trace is still above the zero volt line.
You will also notice that the current continues to flow through the load(the LED) through the complete 70% off time,and never stop's flowing right up until the next on time period. This also shows you that the magnetic field dose not collapse completely before the next on time cycle starts.
So !Bub!
Please show us where exactly the current flow reverses direction?
And please tell us how a current continues to flow(in the same direction ;) ) during the 70% off time if the magnetic field around the inductor collapses instantaneously ?
Show us where this !new! energy is?.
Brad,
That's an awesome posting and my sincere compliments.
I will just throw in some "flywheel" colour commentary:
You can see how there is always current flowing through the inductor -> the flywheel never stops turning.
The flywheel increases in speed when you apply positive torque (positive voltage) to it.
The flywheel decreases in speed when you apply braking to it. So you can imagine the (diode + 10 ohm CVR) acting like a caliper and brake pads putting resistance on the spinning flywheel. When the brakes are applied that is negative torque (negative voltage.)
Bonus round:
What happens when you disconnect the 10-ohm CVR from the circuit? Most of us know that you get a spark in the air gap because the air becomes conductive plasma due to the big negative spike of high voltage. All of the energy in the coil is burned off in the hot plasma. The current stops flowing through the coil.
What is the equivalent for the flywheel?
You clamp down on the brake calipers very fast and with maximum pressure. There is a massive spike of negative torque on the flywheel when the brake pads make contact with the flywheel. All of the energy in the flywheel is burned off in the hot brake pads. The flywheel stops spinning.
Bonus bonus round:
The big spike of negative high voltage happens because the coil has electrical inertia in the form of flowing current. The flowing current pushes its way through the air gap resulting in the generation of the negative spike of high voltage.
The big spike of negative torque happens because the flywheel has rotational inertia. The rotational inertial pushes on the brake pads and that is what causes the spike of negative torque.
Quote from: MileHigh on November 20, 2015, 10:51:46 PM
Brad,
That's an awesome posting and my sincere compliments.
I will just throw in some "flywheel" colour commentary:
You can see how there is always current flowing through the inductor -> the flywheel never stops turning.
The flywheel increases in speed when you apply positive torque (positive voltage) to it.
The flywheel decreases in speed when you apply braking to it. So you can imagine the (diode + 10 ohm CVR) acting like a caliper and brake pads putting resistance on the spinning flywheel. When the brakes are applied that is negative torque (negative voltage.)
Bonus round:
What happens when you disconnect the 10-ohm CVR from the circuit? Most of us know that you get a spark in the air gap because the air becomes conductive plasma due to the big negative spike of high voltage. All of the energy in the coil is burned off in the hot plasma. The current stops flowing through the coil.
What is the equivalent for the flywheel?
You clamp down on the brake calipers very fast and with maximum pressure. There is a massive spike of negative torque on the flywheel when the brake pads make contact with the flywheel. All of the energy in the flywheel is burned off in the hot brake pads. The flywheel stops spinning.
Bonus bonus round:
The big spike of negative high voltage happens because the coil has electrical inertia in the form of flowing current. The flowing current pushes its way through the air gap resulting in the generation of the negative spike of high voltage.
The big spike of negative torque happens because the flywheel has rotational inertia. The rotational inertial pushes on the brake pads and that is what causes the spike of negative torque.
I couldnt agree more MH with your pulsed flywheel analogy 8). The relationship between the two are very similar.
One thing i need to say or would like to discus is this current flow from the inductive kickback in regards to the resistive load value-reference post 51. I am not seeing this on the scope. As i increase the value of the resistive load,the current flowing through the CVR decreases-along with the instantaneous current spike across the CVR. As i increase the load resistance, the current drop's,and the voltage rises. I am not seeing this instantaneous current flow remaining the same as the input current?.
Brad
Brad:
Sure, this is pretty easy to explain using the circuit that you just posted with modifications. I will modify the schematic.
Then just run the same test, but start increasing the value of the resistor. Say, 10 ohms, then 47 ohms, then 100 ohms, then 200 ohms. Something like that.
You can see it's very simple, when the input pulse energizes the inductor at the end of the ON time, let's assume for example that 100 milliamps are going through the coil.
So, when the pulse goes OFF, you have 100 milliamps flowing through the coil. We will ignore the current that also flows through the resistor.
If the resistor is 10 ohms, you should see an exponential decaying waveform that starts at -1 volt in amplitude.
If the resistor is 47 ohms, you should see an exponential decaying waveform that starts at -4.7 volts in amplitude.
If the resistor is 100 ohms, you should see an exponential decaying waveform that starts at -10 volts in amplitude.
Etc, etc.
The higher the value of the resistor, the faster the waveform will decay.
Ideally, your pulse train will have a long enough OFF time to let the current in the coil decay to zero.
So, you notice that this experiment will confirm that as long as you energize the coil for the same amount of time, and as long as there is no current flowing through the coil when you start the pulse, that the initial current flow through the coil is always the same when it discharges through different resistive loads.
MileHigh
And what about the flywheel? Well, if you are getting the flywheel analogy, then figuring out what will happen with your electrical circuit becomes a no-brainer when you do the same experiment in your head with a flywheel.
Recall, that current flow is equivalent to the rotational speed of the flywheel.
So, here is the equivalent experiment done on a flywheel:
You always start with the flywheel spinning at 100 RPM. Then you apply the brakes with different brake pressures. First you apply the brakes gently (10 ohms.) Then you apply the brakes with increased pressure. Then you apply the brakes one more time with even more pressure.
You observe what happens to the flywheel when you try different brake pressures. That's it.
P.S.: Obviously, the higher the brake pressure, the higher the initial negative torque (negative voltage) on the spinning flywheel.
QuoteI couldnt agree more MH with your pulsed flywheel analogy (https://overunityarchives.com/proxy.php?request=http%3A%2F%2Foverunity.com%2FSmileys%2Fdefault%2Fcool.gif&hash=098b4b5d1b0625a21c497652cbb3a23a77c28bc8). The relationship between the two are very similar.
The "secret" is that for all practical intents and purposes they are literally identical. They have identical sets of equations that describe their respective behaviours.
What it means is that Mother Nature decided to keep it simple. It turns out that electrical components act in the same way as physical things that we see and use in our daily lives.
Quote from: MileHigh on November 20, 2015, 11:25:44 PM
Brad:
Sure, this is pretty easy to explain using the circuit that you just posted with modifications. I will modify the schematic.
Then just run the same test, but start increasing the value of the resistor. Say, 10 ohms, then 47 ohms, then 100 ohms, then 200 ohms. Something like that.
You can see it's very simple, when the input pulse energizes the inductor at the end of the ON time, let's assume for example that 100 milliamps are going through the coil.
So, when the pulse goes OFF, you have 100 milliamps flowing through the coil. We will ignore the current that also flows through the resistor.
If the resistor is 10 ohms, you should see an exponential decaying waveform that starts at -1 volt in amplitude.
If the resistor is 47 ohms, you should see an exponential decaying waveform that starts at -4.7 volts in amplitude.
If the resistor is 100 ohms, you should see an exponential decaying waveform that starts at -10 volts in amplitude.
Etc, etc.
The higher the value of the resistor, the faster the waveform will decay.
Ideally, your pulse train will have a long enough OFF time to let the current in the coil decay to zero.
So, you notice that this experiment will confirm that as long as you energize the coil for the same amount of time, and as long as there is no current flowing through the coil when you start the pulse, that the initial current flow through the coil is always the same when it discharges through different resistive loads.
MileHigh
Should the circuit not be like the one below,where we can measure the current flowing through the inductor during the on time,and then measure the current flowing through the VR during the off time? This would show us what current flows through the inductor during the on time,and we can then see what current is flowing through the load during the off time,and how the different resistive loads change the flyback current delivered to those loads. This is what i was referring to,as we are looking into the flyback power delivered to various loads.
What you have depicted in your diagram is a current loop.
Brad
Brad:
You can experiment with that if you want but do you really need it to be that complicated? I think the use of the diode has merit if your signal source is your unamplified signal generator and you will be using low resistance values.
We know that the current going through the coil is going to be an increasing ramp that will eventually level out. Then, when the applied voltage to the coil goes OFF, the voltage across the variable resistor will tell you the current going through the coil provided that you measured the value of the variable resistor beforehand. Do you see what I mean? You can make an argument that the 10-ohm CVR is redundant. The variable resistor is the CVR with the caveat that you cannot see the current increasing through the coil during the energizing cycle. However, do you really need to see the increasing current through the coil while you are energizing it? If you always energize the coil with the same pulse timing, then the final current through the coil at the instant the energizing pulse goes OFF will be the same.
MileHigh
The attached graphic should make everything much clearer. 8)
Quote from: MileHigh on November 21, 2015, 12:40:53 AM
Brad:
You can experiment with that if you want but do you really need it to be that complicated? I think the use of the diode has merit if your signal source is your unamplified signal generator and you will be using low resistance values.
We know that the current going through the coil is going to be an increasing ramp that will eventually level out. Then, when the applied voltage to the coil goes OFF, the voltage across the variable resistor will tell you the current going through the coil provided that you measured the value of the variable resistor beforehand. Do you see what I mean? You can make an argument that the 10-ohm CVR is redundant. The variable resistor is the CVR with the caveat that you cannot see the current increasing through the coil during the energizing cycle. However, do you really need to see the increasing current through the coil while you are energizing it? If you always energize the coil with the same pulse timing, then the final current through the coil at the instant the energizing pulse goes OFF will be the same.
MileHigh
MH
If we do it your way,then the flyback current is not a current source for a load-it is a current loop back onto itself. As we wish to measure the current as a supply source from the flyback,then that source should be dissipated across a load,in this case the VR.
The CVR will show us the current flowing into the storage unit(the inductor),and it will also show us the current of that stored energy being delivered to the load. If we are to look at the current flowing into the inductor,and then the current flowing from the inductor to a load,then the diagram you posted will not show this-it will only show the current flowing through the current loop,and not to a load. For a P/in and P/out measurement,your circuit will not show this,as there is no load,but only a loop. We want the engine to spin up the flywheel,and then for that stored energy in the flywheel to be delivered to a load-not back to the motor.
What point is there in saying the current in and out of the inductor will be the same value,when that current only loops back to where it came from?.
As i said,the current from the flyback is dependent on the load resistance. You then say it is not,and that i am wrong. But then you post a diagram showing a current loop,where the current from the flyback is sent straight back to where it came from. The current then is not a source,as it is a loop. What i was referring to is the current from the flyback being a source of current for a load,and in this case i am correct in saying that the current flowing from the inductor will be dependent on the load resistance.
Brad
Quote from: MileHigh on November 20, 2015, 06:06:42 PM
For an inductor, you can measure the current flowing through it, and the voltage across it.
Look at the picture of the laboratory flywheel. You can see there is a smaller disk so you can use your hands to spin the flywheel.
When you apply voltage across an inductor, the current does not flow instantly. Instead, it slowly rises.
If you have a stationary flywheel, and then you want to make it spin by applying torque to it with your hands, it does not spin at full speed right away. Instead, the rotational speed slowly rises.
This is the first thing you have to understand:
is like "pressure" on the coil.
Applying torque to the flywheel makes the flywheel spin, starting from zero. The torque is a pressure on the flywheel.
Therefore voltage is like torque.
If you can understand that, it's a good start.
Hi MH
Yes very nice analogy with the flywheel.
But i have a question. If i look at the scope shot i have posted earlier, at the end of the pulse , when the voltage stops or in your analogy when the torque stops spinning up the wheel, the current fall down instantly to zero.
So to me in your analogy it would be the same as if something bloqued instantly the wheel. What happen in this case, immediately after the stop, the really high inertia of the spinning wheel will be transformed in a huge and sudden toppling a the complete system which will jump down the table. This is the flybackspike energy.
So question is, is it possible that the "flywheel" stops instantly when the voltage stops ?
Just a supposition
Laurent
Quote from: tinman on November 20, 2015, 10:27:46 PM
Here you go Synchro-test one complete.
What do you see?
1- a 30% on time from the FG(switch closed)
2-a 70% off time(switch open)
During the 30% on time(switch closed) we can see that the voltage(yellow trace) across the inductor(supply voltage) is on the top side of the zero volt line.The current is flowing in a forward direction,as the current trace(blue trace) is also above the zero volt line.
During the 70% off time(switch open) we can now see that the voltage is inverted across the inductor/resistor series circuit-!BUT! the current is still flowing in the same direction as it was during the 30% on time,as the current trace is still above the zero volt line.
You will also notice that the current continues to flow through the load(the LED) through the complete 70% off time,and never stop's flowing right up until the next on time period. This also shows you that the magnetic field dose not collapse completely before the next on time cycle starts.
So !Bub!
Please show us where exactly the current flow reverses direction?
And please tell us how a current continues to flow(in the same direction ;) ) during the 70% off time if the magnetic field around the inductor collapses instantaneously ?
Show us where this !new! energy is?.
@Tinman,
How do you get a 30% percent on time with a Reed switch? You are artificially starving the pulse of sufficient power to even achieve a field collapse!
You need to load the coil to saturation to get a violent field collapse and current reversal, along with peak magnetic field strength for maximum rotor acceleration. Gotoluc is running a secondary coil on flyback. How is your pulse starvation approach supposed to help run his pulse motor and auxiliary Flyback power coil more efficiently? Milehigh has already exposed you as a hoaxer on the initial current fallacy.
Woopy's test was uncontrived and straight forward. You, on the other hand, have intentionally distorted the test conditions to confuse people and appear as some kind of savant. Your results are worthless. I see right through you! You're just a stinking charlatan and a retarded wanker.
@Woopyjump,
"A generator in a power plant produces electromotive force by moving magnets past coils of wire; the relay coil produces electromotive force as the collapsing magnetic field moves past the wires in the coil".
@synchro
KNOCK IT OFF WITH PERSONAL ATTACKS OR OUR DISCUSSION IS OVER!!
@Brad
Would you please rerun your test with a longer off time to allow the current to drop to zero?
Quote from: citfta on November 21, 2015, 07:35:45 AM
@synchro
KNOCK IT OFF WITH PERSONAL ATTACKS OR OUR DISCUSSION IS OVER!!
@Brad
Would you please rerun your test with a longer off time to allow the current to drop to zero?
@Citfta,
Drop dead!
The Ossie moter clips the pulse and recovers a fraction of the Flyback; Igor runs his Reed switch motor at 100 percent duty cycle and recovers proportionally more power. Both motors run near COP 1. The difference is Igor's generating much more rotor flux. This extra flux could power satellite generator coils like Lidmotor's Maggie. These "Scope Junkys" just pedal nothing but Hocus Pocus.
Quote from: synchro1 on November 21, 2015, 06:22:22 AM
@Tinman,
,[/b]
Woopy's test was uncontrived and straight forward. You, on the other hand, have intentionally distorted the test conditions to confuse people and appear as some kind of savant. Your results are worthless. I see right through you! You're just a stinking charlatan and a retarded wanker.
I see you have resorted to insult's.
This normally happens when you cant admit you are wrong-which you are.
It dose not matter if you use a reed switch,transistor,or the FG with a free wheeling diode on the input-->on is on,and off is off. But feel free to post the scope shot of Woopy's primary coil that shows the current flow reversing when the reed switch opens. ::)
QuoteHow do you get a 30% percent on time with a Reed switch?
The reed switch can be moved closer to,or further away from the magnets on the rotor to adjust the on time of the reed switch. I am not sure why you find something so simple to be a difficult task.
QuoteYou are artificially starving the pulse of sufficient power to even achieve a field collapse! You need to load the coil to saturation to get a violent field collapse and current reversal.
There will be no current reversal regardless of input power. And what is the point in saturating the inductor ?. Do you plan on boiling water with the coil?.
Quotealong with peak magnetic field strength for maximum rotor acceleration. Gotoluc is running a secondary coil on flyback. How is your pulse starvation approach supposed to help run his pulse motor and auxiliary Flyback power coil more efficiently?
Luc sends the inductive kickback current to a second inductor,where as i am sending it to an LED.
QuoteMilehigh has already exposed you as a hoaxer on the initial current fallacy.
Did he?
Quote post 51 from MH-->You are right about the current not changing direction but you are not right about the magnitude of the current.
The second part where MH says i am wrong about the magnitude is because he was talking about instantaneous current,and i was talking about average current per cycle. In regards to the instantaneous current,then MH is correct.
What dose MH and Verpies have to say
Verpies-That Igor guy and Synchro are wrong about the current through the inductor reversing when the inductor is opened.
MH-You are right about the current not changing direction.
Im sorry Synchro,but it is you that is wrong--<the current dose NOT reverse direction -regardless of what type of switch you use,or the amount of power you send to the inductor.
And no--you will not get me to join you in your name calling tactics ;)
Brad
@Tinman,
Verpies and Milehigh are both full of shit just like you!
Quote from: citfta on November 21, 2015, 07:35:45 AM
@synchro
KNOCK IT OFF WITH PERSONAL ATTACKS OR OUR DISCUSSION IS OVER!!
@Brad
Would you please rerun your test with a longer off time to allow the current to drop to zero?
I can do that ;)
Synchro
sometimes we post when we shouldn't...a bit too much to drink is one of those times.
maybe you should take a break ...
Chet
@Tinman,
That's how to measure "Flyback" recovery; First we back the Reed switch away from the "Sweet Spot"!
LoL: Look how little "Flyback" we recover when we only charge the power coil part way!
I have asked Stefan to lock this thread as it no longer serves a useful purpose. I have also asked that synchro be put on notice for his childish behavior.
Quote from: synchro1 on November 21, 2015, 08:26:54 AM
@Tinman,
That's how to measure "Flyback" recovery; First we back the Reed switch away from the "Sweet Spot"!
LoL: Look how little "Flyback" we recover when we only charge the power coil part way!
Why lock the thread?-because of synchro?
Anyway,below is a scope shot for the test you requested.
Notice the ringing starts when all the current has been dissipated from the inductor.
Quote from: woopy on November 20, 2015, 10:33:00 AM
Hi all
Youp seems not to be so easy this flyback spike understanding.
For me what i see on the scope is that at the end of the pulse, when the reed sharply opens, the current trace goes to almost instantly (verticaly) to zero. And only AFTER this shut down, begins the flybackspike.
So to me the current who build up the magneticfield is gone , totally dissipated, finished at the end of the pulse. He was totally used to precisely build the magnetic field. So he has no more direction at all.
Than it stays the expanded magnetic field around the coil (??) and what exactly happens at that point is ?? But on the scope , suddenly a strong narrow "high negative voltage" trace appears and also a strong very very narrow and strong current trace (not shown on the pic) also. What does create this event is still puzzling to me.
If it is the collapsing of the magneticfield please explain the process with simple words if possible.
But anyway to me that is a new current who has nothing to do directly with the one that created the magneticfield. It is what i call hes "son"
Youp, please don't shoot me down it's only an idea.
Laurent
Hi Laurent,
Would like to ask which circuit point did you measure the current you show at the yellow trace?
It is the input current going into the drive coil, is this correct?
Here is the schematic, please describe where you inserted the current sensing resistor, was it in series with the drive (5mH) coil? in this case use number 5 as a reference to that point:
http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg466151/#msg466151 (http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg466151/#msg466151)
Thanks,
Gyula
Quote from: synchro1 on November 21, 2015, 08:26:54 AM
@Tinman,
That's how to measure "Flyback" recovery; First we back the Reed switch away from the "Sweet Spot"!
LoL: Look how little "Flyback" we recover when we only charge the power coil part way!
Here is a scope shot from the pulse motor Synchro. Notice the voltage across the coil during the on time-what dose that tell you?. The current trace is now over a 1 ohm 5 watt CVR--enough power in for you now?. So now-show me where the current reverses direction?.
@MH and Verpies.
As i was saying-looking at the scope shot,you can see that the peak current during the on time is more than the instantaneous current during the off time. You can see the current drop in value at switch off.
Great screen shot showing exactly what we have been saying. I have received a PM asking me to not have the thread locked so I guess if you guys want to keep it open I will ask Stefan to not lock it. As far as synchro, guys like him always get theirs in the end.
Hi Guyla
I have redo the measurement so the 1 ohm sensing resistor is placed between the reed and the main coil (05 ohm and 5 mH), the probe for the blue trace is connected (not grounded ) to the output of the main coil (X1) . For the blue trace i had to invert the chanel.
-First pic is a general situation
-Second pic is the trace of the same situation but i tried to expand a max the view to see what happen when the voltage stops and the current fall straight down to zero.
Yep my trace are very different of the Tinman's. Perhaps it is due to very low resistance of my main coil and the low frequency. Tinman is running at 1000Hz ?
Hope this helps
Citfta i don't think that you have to stop this thread.
MH what about the brutal stop of the flywheel ?
Laurent
Quote from: tinman on November 21, 2015, 01:32:51 AM
MH
If we do it your way,then the flyback current is not a current source for a load-it is a current loop back onto itself. As we wish to measure the current as a supply source from the flyback,then that source should be dissipated across a load,in this case the VR.
The CVR will show us the current flowing into the storage unit(the inductor),and it will also show us the current of that stored energy being delivered to the load. If we are to look at the current flowing into the inductor,and then the current flowing from the inductor to a load,then the diagram you posted will not show this-it will only show the current flowing through the current loop,and not to a load. For a P/in and P/out measurement,your circuit will not show this,as there is no load,but only a loop. We want the engine to spin up the flywheel,and then for that stored energy in the flywheel to be delivered to a load-not back to the motor.
What point is there in saying the current in and out of the inductor will be the same value,when that current only loops back to where it came from?.
As i said,the current from the flyback is dependent on the load resistance. You then say it is not,and that i am wrong. But then you post a diagram showing a current loop,where the current from the flyback is sent straight back to where it came from. The current then is not a source,as it is a loop. What i was referring to is the current from the flyback being a source of current for a load,and in this case i am correct in saying that the current flowing from the inductor will be dependent on the load resistance.
Brad
I look forward to seeing the test done the way you want to do it. It will make very little difference, because the initial current will be the same with the 10-ohm resistor or not. In essence, that's the point of the test. I also forgot to acknowledge your point about the average current measurement of the discharge cycle decreasing as you increase the value of the load resistance, which is absolutely true.
QuoteIf we do it your way,then the flyback current is not a current source for a load-it is a current loop back onto itself. As we wish to measure the current as a supply source from the flyback,then that source should be dissipated across a load,in this case the VR.
I think that you might not be seeing the forest for the trees here. If we ignore the diode for a second, you are just looking a the "difference" between the coil discharging through one resistor vs. the coil discharging through two resistors in series. No matter how you look at it, the current will flow in a loop. There can be one resistor, or two resistors in series in the loop, with the discharging inductor as the power source for the loop.
Quote from: woopy on November 21, 2015, 04:52:41 AM
Hi MH
Yes very nice analogy with the flywheel.
But i have a question. If i look at the scope shot i have posted earlier, at the end of the pulse , when the voltage stops or in your analogy when the torque stops spinning up the wheel, the current fall down instantly to zero.
So to me in your analogy it would be the same as if something bloqued instantly the wheel. What happen in this case, immediately after the stop, the really high inertia of the spinning wheel will be transformed in a huge and sudden toppling a the complete system which will jump down the table. This is the flybackspike energy.
So question is, is it possible that the "flywheel" stops instantly when the voltage stops ?
Just a supposition
Laurent
Yes, in your circuit the "secondary flywheel" stops right away after the capacitor finishes discharging. The explanation for this is simple. The secondary coil has a resistance of 210 ohms. So that means that there is "a lot of friction in the ball bearings" for the "secondary flywheel." If you have a coil with a very very low resistance, and you get current going through it and then you short the two terminals of the coil together, then you can assume that with a Hall sensor you will be able to detect the magnetic field around the coil for several tens of seconds or more. Note you do not want to use a current sensing resistor because that would defeat the purpose of seeing how long current will continue to circulate in the coil.
Laurent,
Please see the attached image where I annotated one of your scope traces to illustrate the flywheel analogy.
There is also a very very important lesson to be learned here that I previously mentioned. Let's talk about an electrical coil and the negative voltage spike. What is the source of the negative voltage spike? Is it the coil that is generating the negative voltage spike, or is it the load that is generating the negavive voltage spike as a reaction to the pulse of current from the coil?
Now let's go back to the flywheel analogy. We know that the reason there is a big spike of negative torque on the flywheel is because the brake pads clamp down hard on the flywheel, just like disk brakes in a car. The brake pads are the source of the big spike of negative torque on the flywheel. The brake pads are the mechanical load on the spinning flywheel.
That is telling you that for an electrical coil, the coil itself is not the source of the negative voltage spike. Rather, it is the load that generates the negative voltage spike as a reaction to the pulse of current from the coil.
Quote from: citfta on November 21, 2015, 09:25:17 AM
I have asked Stefan to lock this thread as it no longer serves a useful purpose. I have also asked that synchro be put on notice for his childish behavior.
@Citfta,
Oh! Please don't hurt me boss, I'm jus da poo boss's nigga boss! Big bully Mumzy Jr. Member "Carroll" Lockin the Thread down now, huh!
Quote from: tinman on November 21, 2015, 09:44:50 AM
@MH and Verpies.
As i was saying-looking at the scope shot,you can see that the peak current during the on time is more than the instantaneous current during the off time. You can see the current drop in value at switch off.
Yes, you can see the "step down" that the current takes at switch off. You notice it all makes sense because for a short time you see the voltage trace is near-vertical and negative at that time. That corresponds to a short powerful impulse of negative torque on a flywheel so that the RPM takes a near-instant step down in speed.
Quote from: Erfinder on November 21, 2015, 02:42:11 PM
I'm curious......Are you aware of a method which can be applied to electro mechanical devices which grants one the ability to produce "voltage spikes", a method which does not put the brakes on the prime mover?
Regards
@Erfinder,
"The purpose of a "flyback transformer" in a monitor or television is to generate high voltage about 24 kilovolts dc to picture tube or crt".
Hi Laurent,
Thanks for answering where the 1 Ohm resistor was placed and for the scope shots too. This helps me answer.
I attached a schematic I fabricated from one of tinman's and your drawings. I want to show where is a correct place to insert the 1 Ohm resistor to monitor the current in the coil in both situations i.e. when the reed switch is ON and when the reed switch is OFF.
This is because as you described the placement of the 1 Ohm (i.e. between the reed and the 5 mH coil) it was good for seeing the input current while the reed was ON. And after the reed switch was off, the coil current could not be seen at the same 1 Ohm position because the 1 Ohm was already in an open circuit towards the reed.
If you feel like to repeat the test as per this schematic, yellow trace would show the current in the coil, blue trace would be the voltage across the coil, the Ref point on the negativ battery would be the scope common crocodile grnd.
Thanks,
Gyula
Hi Guyla
here is the scope shot following your shematic, the expanded shot seems really noisy, can we get something out of it ?
MH
thank's for explanation, just a little doubt, it seems on the second scope shot (expanded the one that accopmpanied the shot you commented ) that the "braking " negative voltage seems to happen after that the current stops. Is it perhaps an artifact of my scope ?
Laurent
Hi Laurent,
Thanks. Indeed the expanded shot is noisy, this is probably due to the 1 Ohm which may be small for sensing the lower current the flyback spike would provide towards the 210 Ohm coil plus the 0.3 uF capacitor after the reed is OFF.
If you could use the 10 Ohm instead of the 1 Ohm perhaps we could see the better shot on the current when it starts flowing in the coil after the reed is off.
OF course the 10 Ohm is a much higher value than the 0.5 Ohm coil DC resistance so you may have to increase the input voltage to arrive at the 110-120 mA input current range.
Gyula
Quote from: woopy on November 21, 2015, 04:05:07 PM
thank's for explanation, just a little doubt, it seems on the second scope shot (expanded the one that accopmpanied the shot you commented ) that the "braking " negative voltage seems to happen after that the current stops. Is it perhaps an artifact of my scope ?
Thank you for your comments. I don not think there are any issues with your scope. I believe Gyula is correct, and the scope shot that I marked up did not show the rapidly decaying current in the coil because the CVR was in the "wrong" place. That shows you the problems that can happen when you work without a schematic diagram to backup your scope traces. However, what I marked up on your scope shot still applies. Assuming that you do the test for Gyula you will see the rapidly decreasing current in the coil as the high voltage spike happens. The higher the load resistance, the faster the current will decrease and the higher the negative voltage spike you will see on your scope trace - just like applying the brakes to a spinning flywheel.
P.S.: Vive la France!
Quote from: Erfinder on November 21, 2015, 02:42:11 PM
I'm curious......Are you aware of a method which can be applied to electro mechanical devices which grants one the ability to produce "voltage spikes", a method which does not put the brakes on the prime mover?
Regards
I am honestly drawing a blank here. However, reading between the lines I am guessing that you are concerned about the fact that the voltage spikes represent the loss of useful energy as heat via a resistance. And yes, what that does mean in many cases is that when a beginning experimenter gets excited about the high voltage spikes, he or she probably doesn't realize that that represents the loss of energy.
Switching to my preferred terminology, the current pulses from a discharging inductor don't have to always be dissipated in a resistive load. They are used all the time in switching power supplies to create just about any voltage you want.
Note that if you are going to extract energy from a prime mover like a coil via current pulses, then that will put "the brakes" on the prime mover to a certain extent.
@Tinman,
What do you make of this:
"You should understand that it is the presence of a moving magnetic field that induces a current in a coil winding.
When a battery is connected to a coil, the battery producing D.C., there is no inductive effect. But when you suddenly disconnect the battery, you cause the magnetic field to collapse, and in collapsing, that movement, creates in the coil a reverse current and magnetic field which under certain conditions, coil winding, etc., produces enough voltage to shock you.
Because the falling, magnetic field is reversed in polarity and induces a current reversed in polarity, that effect has become known as reversed electromagnetic force or field or reverse EMF".
Quote from: citfta on November 21, 2015, 10:57:15 AM
Great screen shot showing exactly what we have been saying. I have received a PM asking me to not have the thread locked so I guess if you guys want to keep it open I will ask Stefan to not lock it. As far as synchro, guys like him always get theirs in the end.
Just to let you know, I have moderator privileges of all topics in reactive power usage. So if you want anything done I can also help if Stefan is not available.
Can't believe 8 pages and still going :o
Luc
Quote from: gotoluc on November 21, 2015, 09:14:27 PM
Just to let you know, I have moderator privileges of all topics in reactive power usage. So if you want anything done I can also help if Stefan is not available.
Can't believe 8 pages and still going :o
Luc
@Gotoluc,
We have a wave and particle theory of light that's acceptable. There's a reverse current theory that's used to describe the effect of magnetic field collapse that runs in tandem with the conventional one. I understand the other theory, the reverse current theory has value as well.
Hi luc,
Thanks for the offer of help. We'll see how it goes from here on. But I have no patience for people that resort to name calling and insults when they see they are losing a discussion. That to me indicates they are beginning to realize they are on shaky ground with their idea and are too small to admit they might be wrong.
I guess we should keep the thread open because the scope shots and other info that has been posted are probably a help to some and a good refresher for the rest of us.
Carroll
@Citfta,
We just need to rename "Flyback" "Flyforward"!
Quote from: synchro1 on November 21, 2015, 09:21:37 PM
@Gotoluc,
We have a wave and particle theory of light that's acceptable. There's a reverse current theory that's used to describe the effect of magnetic field collapse that runs in tandem with the conventional one. I understand the other theory, the reverse current theory has value as well.
When it comes to theories and why or how something work I leave that to the scientist to work out.
Using a practical effect that can improve or assist a motor or generator is all that I'm interested in and will put all my energy ;D in that alone.
Luc
Quote from: Erfinder on November 21, 2015, 02:42:11 PM
I'm curious......Are you aware of a method which can be applied to electro mechanical devices which grants one the ability to produce "voltage spikes", a method which does not put the brakes on the prime mover?
Regards
Could that be shorting at a specific time?
For those ignorant of what the term flyback really means I will try to educate them a little. The term flyback originated in the early days of television when all TV's had a CRT or cathode ray tube. After the electron beam was moved from left to right across the screen to paint a line of black and white data the horizontal output tube was switched off very quickly and the inductive spike from the horizontal output transformer was used to drive the horizontal defection coils to cause the electron beam to "flyback" to the left side of the screen. The term originally had nothing to do with the actual discharge of an inductor. Just like many of the terms used by hobbyist the term has taken on a different meaning.
Quote from: MileHigh on November 21, 2015, 01:55:29 PM
Yes, you can see the "step down" that the current takes at switch off. You notice it all makes sense because for a short time you see the voltage trace is near-vertical and negative at that time. That corresponds to a short powerful impulse of negative torque on a flywheel so that the RPM takes a near-instant step down in speed.
MH,Verpies-anyone
When we calculate the dissipated power from the below diagram,during the 70% off time,the circuit should be split in half,and the dissipated power from each half should be equal-correct?.
So if we have say 13,68mW being dissipated across the inductor and resistor,we should also have 13.68mW being dissipated across the LED--?--see pic below.
So should it now be 13.68 + 13.68 X 70% for our power out?
Which would be 19.15mW ?
Quote from: gotoluc on November 21, 2015, 09:41:43 PM
When it comes to theories and why or how something work I leave that to the scientist to work out.
Using a practical effect that can improve or assist a motor or generator is all that I'm interested in and will put all my energy ;D in that alone.
Luc
@Luc,
I suggest someone try increasing rotor magnet strength.
Prototype "Flyback Motor"; Look at the size of the rotor magnet:
https://www.youtube.com/watch?v=1gWZchP8-Hk
Quotes from Tinman's coil shorting circuit thread:
Comment #87 from Erfinder:
"I completely agree with you, Tinman's builds, are awesome".
6 comments later, #93 Tinman replies to Erfinder:
"You are a sick man,and i wish to have nothing more to do with you.
Some told me some time back that you are a couple of cows short in the top paddock,and your words speak volumes".
Laurent, Gyula:
For the zoom in on the switching event (attached), I am quite sure that this is noise associated with the reed switch opening and breaking the voltage supply to the drive coil. To describe that in more detail, it means that when the reed switch opens and breaks contact, it actually makes and breaks the contact multiple times during the switching process which in this case appears to take place over the course of 200 microseconds. It is also called "switch bounce" and with larger automotive-type switches the bouncing can last on the order of a few milliseconds.
The easiest way to stop this would be to go with some sort of a solid state switch configuration that has a conditioned Schmidt trigger input. I don't think it is necessary for this type of investigation.
MileHigh
Quote from: tinman on November 21, 2015, 10:11:43 PM
When we calculate the dissipated power from the below diagram,during the 70% off time,the circuit should be split in half,and the dissipated power from each half should be equal-correct?.
So if we have say 13,68mW being dissipated across the inductor and resistor,we should also have 13.68mW being dissipated across the LED--?--see pic below.
So should it now be 13.68 + 13.68 X 70% for our power out?
Which would be 19.15mW ?
No, there is no reason to assume an even energy split between the two halves. I just did a big measurement treatise on the "Sharing ideas on how to make a more efficent motor using Flyback" thread, you should study it.
Also, why are you back to using an LED? Recently the point that it is no good for power measurements was pounded away at over and over. I suggest that you look up "IV characteristics of a diode" to understand why using an LED is one of the worst things that you can possibly use to make power measurements with.
My suggestion to you is to swap the LED for a resistor and read my measurement treatise and then come up with a logical step-by-step measurement strategy.
Quote from: MileHigh on November 22, 2015, 01:24:17 AM
No, there is no reason to assume an even energy split between the two halves. I just did a big measurement treatise on the "Sharing ideas on how to make a more efficent motor using Flyback" thread, you should study it.
My suggestion to you is to swap the LED for a resistor and read my measurement treatise and then come up with a logical step-by-step measurement strategy.
QuoteAlso, why are you back to using an LED? Recently the point that it is no good for power measurements was pounded away at over and over. I suggest that you look up "IV characteristics of a diode" to understand why using an LED is one of the worst things that you can possibly use to make power measurements with.
It is because of the I/V characteristics(which im well familiar with) of an LED,that i am using an LED
Can you work out why?.
Quote from: synchro1 on November 22, 2015, 12:33:44 AM
Quotes from Tinman's coil shorting circuit thread:
Comment #87 from Erfinder:
"I completely agree with you, Tinman's builds, are awesome".
6 comments later, #93 Tinman replies to Erfinder:
"You are a sick man,and i wish to have nothing more to do with you.
Some told me some time back that you are a couple of cows short in the top paddock,and your words speak volumes".
Nice try synchro.
Now lets have a look at why i said what i said-->in stead of your fumbled effort to discredit me.
Quote Erfinder-->Brad, can I call you Brad, its better to be pissed off than pissed on, well....guess that's relative huh....if getting pissed on is your thing or not.
My reply to that was-->"You are a sick man,and i wish to have nothing more to do with you.
Some told me some time back that you are a couple of cows short in the top paddock,and your words speak volumes".
You really seem to take being wrong to heart synchro,even to go as far as chopping up bit's and pieces of post, and pasting them together to try and make those that show you to be incorrect, look bad.
Nice try,but an epic fail,as people here have the ability to read--do you not know this?.
Hi all
This morning i remembered an experiment i made some time ago, which seems to me as an other analogy , and could be relevant here.
https://www.youtube.com/watch?v=KTluXuMHEkw (https://www.youtube.com/watch?v=KTluXuMHEkw)
A toy balloon is attached to the tap.
In the balloon there is a hole where a semi soft hose is inserted.
The hose make first a U shape maintained by a spring (sandow) and then goes straight down with a certain length.
Then we open the tap so a current of water flows through the balloon and then in the hose. After some short time, the speed of the water falling down in the the down straight part of the hose, due to gravity, is faster than the input water supply, and sucks the the skin of the balloon against the entry of the hose, stopping instantly the current flow. The rest of the system is not important for my understanding of our purpose. What seems important to me in this experiment is the following
Following that brutal stop of the current flow (which is the cause ), then the inertia of the water contained in the hose induces a violent hammer like struck (the effect) which distend the spring (the result).
Mr Uday who proposed this experiment, tried to use this system to rotate a generator. I don't now if he managed.
That is just for fun
Good sunday to all
Laurent
Quote from: tinman on November 22, 2015, 04:14:00 AM
Nice try synchro.
Now lets have a look at why i said what i said-->in stead of your fumbled effort to discredit me.
Quote Erfinder-->Brad, can I call you Brad, its better to be pissed off than pissed on, well....guess that's relative huh....if getting pissed on is your thing or not.
My reply to that was-->"You are a sick man,and i wish to have nothing more to do with you.
Some told me some time back that you are a couple of cows short in the top paddock,and your words speak volumes".
You really seem to take being wrong to heart synchro,even to go as far as chopping up bit's and pieces of post, and pasting them together to try and make those that show you to be incorrect, look bad.
Nice try,but an epic fail,as people here have the ability to read--do you not know this?.
@Tinman,
"An "ideal inductor" has inductance, but no resistance or capacitance, and does not dissipate or radiate energy. However real inductors have side effects which cause their behavior to depart from this simple model".
Sometimes it helps to simplify to increase understanding as in the case of ideal componants like capacitors and inductors without involving all the messy issues that accompany them.
There's nothing I don't understand about the "Magnetic Field Inertia" issue you keep trying to teach me. Additionally, your "concealed coil shorting circuit" has caused me to grow suspicious of you.
On top of that, and more importantly, you seem to fail to understand sufficiently how the rate at which spark gap contacts separate effects the level of Flyback Voltage.
Your attitude has been repugnantly condescending, not only towards me and Erfinder, but others who have challenged your special knowledge on your "Coil Shorting Thread". My opionion of you as a "Mylow" will remain unshaken until such time as you divulge your selfishly "Bogarted" secret!
Quote from: tinman on November 21, 2015, 10:11:43 PM
MH,Verpies-anyone
When we calculate the dissipated power from the below diagram,during the 70% off time,the circuit should be split in half,and the dissipated power from each half should be equal-correct?.
So if we have say 13,68mW being dissipated across the inductor and resistor,we should also have 13.68mW being dissipated across the LED--?--see pic below.
So should it now be 13.68 + 13.68 X 70% for our power out?
Which would be 19.15mW ?
Hi Brad,
I can agree with splitting the circuit in half as you show but I disagree with the assumption that the dissipated power in each half should be equal.
IF your measurements indicate this half-half power split, I think it is only a coincidence in this particular circuit. Just imagine you replace the LED with a normal Si diode, then the voltage and current conditions change in that half part of the circuit and this causes a change in the other half part too: why would the two changes have (again) an equalizing effect?
Under the conditions you show with the measured voltage and current (I suppose you measured them), the voltage drop across the 10 Ohm should be V= 4.8mA*10 Ohm=48 mV, right? (Of course I do not know the DC resistance of the coil.)
If you consider the LED DC resistance it represents in the moment when its current and voltage are 4.8mA and 2.85V, this gives R=593 Ohm. (I neglect LEDs may have even some hundred pF self capacitance.)
It is okay that the total dissipation is approached by considering each components in the closed circuit and adding their dissipations up. Perhaps a close approach would be to consider the coil's DC resistance too and adding up the losses for the 3 components: the LED, the 10 Ohm and the copper loss of the coil and then consider the duty cycle issue.
Gyula
Quote from: tinman on November 22, 2015, 04:04:05 AM
It is because of the I/V characteristics(which im well familiar with) of an LED,that i am using an LED
Can you work out why?.
Offhand I can't. Especially since I believe you stated the math function doesn't really work on your scope. If you explain why it might hit me, not sure. Can you explain why you want to use an LED?
Erfinder,
Thanks for your long reply, not much I can add. Who knows, there may be "deeper meaning" or some kind of deeper insight into some aspects of electronics yet to be discovered. However, I just dreamed up an analogy for my approach. A mirror for most people just works. They look into it and they see a reflection and they understand it and it is tangible. However, "under the hood" you have all of the EM wave mechanics where the electric field in the incident wave bounces of of the flat silvered surface because it is electrically conductive, you can develop some kind of wave equation, discuss the reflection because of the impedance change, bla bla bla. But to the average person it's just a mirror and they take it for what it is, and what it does, on face value - it is what it is.
Electronic circuits and the associated electronics components are like the mirror and can be taken at face value. They do what they do. There is no "radiant energy" with the meaning that you hear on the forums. "Don't split the dipole" means nothing to me. Perhaps the worst offender is the misused and abused Bloch wall. Then there are all the "corkscrew" theories about magnetic fields - when in fact understanding how magnetic fields work are quite enigmatic for the beginner. Hence the discussions on the forums that come up within a certain crowd about the "differences between north and south."
I don't want to get into a big philosophical discussion so I suggest that we leave it at that. I think this thread is serving a good purpose and it's like the clock has been turned back six years and we are trying to master the dynamics of a charging and discharging inductor - and hopefully getting it right so that people have more understanding and more ability to distinguish truth from fiction. One of the problems is that we are swimming in a morass of fiction and you have seen a sprinkling of totally nonsensical postings on this very thread. I can get annoyed with the "Alice in Wonderland just took a huge hit off the bong" postings sometimes. And we can't forget the pervasive criminal element lurking out there scheming up a new pitch to get that GoFundMe hamster wheel turning.
MileHigh
Quote from: tinman on November 21, 2015, 10:11:43 PM
MH,Verpies-anyone
When we calculate the dissipated power from the below diagram,during the 70% off time,the circuit should be split in half,and the dissipated power from each half should be equal-correct?.
So if we have say 13,68mW being dissipated across the inductor and resistor,we should also have 13.68mW being dissipated across the LED--?--see pic below.
So should it now be 13.68 + 13.68 X 70% for our power out?
Which would be 19.15mW ?
Hi Brad,
No disrespect to anyone else's opinion about how to analyze this circuit. Almost any way to analyze could be correct depending on what you are looking for. But I would approach it from a slightly different angle. I would divide the circuit into the source and load. During the off time until the coil's energy is depleted the coil is the source and the rest of the circuit is the load. The source (coil) has to supply the same amount of energy as the load dissipates.
Carroll
I think to resolve all issues use a DPDT relay and external coil and measure coil current then discharge coil separately into capacitor and scope current and voltage.
If you know any such video tutorial on youtube with scope measurement , let me know....
Quote from: synchro1 on November 22, 2015, 05:26:21 AM
@Tinman,
"
Sometimes it helps to simplify to increase understanding as in the case of ideal componants like capacitors and inductors without involving all the messy issues that accompany them.
QuoteYour attitude has been repugnantly condescending, not only towards me and Erfinder,
I have shown you no disrespect synchro,despite your disrespect toward myself-E.G--> post 94 quote: Verpies and Milehigh are both full of shit just like you!
QuoteAn "ideal inductor" has inductance, but no resistance or capacitance, and does not dissipate or radiate energy. However real inductors have side effects which cause their behavior to depart from this simple model".
And i use real inductors ,as i have no ideal inductors.
QuoteThere's nothing I don't understand about the "Magnetic Field Inertia" issue you keep trying to teach me. Additionally, your "concealed coil shorting circuit" has caused me to grow suspicious of you.
Nothing new there synchro. I suspect there are many who think the same way. We humans tend to attack that of what we dont understand.
QuoteOn top of that, and more importantly, you seem to fail to understand sufficiently how the rate at which spark gap contacts separate effects the level of Flyback Voltage.
You do understand synchro that any spark or arc is power being dissipated. There for an ideal switch would have no such arcing.
QuoteMy opionion of you as a "Mylow" will remain unshaken until such time as you divulge your selfishly "Bogarted" secret!
I will keep that in mind, although i doubt i will loose much sleep over it.
Quote from: MileHigh on November 22, 2015, 06:20:11 AM
Offhand I can't. Especially since I believe you stated the math function doesn't really work on your scope. If you explain why it might hit me, not sure. Can you explain why you want to use an LED?
I dont think we need the math (although the math trace works quite fine on my scope)function to work out our P/in and dissipated power when the wave forms are very clean.
I am using an LED for a load for the very reason that the I and V curve is non linear.
It al;pws me to increase or decrease the inductive kickback current while maintaining very close to a set voltage during the off cycle. I can then measure the light output from the LED via way of a light box-->a small solar panel inside a box with the LED,and then we place a resistor across the solar panel output. I can now see if the LED's output rises with the increase of current,while the voltage remains much the same across that LED. For this particular LED,it seems that the optimum voltage is 3.2v,and any additional current makes no difference to the output of the LED. The other thing is i can now see what happens when i increase the voltage on the input--this is how i increase the current on the flyback. Now we have a situation where only the current on both the input and output rises,but both the input and output voltage remain the same,even though i am increasing the voltage on the input. On the flyback side this is expected with the LED in play,but why dose only the current rise on the input,and not the voltage?.
Quote from: gyulasun on November 22, 2015, 05:48:02 AM
Hi Brad,
I can agree with splitting the circuit in half as you show but I disagree with the assumption that the dissipated power in each half should be equal.
IF your measurements indicate this half-half power split, I think it is only a coincidence in this particular circuit. Just imagine you replace the LED with a normal Si diode, then the voltage and current conditions change in that half part of the circuit and this causes a change in the other half part too: why would the two changes have (again) an equalizing effect?
Under the conditions you show with the measured voltage and current (I suppose you measured them), the voltage drop across the 10 Ohm should be V= 4.8mA*10 Ohm=48 mV, right? (Of course I do not know the DC resistance of the coil.)
If you consider the LED DC resistance it represents in the moment when its current and voltage are 4.8mA and 2.85V, this gives R=593 Ohm. (I neglect LEDs may have even some hundred pF self capacitance.)
It is okay that the total dissipation is approached by considering each components in the closed circuit and adding their dissipations up. Perhaps a close approach would be to consider the coil's DC resistance too and adding up the losses for the 3 components: the LED, the 10 Ohm and the copper loss of the coil and then consider the duty cycle issue.
Gyula
Hi Gyula
Yes,the measurements are from the scope shot below.
If we look at the input side of thing's,it is saying our inductor has a resistance of around 1890 ohms at that frequency when we subtract the 10 ohm CVR. The reason i asked that question is that the LED dose not need that amount of current to get the 2.85 volts across it at that 70% duty cycle. So with this circuit,how do we measure the power being dissipated by the LED?.-->circuit below.
Quote from: tinman on November 22, 2015, 08:51:20 AM
Hi Gyula
Yes,the measurements are from the scope shot below.
If we look at the input side of thing's,it is saying our inductor has a resistance of around 1890 ohms at that frequency when we subtract the 10 ohm CVR. The reason i asked that question is that the LED dose not need that amount of current to get the 2.85 volts across it at that 70% duty cycle. So with this circuit,how do we measure the power being dissipated by the LED?.-->circuit below.
Hi Brad,
I can agree with the 4.8mA current (from the scope) for the LED when the voltage across it is 2.85 V, (this 13.68mW is the mean or average comsumption of the LED) and this could be checked with brightness comparison in your "light box": just feed the same pure DC power from your power supply to the same LED to get the same DC output voltage across the box output resistor what the circuit provides to it from the flyback energy. (Be careful to the LED when changing the supply output voltage, better to use a series few hundred Ohm resistor to limit current and measure the voltage across LED with a DMM, of course the current is to be checked too.)
I think the dissipation in the 10 Ohm CVR comes first from the input current of your function generator during the ON times and to this we need to add the dissipation caused by the 4.8 mA current during the OFF times.
The wire resistance of the coil also dissipates during the ON time and the OFF times like the 10 Ohm, while the LED dissipates only during the OFF times, this latter is the 13.68 mW.
EDIT: I think the duty cycle is to be considered for the LED too because the 4.8 mA current through the 10 Ohm includes both the ON and the OFF time currents and I based the LED dissipation on that.
Gyula
Quote from: gyulasun on November 22, 2015, 09:30:18 AM
Hi Brad,
and this could be checked with brightness comparison in your "light box": just feed the same pure DC power from your power supply to the same LED to get the same DC output voltage across the box output resistor what the circuit provides to it from the flyback energy. (Be careful to the LED when changing the supply output voltage, better to use a series few hundred Ohm resistor to limit current and measure the voltage across LED with a DMM, of course the current is to be checked too.)
I think the dissipation in the 10 Ohm CVR comes first from the input current of your function generator during the ON times and to this we need to add the dissipation caused by the 4.8 mA current during the OFF times.
The wire resistance of the coil also dissipates during the ON time and the OFF times like the 10 Ohm, while the LED dissipates only during the OFF times, this latter is the 13.68 mW.
Gyula
QuoteI can agree with the 4.8mA current (from the scope) for the LED when the voltage across it is 2.85 V, (this 13.68mW is the mean or average comsumption of the LED)
This was the very reason i asked the question Gyula,as it dose not take 4.8mA to get 2.85v across the LED-->it only takes around 1.2mA to get the 2.85v across the LED. This i checked simply by using my FG,the LED and CVR. With the FG set to a 70% duty cycle,and set so as i have the 2.85v across the LED,i only had 12mV across the 10 ohm CVR. So it seems that the power dissipated by the LED cannot be measured using the voltage across it,and current flowing through the CVR when in the flyback circuit,as it gives us 4x the actual power required to obtain the 2.85v at a 70% duty cycle-by way of the FG.
Brad
Hi Brad,
Yes and in the meantime I wrote an addition to my previous post.
The CVR should be placed directly in series with the LED and not in series with the coil.
Thanks, Gyula
Quote from: gyulasun on November 22, 2015, 09:49:03 AM
Hi Brad,
Yes and in the meantime I wrote an addition to my previous post.
The CVR should be placed directly in series with the LED and not in series with the coil.
Thanks, Gyula
Gyula
The CVR is directly in series with the LED-see circuit below.
As there is only 3 components in the circuit during the flyback stage,how can they not all be in series?.
Yes it is in series but it is also included in the input circuit current and the scope shot includes that too, no?
EDIT This is why I goofed by saying the 13.68 mW LED dissipation.
Gyula
Quote from: gyulasun on November 22, 2015, 09:57:22 AM
Yes it is in series but it is also included in the input circuit current and the scope shot includes that too, no?
Yes it dose,but we can use the scope shot i posted above with the circuit to separate the two--the on period and off(flyback) period. The power calculations for dissipated power for the LED should be made from the 70% off period. The inductor provides the power for the LED(our current source)during the off period. We measure that power by way of calculating the current flowing through our 10 ohm resistor,and the voltage across the LED--then multiply that result by 70% to get our average power dissipated by the LED over 1 cycle. But when i do the very same with my FG,i find that i only need 25% of that calculated power to get the same voltage across the LED,and for the LED to put out the same light.
This can only mean that !although! the scope show's us the current being delivered to the load,it's actually not there,and this is what i was trying to explain to MH as to why you cannot measure current flowing out from the inductor during the flyback period,as it is a looped system.
This very thing happened not so long ago,when i was posting some measurements from the HTT test,and things didnt add up when the current was calculated across a CVR,and the power dissipated across the load resistor. When we added the two resistor values together,and the voltage across those two resistor's,we ended up with a different power value than we did when we calculated it the other way.
I will do up a video tomorrow night after work,and show you what i mean.
Brad
Okay Brad, thanks and probably using a separate CVR directly in series with the LED too and check the current pulses across it may help to estimate only the LED dissipation. This way the input coil current (during ON time) cannot be sensed by the separate CVR.
Gyula
Hi Gyula and all
I redo the scope measurement you have proposed and i checked all connections and changed some old crocodile pinch.
Now i got a very nice trace where we can clearly see that the current does not fall down vertically as per my first erroneous measurement.
So effectively the "negative voltage" seems to "brake" strongly the current (flywheel) which goes extremely fast to zero in my particular arrangement.
Thank's to all of you for your help.
Hope this helps
Laurent
Hi Laurent,
Very good, thanks for all your efforts. I edited your expanded scope shot to include some lines and explanations, those may be useful for interested readers (schematic is here: http://overunity.com/16203/inductive-kickback/msg466434/#msg466434 (http://overunity.com/16203/inductive-kickback/msg466434/#msg466434) )
Your scope shot nicely shows that from the moment the reed switch switches the current off, the coil current (yellow trace) remains above the zero line i.e. its original polarity is maintained and its amplitude reduces to zero as the flyback pulse with its negative polarity lasts.
Maybe your questions have been answered in connection with this inductive kickback topic? 8)
Gyula
Quote from: tinman on November 22, 2015, 08:43:04 AM
I dont think we need the math (although the math trace works quite fine on my scope)function to work out our P/in and dissipated power when the wave forms are very clean.
I am using an LED for a load for the very reason that the I and V curve is non linear.
It al;pws me to increase or decrease the inductive kickback current while maintaining very close to a set voltage during the off cycle. I can then measure the light output from the LED via way of a light box-->a small solar panel inside a box with the LED,and then we place a resistor across the solar panel output. I can now see if the LED's output rises with the increase of current,while the voltage remains much the same across that LED. For this particular LED,it seems that the optimum voltage is 3.2v,and any additional current makes no difference to the output of the LED. The other thing is i can now see what happens when i increase the voltage on the input--this is how i increase the current on the flyback. Now we have a situation where only the current on both the input and output rises,but both the input and output voltage remain the same,even though i am increasing the voltage on the input. On the flyback side this is expected with the LED in play,but why dose only the current rise on the input,and not the voltage?.
Well, I can see from your scope shots that you always have current flowing in the coil. Personally I find that to be a more complex measurement problem as compared to pulsing the coil with voltage, and then during the OFF time you observe the coil completely discharge its stored energy. That should be a simpler measurement problem to tackle. With current always flowing in the coil during the energizing cycle and during the discharge cycle, two things are happening at the same time during the energizing cycle, - 1) the energizing of the coil, and 2) the discharging of the coil though the LED. I would have to think for a while on how to make the energy measurements in a case like that. Although the LED appears to be clamping the coil discharge through the LED to a fixed voltage, we know it's not truly doing that. I also see how your function generator output (?) droops s bit as the coil current increases which also introduces nonlinearities. It all depends on how accurate you want your measurements to be.
Your observations are all good. You are getting the feel for how you can play with the coil excitation voltage and the ON time for the excitation to literally "dial up" any final current for the coil you want. Likewise you can "dial up" the discharge time through an LED + resistor or just a resistor by choosing whatever components you want for the load. By playing with the inductance of the coil, the ON energizing time, and the ON drive voltage, and the nature of the load, you can dial up just about any type of pulse that you want.
For example, if someone said to you that a certain battery can be pulse charged from a coil, but the maximum initial current flow at the start of the pulse was one-half amp, you can design your pulse to do exactly that.
MileHigh
Quote from: woopy on November 22, 2015, 04:53:17 PM
So effectively the "negative voltage" seems to "brake" strongly the current (flywheel) which goes extremely fast to zero in my particular arrangement.
Laurent,
Think about a very simple thought experiment where you have a coil with some current flowing through it and a variable resistor across the two terminals of the coil. If the variable resistor is a low value then it will take a longer time for the coil to discharge its stored energy. If the variable resistor is a high high value - more RESISTANCE - the coil will discharge its stored energy more quickly. I assume that you would agree that that makes sense.
Now you know that more resistance equals more voltage across the resistor. The initial current is always the same - there must be more voltage across the higher resistance because voltage equals current times resistance.
It's all the same thing as how hard you apply the brakes to a spinning flywheel. The harder you apply the brakes the higher the back-torque acts to stop the flywheel. Most people have an intuitive sense that it is essentially impossible to instantly stop a spinning flywheel because that would take an infinite amount of force. Almost everybody knows that force equals mass times acceleration.
If all this makes sense to you then you should realize that when you see a big spike from a coil when you are working on your bench that that represents the burning off of the stored energy in your coil because the coil discharged through a very high resistance. That's all there is to it. It's no magic, it's no mystery, it's just a discharge of energy and normally that represents the loss of electrical energy being turned into waste heat in your circuit. In that sense it's bad news and nothing to get excited about.
Finally, in theory, any coil no matter what size can manifest a high voltage spike. And if you have been following all of my postings, it's more the
load that generates the high voltage spike, and not the coil itself. It is certainly not "radiant energy" in any way, shape, or form.
A simple rule of thumb: The higher the load resistance across a discharging coil, the higher the initial voltage spike, and the faster it discharges. When in doubt, think of putting the brakes on a spinning flywheel because the behaviour is identical.
MileHigh
Duplicate posting, I will have to find an interesting picture or two.
Again, you need a circuit with coil energized from battery , then disconnected and connected to the capacitor and diode to investigate the problem.
Hi forest,
Brad's post 101 has a scope shot that clearly shows what happens when a coil is charged and discharged. So I am not sure what problem you are referring to. Could you be more specific please? I saw your other post about using a DPDT relay. The problem with that is the noise from the switching of the contacts will make it much harder to see the changeover on a scope.
Carroll
Quote from: citfta on November 23, 2015, 09:31:14 PM
Hi forest,
Brad's post 101 has a scope shot that clearly shows what happens when a coil is charged and discharged. So I am not sure what problem you are referring to. Could you be more specific please? I saw your other post about using a DPDT relay. The problem with that is the noise from the switching of the contacts will make it much harder to see the changeover on a scope.
Carroll
If you look back at the Ossie motor, he used 2 reed switches to disconnect both ends of the source. The purpose of that is that now you can have 2 diodes send the bemf back to the source directly, where with only 1 switch or reed that cannot happen directly.
Mags
Quote from: Magluvin on November 23, 2015, 09:57:12 PM
If you look back at the Ossie motor, he used 2 reed switches to disconnect both ends of the source. The purpose of that is that now you can have 2 diodes send the bemf back to the source directly, where with only 1 switch or reed that cannot happen directly.
Mags
Well now that Im thinking a lot lately ;D , here is a way to overcome the single switch problem.
In a typical situation, putting a diode across the single switch to try to direct bemf back to the source, it does not work, due to the bemf actually being forward emf, opposite the direction of the diodes conductivity function.
But, if we put a tiny cap across the coil and put the reverse diode across the switch, when the switch opens, the bemf/femf will charge the cap. Then the cap reverses the current in the coil and the coil will charge the source through the diode across the switch. ;) ;) ;)
Just ran it on sim. It works. ;)
Mags
The resistor in series with the cap is necessary in sim because when the switch closes the program cannot calculate infinite current flow to the cap. So in real life that res is not necessary.
The cap can be very small to perform the current reversal in the inductor and the process happens much faster.
Mags
Quote from: Magluvin on November 23, 2015, 10:21:53 PM
The resistor in series with the cap is necessary in sim because when the switch closes the program cannot calculate infinite current flow to the cap. So in real life that res is not necessary.
The cap can be very small to perform the current reversal in the inductor and the process happens much faster.
Mags
Correction. The lower right res does the function of current limiting for the sim to process properly. So no res in series with cap necessary in the circuit shown.
Mags
Quote from: MileHigh on November 22, 2015, 09:59:31 PM
Well, I can see from your scope shots that you always have current flowing in the coil. Personally I find that to be a more complex measurement problem as compared to pulsing the coil with voltage, and then during the OFF time you observe the coil completely discharge its stored energy. That should be a simpler measurement problem to tackle. With current always flowing in the coil during the energizing cycle and during the discharge cycle, two things are happening at the same time during the energizing cycle, - 1) the energizing of the coil, and 2) the discharging of the coil though the LED. I would have to think for a while on how to make the energy measurements in a case like that. Although the LED appears to be clamping the coil discharge through the LED to a fixed voltage, we know it's not truly doing that. I also see how your function generator output (?) droops s bit as the coil current increases which also introduces nonlinearities. It all depends on how accurate you want your measurements to be.
Your observations are all good. You are getting the feel for how you can play with the coil excitation voltage and the ON time for the excitation to literally "dial up" any final current for the coil you want. Likewise you can "dial up" the discharge time through an LED + resistor or just a resistor by choosing whatever components you want for the load. By playing with the inductance of the coil, the ON energizing time, and the ON drive voltage, and the nature of the load, you can dial up just about any type of pulse that you want.
For example, if someone said to you that a certain battery can be pulse charged from a coil, but the maximum initial current flow at the start of the pulse was one-half amp, you can design your pulse to do exactly that.
MileHigh
I agree, and that's what I'm refering to. Investigate the situation where there is no current flowing into coil, simply because coil is disconnected from power source and connected to the led.
"The direction of *current* flow is opposite to the direction of *electron* flow".
"First of all, current doesn't flow. Charge flows. The conventional direction of the current is from + to -, and this direction can be the same as the direction of actual charge flow or the opposite, depending on the type of the charge carrier. As you know, the formula of definition for the electric current is I = dq/dt. Now, if we consider a conductor, in which the charge carrier is the electron which is a negative charge, the dq term will also be negative and so will be I in the direction of the electron flow. That's why the direction in which I will be positive is the opposite direction of actual electron flow, meaning that the electrons actually flow from - to +. If we consider a semiconductor, in which there are two charge carriers - the electron and the hole (which is a positive charge) - the electric current will be given by the flow of both (electrons and holes). While the direction of the actual electrons flow will be from - to +, the direction of actual holes flow will be from + to - (the same as the conventional current direction) because for the hole current the dq term is positive. So, yes, charge can flow from + to - and from - to + in the same time (for example in a PN junction) but the direction of the electric current is from + to -".
Well, I got as far as page six in this very interesting discussion.
Has it yet been concluded that both voltage and current which
are induced within the inductor during charge and discharge
do indeed change direction?
When pulsing an inductor with DC pulses and monitoring the
voltage across the inductor the polarity change cannot be
seen directly but its effect will be observed.
As current from a source pulse increases from zero to its
maximum the induced Counter ElectroMotive Force opposes
the source voltage and current. The result is the charging
time constant. The source current through the inductor
is stronger than the induced Counter ElectroMotive Force
current which opposes it or pushes back against it with the
result that the charging current increases from zero to
maximum (at inductor saturation) in the characteristic time
constant curve; gradually increasing as it overcomes the
induced Counter ElectroMotive Force.
Then when the source current is discontinued and the
charged magnetic field collapses the induced voltage
and current will reverse as the inductor itself becomes
a source of current which will sustain current in the
same direction as the current from the source which
charged it, thereby effectively aiding that source.
While charging; the inductor will oppose the source.
While discharging; the inductor will aid the source.
Yes, the induced voltage within the inductor does
change polarity as well as the induced current flow.
A parallel winding on the same core as the inductor
under test will demonstrate the polarity reversals as
the inductor is charged and discharged. There the
reversals will be clearly seen as it is separated from
the overpowering influence of the source current in
the driven winding.
This is how Inductors were taught in Naval Electronics
Technician "A" School, Treasure Island, California in the
'60s. At that time Navy Electronics Schools were the
absolute best in the whole nation. ;)
Hi SeaMonkey,
Yes I think we pretty well established with some good scope shots that while the voltage reverses when the power is removed from an inductor the current continues to flow in the same direction until the magnetic field has collapsed. One of our senior members explained that paradox this way. When the applied voltage is removed the inductor which was the load becomes a source instead of a load. Since it is now a source of current the polarity has to be the opposite of the direction of current flow.
Another way to look at it is which way does the current flow inside a DC generator? If the electron flow (conventional way of thinking) flows from negative to positive outside the generator then which way must it flow inside the generator? Obviously it has to flow from positive to negative which of course is because of the generating forces being applied internally. And that is exactly what happens as the magnetic field collapses in an inductor when the source voltage is removed.
So it appears what you were taught is still valid.
Regards,
Carroll
@cifta
QuoteYes I think we pretty well established with some good scope shots that while the voltage reverses when the power is removed from an inductor the current continues to flow in the same direction until the magnetic field has collapsed. One of our senior members explained that paradox this way. When the applied voltage is removed the inductor which was the load becomes a source instead of a load. Since it is now a source of current the polarity has to be the opposite of the direction of current flow.
In a generator or transformer the external magnetic field change which may expand into or contract from the coil region determines the induced Electro-Motive Force direction thus the direction of current flow. When the field change is expanding the current flows in one direction and when it contracts the current reverses flowing in the other direction, an Alternating Current.
In an inductor the applied Emf from let's say a battery "produces" a current flow which then produces an expanding magnetic field which then "induces" a Cemf against the applied Emf from the battery. The Cemf is "induced" due to the expanding magnetic field and has the same direction as if an external magnetic field had induced it as in a generator or transformer. However the battery Emf maintains the current flow forward against the Cemf.
Then when we remove the battery Emf/current the contracting magnetic field induces an Emf thus a current in the same direction as the battery Emf/current. The inductor acts just like a coil in a generator or transformer and the magnetic field change induces an Emf in the same direction. However the source battery applied Emf/current dominates over the the induced Cemf from the expanding magnetic field.
Let's apply this thought, when a magnet passes a coil the expanding magnetic field into the coil induces an Emf and current in one direction and as the magnetic field contracts from the coil it induces an Emf and current in the opposite direction... an Alternating Current. However if we applied an external Emf from let's say a battery or another coil to exactly "Counter" the induced Emf from the magnets expanding magnetic field then no current would flow. The only output would be from the contracting magnetic field as the magnet leaves the coil inducing a current in one direction only... a Direct Current. We could say the external applied Emf opposing the induced Emf has rectified the output current. Now you know how to produce DC from an AC generator without rectifying the current with diodes.
In fact we can have any number of applied or induced Emf's in a circuit acting in different directions however the sum of the Emf's will determine the direction of the current flow.
AC
I propose a new way of defining the voltage apparent reversal which I think will make things less confusing. I'll explain.
The way I see it is very much the same as Citfa, MileHigh and others except that rather than just stating that the voltage reverses which is vague.
I propose that
when the coil is switched off the applied voltage can be forgotten and a new voltage is caused due to the energy in the magnetic field, the current and the resistance if faces.
It is the new voltage which is produced that has an opposite polarity across the coil when compared to the initial applied voltage. The initial applied voltage from the initial supply remains the same.
The opposite voltage is a different voltage, emf.
It is the voltage produced at the coil when the coil becomes the separate supply that is opposite to the initial voltage applied from the supply "battery for example".
To put it another way.
The voltage applied to the coil from the initial supply does not reverse at all, the new voltage that the coil applies to the load resistance is opposite polarity to the initial voltage applied to the coil.
The discharge current can be thought of as new as well, the new discharging current is in the same direction as the charging current.
The one thing that is not new is the energy. The same energy in both, some energy is lost as heat and radiation ect. so the
total output power is always less than the
total input power per cycle on a whole system basis.
..
QuoteQuote from: allcanadian on July 26, 2016, 05:12:08 PM
In fact we can have any number of applied or induced Emf's in a circuit acting in different directions however the sum of the Emf's will determine the direction of the current flow.
AC
As clearly stated by AC and probably others, the above would appear to be true of course. And that is very relevant.
Generally when thinking of this kind of situation the coil is switched off, the initial applied voltage is removed until the coil is discharged fully. Sometimes a new current path is enabled by whichever means to discharge back to the supply battery, sometimes to another load, so the voltage simply has to be a new and different voltage, this is reinforced by the fact that the discharge voltage can be very high if the coil discharge is through a high resistance.
..
I can well remember at the age of 11 onward beginning
my education in electricity/electronics.
My dad was a railroad switchman who had a supply of
six volt lantern batteries. I had a good supply of discarded
radios which yielded several very useful transformers.
I discovered that connecting the low voltage winding of a
filament transformer to the six volt battery briefly caused
a large, flamelike spark to be produced. It seemed almost
magical at the time.
Once while creating those large flaming sparks I held the
bare ends of the wires to get a firm grip while sparking
the connection and got a tremendous jolt. It was a
strong and painful jolt of high voltage electricity.
In those days the only way to find answers to questions
such as "What causes the large flamelike spark?" or
"Why did I get shocked from the six volt battery and
filament transformer combination?" was to go to a
library and find books on electricity. Fortunately, most
libraries then had a good selection of books on electricity
for experimenters which contained instructions for lots of
projects. Another very useful resource was the encyclopedia.
The answers didn't come quickly but they were eventually
found. Of course, playing with inductors at an early age
assures that one is "bitten by the bug" which causes the
life-long addiction to the study of all things electrical and
electronic.
Then a bit later Navy Electronics Schools played a very big
role in my ongoing education. The types of equipments that
the Navy Electronics Technician is able to maintain are incredible.
The subject matter of the schools where technicians are trained
was fascinating and diverse. And of course life aboard ship which
plies the seas to numerous foreign destinations was 'icing on the cake'
of Adventure. It was a good life!
I think we should first concentrate on simple boost converter.
I think, when the inductor is charged from battery it creates cemf against battery emf due to magnetic field expanding. When connection is changed then battery emf passed through the inductor charging capacitor via diode very quickly to the battery voltage and then capacitor is acting like a break in the circuit. Magnetic field of coil collapse and create emf which add to the emf of battery and capacitor voltage is rised to the sum of battery voltage and induced emf of coil.
Is there any current generated when coil field is collapsing or it's just the combined emf's of coil and battery which can charge capacitor to the higher voltage then battery ?
Please explain how you see it operates. Please correct me if I'm wrong.[/size]
Hi forest,
The answer to your question is yes. Yes there is a current being supplied by the coil field collapse. Your second part where you ask if it is the combined effect of the coil's emf and the batteries emf is also correct. Current is just the movement of electrons. So when the coil field collapses then it induces a voltage (emf) in the coil. If there is a path for the current to flow then current will flow because of the emf being applied. As you correctly thought this voltage is added to the voltage of the battery to charge the cap to a higher voltage than the battery voltage. The higher voltage also means that more current will flow into the cap. So you see the coil collapse adds voltage and because of the added voltage also adds current. Here is a nice tutorial all about switching power supplies. It includes boost, buck and inverting power supplies.
https://www.maximintegrated.com/en/app-notes/index.mvp/id/2031
That was a good question. Thanks for asking.
Carroll
This has been my position the entire time:
"In fact we can have any number of applied or induced Emf's in a circuit acting in different directions however the sum of the Emf's will determine the direction of the current flow".
Excellent suggestion Carroll.
Also, there are a number of videos on the subject
as well:
How to select a Boost Converter
(https://www.youtube.com/watch?v=-V_p1GBH4pk)
How does the Boost Converter work?
(https://www.youtube.com/watch?v=wJU7AJgERG8)
There are lots of videos on YouTube about
Boost Converters. Learning has never been
easier!
Some interesting channels:
Mr. Carlson's Lab (https://www.youtube.com/user/MrCarlsonsLab/videos) (how to fix/restore things)
Eric Goodchild (https://www.youtube.com/user/crazeman5522/videos)
EEVblog (https://www.youtube.com/user/EEVblog/videos) (very lively)
z1power (https://www.youtube.com/user/z1power/videos)
TRX Bench (https://www.youtube.com/user/TRXBench/videos) (for those who like to fix things)
I know I know, but don't you think that if there is additional current generated due to magnetic field collapse of coil in boost inverter then it could be used to create overunity converter ? If this is only due to combined higher emf "squeezing" electrons in output capacitor then obviously we end up with some 99.99% efficiency.
I am not going to say you are wrong about that. There are still a lot of things about electromagnetics we don't totally understand. I can only say that in all my years of working with coils and pulse motors I have not seen any evidence that would lead me to believe any extra current is produced when the magnetic field of a coil collapses.
It seems to me if there is no advantage in letting the current out of the coil, then don't. Lock the current in there and let if flow while retaining the magnetic field it generated, much like a permanent magnet. And since you already have current in there, try adding some more. As long as you don't saturate whatever core material you are using, one should be able to push the field strength up, up and up. The lower the resistance in your coil windings, the longer you can hold each additional surge of current.
When you do finally desire to unleash all this stored magnetic energy, why not just let it go back to a capacitor to be reused again. You control the change in current via resistance, so you can set to what voltage you want the capacitor to charge up to--big cap / small voltage or little cap / high voltage. The loss you incur from the transfer can be supplemented by your power source so the cap is ready for the next cycle. Use the cap to energize the coil in the first place and only use the power supply to keep it topped off.
My hunch is if you do all this properly, you can create serious magnetic force to drive an armature and perform work without spending the classical 746 watts per horsepower. Supposedly Paul Babcock has done this and I wouldn't be too surprised to learn that others have also.
Easy?
No, which is why we are all struggling. I just think if we get the concepts right and decide what it is we want, there's a way to achieve it.
Myself, I think there is a way to create constant linear or rotational force in such a way there is no need to let the coil windings collapse--they will regardless unless you are using superconductors, but you only intervene when the current has become so weak that it would make sense to pump them back up again. That's when you open the switch and only when you open the switch. Just a crazy idea, but since I thought it, it would be silly not to pursue it.
Hi all this might be a different type of inductive kick back but very interesting..
https://www.youtube.com/watch?v=vyVJ-lOO-50
"Let us consider another experiment. In Fig. 3 (below) a large wooden spool wound with a large number of turns of fine, insulated wire is connected to a galvanometer. A small spool wound with a few turns of insulated wire is connected in series with a dry cell and a contact key. Let us put the small spool down inside the large spool and press the contact key, closing the circuit. The galvanometer needle will be deflected, showing that a current has been induced in the large coil. What has happened? When we closed the switch and current started flowing in the small coil, that current caused a magnetic field to develop around it, and that developing magnetic field induced an emf and current in the large coil. If the key is kept closed, the induced current in the large coil soon stops. When the circuit is broken, the strength of the magnetic field quickly falls to zero and an induced current flowing in the opposite direction occurs. In both cases the induced current stops flowing when the magnetic field stops changing".
Faraday's induction:
"The direction of movement of the magnet altered the direction of the induced current, and similar results were achieved with the South pole of a magnet although the induced current was always opposite to current induced with the North pole of a magnet".
Viewing the two diagrams below demonstrates the correctness of the current reversal claims in each case. The movement of the primary field into and out of the secondary coil on the left causes the same current reversal as the magnet causes in the coils on the right when the magnet reverses direction.
Quote from: tinman on November 20, 2015, 09:25:40 AM
2
When the current source to an inductor is interrupted, the current will continue to flow through the inductor in the same direction,but the voltage across that inductor will invert. This is how the simple circuit below is able to work,and the LED can be lit from the flyback<--im guessing you guys want to use the term !flyback! so as to keep it simple?.
Diagram 1 shows the switch closed,and current flows from the battery into the top of the inductor,and out the bottom of the inductor,and into the negative of the battery. At this point in time,i also assume that we will stick to conventional current flow throughout this thread-so as to keep it simple?.
Diagram 2 shows the current flow through the inductor,and voltage polarity across that inductor the moment the switch becomes open. This current flow,and voltage polarity will remain the same until such time as the magnetic field around the inductor has fully collapsed-all stored energy has been depleted. We know the current flowing through the inductor must be in the same direction,and the voltage must have inverted in order for the LED to light.
Why you turn the battery round to confuse things ? Master Po ?
The battery's the source; The inductor the destination. Power moves away from the battery toward the inductor. The inductor stores the power from the battery in a magnetic field that expands Outward from the center of the inductor. When the current is interrupted, the magnetic field collapses Inward toward the center of the inductor and the inductor becomes the source. The flyback current then moves away from the inductor. Current moving away is in the opposite direction as current moving toward.
Struth thats more confusing for my brain this time of night.
lets suppose we have a relay in the collector of an NPN transistor when we engage the relay
by turning the NPN trasistor on, energy will build up in a race to saturate the coil.
All is fine till we turn the device off thus disconnecting the supply we now have an inertia problem
where the electrons build up in the same direction, as far as i know if we have way to dissipate the energy
the voltage will reverse and unless you have a protection diode between the + power rail and the transistors collector
(across the coil in the revers direction the fly back will destroy your transistor in its attempt to escape.
@Aliengray,
This inductor discharge curve is the only thing you need to concern yourself with: Flyback takes the path of least resistance. The notion that there's some kind of inertia in the inductor is complete hogwash.
The flyback is at a higher voltage potential then the source voltage; so if the resistance between the switch contacts is less then the resistance to ground, the flyback current will spark jump the gap to return to source across the open switch.
Everyone's seen the way current flows from higher to lower positive poles in our split positive tests.
Igor's reed switch spinner 2 sends flyback power back to the positive pole of the source battery through his reverse biased LED:
https://www.youtube.com/watch?v=vWvI7T7h3tk (https://www.youtube.com/watch?v=vWvI7T7h3tk)
Quote from: synchro1 on February 23, 2018, 07:55:37 PM
@Aliengray,
The notion that there's some kind of inertia in the inductor is complete hogwash.
Maxwell says otherwise
The inertia is defined as:
9.10938356x(10^-31)Kg * 0.0002m/s * [6.25x(10^18)*# of coulombs]
and the vector is along the wire, in the direction of the current.
This is the state of motion that has to be overcome when you switch the current off
Quote from: sm0ky2 on February 24, 2018, 01:30:50 AM
Maxwell says otherwise
The inertia is defined as:
9.10938356x(10^-31)Kg * 0.0002m/s * [6.25x(10^18)*# of coulombs]
and the vector is along the wire, in the direction of the current.
This is the state of motion that has to be overcome when you switch the current off
@smOky2,
That's all included in the discharge curve.
Allen,
No matter how many times you insist your false ideas are true those of us that actually understand how to use our test equipment properly are not fooled by your false claims. I have purposely not gone to your foolish thread and tried to correct all your mistaken ideas. Now please do the same and keep them in YOUR thread. If you continue to spread your false ideas in this thread I will go to YOUR thread and prove you wrong again just like I did a couple of years ago. You got yourself banned then because your only reply when proven wrong was to attack with foul language and personal insults. I had hoped you learned from that but we will see.
Respectfully,
Carroll
Quote from: citfta on February 24, 2018, 01:37:25 PM
Allen,
No matter how many times you insist your false ideas are true those of us that actually understand how to use our test equipment properly are not fooled by your false claims. I have purposely not gone to your foolish thread and tried to correct all your mistaken ideas. Now please do the same and keep them in YOUR thread. If you continue to spread your false ideas in this thread I will go to YOUR thread and prove you wrong again just like I did a couple of years ago. You got yourself banned then because your only reply when proven wrong was to attack with foul language and personal insults. I had hoped you learned from that but we will see.
Respectfully,
Carroll
@Citfa,
I never instigate attacks.
Allen
you are so use to attacking people its become second nature to you,
But Nowadays
you stick out like a sore thumb here.
and Carroll is correct
attacks are no way to move forward...
Nor are racially derogatory innuendos/insinuations [most find those disgusting ESPECIALLY ADMINISTRATION HERE !!
the last few pages of your thread show your behavior here ,and No I have not bothered Stefan because I keep hoping you'll
show something which proves your point
@ Allen
I see your comment below
,here all the builders work together to actually seek out anomalies.
and they don't take wooden Nickles or fancy words and theories for currency.
The bench and results Allen
your whistling the wrong tune with your childish attacks .
Quote from: ramset on February 24, 2018, 02:28:43 PM
Allen
you are so use to attacking people its become second nature to you,
But Nowadays
you stick out like a sore thumb here.
and Carroll is correct
attacks are no way to move forward...
Nor are racially derogatory innuendos/insinuations [most find those disgusting ESPECIALLY ADMINISTRATION HERE !!
the last few pages of your thread show your behavior here ,and No I have not bothered Stefan because I keep hoping you'll
show something which proves your point
@ramset,
Try and find someone to help dislodge that pea whistle that's stuck down your throat.
@Ciitfa,
Power moves from the power source through the top of the coil. When the current's interrupted from the power source, power can no longer flow into the coil through the top of the coil. How can current be moving in the same direction when current no longer has any way to channel in through the top? Current moves away from the top of the coil not toward it from the battery when the coil turns around and inverts it's relationship as the source and not the destination of the power.
Still wrong as always. This has been explained to you many many times in this thread and now smoky2 has also tried to explain it to you in your own thread. But you still continue to deny the truth. Please go back to your own thread and continue your discussion of your mistaken ideas. I really like the idea of negative inductance. It reminds me of someone that claims to have a glass that has less water in it than zero. Your ideas are completely wrong so just keep them in your own thread and I won't have to go there and correct you. If you continue in this thread I will go there. That is a promise.
Quote from: citfta on February 24, 2018, 03:55:00 PM
Still wrong as always. This has been explained to you many many times in this thread and now smoky2 has also tried to explain it to you in your own thread. But you still continue to deny the truth. Please go back to your own thread and continue your discussion of your mistaken ideas. I really like the idea of negative inductance. It reminds me of someone that claims to have a glass that has less water in it than zero. Your ideas are completely wrong so just keep them in your own thread and I won't have to go there and correct you. If you continue in this thread I will go there. That is a promise.
@Citfta,
What causes the flyback current to reverse direction and exit through the top of the coil as it has a strong tendency to do routinely?
@Citfta,
Alright, you're on over at my "Negative Inductance" thread.
Quote from: citfta on February 24, 2018, 03:55:00 PM
Still wrong as always. This has been explained to you many many times in this thread and now smoky2 has also tried to explain it to you in your own thread. But you still continue to deny the truth. Please go back to your own thread and continue your discussion of your mistaken ideas. I really like the idea of negative inductance. It reminds me of someone that claims to have a glass that has less water in it than zero. Your ideas are completely wrong so just keep them in your own thread and I won't have to go there and correct you. If you continue in this thread I will go there. That is a promise.
Dear MR citfa Carol I'm really getting confused in this thread ;D Who is Allen me ? what have i said ?
I came on here (this tread) to learn, perhaps unlearn what i already know, the problem is i'm now totally confused as I think most people are in here with the abusive terminology used in here (idiots hogwash ect ect) and not to mention the aggressive comments.
Please explain the difference between Fly back, Bemf, and inductive kick back, i'm asking you because i know there is a good chance of getting a respectable answer from you good self.
Many thanks AG (Allen)
AG - I think they are referring to the other Allen posting in this thread.
Things are much simpler (and less personal) when we stick to our self-assigned pseudonyms
This should help explain inductive kickback.
https://youtu.be/IGRt0Kg55Sc
Please note my camera was not fast enough to always catch the flash of the Led that is across the coil. It did always flash.
Sorry AG,
I was not referring to you. I was referring to Allen Burgess known as Synchro1 on this forum. I am sure you have read many of his posts in this thread where he has attacked those that didn't agree with him. Please watch the video and then if you have questions about inductive kickback or BEMF or whatever I will try to answer if I can.
Respectfully,
Carroll
Quote from: citfta on February 24, 2018, 09:02:26 PM
This should help explain inductive kickback.
Please note my camera was not fast enough to always catch the flash of the Led that is across the coil. It did always flash.
Don't you find it at all interesting that this graph below,
posted by Synchro himself, refutes him totally? Note the current and voltage curves after disconnect.
I don't think the fellow even knows how to interpret graphs.
And, by the way....
https://www.youtube.com/watch?v=8vyg0NX0zh4
https://www.youtube.com/watch?v=X66foYixu9Y
I made these videos back in late March and early April of last year, for the same reason that you've made yours. Did it do any good? Well, for some people it did, but obviously the "target audience" was too busy picking up paper clips to pay attention.
Quote from: AlienGrey on February 24, 2018, 08:36:58 PM
Dear MR citfa Carol I'm really getting confused in this thread ;D Who is Allen me ? what have i said ?
I came on here (this tread) to learn, perhaps unlearn what i already know, the problem is i'm now totally confused as I think most people are in here with the abusive terminology used in here (idiots hogwash ect ect) and not to mention the aggressive comments.
Please explain the difference between Fly back, Bemf, and inductive kick back, i'm asking you because i know there is a good chance of getting a respectable answer from you good self.
Many thanks AG (Allen)
When i said !idiots!,i was referring to those that post articles on the net,that do nothing more than tell untruths about how inductors work,which just further confuses those trying to learn the truth.
I was not referring to anyone here on this forum.
Brad
Quote from: tinman on February 24, 2018, 10:30:22 PM
When i said !idiots!,i was referring to those that post articles on the net,that do nothing more than tell untruths about how inductors work,which just further confuses those trying to learn the truth.
I was not referring to anyone here on this forum.
Brad
Brad did you say that? sorry no offence I couldn't have read your comment on the tread properly in the tread.
To be honest I had to make up a circuit and re test the theory again for my self just to make sure.
I think TK has a video showing this phenomena ! i will try to re post it if I can find it i'm sure he wont mind 8)
https://www.youtube.com/watch?v=8vyg0NX0zh4
Hmm I wonder if any one will notice :D
Happy days
AG Allen
Quote from: citfta on November 20, 2015, 09:56:25 AM[/font]A link to get you started: https://en.wikipedia.org/wiki/Inductor (https://en.wikipedia.org/wiki/Inductor)I had to leave for a while but now I am back.To answer the question about current flow and opposite polarity, answer my question about which direction does current flow inside a battery. Does it flow from positive to negative or the other way round? And which way does it flow outside the battery? When you get the answer to that you will understand why polarity can switch on an inductor but the current still flows the same way.Also do you have a way to conduct the test with a 2 channel scope?Carroll
[/font]
In a battery the current flows from the positive pole over a conductor to the negative pole. However, if you attach a battery charger instead of a conductor, then current flows in at the batteries positive pole and out of the negative pole. It requires the charger voltage to be higher than the battery voltage.
Current and voltage have this relationship that there can be voltage without current, like an empty riverbed, but no current without voltage, like a river without a bed.
Therefor I do not agree with your initial statement. Inductive Kickback has reverse Voltage and therefor reverse current flow.At any point in a curcuit, current will always flow from a higher to a lower voltage. Even if the current measured accross the coil seems time delayed, it still holds true.
Oh, btw, Carroll, there is no current flow inside the battery, only ion exchange.
Therefor this allows us to say, that every electron that goes out at plus, must come in at minus, in order to make this battery work at all. It does not have to be the same ekectron tho. It could be one from china or from mars. The battery doesn't care.
Internally it's a voltaic or capacitive Action, eg. oxidation of one of two metals by ion exchange that forces and requires the outside electron transfer. But you knew that. (And please no physicuans vs electricians electron direction talks... you know what I mean)
And since we are talking about inductive kickback,a few more words.
The voltage of the Kickback must be higher than the supply to make the current flow in reverse.
That is, if you did not direct it to flow to ground, or to ring/afterburn the coil in freerunning diode. No, I mean back to the positive pole.
Assuming the back spike is high, tho short, even with no load in it, which is the basic condition in this experiment, (big coil, big pulse), then you can imagine, that any high resistor would only hinder the Spike to flow back to the supply.
But one can easily use the wrong coil, that does not reach flowback condition, which then just locks or parasiticly dampens the coil state.
Quote from: dieterTherefor I do not agree with your initial statement. Inductive Kickback has reverse Voltage and therefor reverse current flow.
Wrong yet again. Study the graph AGAIN posted below. When the inductor is switched off, the inductor now acts as a source, the voltage V
L across it is reversed-- and hence the current I
L continues to flow from the inductor IN THE SAME DIRECTION as before.
Nature doesn't care about what you agree with or not, but the facts of the matter are clear. Perhaps you would like to try to explain the results of the LED demonstrations that Citfta and I have published.
It seems that what is being missed here is that the coil
is Conductive
current flows from + to - THROUGH the coil!
in both conditions
First from source
Second from coil (reverse polarity)
This current is in the Same direction
if you want to route this "back to source" you do it from the - end
back to the + of the battery
You still aren't changing the direction of current Through the coil.
Battery has an internal resistance that prevents this
The coil (being a conductor) does not
Quote from: sm0ky2 on February 25, 2018, 07:01:40 PM
It seems that what is being missed here is that the coil
is Conductive
current flows from + to - THROUGH the coil!
in both conditions
First from source
Second from coil (reverse polarity)
This current is in the Same direction
if you want to route this "back to source" you do it from the - end
back to the + of the battery
You still aren't changing the direction of current Through the coil.
Battery has an internal resistance that prevents this
The coil (being a conductor) does not
@smOky2,
Wrong! Power travels away from the positive electrode then power moves toward the positive electrode. Toward and away are different directions.
The current turns around; It does not continue in the same direction!
MAKE A VIDEO AND PROVE IT! You have been shown two videos that prove you are wrong. Now make one and prove you are right. You and I both know you can't do it. The video you keep referring to where some guy claims he is sending power back to the battery is wrong. As someone has pointed out in the comments he has the Led that he claims is sending the power back hooked up backwards. TK and I have both shown with videos that current cannot and does not flow back to the source.
@Citfta,
What do you mean "some guy"? Igor Moroz is much more then just some guy. You are the "Just some guy"!
Here's a video where I'm doing it:
https://www.youtube.com/watch?v=COLDfC6aPFM (https://www.youtube.com/watch?v=COLDfC6aPFM)
I may be just some guy also but I do know how to connect a Led up properly. NOW MAKE A VIDEO AND PROVE ALL THE REST OF US ARE WRONG!! Stop changing the subject.
Quote from: citfta on February 26, 2018, 05:51:03 AM
I may be just some guy also but I do know how to connect a Led up properly. NOW MAKE A VIDEO AND PROVE ALL THE REST OF US ARE WRONG!! Stop changing the subject.
@Citfta,
I just posted a video above. Take three AA batteries. Connect two AA batteries in series, then connect the negative electrodes. Now attach an LED between the positive poles and measure the voltage in the single battery. Let the LED shine for a few minutes then re-check the voltage in the single AA.
The long prong on the LED is the positive. The positive or long prong goes to the positive electrode of the top AA in the two AA's in series, that's the nipple end. The negative or short prong to the positive electrode of the single AA. The negative electrodes, thats the flat sides of the AA's are connected.
@Citfta,
Do you understand what's going on in the video? The combined voltage on the left is 13.5 volts. The voltage on the AA on the right is 1.5 volts. The positive electrode of the LED is connected to the positive electrode of the AA on the left, and the negative electrode of the LED is connected to the positive electrode of the AA on the right. The negative electrodes are connected through the 12 volt battery. The voltage on the AA on the right is increasing as the LED illuminates.
@Citfta,
Look, it takes a diode to loop the inductive kickback. The flyback is higher in voltage, so it wants to go home through the positive electrode of the battery from whence it came. The flyback Will travel in the same direction if you send it down the ground drain, I'll grant you that, but you just have it wired wrong.
Allen,
You are like a Pre-schooler trying to explain to an engineer how to connect a circuit. What a joke you are. David Bowling, Matt Jones and I have been working with the 3 battery generating system for several years now and you think you can explain it to ME.
STOP CHANGING THE SUBJECT AND SHOW A VIDEO THAT PROVES THE CURRENT REVERSES WHEN POWER IS DISCONNECTED FROM A COIL!!!!!!
I see you posted again while I was typing this. Explain what is going on with the video I posted. How is it connected wrong? It clearly shows which direction the current is flowing.
NOW MAKE A VIDEO THAT PROVES CURRENT REVERSES WHEN POWER IS DISCONNECTED FROM THE COIL!!!!
Stop changing the subject and just do it.
Perhaps trouble makers on this site should have 'Names' in colors so as to warn others to be aware of misleading information or deception after a warning by admin!' no one with intelligence can be that much of a minion, surely !
Just a thought ! the pages are getting filled with totally misleading ''rabbit hole fodder probably deliberately'','' probably laughing him self stupid on the wind up'', no one can be that stupid, can they?
AG.
Quote from: synchro1 on February 26, 2018, 06:35:13 AM
@Citfta,
Look, it takes a diode to loop the inductive kickback. The flyback is higher in voltage, so it wants to go home through the positive electrode of the battery from whence it came. The flyback Will travel in the same direction if you send it down the ground drain, I'll grant you that, but you just have it wired wrong.
You have posted another one of your mistaken ideas. The flyback voltage will only go as high as it needs to go in order to return back to the coil. If there is a high resistance in the path then yes the flyback voltage can be higher than the applied voltage. But it is only higher in relation to the other end of the coil. Not some imaginary connection on the other side of the supply. If the return path is low resistance then the flyback voltage will be lower than the supply voltage.
It is amazing to me that you have wasted at least two years of you life trying to convince the rest of the world that you are the only one that understands how a coil works.
Quote from: citfta on February 26, 2018, 06:48:16 AM
You have posted another one of your mistaken ideas. The flyback voltage will only go as high as it needs to go in order to return back to the coil. If there is a high resistance in the path then yes the flyback voltage can be higher than the applied voltage. But it is only higher in relation to the other end of the coil. Not some imaginary connection on the other side of the supply. If the return path is low resistance then the flyback voltage will be lower than the supply voltage.
It is amazing to me that you have wasted at least two years of you life trying to convince the rest of the world that you are the only one that understands how a coil works.
@Citfta,
You're the one that needs an elementary text book pal. The flyback wants to return home without the diode, thats why we need tiny capacitors on our switches.
Hi AG,
Yes it is unfortunate that anyone can post anything they want on most forums, even when their ideas are clearly wrong. On a couple of other forums I have a signature that says" Just because it is on Youtube does not make it real". And of course the same thing applies to most forums. There are no forum police to check all the things posted so anyone can post anything and get by with it unless someone else decides to challenge them on it. The threads that are clearly foolishness I tend to just ignore. But sometimes I feel the claims are so misleading they need to be corrected. I have seen several times on these forums where people have spent a lot of time and money chasing some idea that was false from the start. I feel like if I know for certain an idea is false I should at least explain why I think so in hopes of saving someone many hours of time and a lot of money.
I hope my explanations about inductive kickback have been helpful to you.
Respectfully,
Carroll
Quote from: citfta on February 26, 2018, 07:01:35 AM
Hi AG,
Yes it is unfortunate that anyone can post anything they want on most forums, even when their ideas are clearly wrong. On a couple of other forums I have a signature that says" Just because it is on Youtube does not make it real". And of course the same thing applies to most forums. There are no forum police to check all the things posted so anyone can post anything and get by with it unless someone else decides to challenge them on it. The threads that are clearly foolishness I tend to just ignore. But sometimes I feel the claims are so misleading they need to be corrected. I have seen several times on these forums where people have spent a lot of time and money chasing some idea that was false from the start. I feel like if I know for certain an idea is false I should at least explain why I think so in hopes of saving someone many hours of time and a lot of money.
I hope my explanations about inductive kickback have been helpful to you.
Respectfully,
Carroll
@Citfta,
I don't have to respect you and that other crackpot for acting like important authorities because you show one isolated effect in your tedious videos with your inane scope shots. I don't have to upload any videos of getting bit by flyback. That's why we call it flyback.
@Citfta,
I'm going to show you an over unity circuit. We can replace the AA battery in this Ludic Science boost converter video posted below, with a high inductance Energex rechargeable AA. We wrap the AA battery with a copper coil to use the battery as a substitute for a ferrite core. Next we connect the negative end of the LED to the positive end of the battery, and connect the tail end of the coil to the negative electrode of the AA. This circuit will charge the battery and light the LED at the same time.
https://www.youtube.com/watch?v=iHmTc0PwiyY&t=214s (https://www.youtube.com/watch?v=iHmTc0PwiyY&t=214s)
BORING, BORING, BORING.
STOP TRYING TO CHANGE THE SUBJECT AND SHOW SOME PROOF THAT CURRENT REVERSES WHEN POWER IS REMOVED FROM A COIL!!!!
I will not waste any more time on this. You refuse to face the reality of how an inductor works so we have no more to discuss.
BYE BYE
Quote from: citfta on February 26, 2018, 08:07:07 AM
BORING, BORING, BORING.
STOP TRYING TO CHANGE THE SUBJECT AND SHOW SOME PROOF THAT CURRENT REVERSES WHEN POWER IS REMOVED FROM A COIL!!!!
I will not waste any more time on this. You refuse to face the reality of how an inductor works so we have no more to discuss.
BYE BYE
@Citfa,
You're just an old fuddy duddy fart.
I am yet to see anyone here so far explain the complete series of events of what takes place within an inductor,from the time of connection to the source,to the end of the inductive kickback cycle.
Synchro1--you are just much further behind than most others here,except for your mate dieter-->he's just as lost.
Len'z law explains it. The magnetic field collapse and the rule is it should reinforce the current which produced it. Maybe someone explain it details.
Quote from: tinman on February 26, 2018, 09:27:34 AM
I am yet to see anyone here so far explain the complete series of events of what takes place within an inductor,from the time of connection to the source,to the end of the inductive kickback cycle.
Synchro1--you are just much further behind than most others here,except for your mate dieter-->he's just as lost.
@Tinman,
I'd trust you to inflate a low pressure tire.
@Citfta,
You know very well along with Tinselkoala and most of the rest of the group with the exception of the least enlightened among us that the direction of flyback depends entirely on the spacing of the interrupter contacts . If either of you had a shred of decency you would grab a video with scope shots of the inductive kickback traveling in both directions, and move forward with an informative discourse.
Instead, you lean on a worn out video that would take me five minutes to set up and capture on digital format. Have neither of you any sense of decency? Both you and Tinselkoala milk those deceptive videos to death to appear like bulwarks of the Overunity site when you're just preying on the gullibility of our beginners.
There was some discussion some years ago regarding
Inductive Kickback on several of the forums where it
was acknowledged by the "beginners" that inductive
discharge is characterized by a voltage reversal
across the coil. What was difficult for many of them
to accept, however, was that the inductive discharge
sustained current flow in the same direction as that
which initially charged it.
It seems the confusion still exists today. For some,
once an idea is accepted as truth it becomes so
ingrained that it is near impossible to admit that the
idea is incorrect.
Once the "beginner" comprehends the function of the
Inductor in a DC Buck Converter then the confusion
tends to evaporate.
One is then able to comprehend the function of the
Inductor in the DC Boost Converter.
Inductive Kickback isn't necessarily greater than the
source voltage, but it can be when desired. It is
dependent upon how the circuit is configured.
If the voltage flips polarity, how is this possible:
circuit:
9V battery -> diode -> coil -> cap -> back battery
I charge the cap to 7-8~V through the coil and cut the battery off. Magnetic field collapsing in the coil will push the cap to 14-16V and I can keep doing this. With a mosfet I got the cap all the way to 200V. How is this possible, if the voltage going from coil to cap is of different sign? What is the cap seeing and negative voltage should not pass that diode?
Here is the same with a signal generator and mosfet
http://tinyurl.com/yc7ressj
Quote from: Belfior on February 27, 2018, 08:45:55 AM
If the voltage flips polarity, how is this possible:
circuit:
9V battery -> diode -> coil -> cap -> back battery
I charge the cap to 7-8~V through the coil and cut the battery off. Magnetic field collapsing in the coil will push the cap to 14-16V and I can keep doing this. With a mosfet I got the cap all the way to 200V. How is this possible, if the voltage going from coil to cap is of different sign? What is the cap seeing and negative voltage should not pass that diode?
Here is the same with a signal generator and mosfet
http://tinyurl.com/yc7ressj
Nice video but why not go positive in to the inductor and then into an FET switch IRF 940 P channel and drive it with a 555 ect you will get a nice positive voltage post with little current draw.
it is not a video. You can design circuits with that and run them. Could you use that to draw the circuit with the 555 and run it in the simulation?
Not sure why this same problem keeps arisin in 3-4 different threads.....
Look, current is like water.
Water runs through the largest/less constricted pathway
The same is true with electrical current
Where is the lowest R?
The inductor
Why?
The inductor is also a Conductor
Current flows from the + of the inductor to the - of the inductor
Through the inductor
It doesn't matter if it is the induction event,
Or the subsequent kickback
Current is in the same direction
R.I.P. George Simon Ohm
Quote from: Belfior on February 27, 2018, 10:31:49 AM
it is not a video. You can design circuits with that and run them. Could you use that to draw the circuit with the 555 and run it in the simulation?
I don't have that package, can you try it ? you need to find the best pulse width in for the best output i expect
the HS 555 will work best.
I can do it with a scope the hard way
Smokey the trick here is to slow the current down !!! and just grab the voltage, think you can handle that idea ?
It works best with IRON ALUMINUM coated wire so i'm told but where are you going to get it ?
Quote from: Belfior on February 27, 2018, 08:45:55 AM
If the voltage flips polarity, how is this possible:
circuit:
9V battery -> diode -> coil -> cap -> back battery
I charge the cap to 7-8~V through the coil and cut the battery off. Magnetic field collapsing in the coil will push the cap to 14-16V and I can keep doing this. With a mosfet I got the cap all the way to 200V. How is this possible, if the voltage going from coil to cap is of different sign? What is the cap seeing and negative voltage should not pass that diode?
Here is the same with a signal generator and mosfet
http://tinyurl.com/yc7ressj (http://tinyurl.com/yc7ressj)
Hi Belfior,
It doesn't seem like anyone has actually answered your question yet so I will try to answer it. First I need to warn you that you should totally ignore any of the foolishness posted by Synchro1. I don't know if he is deliberately being misleading or he just is really that confused about how inductors work. But if you have read the whole thread you have seen several try to convince him he is wrong about inductors.
The first thing you need to understand is actually how inductors work. A mechanical equivalent is the flywheel. An inductor resists a CHANGE in current flow just like a flywheel resists CHANGE in speed. A former member of this forum (Milehigh) taught me that equivalent many years ago. So what is happening in your circuit?
When you first turn on your circuit the voltage from the battery is flowing through the inductor to the cap and charging the cap up to close to the battery voltage. Then the charging of the cap stops and all current flow stops. Now you turn on the fet. Current again begins to flow from the battery through the coil and fet and back to the battery. What you have to remember is the current flow is not instantaneous. Just like when you first turned on the circuit you saw at the bottom the current go up at an angle. This is how an inductor works, it resists current CHANGE. So depending on the voltage applied and the inductance of the coil it may take a few milliseconds to several milliseconds for the current to reach its peak value. As the current reaches the peak value the voltage across the coils drops. You can also see that in the bottom trace.
So you now have full current flowing through the coil and back to the battery through the fet. So what happens when you turn the fet off? Remember the coil or inductor resists current CHANGE. The goes the same for whether we try to increase the current or decrease the current. When you turn off the fet that current flowing through the coil wants to keep going just like a flywheel wants to keep spinning after you remove power. But it now can't keep going through the fet because it is now turned off. One of the interesting things about inductors is they will increase the voltage during inductive kickback until that current can go somewhere. Rather strange but not really if you continue to compare them to a flywheel. What happens if you try to stop a flywheel from spinning? The energy of the flywheel will be transferred to whatever you are using to stop the flywheel.
So where can the stored energy in your coil go? It wants to go back to itself. The coil which was the load has now become a source of energy instead of a load. That means that if the current is still wanting to go the same direction as it was going, then the bottom of the coil has now become positive in relation to the top of the coil. Or another way of looking at is to replace the coil with at a battery at the time of turning off the power. And you connect the battery so that it wants to keep the current flowing the same way. So the positive of the battery would be at the bottom to keep the current going the same way.
Now since the current can't continue through the fet it will have to go through the blocking diode and into the cap and back through the battery and then back to itself. Of course this depletes the magnetic field and the coil discharges. You will notice that the voltage from the discharging coil gets added to the voltage from the battery. But in addition as the cap charges and the internal impedance of the cap goes up the discharging coil keeps raising the voltage to overcome the resistance of the return path. This is why coils are used in boost converters. You need to study them to get a better understanding of what is going on when charging and discharging coils.
I hope this has been some help to you. The main thing and most confusing to those new to electronics is to remember that the coil switches from being a load to being a source of current when the power is removed from the coil. And that explains why the current can continue in the same direction while the polarity of the coil changes.
Respectfully,
Carroll
The main problem is that scientists do not experiment with electronic and engineers do not go out of the scope of handbooks with equations how to construct DC-DC converters.
What is missing is fundamental scope on what is going on at the SINGLE impulse stage.The most impostant not answered question is : what is the source of current in case of simple boost converter ?
Id call it Inductive Kick Forward. ;D ;D
Mags
Quote from: citfta on February 27, 2018, 11:26:25 AM
But in addition as the cap charges and the internal impedance of the cap goes up the discharging coil keeps raising the voltage to overcome the resistance of the return path.
Thx Carroll!
What If I put a 10MOhm resistor on the return path? Can I get the voltage to go into thousands and then have it jump a gap to create another return path?
Quote from: citfta on February 27, 2018, 11:26:25 AM
Hi Belfior,
It doesn't seem like anyone has actually answered your question yet so I will try to answer it. First I need to warn you that you should totally ignore any of the foolishness posted by Synchro1. I don't know if he is deliberately being misleading or he just is really that confused about how inductors work. But if you have read the whole thread you have seen several try to convince him he is wrong about inductors.
The first thing you need to understand is actually how inductors work. A mechanical equivalent is the flywheel. An inductor resists a CHANGE in current flow just like a flywheel resists CHANGE in speed. A former member of this forum (Milehigh) taught me that equivalent many years ago. So what is happening in your circuit?
When you first turn on your circuit the voltage from the battery is flowing through the inductor to the cap and charging the cap up to close to the battery voltage. Then the charging of the cap stops and all current flow stops. Now you turn on the fet. Current again begins to flow from the battery through the coil and fet and back to the battery. What you have to remember is the current flow is not instantaneous. Just like when you first turned on the circuit you saw at the bottom the current go up at an angle. This is how an inductor works, it resists current CHANGE. So depending on the voltage applied and the inductance of the coil it may take a few milliseconds to several milliseconds for the current to reach its peak value. As the current reaches the peak value the voltage across the coils drops. You can also see that in the bottom trace.
So you now have full current flowing through the coil and back to the battery through the fet. So what happens when you turn the fet off? Remember the coil or inductor resists current CHANGE. The goes the same for whether we try to increase the current or decrease the current. When you turn off the fet that current flowing through the coil wants to keep going just like a flywheel wants to keep spinning after you remove power. But it now can't keep going through the fet because it is now turned off. One of the interesting things about inductors is they will increase the voltage during inductive kickback until that current can go somewhere. Rather strange but not really if you continue to compare them to a flywheel. What happens if you try to stop a flywheel from spinning? The energy of the flywheel will be transferred to whatever you are using to stop the flywheel.
So where can the stored energy in your coil go? It wants to go back to itself. The coil which was the load has now become a source of energy instead of a load. That means that if the current is still wanting to go the same direction as it was going, then the bottom of the coil has now become positive in relation to the top of the coil. Or another way of looking at is to replace the coil with at a battery at the time of turning off the power. And you connect the battery so that it wants to keep the current flowing the same way. So the positive of the battery would be at the bottom to keep the current going the same way.
Now since the current can't continue through the fet it will have to go through the blocking diode and into the cap and back through the battery and then back to itself. Of course this depletes the magnetic field and the coil discharges. You will notice that the voltage from the discharging coil gets added to the voltage from the battery. But in addition as the cap charges and the internal impedance of the cap goes up the discharging coil keeps raising the voltage to overcome the resistance of the return path. This is why coils are used in boost converters. You need to study them to get a better understanding of what is going on when charging and discharging coils.
I hope this has been some help to you. The main thing and most confusing to those new to electronics is to remember that the coil switches from being a load to being a source of current when the power is removed from the coil. And that explains why the current can continue in the same direction while the polarity of the coil changes.
Respectfully,
Carroll
This is just a bunch of baffle gouge Carroll in his limited knowledge simply parroted from Milehigh to pass for a savant. He doesn't answer the question put to him, he just skillfully mis-directs the answer to a different subject.
I asked Carroll which direction the inductive kickback would take in Igor's "Reed switch spinner 2" circuit if the resistance to the battery from the top of the pulse coil was lower then the resistance to the ground, and he vanished off the thread.
Quote from: Belfior on February 27, 2018, 04:37:01 PM
Thx Carroll!
What If I put a 10MOhm resistor on the return path? Can I get the voltage to go into thousands and then have it jump a gap to create another return path?
Hi Belfior,
You have to remember what else is in the circuit. If you raise the voltage too high it will find another path. In the circuit you have shown it will most likely break down the fet and go through it. Repeatedly doing that will destroy the fet. How high the voltage can go is limited by how high the applied voltage is and how large the inductor is. And of course by what other path it can take. But I have easily gotten caps to charge as high as 300 volts from a 12 volt supply. I have used a slightly different circuit than what you have shown. In my circuit I just put a diode in series with a cap and the return of the cap back to the top side of the coil instead of to the battery. Put that circuit on your simulator and try it for a while.
Glad I could help some. Keep studying and learning. Electronics can be a lot of fun.
Take care,
Carroll
Quote from: citfta on February 27, 2018, 05:15:15 PM
Hi Belfior,
You have to remember what else is in the circuit. If you raise the voltage too high it will find another path. In the circuit you have shown it will most likely break down the fet and go through it. Repeatedly doing that will destroy the fet. How high the voltage can go is limited by how high the applied voltage is and how large the inductor is. And of course by what other path it can take. But I have easily gotten caps to charge as high as 300 volts from a 12 volt supply. I have used a slightly different circuit than what you have shown. In my circuit I just put a diode in series with a cap and the return of the cap back to the top side of the coil instead of to the battery. Put that circuit on your simulator and try it for a while.
Glad I could help some. Keep studying and learning. Electronics can be a lot of fun.
Take care,
Carroll
@Citfta,
Maybe Igor, in his infernal genius, simply tapered his ground lead.
Something I had thought of trying in the past...
If we have an inductor and we energize it then take away the input, the output of the coil doesnt really do much if the load is another inductor as the receiving inductor would impede the output dump of the first coil. But, if the receiving coil were bifi, its capacitance would accept the field collapse currents from the first coil. I wonder....
Mags
Quote from: Magluvin on February 27, 2018, 06:07:03 PM
Something I had thought of trying in the past...
If we have an inductor and we energize it then take away the input, the output of the coil doesnt really do much if the load is another inductor as the receiving inductor would impede the output dump of the first coil. But, if the receiving coil were bifi, its capacitance would accept the field collapse currents from the first coil. I wonder....
Mags
@Magluvin,
Genius shining Mags! A fat switching diode might help.
As long as the inductance of the primary is
equally divisible by the capacitance of the secondary bi-filar
Quote from: Erfinder on February 28, 2018, 02:02:36 AM
Have you built anything, ever? Something that relates to the subject? I ask because with each post you make it's becoming clearer that you are out of your element.....
If you don't understand the implications of what I stated
Perhaps it is not my element that I am out of....
Rethink your intention, and consider it carefully
What exactly will happen
If let's say our inductor were 80mH
and our bifilar secondary had a capacitance of 20uF
Or perhaps 120mH and 30uF
Or 100 and 10
And we release it from a saturated state???
And why would I suggest such a thing?
Or you could skip this information
Refuse to look at it on a scope and watch a Visualization of
the effect you spend pages talking about
ignore my input.
It's just one scenario you may happen upon
as you play in the dark making random coils feeding
other random coils and theorizing what happens to the energy.
there's a whole set of less effective inductor to inductor feedback
scenarios on either side of these nodes that will give you a cascade
of "textbook" values you can talk about for days on end.
you can scope them and calculate the inefficiency of the design.
watch how one coil stores one amount of energy and the other stores another
and talk more about why.
and one day everyone can know what they already know.
and your view of the energy can remain as it were before we started.
And we can go in circles and do it again on the other 3 threads.
And go hmm.... the energy goes in, now what happens?
I'm surprised no ones mentioned nickel-cobalt-iron wire
don't leave it to me to store it somewhere else in a qualitative manner
for it to be measured, scoped, and trace the losses through known scientific
values of inductance, impedance, reluctance, and permeance
as we watch it cycle back and forth on our scopes.
after all, if we had the data of the collapsing magnetic field
what would we have left to talk about?
Please, continue with your lecture Dr, erfinder.
I will try to pay attention as you teach us about the magnetism in the inductive "kickback"
and how to quantify it.
You all seem to consider there is overunity in inductive kickback and some weird coil setup.Why?
Quote from: Magluvin on February 27, 2018, 06:07:03 PM
Something I had thought of trying in the past...
If we have an inductor and we energize it then take away the input, the output of the coil doesnt really do much if the load is another inductor as the receiving inductor would impede the output dump of the first coil. But, if the receiving coil were bifi, its capacitance would accept the field collapse currents from the first coil. I wonder....
Mags
Ok Mags
The circuit below.
The current through L2 will be the same as the current through L1 !so they say!
The current through L1,the LED,and L2,must be the same as provided by the source !so they say!
L2 has 1/2 as many turns as L1,and so we have a 2:1-(L2 to L1) transformer,and so the voltage across L1 is twice that of the voltage across L2 !so they say!
And the current through L1 is 1/2 that of the current through L2 ??? !so they say!
Think about it ;)
Quote from: forest on February 28, 2018, 08:42:25 AM
You all seem to consider there is overunity in inductive kickback and some weird coil setup.Why?
I think it is a good place to start. If there is going to be be any OU found it is going to be a simple solution. It is so simple, that it is easy to hide.
It is going to be something you don't think of looking into. That is why people find it by accident. Good information can be found on text books, that are paid by energy companies and taught to engineers. Then PhDs and engineers become the next set of priests and they will defend this textbook to death, because they have a piece of paper on the wall saying "you are now a PhD, because you read and believe in this book".
So you can start by opening a textbook and find every sentence that goes like "Thou shall not put DC into a trafo" and experiment with it. Or "operational amplifier must always return to unity, even if you feed it back to itself". So when they say "no" or "you cannot" you experiment with it. Is there s special case that you CAN do it?
What are the things you ALWAYS do without even thinking? You start experimenting with a circuit with 2 poles. Is that the reason for failure? Do you need 2 poles?
Egyptians moved rocks that we cannot move in 2018, and they did not have transistors. Makes me think the answer is a simple one and the reason we don't get it is our training
author=Belfior link=topic=16203.msg517466#msg517466 date=1519827274]
QuoteI think it is a good place to start. If there is going to be be any OU found it is going to be a simple solution. It is so simple, that it is easy to hide.
It is going to be something you don't think of looking into.
Indeed. ;)
Quote from: Magluvin on February 27, 2018, 06:07:03 PM
Something I had thought of trying in the past...
If we have an inductor and we energize it then take away the input, the output of the coil doesnt really do much if the load is another inductor as the receiving inductor would impede the output dump of the first coil. But, if the receiving coil were bifi, its capacitance would accept the field collapse currents from the first coil. I wonder....
Mags
Mags,
Actually, a source inductor directly dumping it's energy into a target inductor without any capacitance involved, will nearly equal the same result as if the target inductor was not present. If however a capacitance is used for temporary storage of energy, then energy from a source inductor can be shuttled into a target inductor.
This makes your suggestion of using the distributed capacitance between two tightly coupled windings rather significant IMO and worthy of investigation!
I've attached a paper below of my research into possible gain in powdered iron toroids using circuitry that transfers energy from a linear inductor via a capacitor into a biased non-linear inductor.
Regards,
Pm
Thanks all for the replies to my idea.
I have all I need to do this other than time, of which I hope to have that here soon.
The question remains as to what can we do with that? What will happen? Do we use its magnetic field to drive a motor or a secondary? Do we keep on kicking the bifi to keep it in a high resonant state? Etc, etc. I think the idea will work, but we need to come up with how we can use it to our benefit.
It seems we can get, say 'inductive kickback', very simply and in many ways very efficiently. So like Partsman picked up on and stated, the capacitance of the bifi coil should be able to take on the inductors field collapse currents very well compared to a normal inductor.
I dont know if it would be any different than just discharging a hv cap into a bifi, but it might be...
Havnt seen it done before. So if anyone wants to dig in, be my guest. My bench is loaded with work at the moment.
Mags
Mags
Can you post a simple Block schematic of how you would like this done ?
Thanks
Chet
Quote from: ramset on February 28, 2018, 12:47:52 PM
Mags
Can you post a simple Block schematic of how you would like this done ?
Thanks
Chet
Ok. After work.
Mags
My first try would be to use a diode to capture the collapse currents and direct them into the bifi as if it were a cap. If what Tesla claimed in the pat to be correct, the bifi should take on the currents as fast as the bifi coil resistance and cap value would allow. If so, would there be a mag field built as fast as compared to a similar normal inductor with input applied? Dunno yet....
Mags
Let me rephrase my question , still not answered and nobody interested in : the collapsing field of disconnected coil is the source of EMF or EMF and current ? For example if the coil is part of boost converter does it only provide additional emf which is added to emf from battery or also produce the current ? Or maybe it depends on coil type or other factor ? Because boost converters are not OU I suppose coil is not producing current, but that's jsut my blind guess....
Quote from: forest on February 28, 2018, 01:38:11 PM
Let me rephrase my question , still not answered and nobody interested in : the collapsing field of disconnected coil is the source of EMF or EMF and current ? For example if the coil is part of boost converter does it only provide additional emf which is added to emf from battery or also produce the current ? Or maybe it depends on coil type or other factor ? Because boost converters are not OU I suppose coil is not producing current, but that's jsut my blind guess....
@forest,
When the current is interrupted from the source, the magnetic field in the inductor begins to collapse from the outside in. This collapse generates power, both current and voltage, that together are both
an entirely new event.
Quote from: synchro1 on March 01, 2018, 05:26:46 PM
@forest,
When the current is interrupted from the source, the magnetic field in the inductor begins to collapse from the outside in. This collapse generates power, both current and voltage, that together are both an entirely new event.
Mr Syncro have a look at his series off lectures MIT 8,02 try and re program your self, good luck https://www.youtube.com/channel/UCiEHVhv0SBMpP75JbzJShqw lecture 1
https://youtu.be/nGQbA2jwkWI (https://youtu.be/nGQbA2jwkWI)
https://youtu.be/MzAPu_p2wI4 (https://youtu.be/MzAPu_p2wI4)
https://youtu.be/3sP9kh4xtKo (https://youtu.be/3sP9kh4xtKo)
#16, 17, 18 are a good place to start
Quote from: AlienGrey on March 01, 2018, 06:20:01 PM
Mr Syncro have a look at his series off lectures MIT 8,02 try and re program your self, good luck https://www.youtube.com/channel/UCiEHVhv0SBMpP75JbzJShqw (https://www.youtube.com/channel/UCiEHVhv0SBMpP75JbzJShqw) lecture 1
@AlienGray,
Maybe I should read the entire "Encyclopedia Britannica" from A to Z while I'm at at too for you? How would you like it if I gave you seven Herculean chores for homework? Can you make a specific point please?
Quote from: sm0ky2 on March 01, 2018, 07:00:33 PM
https://youtu.be/nGQbA2jwkWI (https://youtu.be/nGQbA2jwkWI)
https://youtu.be/MzAPu_p2wI4 (https://youtu.be/MzAPu_p2wI4)
https://youtu.be/3sP9kh4xtKo (https://youtu.be/3sP9kh4xtKo)
#16, 17, 18 are a good place to start
@smoKy2,
There's nothing in any of these videos that contradicts what I stated as fact above.
Current flows to the coil from the power source until it's interrupted by the separation of the contacts. Next, comes the beginning of an entirely new event; Current begins to flow away from the coil. The away event is an entirely new and uniquely different event from the closed contact power event.
Look at the question I answered for forest above. The guy's really confused about two events happening simultaenously, and asking why there's no Overunity from overlapping. This really amounts to complete nonsense. Everyone's better off viewing Inductive kickback as simply as I have it described.
Quote from: Magluvin on February 28, 2018, 01:03:25 PM
My first try would be to use a diode to capture the collapse currents and direct them into the bifi as if it were a cap. If what Tesla claimed in the pat to be correct, the bifi should take on the currents as fast as the bifi coil resistance and cap value would allow. If so, would there be a mag field built as fast as compared to a similar normal inductor with input applied? Dunno yet....
Mags
Below is the diagram of what I said above. It also gives reference to which way current flows in the inductor L1 when SW! switch is opened. Now just replace C1 with your bifi coil. As simple as it gets.
Mags
Please. Talk is cheap.
PROVE that inductor is the source of current and voltage (EMF) once interrupted.
Prove it in electronic circuit demo using one single step , not a talk about duty cycle and DC-DC inverters equations. Like in Magluvin simple circuit, but with additional electronic switch to disconnect power source once coil is "magnetically charged".
Quote from: forest on March 02, 2018, 05:34:35 AM
Please. Talk is cheap.
PROVE that inductor is the source of current and voltage (EMF) once interrupted.
Prove it in electronic circuit demo using one single step , not a talk about duty cycle and DC-DC inverters equations. Like in Magluvin simple circuit, but with additional electronic switch to disconnect power source once coil is "magnetically charged".
Forest
You have been around here as long as I have and maybe longer. You dont know this yet??
Im surprised. ??? We could reverse the positions of the switch and the inductor, then reverse the polarity of the diode and get just about the same thing into the cap, just opposite polarity due to the configuration.
The previous pic shows the input in series with the inductor. This one isolates the inductor from the input. So in this pic below, where does the cap get its charge from??? ;) Do you think the cap wont get charged from the collapse currents of the inductor once the input switch is opened? ??? ???
These circuits in the pics show conventional current flow. I particularly think in electron flow from neg to pos. But this is what I found on the web and I modified the original pic to make the second pic example
Mags
Mags
seemed to me as If Forest was questioning the "other" fellow's input...
regardless , a fair analysis needs doing with as simple as possible a circuit [as you presented]
and perhaps whatever other ideas [for testing] can be presented to bolster claims ?
this needs a very simple benchmark moment for future reference ....one way or the other.
Question below
Are we to look for heretofore unknown excess energy [magnetic or other] being available due to a Bifi...Pancake coil ?
{in addition to the energy flow analysis}
perhaps a quick summary of your thoughts or possible claim ?
respectfully
Chet
Quote from: ramset on March 02, 2018, 02:12:47 PM
Mags
seemed to me as If Forest was questioning the "other" fellow's input...
regardless , a fair analysis needs doing with as simple as possible a circuit [as you presented]
and perhaps whatever other ideas [for testing] can be presented to bolster claims ?
this needs a very simple benchmark moment for future reference ....one way or the other.
Question below
Are we to look for heretofore unknown excess energy [magnetic or other] being available due to a Bifi...Pancake coil ?
{in addition to the energy flow analysis}
perhaps a quick summary of your thoughts or possible claim ?
respectfully
Chet
@ramset,
Are you referring to me Mr. Snodgrass? Take a look at my latest video "Poet of the Pickwick Papers":
https://www.youtube.com/watch?v=9K75ALn6__Y
Mags
Perfect example. Thanks. Now you can easily proof overunity. Just replace coil with transformer with secondary connected to diode bridge and to the capacitor. Single ON pulse will charge capacitor by induction, OFF by drop of magnetic field of coil. IS that OU ? Why the hell nobody tested it yet for single pulse ? It jsut require moderate equipment, just fast mosfets and digital scope.
@Magluvin,
I'm really stoked about the possibility of running a second spinner with the serial bifilar coil off the inductive kickback from the reed switch power coil. This setup is blindingly simple compared to Gotoluc's capacitor and horseshoe core coils arrangement. Thanks for the put together idea.
Quote from: ramset on March 02, 2018, 02:12:47 PM
Mags
seemed to me as If Forest was questioning the "other" fellow's input...
regardless , a fair analysis needs doing with as simple as possible a circuit [as you presented]
and perhaps whatever other ideas [for testing] can be presented to bolster claims ?
this needs a very simple benchmark moment for future reference ....one way or the other.
Question below
Are we to look for heretofore unknown excess energy [magnetic or other] being available due to a Bifi...Pancake coil ?
{in addition to the energy flow analysis}
perhaps a quick summary of your thoughts or possible claim ?
respectfully
Chet
Wont be long now,and we'll all be building bedini motors with another loss placed between the collector diode and charge battery.
Bound to be overunity there ::)
Brad
@Tinman,
Good point; However, running a neutralization pulse oscillator as demonstrated by me, may generate a secondary output from the bifilar. Those videos are on the "Negative Inductance" thread, but they feature a commutator that turns the oscillator coil into a hybrid output coil that generates power from the attraction of the overhead spring powered magnet to the ferrite coil core.
Quote from: synchro1 on March 02, 2018, 07:19:44 PM
@Tinman,
Good point; However, running a neutralization pulse oscillator as demonstrated by me, may generate a secondary output from the bifilar. Those videos are on the "Negative Inductance" thread, but they feature a commutator that turns the oscillator coil into a hybrid output coil that generates power from the attraction of the overhead spring powered magnet to the ferrite coil core.
Well i thought Mags was on the right track-kind of,but then went back to the same old same old.
I posted a bifilar transformer configuration a few pages back,but no one was interested--oh well :o
Anyway,i am having fun with it--i call it the !transphasic converter! :D
So ,the scope shot shows something that should never happen !apparently! ::) -schematic attached.
Due to my common ground issue with my scope,i had to scope one CVR at a time.
I saved the CVR1 scope shot(blue trace),and then inverted and overlaid the CVR 2 (yellow trace) on top of the blue trace.
The 962Hz Ac signal is supplied via a battery operated SG,removing any type of ground loop issues.
Both CVRs were swapped around,to eliminate any error in there measured power dissipation.
Take note of the current value flowing through the two resistors,where the !cant! happen- happens.
Quote from: tinman on March 02, 2018, 08:29:20 PM
Well i thought Mags was on the right track-kind of,but then went back to the same old same old.
I posted a bifilar transformer configuration a few pages back,but no one was interested--oh well :o
Anyway,i am having fun with it--i call it the !transphasic converter! :D
So ,the scope shot shows something that should never happen !apparently! ::) -schematic attached.
Due to my common ground issue with my scope,i had to scope one CVR at a time.
I saved the CVR1 scope shot(blue trace),and then inverted and overlaid the CVR 2 (yellow trace) on top of the blue trace.
The 962Hz Ac signal is supplied via a battery operated SG,removing any type of ground loop issues.
Both CVRs were swapped around,to eliminate any error in there measured power dissipation.
Take note of the current value flowing through the two resistors,where the !cant! happen- happens.
The right track to what? What you have shown is the right track and what I have suggested is same old same old?? :P
Apparently Partsman and syncro get what I am suggesting and find it new and interesting. And if what Im suggesting is soo same old same old, then show me an example of someone doing just what I have suggested. ;) Doubt you will be able to find that anywhere. ;)
In fact I am going to steal my own time tonight after laundry to put it on the table. What I had originally suggested is nothing like what you are showing and I 'dont' see that what you are showing shouldnt happen in the least. Its just one big series circuit driven with an ac input. Woopty doo. ;D Well you keep on having fun with all that. Enjoy. ;)
Mags
Quote from: Magluvin on March 02, 2018, 09:15:16 PM
The right track to what? What you have shown is the right track and what I have suggested is same old same old?? :P
Apparently Partsman and syncro get what I am suggesting and find it new and interesting. And if what Im suggesting is soo same old same old, then show me an example of someone doing just what I have suggested. ;) Doubt you will be able to find that anywhere. ;)
In fact I am going to steal my own time tonight after laundry to put it on the table. What I had originally suggested is nothing like what you are showing and I 'dont' see that what you are showing shouldnt happen in the least. Its just one big series circuit driven with an ac input. Woopty doo. ;D Well you keep on having fun with all that. Enjoy. ;)
Mags
You know of another circuit,where the current on one leg of the AC input is higher than the current on the other leg of the AC input ?--would love to see that--bet you cant find one example.
QuoteIf we have an inductor and we energize it then take away the input, the output of the coil doesnt really do much if the load is another inductor as the receiving inductor would impede the output dump of the first coil. But, if the receiving coil were bifi, its capacitance would accept the field collapse currents from the first coil.
The capacitance value of a bifi coil is near to nothing--in the pF range,and so will offer very little storage capacity.
QuoteIt seems we can get, say 'inductive kickback', very simply and in many ways very efficiently. So like Partsman picked up on and stated, the capacitance of the bifi coil should be able to take on the inductors field collapse currents very well compared to a normal inductor.
How is this any different to a transformer?,such as used in boost converters.
Quotethen show me an example of someone doing just what I have suggested
Have you forgotten about the weeks and weeks of research i(and others) done on the bifi pancake coil,where we bought rolls of flat copper tape to make our bifi coil,so as we could get the capacitance up as high as we could-far higher than you will ever get with a wire wound bifi coil--and nothing special was ever found by any of us.
We put all kinds of signals into those coil's--ac sine,square wave,HV pulses--and nothing.
So yes,others have already been down this road Mags,and had no luck.
Maybe you will find what we did not,so good luck :)
All i was doing,was offering what i had found to be interesting,when a bifi coil is wound and configured in a certain way-->which has not been done before.
But anyway,i will leave it to you,and wont hassle you anymore
Brad.
Quote from: tinman on March 02, 2018, 10:52:33 PM
You know of another circuit,where the current on one leg of the AC input is higher than the current on the other leg of the AC input ?--would love to see that--bet you cant find one example.
The capacitance value of a bifi coil is near to nothing--in the pF range,and so will offer very little storage capacity.
How is this any different to a transformer?,such as used in boost converters.
Have you forgotten about the weeks and weeks of research i(and others) done on the bifi pancake coil,where we bought rolls of flat copper tape to make our bifi coil,so as we could get the capacitance up as high as we could-far higher than you will ever get with a wire wound bifi coil--and nothing special was ever found by any of us.
We put all kinds of signals into those coil's--ac sine,square wave,HV pulses--and nothing.
So yes,others have already been down this road Mags,and had no luck.
Maybe you will find what we did not,so good luck :)
All i was doing,was offering what i had found to be interesting,when a bifi coil is wound and configured in a certain way-->which has not been done before.
But anyway,i will leave it to you,and wont hassle you anymore
Brad.
"You know of another circuit,where the current on one leg of the AC input is higher than the current on the other leg of the AC input ?--would love to see that--bet you cant find one example."
Actually I could care less. This thread is about inductive kickback, of which your circuit does not contain. And actually I believe I have seen the likes of what you state above. For some strange reason I think it was you that had shown something like this before. If not then it was someone else. Just some reactive anomaly as far as Im concerned. No inclination to OU possibilities? Not interested.
"The capacitance value of a bifi coil is near to nothing--in the pF range,and so will offer very little storage capacity."
That sounds like a MileHigh quote. Sorry if you only get picofarads in your bifi coils. Mine are mostly in the nanofarads, as was my pancake coil back in that thread. Tough break kid. ;) And if you remember so much about the bifi pancake experiments as you state in your post here, then you should have remembered that some had gotten more than just picofarads at the time. But, I guess you were not paying that close of attention. Dunno.
"How is this any different to a transformer?,such as used in boost converters."
Maybe you should reread all I have given on the idea I presented again before you question me with such, as I never suggested any mutual inductance between the charged inductor and the bifi coil. Only you have presented anything like a transformer with the 2 windings having mutual inductance and zero inductive kickback as this thread is based on. So your questioning me on that is out of order and without any real basis. ;)
"Have you forgotten about the weeks and weeks of research i(and others) done on the bifi pancake coil,where we bought rolls of flat copper tape to make our bifi coil,so as we could get the capacitance up as high as we could-far higher than you will ever get with a wire wound bifi coil--and nothing special was ever found by any of us.
We put all kinds of signals into those coil's--ac sine,square wave,HV pulses--and nothing."
I absolutely have not. But we didnt try THIS idea that I have presented. So all of that had shed light on what not to do again. And this idea is not one of those. So I can see that you think that those exercises had covered everything that could be done. Well Im saying here it did not. Sorry you have given up looking for more that could be done beyond those things you mention here. I have not. Im an idea guy. ;D
"So yes,others have already been down this road Mags,and had no luck."
Well this idea I presented was not on those roads already ridden. And maybe there are more roads to explore beyond this idea. :o ;)
"Maybe you will find what we did not,so good luck"
Yes. Maybe...
"All i was doing,was offering what i had found to be interesting,when a bifi coil is wound and configured in a certain way-->which has not been done before."
Well, firstly it was off topic, as we, partsman, ramset, syncro and forest were into my idea here on trying to take some advantage of 'Inductive Kickback' that may have not been tried before, except for maybe Tesla. ;) Sorry if others did not take interest in what you presented, as you stated in your second posting of your circuit. Again, it was way off topic anyway. Whether there is more current through one winding than the other or not, unless there is some possibility of OU there, Im not too concerned with it, as I am here for what this site was meant to try and find, overunity. Not parlor tricks.
"But anyway,i will leave it to you,and wont hassle you anymore"
Its not my thread. Oh, were you hassling me? Well Im still waiting for an answer to the first questions in my last post. You said "Well i thought Mags was on the right track-kind of,but then went back to the same old same old." ......
What was the right track? ??? What is same old same old about what I have presented? ??? Show me what I have presented is same old instead of just saying I have never seen what YOU have shown before! Bs dude. Bait and switch is all that was. Personally I just found that all to be just insults really. And I dont think you will answer even though I have asked twice now. ::) Pretty much because it was just a blind insult statement towards me, because so far you dont really know what my idea really is, as far as your replies here truly show. ::)
Mags
Quote from: Magluvin on March 03, 2018, 12:42:05 AM
"You know of another circuit,where the current on one leg of the AC input is higher than the current on the other leg of the AC input ?--would love to see that--bet you cant find one example."
Actually I could care less. This thread is about inductive kickback, of which your circuit does not contain. And actually I believe I have seen the likes of what you state above. For some strange reason I think it was you that had shown something like this before. If not then it was someone else. Just some reactive anomaly as far as Im concerned. No inclination to OU possibilities? Not interested.
"The capacitance value of a bifi coil is near to nothing--in the pF range,and so will offer very little storage capacity."
That sounds like a MileHigh quote. Sorry if you only get picofarads in your bifi coils. Mine are mostly in the nanofarads, as was my pancake coil back in that thread. Tough break kid. ;) And if you remember so much about the bifi pancake experiments as you state in your post here, then you should have remembered that some had gotten more than just picofarads at the time. But, I guess you were not paying that close of attention. Dunno.
"How is this any different to a transformer?,such as used in boost converters."
Maybe you should reread all I have given on the idea I presented again before you question me with such, as I never suggested any mutual inductance between the charged inductor and the bifi coil. Only you have presented anything like a transformer with the 2 windings having mutual inductance and zero inductive kickback as this thread is based on. So your questioning me on that is out of order and without any real basis. ;)
"Have you forgotten about the weeks and weeks of research i(and others) done on the bifi pancake coil,where we bought rolls of flat copper tape to make our bifi coil,so as we could get the capacitance up as high as we could-far higher than you will ever get with a wire wound bifi coil--and nothing special was ever found by any of us.
We put all kinds of signals into those coil's--ac sine,square wave,HV pulses--and nothing."
I absolutely have not. But we didnt try THIS idea that I have presented. So all of that had shed light on what not to do again. And this idea is not one of those. So I can see that you think that those exercises had covered everything that could be done. Well Im saying here it did not. Sorry you have given up looking for more that could be done beyond those things you mention here. I have not. Im an idea guy. ;D
"So yes,others have already been down this road Mags,and had no luck."
Well this idea I presented was not on those roads already ridden. And maybe there is more roads to explore beyond this idea. :o ;)
"Maybe you will find what we did not,so good luck"
Yes. Maybe...
"All i was doing,was offering what i had found to be interesting,when a bifi coil is wound and configured in a certain way-->which has not been done before."
Well, firstly it was off topic, as we, partsman, ramset, syncro and forest were into my idea here on trying to take some advantage of 'Inductive Kickback' that may have not been tried before, except for maybe Tesla. ;) Sorry if others did not take interest in what you presented, as you stated in your second posting of your circuit. Again, it was way off topic anyway. Whether there is more current through one winding than the other or not, unless there is some possibility of OU there, Im not too concerned with it, as I am here for what this site was meant to try and find, overunity. Not parlor tricks.
"But anyway,i will leave it to you,and wont hassle you anymore"
Its not my thread. Oh, were you hassling me? Well Im still waiting for an answer to the first questions in my last post. You said "Well i thought Mags was on the right track-kind of,but then went back to the same old same old." ......
What was the right track???? What is same old same old about what I have presented???? Show me what I have presented is same old instead of just saying I have never seen what YOU have shown before! Bs dude. Bait and switch is all that was. Personally I just found that all to be just insults really. And I dont think you will answer even though I have asked twice now. ::) Pretty much because it was just a blind insult statement towards me, because so far you dont really know what my idea really is, as far as your replies here truly show. ::)
Mags
Hey-you enjoy your self Mag's
Good luck.
Brad
Quote from: tinman on March 03, 2018, 01:10:51 AM
Hey-you enjoy your self Mag's
Good luck.
Brad
Yep. Wont answer the 2 questions. Because there are no real answers. ;)
Mags
Quote from: Magluvin on March 03, 2018, 01:19:33 AM
Yep. Wont answer the 2 questions. Because there are no real answers. ;)
Mags
It would seem you have gotten your self all mixed up here Mags
Quote"Well i thought Mags was on the right track-kind of,but then went back to the same old same old."
To use a bifi coil in a different manner to that which has been done before.
The same old same old,is returning back to inductive kickback,thinking that there is magic to be found,even after years of experimenting,and absolute proof that inductive kickback always yields less energy than what it took to create it.
QuoteThis thread is about inductive kickback, of which your circuit does not contain.
Are you blind?
Look at the scope shot Mag's--the whole bloody AC signal is loaded with inductive spikes.
Do you think the only way to create inductive kickback spikes is the way you say?.
QuoteSorry if you only get picofarads in your bifi coils. Mine are mostly in the nanofarads, as was my pancake coil back in that thread. Tough break kid.
Well,as you have gone from wire wound coils(as i stated in my post),to tape wound coils(which i referred to only in reference to those particular experiments),my bifi tape coil was in the microfarad range--tough break kid ::)
QuoteAnd if what Im suggesting is soo same old same old, then show me an example of someone doing just what I have suggested. ;) Doubt you will be able to find that anywhere.
You mean-sending the inductive kickback current spike to one of the windings of your bifi coil,and then using the second winding to power a device--like a motor,as i believe you stated.
How will your bifi coil be any different to a single wound coil,with some added capacitance across it?
How is this any different than a toroid wound transformer?
How many people you know that have sent inductive kickback spikes to one winding of a toroid transformer,and then tried to use a second winding to power something?
Have you seen a boost converter with a toroid transformer in it Mag's ?
So,if you are going to be getting your knickers in a twist,even after i wished you good luck-->then SAME OLD SAME OLD.
Brad
Hi Tinnman,
I'm confused now you said that the kickback always yields less energy than what it took to create it.
Didn't the video that TK showed prove the spike is higher than the input?
He had the supply set at 1.8 volts, the forward bias diode wasn't lit, but when he disconnected , the back spike lit the led that was reverse biased.
just asking
Thanks artv
Quote from: shylo on March 03, 2018, 04:12:08 AM
Hi Tinnman,
I'm confused now you said that the kickback always yields less energy than what it took to create it.
Didn't the video that TK showed prove the spike is higher than the input?
He had the supply set at 1.8 volts, the forward bias diode wasn't lit, but when he disconnected , the back spike lit the led that was reverse biased.
just asking
Thanks artv
The inductive kickback can be either high current and low voltage,or high voltage and low current-depends on the load.
When i say high and low current,i am referring to the value of current reached for the duration of the inductive kickback part of the cycle.
For example-
If we have a high resistance load,we would see a high voltage with low current value.
If we have a low resistance load,we would see a low voltage with a higher current value.
QuoteDidn't the video that TK showed prove the spike is higher than the input?
It proved that the voltage from the inductive kickback was higher than the source voltage--as per the joule thief.
But the available overall current value will be far less than that which created it.
As soon as you add copper wire(or any resistive conductor),you loose some of the input power to resistive losses(heat),and so,the inductive kickback energy will be less than what it took to create it.
Mags will find that he will incur further losses by adding another bifilar coil to the mix.
If i am wrong,then hats off to Mags.
Brad
Quote from: Erfinder on March 03, 2018, 04:34:14 AM
Excuse the intrusion, in my opinion it's not enough to end the argument with this statement in red....... proof is relative..... the question of why inductive kickback yields less energy (in your opinion, based on your research and those of like mind with you) than it took to create it demands a satisfactory explanation of why one gets back less. Following this line of thought back far enough one discovers ones folly in logic, and introduces one to the means through and by which we can approach what you are dictating (in so many words) is impossible.
Just explained that in my last post.
author=Erfinder link=topic=16203.msg517624#msg517624 date=1520070858]
QuoteWe are going to have to agree to disagree, and that's fine with me...... See in my opinion, you haven't "explained" anything. You sight examples, you share personal experience, and its here where the bias comes in. You cannot speak from that which you have not experienced, and yet your posture is likened to that of an individual who "knows". You do not offer proof of that you get back less, you present information you've gathered which supports your position, nothing more.
What i have to offer,is the same as everyone else has to offer--fact's,backed by the laws of physics.
Not only is it my experience,but the experience of the scientific community-->and that is,when ever a current is passed through a conductor,some of the energy is dissipated as heat--unless you are using super conductor's. But even then,there are losses,in way of the energy required to keep the temperature low enough,so as the conductor maintains it's super conductive state.
QuoteMy suggestion is to ask and answer the question of why you get back less than you put in. It's the only question that matters at this point, it's the only question which hasn't been properly addressed in all these years of back and forth debating this issue.
I have answered that question,or presented one of the energy losses. There are other losses also,but they also convert to a heat loss in the end.
Now,i know you are going to say that you know different,as we have heard for years now.
You have shared some things with some people--such as Mag's,and as expected,nothing ever came of it.Mags even said to me in a post--Erfinder is on to something--we shall see he said.
And yes,we are still waiting to see ::)
One would think that if you !Erfinder! could get more energy from the inductive kickback than it took to create it,you would not be here now.
You would be so busy cutting deals with investors on your free energy device,that you would not have the time to frequent these forums. We may catch a glimpse of your private jet cruising over head,but that would be it.
Yes yes,i know how it go's Erfinder--no need to explain-->we are not worthy of your higher IQ knowledge.
QuoteI traveled thousands of miles, and spent money I didn't have to take part in something that was in hindsight beneath me... I'd do it again in a heartbeat. They have nothing on me, didn't then and are galaxies behind me today..
One day Erfinder,hopefully your overunity percentage will be as big as your ego ;)
Brad
It's too silly for anyone to pretend it's possible to get more from the kickback. However, directed through a diode to a capacitor that is connected to a bifilar coil with a ferrite core, the accumulated back spike can neutralize the attraction to a spring tensioned magnet periodically as the capacitor fills and discharges into the coil. The over unity would come from the 3.33 sheer to push pull ratio. The only adjunct would be the microwave carousel and latching Reed switch commutator needed to direct the power induced by the magnet attracted back to the ferrite core, into a final storage capacitor destination.
This would deliver an over unity output from the the BEMF. Let's say the back spike is 70% of the input, multiplied by the sheer to push pull ratio of 3.33 would deliver 2.33 times the input from the primary pulse coil.
I spotted an experimenter over at the Energetic Forum site who was pulsing a magnet on a swinging leggo pendulum with collected BEMF discharging periodically from a capacitor. Does anyone recall this guy's user name?
Quote from: Erfinder on March 03, 2018, 07:00:36 AM
you state what is impossible, and back this up with what you consider as support from mainstream, and personal bias. We are supposed to accept what you say and what you demonstrate as being the last word on the subject... Not I.....
I suggested very simply that one ask why we don't get back more! I know why, and I am sure you think you know why, but for whatever reason, you don't entertain the question..... I am not here to show you anything, prove nor disprove anything. I am asking why are we supposed to take your word for it that we cannot get more back, motivated in part through considering the information you offer as "proof" for why we cannot. So I say it again, you haven't explained anything, you site examples which support your biased view. You lack the necessary integrity and courage to really look at what's going on, from this posture, you will never recognize and face your folly.
I told you time and time again, the search for overunity is ignorance.... You play that game, I am not interested in overunity, I am after that which is there... unfortunately for you, you are oblivious as to what that is.
I know we live and breath in a sea of energy,but i also know that that sea will not be sailed in a leaky boat.
The high tension electric field holds the answers we seek,where the restrictions of the magnetic field are broken.
We need only energy,not force nor mass to achieve our goal.
QuoteI am not here to show you anything, prove nor disprove anything.
Yes,we all know that.
Quote from: tinman on March 03, 2018, 03:11:41 AM
It would seem you have gotten your self all mixed up here Mags
To use a bifi coil in a different manner to that which has been done before.
The same old same old,is returning back to inductive kickback,thinking that there is magic to be found,even after years of experimenting,and absolute proof that inductive kickback always yields less energy than what it took to create it.
Are you blind?
Look at the scope shot Mag's--the whole bloody AC signal is loaded with inductive spikes.
Do you think the only way to create inductive kickback spikes is the way you say?.
Well,as you have gone from wire wound coils(as i stated in my post),to tape wound coils(which i referred to only in reference to those particular experiments),my bifi tape coil was in the microfarad range--tough break kid ::)
You mean-sending the inductive kickback current spike to one of the windings of your bifi coil,and then using the second winding to power a device--like a motor,as i believe you stated.
How will your bifi coil be any different to a single wound coil,with some added capacitance across it?
How is this any different than a toroid wound transformer?
How many people you know that have sent inductive kickback spikes to one winding of a toroid transformer,and then tried to use a second winding to power something?
Have you seen a boost converter with a toroid transformer in it Mag's ?
So,if you are going to be getting your knickers in a twist,even after i wished you good luck-->then SAME OLD SAME OLD.
Brad
"It would seem you have gotten your self all mixed up here Mags"
Lol. Oh sure. Its only been a few pages. Yeah Im all mixed up. Lets see......
Ok... Question 1 What is the right track?
Your answer finally is....
"To use a bifi coil in a different manner to that which has been done before."
Ok then. But that still does not define what the right track is that you thought I was on. Just because now you say it IS a different manner of use of the bifilar does not explain the right track comment. So still no real answer. Why was it the right track Brad? Show me that you understand the idea presented, as so far you havnt shown that you do at all.
"The same old same old,is returning back to inductive kickback,thinking that there is magic to be found,even after years of experimenting,and absolute proof that inductive kickback always yields less energy than what it took to create it. "
So let me get this straight Brad. You initially thought I was on the right track by me saying to send the inductive kickback of a charged inductor to a bifilar coil, because that was my initial presentation of the idea, but now the part of the Inductive Kickback use is an issue for you. Lol. And you say Im mixed up? Well that is a terrible answer Brad. You are just trying to climb out the hole you have dug yourself here. Its not me that is mixed up Brad, clearly. Again you try to insult me further. ;)
"Are you blind?
Look at the scope shot Mag's--the whole bloody AC signal is loaded with inductive spikes.
Do you think the only way to create inductive kickback spikes is the way you say?."
Definition of Inductive Kickback - the pulses of high voltage produced when direct current through an inductor is interrupted ...
Your circuit does not fall under that very common definition Brad. You really must think I am that gullible to accept what you are saying here. But maybe you were mixed up when you wrote that. Show me a definition on the internet that backs up your claim on inductive kickback in your circuit Brad. I have reposted your circuit from this post from you below where you clearly state an ac input.
Ok Brad. From your post http://overunity.com/16203/inductive-kickback/msg517605/#msg517605 (http://overunity.com/16203/inductive-kickback/msg517605/#msg517605)
"The 962Hz Ac signal is supplied....."
Just where in that AC supplied signal is the direct current through the inductor interrupted Brad??? So no Brad, there is no inductive kickback shown in your scope shot as you claim above. But maybe your just mixed up here. ;) Again. And you ask if I am blind. ::) Well I suppose you wish I was in this case Brad. ;)
"Well,as you have gone from wire wound coils(as i stated in my post),to tape wound coils(which i referred to only in reference to those particular experiments),my bifi tape coil was in the microfarad range--tough break kid"
WOW!!! So you are the man now!!! Hey? Did you not just previously state "The capacitance value of a bifi coil is near to nothing--in the pF range,and so will offer very little storage capacity." http://overunity.com/16203/inductive-kickback/msg517611/#msg517611 (http://overunity.com/16203/inductive-kickback/msg517611/#msg517611)
Hmmm. Lol Well maybe you were just mixed up. Again.
"You mean-sending the inductive kickback current spike to one of the windings of your bifi coil,and then using the second winding to power a device--like a motor,as i believe you stated."
Ha! Show me the post where I said that Brad. I have never said that. I havnt said anything about doing anything with one winding of a bifi coil here in this discussion. I have not said anything about using the second winding to power a load. I have only talked about the bifi winding as a whole here, period. Show me that I stated that Brad. Dont you ever put words in my mouth that you cannot back up brad. Now you are really pissing me off just so you can save face here. Or maybe you are just mixed up.
"How will your bifi coil be any different to a single wound coil,with some added capacitance across it?
How is this any different than a toroid wound transformer?
How many people you know that have sent inductive kickback spikes to one winding of a toroid transformer,and then tried to use a second winding to power something?"
Again, you clearly have not read what my idea is all about. CLEARLY!. You are just injecting falsehoods here. What the hell is wrong with you man?
Go ahead and try to send an inductive kickback spike through a single winding inductor. It will impede and reject that spike, as Partsman stated as he clearly read what I presented and understood it very clearly and you clearly have not. And now as you say in the above quote "How will your bifi coil be any different to a single wound coil,with some added capacitance across it?" Do you not believe that the inductive kickback spike of an inductor will be accepted by that capacitance across the other coil as you have just stated?? ::) Your losing it man.
"Have you seen a boost converter with a toroid transformer in it Mag's ?"
The only one here talking about transformers and mutual inductance AND using the windings of a bifi coil separately here is you. So your question is not relevant here.
"So,if you are going to be getting your knickers in a twist,even after i wished you good luck-->then SAME OLD SAME OLD.
No Brad. You are making it seem like what I have presented is same old same old, when really you clearly do not get what the idea is about at all. But at least these other guys do. We are all ten steps ahead of you on this.
Mags
author=Magluvin link=topic=16203.msg517639#msg517639 date=1520084349]
QuoteOk... Question 1 What is the right track?
Your answer finally is....
"To use a bifi coil in a different manner to that which has been done before."
Ok then. But that still does not define what the right track is that you thought I was on. Just because now you say it IS a different manner of use of the bifilar does not explain the right track comment. So still no real answer. Why was it the right track Brad?
Like i said Mag's--your lost,or missed the first circuit i presented.
This is the right track in regards to using a bifi coil that i was talking about.
See pic below !again!-->dose that look like an inductive input spike,or quick input pulse ? ;)
QuoteSo let me get this straight Brad. You initially thought I was on the right track by me saying to send the inductive kickback of a charged inductor to a bifilar coil, because that was my initial presentation of the idea,
Yes
Quotebut now the part of the Inductive Kickback use is an issue for you. Lol. And you say Im mixed up? Well that is a terrible answer Brad. You are just trying to climb out the hole you have dug yourself here. Its not me that is mixed up Brad, clearly. Again you try to insult me further. ;)
I presented a bifi coil arrangement that i have been experimenting with,that showed some unusual effect's (post 253),but then you mixed it up with my second post of the same bifi coil running on an AC --so yes,your mixed up Mags.
The inductive kickback will yield less energy through your bifi coil,than that of just using the inductive kickback energy straight from the source.
QuoteDefinition of Inductive Kickback - the pulses of high voltage produced when direct current through an inductor is interrupted ...
Your circuit does not fall under that very common definition Brad. You really must think I am that gullible to accept what you are saying here. But maybe you were mixed up when you wrote that. Show me a definition on the internet that backs up your claim on inductive kickback in your circuit Brad. I have reposted your circuit from this post from you below where you clearly state an ac input.
As i stated Mag;s,i guess you missed the first one-post 253
QuoteJust where in that AC supplied signal is the direct current through the inductor interrupted Brad??? So no Brad, there is no inductive kickback shown in your scope shot as you claim above. But maybe your just mixed up here. ;) Again. And you ask if I am blind. ::) Well I suppose you wish I was in this case Brad. ;)
See previous comment--post 253
QuoteWOW!!! So you are the man now!!! Hey? Did you not just previously state "The capacitance value of a bifi coil is near to nothing--in the pF range,and so will offer very little storage capacity."
Yep,in a wire wound bifi coil,the capacitance value will be low,and in a copper tape wound bifi coil,the capacitance value will be higher.
As i said,you are mixed up.
QuoteHa! Show me the post where I said that Brad. I have never said that. I havnt said anything about doing anything with one winding of a bifi coil here in this discussion. I have not said anything about using the second winding to power a load. I have only talked about the bifi winding as a whole here, period. Show me that I stated that Brad. Dont you ever put words in my mouth that you cannot back up brad. Now you are really pissing me off just so you can save face here. Or maybe you are just mixed up.
Quote: The question remains as to what can we do with that? What will happen? Do we use its magnetic field to drive a motor or a secondary? Do we keep on kicking the bifi to keep it in a high resonant state?
Maybe your just going to let it dissipate in the bifi coil Mag's ::)
QuoteAgain, you clearly have not read what my idea is all about. CLEARLY!. You are just injecting falsehoods here. What the hell is wrong with you man?
Well ,reading all your post's,you are all over the show,with no clear path to follow.
Your circuits and your words say--swap out the capacitor for a bifi coil--then what?
QuoteGo ahead and try to send an inductive kickback spike through a single winding inductor. It will impede and reject that spike,
No it wont.
Quoteas Partsman stated as he clearly read what I presented and understood it very clearly and you clearly have not. And now as you say in the above quote "How will your bifi coil be any different to a single wound coil,with some added capacitance across it?"
And i ask again.
I answered your questions,so now answer mine.
How will it be different ?.
QuoteThe only one here talking about transformers and mutual inductance AND using the windings of a bifi coil separately here is you. So your question is not relevant here.
Fair enough.
So,how are you going to use the bifi coil ?.
QuoteNo Brad. You are making it seem like what I have presented is same old same old, when really you clearly do not get what the idea is about at all. But at least these other guys do. We are all ten steps ahead of you on this.
Breaking new ground on the !never tried before! are we ;)
Well,guess we'll see soon enough how far ahead of me you are.
Once you have the best you can get,i will show you what i have--unless of course you do not wish to see it,which is fair enough.
Quote from: tinman on March 03, 2018, 10:25:33 AM
author=Magluvin link=topic=16203.msg517639#msg517639 date=1520084349]
Like i said Mag's--your lost,or missed the first circuit i presented.
This is the right track in regards to using a bifi coil that i was talking about.
See pic below !again!-->dose that look like an inductive input spike,or quick input pulse ? ;)
Yes
I presented a bifi coil arrangement that i have been experimenting with,that showed some unusual effect's (post 253),but then you mixed it up with my second post of the same bifi coil running on an AC --so yes,your mixed up Mags.
The inductive kickback will yield less energy through your bifi coil,than that of just using the inductive kickback energy straight from the source.
As i stated Mag;s,i guess you missed the first one-post 253
See previous comment--post 253
Yep,in a wire wound bifi coil,the capacitance value will be low,and in a copper tape wound bifi coil,the capacitance value will be higher.
As i said,you are mixed up.
Quote: The question remains as to what can we do with that? What will happen? Do we use its magnetic field to drive a motor or a secondary? Do we keep on kicking the bifi to keep it in a high resonant state?
Maybe your just going to let it dissipate in the bifi coil Mag's ::)
Well ,reading all your post's,you are all over the show,with no clear path to follow.
Your circuits and your words say--swap out the capacitor for a bifi coil--then what?
No it wont.
And i ask again.
I answered your questions,so now answer mine.
How will it be different ?.
Fair enough.
So,how are you going to use the bifi coil ?.
Breaking new ground on the !never tried before! are we ;)
Well,guess we'll see soon enough how far ahead of me you are.
Once you have the best you can get,i will show you what i have--unless of course you do not wish to see it,which is fair enough.
"I presented a bifi coil arrangement that i have been experimenting with,that showed some unusual effect's (post 253),but then you mixed it up with my second post of the same bifi coil running on an AC --so yes,your mixed up Mags.
The inductive kickback will yield less energy through your bifi coil,than that of just using the inductive kickback energy straight from the source."
No I did not mix it up... You said in post 283 this...."
"Look at the scope shot Mag's--the whole bloody AC signal is loaded with inductive spikes.
Do you think the only way to create inductive kickback spikes is the way you say?."
In the post 277 with the scope shot you said.....
"The 962Hz Ac signal is supplied via a battery operated SG,"
So clearly the scope shot that you claim has all the inductive kickback spikes is not the circuit you had first shown using input pulses, as you are referring to here to cover your ass. Im not missing anything Brad. You are full of it. If it was your first circuit that was so interesting then why the second circuit using ac input to show a scope shot? Im not mixing anything. You are mixing everything. I know the first circuit had a pulse input. I know the scope shot is with ac input and then you claim inductive kickback in the scope shot, but then say it was the first circuit. You are nuts dude. Totally nuts.. Its only 4 pages where all this goes down from the time you posted the first circuit. Anyone here can reread it all for themselves and see you are playing games. Shameful dude, shameful.
Your circuit has nothing to do with what I had suggested in any way.. It is just some distraction from what WE were talking about. Thats all it is. Some useless distraction.
As to the rest of your post, its not worth the time.
Mags
Quote from: Magluvin on March 03, 2018, 11:57:13 AM
"I presented a bifi coil arrangement that i have been experimenting with,that showed some unusual effect's (post 253),but then you mixed it up with my second post of the same bifi coil running on an AC --so yes,your mixed up Mags.
The inductive kickback will yield less energy through your bifi coil,than that of just using the inductive kickback energy straight from the source."
No I did not mix it up... You said in post 283 this...."
"Look at the scope shot Mag's--the whole bloody AC signal is loaded with inductive spikes.
Do you think the only way to create inductive kickback spikes is the way you say?."
In the post 277 with the scope shot you said.....
"The 962Hz Ac signal is supplied via a battery operated SG,"
So clearly the scope shot that you claim has all the inductive kickback spikes is not the circuit you had first shown using input pulses, as you are referring to here to cover your ass. Im not missing anything Brad. You are full of it. If it was your first circuit that was so interesting the why the second circuit using ac input to show a scope shot? Im not mixing anything. You are mixing everything. I know the first circuit had a pulse input. I know the scope shot is with ac input and then you claim inductive kickback in the scope shot, but then say it was the first circuit. You are nuts dude. Totally nuts.. Its only 4 pages where all this goes down from the time you posted the first circuit. Anyone here can reread it all for themselves and see you are playing games. Shameful dude, shameful.
Your circuit has nothing to do with what I had suggested in any way.. It is just some distraction from what WE were talking about. Thats all it is. Some useless distraction.
As to the rest of your post, its not worth the time.
Mags
@Mags,
Watch this video: You guy's need to do less kibitzing and more building.
https://www.youtube.com/watch?v=z8qFTKWIdCQ (https://www.youtube.com/watch?v=z8qFTKWIdCQ)
Quote from: synchro1 on March 03, 2018, 12:19:48 PM
@Mags,
Watch this video: You guy's need to do less kibitzing and more building.
https://www.youtube.com/watch?v=z8qFTKWIdCQ (https://www.youtube.com/watch?v=z8qFTKWIdCQ)
Hey Sync
Im just getting to my simple setup. Im using the coils I had made and shown some months ago that are as identical as possible where one is bifi and the other is single wire. I had made the single wire in 2 halves as to do similar parallel and series experiments with each to compare tests.
Just trying some other coils to use as the kickback coil. And then finally another coil as a pickup coil to measure any magnetic output from the single wire and bifi coils to be tested.. The single wire should show very little to nothing when it receives the kickback electrical spike from the kicker coil electrical output across its leads, and then we will see what the bifi coil will do after that test. Simple simple simple.
Hope to see what you have come up with soon. ;) I should be done in an hour or so. Will post the vid when done. The pickup coil will be on the scope to see what happens. Going to rig it to a switch on a bread board to easily switch between the single wire coil and the bifi to make it easy to show each test during the vid and then all I have to do is move the test coils in place for each test as the vid goes along.
Mags
Ok. Finally got it set up. Had problems with the breadboard. Dunno what but connections sucked. So dug out a new one. Have to go home for a bit and then ill come back and do a vid. ;D
Mags
Quote from: Magluvin on March 03, 2018, 11:57:13 AM
"I presented a bifi coil arrangement that i have been experimenting with,that showed some unusual effect's (post 253),but then you mixed it up with my second post of the same bifi coil running on an AC --so yes,your mixed up Mags.
The inductive kickback will yield less energy through your bifi coil,than that of just using the inductive kickback energy straight from the source."
No I did not mix it up... You said in post 283 this...."
"Look at the scope shot Mag's--the whole bloody AC signal is loaded with inductive spikes.
Do you think the only way to create inductive kickback spikes is the way you say?."
In the post 277 with the scope shot you said.....
"The 962Hz Ac signal is supplied via a battery operated SG,"
So clearly the scope shot that you claim has all the inductive kickback spikes is not the circuit you had first shown using input pulses, as you are referring to here to cover your ass. Im not missing anything Brad. You are full of it. If it was your first circuit that was so interesting then why the second circuit using ac input to show a scope shot? Im not mixing anything. You are mixing everything. I know the first circuit had a pulse input. I know the scope shot is with ac input and then you claim inductive kickback in the scope shot, but then say it was the first circuit. You are nuts dude. Totally nuts.. Its only 4 pages where all this goes down from the time you posted the first circuit. Anyone here can reread it all for themselves and see you are playing games. Shameful dude, shameful.
Your circuit has nothing to do with what I had suggested in any way.. It is just some distraction from what WE were talking about. Thats all it is. Some useless distraction.
As to the rest of your post, its not worth the time.
Mags
Inductive kickback is caused by the interruption of current flow through the inductor.
Only you mags would think that the interruption has to be a disconnection.
Quote from: Erfinder on March 03, 2018, 10:48:54 AM
QuoteFor the record, I was being sarcastic
Of course,hence the cheesy grin-as you stated.
QuoteI see you have found value in impedance mismatching.....
Oddly enough-yes.
QuoteI see you "appear" to recognize where one is to, assuming one is desirous of such, position ones load at the junction between the differing impedance....
I see great value-so far,but much left to look at yet.
The larger unit i made from a MOT core,is playing havoc with my switching transistor,but i can see any high voltage spikes,or large current values across the transistor anywhere,so not sure what is going on there. Maybe will have to switch to mechanical switching.
Quoteinteresting choice for a load...cheesy, but interesting nonetheless, a diode, light emitting, what's powering it
The small unit is powered by my FG ATM.
Quote.....were it my system, the load would be powered by the reflected.....
The reflected what?
QuoteA broken watch indicates the correct time twice a day....
A good watch indicates the correct time 24 hours a day. ;)
Ok, sorry it took longer than expected. Spent more time than I wanted to, but sometimes you have to try and test things and run into issues, etc. But its all setup.
The input is 1v from the power supply. Im just using a push button switch to energize the kicking inductor. That coil is 24.8ohm and 10mh
The identical(practically) single wire and bifi coils, left and right respectively, are 43 ohm and 35.7mh. The pickup coil between the test coils will be the yellow trace on the scope and the blue trace will be monitoring the kicker coil.. Doing the vid now to show results....
Mags
Here's a test of inductive kickback illuminating 9 LEDs. The voltage off the end of the LED rail is 6 volts. Tinman stated that Igor's "Reed Switch Spinner 2" is not returning any backspike power back to his source battery. This is wrong! The bifilar is next.
https://www.youtube.com/watch?v=69diZuFSI4o (https://www.youtube.com/watch?v=69diZuFSI4o)
Here is the vid.
https://www.youtube.com/watch?v=4zSjnNdeZ-Q
Mags
Quote from: Magluvin on March 04, 2018, 06:59:41 AM
Here is the vid.
https://www.youtube.com/watch?v=4zSjnNdeZ-Q (https://www.youtube.com/watch?v=4zSjnNdeZ-Q)
Mags
Youtube says video not there.
Norman
Quote from: Magluvin on March 04, 2018, 04:49:52 AM
Ok, sorry it took longer than expected. Spent more time than I wanted to, but sometimes you have to try and test things and run into issues, etc. But its all setup.
The input is 1v from the power supply. Im just using a push button switch to energize the kicking inductor. That coil is 24.8ohm and 10mh
The identical(practically) single wire and bifi coils, left and right respectively, are 43 ohm and 35.7mh. The pickup coil between the test coils will be the yellow trace on the scope and the blue trace will be monitoring the kicker coil.. Doing the vid now to show results....
Mags
@Magluvin,
Can you measure the inductances on the two coils please? Remember the 5% greater inductance measurement of the bifilar?
Quote from: norman6538 on March 04, 2018, 07:43:56 AM
Youtube says video not there.
Norman
Hmm, Im watching it again right now. Ill look and see what might be the issue.
Mags
Check now. For some reason it was on private, but that should still let you see it if I give the link...
https://www.youtube.com/watch?v=4zSjnNdeZ-Q
Mags
Quote from: synchro1 on March 04, 2018, 07:49:56 AM
@Magluvin,
Can you measure the inductances on the two coils please? Remember the 5% greater inductance measurement of the bifilar?
Not true. Both are series connected and both measure 35.7mh
Mags
Strange. When i got to wath the vid, it starts at 48 sec in. Does that happen to anyone else?
Mags
Rebooted firefox and is ok now. Having internet issues also. Keeps bleeping with a message at the bot right to do a troubleshoot and when I do it says no issues.
Never seen that before.
Mags
Current reading of inductive kickback on the "Reed Switch Spinner": The kickback voltage measures 6 volts. Naturally the input is 12. The BEMF output measures around 50 milli watts:
https://www.youtube.com/watch?v=eT2VcWNIKA0 (https://www.youtube.com/watch?v=eT2VcWNIKA0)
The video is there now. thanks. Norman
Quote from: Magluvin on March 04, 2018, 08:01:00 AM
Check now. For some reason it was on private, but that should still let you see it if I give the link...
https://www.youtube.com/watch?v=4zSjnNdeZ-Q (https://www.youtube.com/watch?v=4zSjnNdeZ-Q)
Mags
@Magluvin,
Congratulations on the definitive and complete obliteration of the false Milehigh position that there's no difference between the bifilar and single wire coil. Excellent video. I urged Tinselkoala repeatedly to test the two coils with pulse power for years, and the only tests he ever performed were tests with constant current, both D.C. and A.C. to arrive at the erroneous conclusion that Tesla was just a dummy.
Liar.
Quote from: synchro1 on March 04, 2018, 12:56:02 PM
Current reading of inductive kickback on the "Reed Switch Spinner": The kickback voltage measures 6 volts. Naturally the input is 12. The BEMF output measures around 50 milli watts:
https://www.youtube.com/watch?v=eT2VcWNIKA0 (https://www.youtube.com/watch?v=eT2VcWNIKA0)
Well syncro,your video just proved i was right,and you are wrong.Inductive kickback is a direct current (DC),and what i said was that the generating effect of the spinning magnet is what was sending current back to the source,as the spinning magnet would produce a alternating current(AC) across the coil.
You just showed a video,where you were measuring alternating current,and trying to pass it of as the inductive kickback current, lol--whats the chances of that.
Enjoy.
Quote from: tinman on March 05, 2018, 01:53:56 AM
Well syncro,your video just proved i was right,and you are wrong.Inductive kickback is a direct current (DC),and what i said was that the generating effect of the spinning magnet is what was sending current back to the source,as the spinning magnet would produce a alternating current(AC) across the coil.
You just showed a video,where you were measuring alternating current,and trying to pass it of as the inductive kickback current, lol--whats the chances of that.
Enjoy.
@Tinman,
The inductive kickback measures around the same on the D.C. scale. I just measured the input current, and I'm processing a video of that right now. I'll upload it soon. The input current is around 200 mA D.C. at 12 volts. This gives us an input power of around 2.4 watts and an inductive backspike of 50 milli watts. That's right around a 20% recovery figure. I 've measured this COP in the past with my Bedini circuits and the ratio is the same, so I certain my figures are correct.
The good news is that the Inductive kickback directed to the ferrite core series bifilar coil through my 70uF 25 volt capacitor with no diode (The diode kills the effect) is generating a very powerful oscillation in the overhead ferrite magnet elastic band piston. The problem is, that it's at the same frequency as the R.P.M. of the spinner which measures around 3300 R.P.M. with my frequency meter. This oscillation frequency is too great for my commutator to keep up with; However, this amounts to real progress in my opinion. We have Magluvin to thank for this ingenious path of discovery.
Quote from: synchro1 on March 05, 2018, 06:48:54 AM
@Tinman,
I just measured the input current, and I'm processing a video of that right now. I'll upload it soon. The input current is around 200 mA D.C. at 12 volts. This gives us an input power of around 2.4 watts and an inductive backspike of 50 milli watts. That's right around a 20% recovery figure. I 've measured this COP in the past with my Bedini circuits and the ratio is the same, so I certain my figures are correct.
The good news is that the Inductive kickback directed to the ferrite core series bifilar coil through my 70uF 25 volt capacitor with no diode (The diode kills the effect) is generating a very powerful oscillation in the overhead ferrite magnet elastic band piston. The problem is, that it's at the same frequency as the R.P.M. of the spinner which measures around 3300 R.P.M. with my frequency meter. This oscillation frequency is too great for my commutator to keep up with; However, this amounts to real progress in my opinion.
QuoteThe inductive kickback measures around the same on the D.C. scale.
So,your meter reads the same amount of current on both the DC and AC setting--thats great :D
We have Magluvin to thank for this ingenious path of discovery.
Fantastic
You should both be up for your nobel prize any time soon.
Quote from: tinman on March 05, 2018, 07:34:43 AM
So,your meter reads the same amount of current on both the DC and AC setting--thats great :D
We have Magluvin to thank for this ingenious path of discovery.
Fantastic
You should both be up for your nobel prize any time soon.
@Tinman,
The inductive kickback is a "Low ripple A.C. or weak pulsed D.C." all depends how you look at it. You know what it's scope signature looks like. How's your flat tire repair business doing?
Here's a video of a D.C. input measured on the A.C. scale. I have another one that measures the same input on the D.C. scale that reads the same. The very powerful Neo sphere strongly influences the measurements.
https://www.youtube.com/watch?v=pMA1X-sjlz4 (https://www.youtube.com/watch?v=pMA1X-sjlz4)
Quote from: synchro1 on March 05, 2018, 07:40:55 AM
@Tinman,
The inductive kickback is a "Low ripple A.C. or weak pulsed D.C." all depends how you look at it. You know what it's scope signature looks like. How's your flat tire repair business doing?
Here's a video of a D.C. input measured on the A.C. scale. I have another one that measures the same input on the D.C. scale that reads the same. The very powerful Neo sphere strongly influences the measurements.
https://www.youtube.com/watch?v=pMA1X-sjlz4 (https://www.youtube.com/watch?v=pMA1X-sjlz4)
Syncro
There is no point in going any further in trying to explain what is going on with your DUT,as you have your mind made up.
Quote from: tinman on March 05, 2018, 08:22:56 AM
Syncro
There is no point in going any further in trying to explain what is going on with your DUT,as you have your mind made up.
@Tinman,
Here's a video of the input current measured on the D.C. scale:
https://www.youtube.com/watch?v=jkWTy_PyDaY (https://www.youtube.com/watch?v=jkWTy_PyDaY)
The value's fluctuating due to the erratic nature of the spinner, so it requires a "Round Off". The round number is the same with A.C. or D.C.! You see the OL on the 200mA setting, so I'm calling it 200.
We're dealing with
roughly 20% kickback return on input power with this setup.
There is probably a fairly 'good' reason there is the word 'Kick' in Inductive Kickback, otherwise it might be called inductive flowback, or just simply pushback. Kick has impact. A kick would not be something I would expect from an ac signal. There is nothing out there on that other than here in this thread for some 'odd' reason.
But, if sync is getting ac and has a bifi coil taking on an inductive kickback from the pulse motor, then that kick will get the bifi oscillating. Just like in my scope shots. Once that kick current charges the bifi capacitance, then that is the end of the kick and it is dissipated, leaving the bifi to oscillate. I did some other things that I have found to be pretty unique also. The bifi will go into oscilation with only 1 wire from the kicker coils output through the diode. And not just a little weak oscillation as one might expect in comparison to 2 wire connection loop. That was fairly loose coupling I had between the test coils and the pickup coil. So now is time to make a transformer with a bifi primary and drive that primary with the inductive kickback of another coil and see if things look promising. And a couple other weird things that I will get a better grip on and then show here.
Using the inductive kickback as a test of the single wire coil and the bifi was a very straight forward way of showing this effect as it either blocks the spike or takes it on. There is no way to dispute it. Its proven. Like Sync said, this one thing about bifi coils is part of a long ongoing argument between MH(not just him either) and me. There are differences between the 2 types of coils that are more than just a difference in resonant freq between the 2. So from here we move forward and see what we can do with this facet of the bifi coils. One thing Im thinking now, going for the most capacitance with your bifi may not always be the best bet. If we had a bifi made of thicker wire and less capacitance, the kickback spike could charge that lower capacitance to a much higher voltage. Trying some things on that. Will see..
What I am finding is that the bifilar does not impede the input 'like' a single wire coil does. If you have a circuit that uses a choke coil to suppress pulses to keep them from other parts of the circuit, and then you replaced that choke coil with a bifilar coil of the same wire, turns, etc, then I believe that it will no longer choke those problematic pulses, spikes, etc. So far it does things as tesla says.
Whether I choose to say disconnect the input, as I am doing in the circuit I had shown, none the less it is an interruption of the input to the kicker coil. If we wanted to stop a water source from a house, we could just shut the valve, without completely disconnecting the pipe from the house. In both cases we have interrupted the water from getting to the house. Show me just 1 article on the net that uses the words inductive kickback in a circuit that has an AC input to a circuit of a transformer, light bulb and a resistor. There are none. Just like there are none in that scope shot.
Mags
Quote from: tinman on March 05, 2018, 07:34:43 AM
So,your meter reads the same amount of current on both the DC and AC setting--thats great :D
We have Magluvin to thank for this ingenious path of discovery.
Fantastic
You should both be up for your nobel prize any time soon.
Sure as hell wont be you. ;)
Mags
My serial bifilar setup is accepting kickback and the magnetic oscillation is powering an overhead magnet piston GAP appendage that together act as a current amplifier. I tried measuring overall output, but my input current mesurements killed my Reed switch, so I'm out of business for the time being.
Quote from: Magluvin on March 05, 2018, 08:53:29 AM
Sure as hell wont be you. ;)
Mags
QuoteShow me just 1 article on the net that uses the words inductive kickback in a circuit that has an AC input to a circuit of a transformer, light bulb and a resistor. There are none. Just like there are none in that scope shot.
As this forum is on the net,and my AC circuit showing inductive spikes is on this forum,then you have been shown 1 article on the net showing inductive spikes in an AC circuit.
Once again,see scope shot below,which is full of inductive spikes.
I see you are also avoiding the original circuit i posted,as that shows a spike being sent to a bifi coil.
Inductive kickback is a term,where as inductive spike is a description.
QuoteUsing the inductive kickback as a test of the single wire coil and the bifi was a very straight forward way of showing this effect as it either blocks the spike or takes it on. There is no way to dispute it. Its proven.
That video was no proof of anything.
Once again i ask--what is the difference between a bifi coil,and a single wound coil with added capacitance across it?
How much energy was sent to each coil?
How much energy was wasted as heat in each coil?
What was the load on the pickup coil?--none as far as i could see.
So Mags,your video proves nothing at all--maybe the next one will be more informative.
I plan to generate inductive kickback with my battery core mechanical contact oscillator, then channel it into the serial bifilar and tiny capacitor to try and oscillate my GAP appendage that way. I hope to measure gain in a storage capacitor through a diode that way.
Tinman has to be right about Igor's "Reed Switch Spinner". Power from the magnet has to account for the near unity COP. There's no way the inductive kickback alone could supply that much power back to the source battery.
Quote from: tinman on March 05, 2018, 09:40:50 AM
As this forum is on the net,and my AC circuit showing inductive spikes is on this forum,then you have been shown 1 article on the net showing inductive spikes in an AC circuit.
Once again,see scope shot below,which is full of inductive spikes.
I see you are also avoiding the original circuit i posted,as that shows a spike being sent to a bifi coil.
Inductive kickback is a term,where as inductive spike is a description.
That video was no proof of anything.
Once again i ask--what is the difference between a bifi coil,and a single wound coil with added capacitance across it?
How much energy was sent to each coil?
How much energy was wasted as heat in each coil?
What was the load on the pickup coil?--none as far as i could see.
So Mags,your video proves nothing at all--maybe the next one will be more informative.
"As this forum is on the net,and my AC circuit showing inductive spikes is on this forum,then you have been shown 1 article on the net showing inductive spikes in an AC circuit.
Once again,see scope shot below,which is full of inductive spikes.
I see you are also avoiding the original circuit i posted,as that shows a spike being sent to a bifi coil.
Inductive kickback is a term,where as inductive spike is a description."
I see. Lol. So you here in this thread on the net are the only one. Like that counts. ::) ::) ::) ROTF
Avoiding the first circuit? I see no scope shot for that circuit. No vid, no nothing. Led lights up? Sweet and Yay. Good job. Hmm, no measurements? Well then you have shown nothing, as your basis for 'proving something' shows you are showing nothing there pretty much. Yet when I see the scope shot that is in reference to your AC input circuit, you keep jumping me back to the first pulsed circuit, as if that is the circuit that the scope shot you have shown is from that circuit. ??? ??? ??? ::) Dude its just getting weird man. Weird indeed.
"That video was no proof of anything."
Well then you got nothing from it. Some do, some dont. Thats what happens with age. Its happens to some. They may have a pill for that. ;) What it does show is something that you never knew of, and something other than you have ever shown. If you dont get that this is something that Tesla tells us in the patent that this coil can do compared to an single wire coil of the same wire, same total wire length , coil width, inner outer dia, then I guess thats your problem. That was the whole point of the vid was to show this one major difference between the 2 test coils that nobody here has shown yet. There will be more vids showing more of the details of what we can do with this.
Soo what? Do you think I am faking what I have claimed and have shown? Heck, your scope shot is so crammed together, how could we tell if those are inductive kickback spikes or just a mix of freq that the circuit provides from the ac input? Yet you say its there, hiding in the thickness of it all. Cool. Oh, sweet 21.3mv rms and 46mv rms. That must be your proof of something going on there. Your the winner! Sit back and have a beer! Your proof of something, what ever that is, has more merit than what I have shown! lol Whatever dude. Make yourself a trophy and sit on it.
"Once again i ask--what is the difference between a bifi coil,and a single wound coil with added capacitance across it?
How much energy was sent to each coil?
How much energy was wasted as heat in each coil?
What was the load on the pickup coil?--none as far as i could see."
Where is yours??? Hypocrite are you? ::) If you cannot understand that for this first vid it was just to visually show an effect that hasnt been shown here before, then I say its just you being troll. That vid demonstrates what tesla claimed with the bifilar coils. This is just you here 'trying' to one up me with your circuit and a scope shot that you have not even explained yet, just a distraction, and you have to come back after the first circuit and ask why hasnt anyone talked about your circuit yet. lol Thats funny. I suppose you are not getting the attn you wanted. Well I guess you need to do a bit better than what you have so far and maybe you wont have to come back and ask again, to make sure people are paying attn to you and your exciting circuit. ;) You have the good luck with those tactics to get people interested.....
"So Mags,your video proves nothing at all--maybe the next one will be more informative."
Well then it is your loss. And your showing of your circuits and scope shot 'prove' and show even less! If you cannot see that the single wire coil rejected the kicking coil spike as a normal coil should, and the bifilar coil accepted the spike and created a much better magnetic field output to the pickup coil than the single wire coil, then I guess you just cant. Cool beans. I will continue on with it and you go and continue on with your, whatever it is your trying to show.
You open a thread on your circuit and Ill open a thread on my idea and lets see whose thread gets more interest. I mean you are all 'Here, Here, look at my circuit and look at my scope shot!!' Well then open a thread and do the do. You stay in your thread and Ill stay in mine. If you want to comment on me in your thread or anywhere else on thins forum, fine, just stay on your side of the fence and Ill stay on mine. This all started with you sayin you thought I was on the right track and me asking you what that meant exactly. And here we are. You cant post a thing without poking me with nearly every post. Meanwhile you have shown didly, according to your terms of what proves something or not. I have explained everything that I intended to show 'so far' with what I have presented. You show 3 pics and zero explanation, no load or input measurements, etc. and I suppose we are just all suppose to guess at what you are getting at and expect commentary and praise.. Well so far Im at a total loss along with many others with your circuit as you have explained nothing... I see Erfinder seems to know what your circuit may be doing and I bet he knows more about it than you. ;) Id bet on that. He and I have talked for some time and he IS above you(including me and others here) in nearly all this stuff ;D Maybe he can shed some light on it, being you dont. He has opened my mind up to certain things that has helped me figure out a thing or two about a thing or two. :-X ;D And I had thanked him for just about every conversation we had. ;)
Mags
@Tinman,
Here's a quote from Tesla's patent: It states that the capacitive energy stored in the serial bifilar coil is 250,000 times greater then a single wire coil of the same inductance. I fail to see how adding a capacitor to a single wire coil can achieve this kind of result.
"Consider a single wire pancake coil with 1000 turns and 100 volts applied. There will be 0.1 volt difference between each consecutive layer in the coil. The capacitance of the coil will be mainly determined by the geometry (plate spacing) between each consecutive coil layer and is constant for a given coil. The capacitive energy stored in such a coil will be ~ E1 = 1/2 C V^2, or 1/2 C (0.1 ^2). On the other hand if you wind a similar bifilar pancake coil with 1000 turns (same total length of wire) and again apply 100 volts then the midpoint solder connection will be at 50 volts. Now the voltage difference between each consecutive layer in the coil has been increased from 0.1 volt to 50 volts. The capacitance of the coil determined by the geometry hasn't changed; However, the capacitive energy stored in such a bifilar wound coil will be ~ E2 = 1/2 C V^2, or 1/2 C (50 ^2). Or 250,000 times greater than the single wire wound coil of the same inductance".
@Tinman,
I'm involved in serious testing. I connected a 70uF 25 volt capacitor in series between the positive output of my reed switch pulse coil and the beginning of my ferrite core serial bifilar coil and achieved a strong shudder in the elastic spring tensioned magnet piston held inside the attraction zone from the inductive kickback. I can't imagine how I would accomplish this with a second capacitor in the way.
The "shudder" I'm referring to is generating current inside the serial bifilar coil that is a direct result of the amplification of a resonant oscillation between the serial bifilar windings. Nothing like this is going on inside the single wire coil. It's dead, not alive and singing like Tesla's bifilar. When are you gonna get off the rag, huh Bub?
Quote from: synchro1 on March 05, 2018, 03:06:57 PM
@Tinman,
Here's a quote from Tesla's patent: It states that the capacitive energy stored in the serial bifilar coil is 250,000 times greater then a single wire coil of the same inductance. I fail to see how adding a capacitor to a single wire coil can achieve this kind of result.
"Consider a single wire pancake coil with 1000 turns and 100 volts applied. There will be 0.1 volt difference between each consecutive layer in the coil. The capacitance of the coil will be mainly determined by the geometry (plate spacing) between each consecutive coil layer and is constant for a given coil. The capacitive energy stored in such a coil will be ~ E1 = 1/2 C V^2, or 1/2 C (0.1 ^2). On the other hand if you wind a similar bifilar pancake coil with 1000 turns (same total length of wire) and again apply 100 volts then the midpoint solder connection will be at 50 volts. Now the voltage difference between each consecutive layer in the coil has been increased from 0.1 volt to 50 volts. The capacitance of the coil determined by the geometry hasn't changed; However, the capacitive energy stored in such a bifilar wound coil will be ~ E2 = 1/2 C V^2, or 1/2 C (50 ^2). Or 250,000 times greater than the single wire wound coil of the same inductance".
No... that is not a quote from Tesla's Patent 512,340. Yet Again, we see how you misrepresent, misquote, misinterpret and even flat-out lie about the work of others. So we may only presume that this applies even more to your own work.
Here is the exact quote FROM TESLA'S PATENT #512,340 (emphasis mine):
QuoteLet A, Fig. 1, designate any given coil the spires or convolutions of which are wound upon and insulated from each other. Let it be assumed that the terminals of this coil show a potential difference of one hundred volts, and that there are one thousand convolutions; then considering any two contiguous points on adjacent convolutions let it be assumed that there will exist between them a potential difference of one-tenth of a volt. If now, as shown in Fig. 2, a conductor B be wound parallel with the conductor A and insulated from it, and the end of A be connected with the starting point of B, the aggregate length of the two conductors being such that the assumed number of convolutions or turns is the same, viz., one thousand, then the potential difference between any two adjacent points in A and B will be fifty volts, and as the capacity effect is proportionate to the square of this difference, the energy stored in the coil as a whole will now be two hundred and fifty thousand as great.
That large number applies ONLY to a coil with the input parameters as Tesla stated: A 1000 turn flat pancake coil with a voltage drop of 100 volts across it. Tesla is here just pulling the input numbers out of his hat, in order to illustrate his point. To claim from this that "any" series connected bifilar pancake coil has "250,000 times more energy storage" when that only applies to the specific parameters Tesla stated here -- is simply false, and in fact Tesla could have used different input numbers and come up with a wildly different final result.
Please construct such a coil, apply whatever voltage you need to get a 100 volt drop across it, and report your results.
Now apply the same calculation to a set of practical coils of, say, 200 turns with an _actual_ voltage drop across them of 2 volts and see what you get.
"The capacitance of the coil determined by the geometry hasn't changed" -- this language does not appear at all in the actual patent, and in fact is false. The whole point of Tesla's patent is that the "capacitance determined by the geometry" HAS changed.
Furthermore, another exact quote from Tesla's patent is as follows:
QuoteI have found that in every coil there exists a certain relation between its self-induction and capacity that permits a current of given frequency and potential to pass through it with no other opposition than that of ohmic resistance, or, in other words, as though it possessed no self-induction. This is due to the mutual relations existing between the special character of the current and the self-induction and capacity of the coil, the latter quantity being just capable of neutralizing the self-induction for that frequency. It is well-known that the higher the frequency or potential difference of the current the smaller the capacity required to counteract the self-induction; hence, in any coil, however small the capacity, it may be sufficient for the purpose stated if the proper conditions in other respects be secured. In the ordinary coils the difference of potential between adjacent turns or spires is very small, so that while they are in a sense condensers, they possess but very small capacity and the relations between the two quantities, self-induction and capacity, are not such as under any ordinary conditions satisfy the requirements herein contemplated, because the capacity relatively to the self-induction is very small.
In every coil... and in any coil.
Quote from: synchro1 on March 05, 2018, 04:10:27 PM
@Tinman,
I'm involved in serious testing.
(snip)
When are you gonna get off the rag, huh Bub?
You earned another ROFL for that one.
Quote from: TinselKoala on March 05, 2018, 05:29:51 PM
You earned another ROFL for that one.
@Tinselkoala,
So what if it's not an exact quote? It's the quote rephrased in modern English. Why not run another one of your banal constant current tests? You never once tried the kind of current interruption test I suggested to demonstrate the power of "Impulse Magnetization" smOky2 demonstrated with his PMH video links on the SMOT thread and Mags just got through showing us with his scope shots. What do you make of Magluvin's latest test results Mr. Webster?
Quote from: Magluvin on March 05, 2018, 02:07:27 PM
"As this forum is on the net,and my AC circuit showing inductive spikes is on this forum,then you have been shown 1 article on the net showing inductive spikes in an AC circuit.
Once again,see scope shot below,which is full of inductive spikes.
I see you are also avoiding the original circuit i posted,as that shows a spike being sent to a bifi coil.
Inductive kickback is a term,where as inductive spike is a description."
I see. Lol. So you here in this thread on the net are the only one. Like that counts. ::) ::) ::) ROTF
Avoiding the first circuit? I see no scope shot for that circuit. No vid, no nothing. Led lights up? Sweet and Yay. Good job. Hmm, no measurements? Well then you have shown nothing, as your basis for 'proving something' shows you are showing nothing there pretty much. Yet when I see the scope shot that is in reference to your AC input circuit, you keep jumping me back to the first pulsed circuit, as if that is the circuit that the scope shot you have shown is from that circuit. ??? ??? ??? ::) Dude its just getting weird man. Weird indeed.
"That video was no proof of anything."
Well then you got nothing from it. Some do, some dont. Thats what happens with age. Its happens to some. They may have a pill for that. ;) What it does show is something that you never knew of, and something other than you have ever shown. If you dont get that this is something that Tesla tells us in the patent that this coil can do compared to an single wire coil of the same wire, same total wire length , coil width, inner outer dia, then I guess thats your problem. That was the whole point of the vid was to show this one major difference between the 2 test coils that nobody here has shown yet. There will be more vids showing more of the details of what we can do with this.
Soo what? Do you think I am faking what I have claimed and have shown? Heck, your scope shot is so crammed together, how could we tell if those are inductive kickback spikes or just a mix of freq that the circuit provides from the ac input? Yet you say its there, hiding in the thickness of it all. Cool. Oh, sweet 21.3mv rms and 46mv rms. That must be your proof of something going on there. Your the winner! Sit back and have a beer! Your proof of something, what ever that is, has more merit than what I have shown! lol Whatever dude. Make yourself a trophy and sit on it.
"Once again i ask--what is the difference between a bifi coil,and a single wound coil with added capacitance across it?
How much energy was sent to each coil?
How much energy was wasted as heat in each coil?
What was the load on the pickup coil?--none as far as i could see."
Where is yours??? Hypocrite are you? ::) If you cannot understand that for this first vid it was just to visually show an effect that hasnt been shown here before, then I say its just you being troll. That vid demonstrates what tesla claimed with the bifilar coils. This is just you here 'trying' to one up me with your circuit and a scope shot that you have not even explained yet, just a distraction, and you have to come back after the first circuit and ask why hasnt anyone talked about your circuit yet. lol Thats funny. I suppose you are not getting the attn you wanted. Well I guess you need to do a bit better than what you have so far and maybe you wont have to come back and ask again, to make sure people are paying attn to you and your exciting circuit. ;) You have the good luck with those tactics to get people interested.....
"So Mags,your video proves nothing at all--maybe the next one will be more informative."
Well then it is your loss. And your showing of your circuits and scope shot 'prove' and show even less! If you cannot see that the single wire coil rejected the kicking coil spike as a normal coil should, and the bifilar coil accepted the spike and created a much better magnetic field output to the pickup coil than the single wire coil, then I guess you just cant. Cool beans. I will continue on with it and you go and continue on with your, whatever it is your trying to show.
You open a thread on your circuit and Ill open a thread on my idea and lets see whose thread gets more interest. I mean you are all 'Here, Here, look at my circuit and look at my scope shot!!' Well then open a thread and do the do. You stay in your thread and Ill stay in mine. If you want to comment on me in your thread or anywhere else on thins forum, fine, just stay on your side of the fence and Ill stay on mine.
This all started with you sayin you thought I was on the right track and me asking you what that meant exactly. And here we are. You cant post a thing without poking me with nearly every post. Meanwhile you have shown didly, according to your terms of what proves something or not. I have explained everything that I intended to show 'so far' with what I have presented. You show 3 pics and zero explanation, no load or input measurements, etc. and I suppose we are just all suppose to guess at what you are getting at and expect commentary and praise.. Well so far Im at a total loss along with many others with your circuit as you have explained nothing... I see Erfinder seems to know what your circuit may be doing and I bet he knows more about it than you. ;) Id bet on that. He and I have talked for some time and he IS above you(including me and others here) in nearly all this stuff ;D Maybe he can shed some light on it, being you dont. He has opened my mind up to certain things that has helped me figure out a thing or two about a thing or two. :-X ;D And I had thanked him for just about every conversation we had. ;)
Mags
Hey Mags,your right,i shouldnt be making any negative comments towards your work.
So i here by apologise.
Brad
Quote from: Erfinder on March 06, 2018, 04:55:41 AM
When considering Tesla's coil for electromagnets and contemplating its "intended" use, one must consider the quality of the current directed into the coil, the geometry of the coil into which the current is to being directed, and ultimately, the relation the one has to the other.
What we see is not what we get. What I mean is, what we read in the patent about the coil is not the message.... the coil and everything said about it is a red herring. What is the message?
Regards
This is not the movie Natiolan Tresure (https://en.wikipedia.org/wiki/National_Treasure_(film)). We don't have time to fly all over the globe finding secret messages behind paintings. I get that you are trying make people to think and "teach them how to fish", but we are in the boat with our lines and hooks already. If you are the guy with worms, it does not help to keep the worms in your pants and go "Where is the worm? What does the worm mean exactly? What did Tesla mean when he said 'put the worm in the hook'? "
And it is not like there is a riddle or two, but it is every freaking message in every thread!
Quote from: Erfinder on March 06, 2018, 08:23:03 AM
You are not the voice of the body of seekers.
Yeah but I kinda am. I've seen people like you in every walk of life. Arrogant, condescending and full of shit. There is just the outer shell that fools somebody, but no content inside. Just a show, but no dinner. It just cries out "mother & abandonment issues" every time I see it.
You prance around like a Greek poet and some blind follow you around and take a deep bow, if they happen to accidentally cross your path.
Where is the input that you are giving this thread? I mean actual input and not a post where you state that you are great and all others are shit.
"It will be
too late when you finally realize that you cannot
drink from the sea, nor receive sustenance
therefrom. Mutiny is inevitable, followed by
cannibalism. In this scenario, the captain doesn't
have the option of going down with the ship, nor does he have a choice in the matter of his
placement on the menu."
like.^
Quote from: Erfinder on March 06, 2018, 09:11:07 AM
I am going to answer this question with a question.... can you read?
Got you there game, set and match.
And 1 by 1,they start to learn
;)
Hey Erfinder.
If you can, maybe we can skype later.
Mags
Not you EF or synchro1 -
The rest sad bunch - I posed this some time ago placing a cap across a coil and even began instructions as to how I did it what to balance in the coil, cap, and spike as well as several methods that can be used to harness... only to be laughed off the stage - hahaha
you all go at your own pace...
here is one of the first vids in the bunch:
https://www.youtube.com/watch?v=41cfHcb7qd8
Quote from: minoly on March 06, 2018, 05:28:08 PM
Not you EF or synchro1 -
The rest sad bunch - I posed this some time ago placing a cap across a coil and even began instructions as to how I did it what to balance in the coil, cap, and spike as well as several methods that can be used to harness... only to be laughed off the stage - hahaha
you all go at your own pace...
here is one of the first vids in the bunch:
https://www.youtube.com/watch?v=41cfHcb7qd8
I have asked twice now in this thread-what is the difference between having a bifi coil collect the inductive spike,and a single wire coil with a cap across it ?.
All declined to answer.
Quote from: minoly on March 06, 2018, 05:28:08 PM
Not you EF or synchro1 -
The rest sad bunch - I posed this some time ago placing a cap across a coil and even began instructions as to how I did it what to balance in the coil, cap, and spike as well as several methods that can be used to harness... only to be laughed off the stage - hahaha
you all go at your own pace...
here is one of the first vids in the bunch:
https://www.youtube.com/watch?v=41cfHcb7qd8 (https://www.youtube.com/watch?v=41cfHcb7qd8)
@minoly,
You're the guy I was talking about. I searched high and low for your videos and finally gave up. I'm glad you posted them for us over here. Thank you.
Quote from: tinman on March 06, 2018, 06:21:30 PM
I have asked twice now in this thread-what is the difference between having a bifi coil collect the inductive spike,and a single wire coil with a cap across it ?.
All declined to answer.
Assuming you kept the impedance the same on both
and the input was the same
you would have an increase in magnetic field in one direction
and would lose an equivalent in electricity on the way out
input would be similar to the output, minus losses
In a transformer, this can have technical advantages
a secondary with a lower impedance would harness more effieicntly
than it would from the field of the monofilar coil
A capacitor performs a similar function with the charges
on its plates.
The simple answer is [sqrt](resistance^2 + reactance^2)
In both cases, reactance is 1/2(pi)(freq)(capacitance)
In the capacitor, R is very small, so it is mostly the capacitance
that governs the equation.
Ok, the bifi measures 15.7nf between turns but the cap needed for the normal coil is 4.3nf and the resonant freq foir each comes out to 12.37khz and 12.88khz for each. As close as I can get for now with the bunch of little caps I have on hand. It was just a small disk cap for the 4.3nf to get the freq that close. So Im expecting there to be similar results with the kicker circuit like Tk says. But I got tied up into some things with getting the 2 coils to be a similar as possible with at least the resonant freq with the right cp on the normal coil, so didnt get to doing that yet.
This was just energizing the test coils directly with everything being even simpler, and they both look very similar in the 1st 2 scope shots with .5v input and the bifi tends to be a bit higher amplitude. The normal coil shot is a very uniform trumpet ring down, but the bifi has a bit of a swing to it and the second down stroke of the wave seems to be clipped earlier than the first swing down. Now when i go to 5v input, things change dramatically as seen in the second set of shots below. The normal coil with the disk cap shot seems to show lots of issues, yet the bifi seems to be that same shape as the first shot and a lot higher p-p than the normal coil. The cap is 500v. Cant see that as the issue for the bad performance of the normal coil as I have a few of those caps and they all perform the same in the circuit.. I did it over again and again. The higher the input, the worse the normal coil acted in comparison to the bifi. Could the windings insulation be breaking down? If so, Id think possibly if there were sparking between windings that there might be noise. Will have to go over it all again tomorrow.
Planned to do a lot more than this tonight, but Im beat now. I have to see what is the problem or reasoning for these results of what should be simple things. Tomorrow Ill have time to do a vid demonstrating the use of the switch flicking and see what might be going on with this setup. The switch flick is still working well this evening. The kicker coil and the diode are off the board and out of the circuit. so its not anything else affecting the circuit. Even reversed the dpdt switch around the other way and the same results. Doesnt make sense. .5v in pretty close results. 5v in, pretty crappy results. The only thing that changed was the voltage in. Like the diode cant be breaking down, the voltage across the coil never gets as high as the bifi shot, of which it should cause the same issue if the diode were bad.
Mags
Quote from: Erfinder on March 07, 2018, 01:59:12 AM
Mags,
If I may make a suggestion, go back to your original experiment..... Input from others is well....input from others.... you were onto it, I wanted to comment but stayed in my place.... allowing you to do your thing without interference... don't let anyone take your off your path.
Thanks. ;)
Watch this PMH video
https://youtu.be/mLd_py1xTO8 (https://youtu.be/mLd_py1xTO8)
Under fairly controlled conditions,i compared a bifilar coil against a mono coil.
As i first expected,i found no difference between the two. In fact,the mono performed a little better in this case.
https://www.youtube.com/watch?v=lgTmKrLCWI4
Excellent demo on how NOT to see the effect... people should pay close attention as Edison found thousands of ways NOT to make a lightbulb...
Quote from: minoly on March 07, 2018, 11:16:57 AM
Excellent demo on how NOT to see the effect... people should pay close attention as Edison found thousands of ways NOT to make a lightbulb...
Hi Minoly, i must agree with you , not a good way to show the difference between the two types of coil .
It seems to me a bit ambiguous say that the single coil perform better than bifilar coil, based on the configuration that Tinman provide in their video demo.
Other points should must considered, like the shape of pulse used to pulse the coils, the frequency used, geometry of coil and so on.
What does this video try to prove? I really can not understand the purpose of the configuration used in the video .
Tinman can you clarify how you are pulsing the coil just to i try follow you ? Are you using a function generator directly in the transistor pulsing a primary coil ?
Are you using any kind of diode to catch the collapse field in the test coils ?
I Will appreciate if you can clarify myself .
Thanks
Quote from: nelsonrochaa on March 07, 2018, 01:55:52 PM
Hi Minoly, i must agree with you , not a good way to show the difference between the two types of coil .
It seems to me a bit ambiguous say that the single coil perform better than bifilar coil, based on the configuration that Tinman provide in their video demo.
Other points should must considered, like the shape of pulse used to pulse the coils, the frequency used, geometry of coil and so on.
What does this video try to prove? I really can not understand the purpose of the configuration used in the video .
Tinman can you clarify how you are pulsing the coil just to i try follow you ? Are you using a function generator directly in the transistor pulsing a primary coil ?
Are you using any kind of diode to catch the collapse field in the test coils ?
I Will appreciate if you can clarify myself .
Thanks
I feel the same. If the coils are as identical as he says, then there should be an equal outcome. And whats with the 20v on the scope he keeps pointing to, but it never changes. Always 20.00v dead nuts it seems whether it is testing the normal coil or the bifi. Watched it twice and Im at a somewhat of a loss here with you guys. ??? i take the scope shots and zoom into ground zero to see things close up and in detail.
It seems he is posting it as a counter to my vid, as he keeps mentioning me in the vid. Personally I dont believe his results either. My simplistic way of approaching and showing make it very easy to see what is going on. Anyone can put it together and play. Need the scope at least. I had shown the very basic circuit successfully with the simplest of switching. I had gone over the issues of using the manual switch when the input voltages reach higher levels than I started with. At 20v the noisy switching that happens more often was seen on the scope and I pointed it out and also indicated by my stereo amplifier popping the speakers each time it occurred. So I suppose if I do a loaded pickup coil, and my results oppose his, then it is mine that is still wrong again I will venture to say..... And for the most part, not all or even most of my switching is noisy. Even TK found it remarkable. ;D ;D ;D But will be getting a transistor circuit going here this evening, not just to satisfy others. More to go into this deeper..... ;) I think my very primitive setup did quite well and has shown that there are differences between a bifi coil and a similar 1 wire coil, and especially a 1 wire coil with the added capacitance, as TK requested I do in comments on my first vid, caveman style. ;D At least we 'should not' have to hear that there is no difference any longer. But I have a growing feeling we will. ;)
Was on the phone with someone and they seemed to mostly want to see in vs out. For this first part Im just showing what I believe are interesting things going on here with this stuff that so far may be useful to us. New things to work into our projects is always a good thing. How can we even approach trying to put something together for complete testing if we dont know or fully understand the basic workings and differences of these particular components? like if i had only gone and compared the test coils at the initial .5v in, then I may have just walked away thinking, yeah, there really seems to be no difference and just conclude that they were right. but it didnt happen that way. ;) I am hoping these things Im showing will help others. I dont hide it or fake it. I show and I divulge info freely. Im not looking for a trophy to sit on. I just want to see happy people. ;D
More to come.
Mags
These guy's kill me. I've been watching all these failed attempts to get the bifilar to work for five or six years now. They always make the same mistake. They test the bifilar dead. The bifilar coil has a quarter million times the capacitance as the single wire coil between the windings. How many times has everyone listened to me repeat myself.
The serial bifilar coil needs to be charged before it will work.
Tinman needs to shock charge his serial bifilar coil by shorting the coil electrodes across two hot battery leads, like we do when we lock a Leedskalnin PMH, causing a large spark to jump. This brings it's enormous capacitance to high potential. The coil begins to audibly ring and spontaneously generates a powerful magnetic field.
Magluvin ran a series of tests on his bifilar, shock charging it with BEMF before he grabbed his video. Tinman just starts off cold with an inert coil and gets the same tedious no results all these other failed testers show. No difference! I get angry with them and grow insulting and Augustus Snodgrass rings off his there district fire alarm. Same bullshit routine.
@Magluvin,
I uploaded a capacitor charge curve last year on Evostar's thread and compared the 66.33 bottle neck to a water jug setting a flame war off with TK that led to the negative inductance feud.
We're discussing "Inductor Capacitance" which is different from Henries of inductance. Here's another statement for you "Seventh Level Druids" to puzzle through:
Shock charge the bifilar to 33 percent "Negative Capacitance", the deep well zone for acceptance of BEMF.
Do you guys have any idea how far ahead of you I am with my latest invention?
The "Negative Henry" is a measure of impedance, which indicates the presence of magnetic energy. "Negative Capacitance" is a measure of "Inductor Capacitance Charge". Inductor negative capacitance also has a magnetic field equivalency.
Quote from: minoly on March 07, 2018, 11:16:57 AM
Excellent demo on how NOT to see the effect... people should pay close attention as Edison found thousands of ways NOT to make a lightbulb...
Oh,i see.
So clean switching,and controlled experiments are out of the question here?
We also have to make sure that the receiving bifi coil is doing no work,in order to have a successful test,and see the wonderful high amplitude self oscillating bifi coil.
Let me guess--
If my results were the same as those that Mag's got,my test would have been an excellent demonstration of how much better a bifi coil is than a single wound coil--right? ::)
Isnt it odd that no one seems to use this wonderful bifi coil in any application today. I mean,if it is so much better,then why not use them in transformers,or maybe even winding configurations in electric motors.
Even in the wonderful Tesla coils,you do not see bifi coils used--both the primary and secondary coils are single wound coils.
Even Bedini him self never used a bifi configuration in his !ever famous! pulse motors.
Last but not least--some here need to understand the difference between an electric field induced EMF,and a magnetically induced current,and it would seem that you are one of those people minoly.
Brad
author=Magluvin link=topic=16203.msg517930#msg517930 date=1520455108]
QuoteI feel the same. If the coils are as identical as he says, then there should be an equal outcome.
Are yours not the same Mag's?
Quote post 305: The identical(practically) single wire and bifi coils, left and right respectively, are 43 ohm and 35.7mh.
So,with my identical coil's,i should get an equal outcome,but with your identical coil's(your own words),you get a different outcome--how wonderful ::)
QuoteAnd whats with the 20v on the scope he keeps pointing to, but it never changes.
That is the RMS value of the inductive kickback,which remains the same with both coils.
QuoteWatched it twice and Im at a somewhat of a loss here with you guys. ??? i take the scope shots and zoom into ground zero to see things close up and in detail.
Your scope shot's show nothing more than an open oscillating coil,where as mine show the actual receiving coil doing work.
So unless you just plan on your coil doing no work,then your scope shots mean nothing much at all.
QuoteIt seems he is posting it as a counter to my vid, as he keeps mentioning me in the vid.
In actual fact Mag's,i spent the time,and carried out the test because i wanted to see your results for my self.
I only mentioned you because this was my replication of your test,but where i place a load on the receiving coil.
QuotePersonally I dont believe his results either.
I see. So you think that some how(even though i kept the camera rolling throughout the whole test),that i faked the test?.
Let me guess--if my results turned out to be the same as yours,then my test would have been a great success ? ::)
QuoteMy simplistic way of approaching and showing make it very easy to see what is going on.
And my test was more controlled,and applied in real world applications,where work was being done by each coil,where as yours was using a contact switch,where arcing would have been taking place,and your receiving coil was doing no work,and there for neither of your coils(bifi and single wound) were doing any work.
QuoteNew things to work into our projects is always a good thing. How can we even approach trying to put something together for complete testing if we dont know or fully understand the basic workings and differences of these particular components?
What you first need to understand,is the difference between an electric field induced EMF across a receiving coil,and a magnetically induced current through a receiving coil.
QuoteSo I suppose if I do a loaded pickup coil, and my results oppose his, then it is mine that is still wrong again I will venture to say.....
Is that not what i am getting now?--my results are opposite your's,and so you say that- Quote: Personally I dont believe his results either
So you suggest that i will do exactly what you have done ::)
QuoteI think my very primitive setup did quite well and has shown that there are differences between a bifi coil and a similar 1 wire coil
And my controlled test,where work is being done,was junk--right? ::)
Quoteand especially a 1 wire coil with the added capacitance, as TK requested I do in comments on my first vid, caveman style. ;D At least we 'should not' have to hear that there is no difference any longer. But I have a growing feeling we will. ;)
The question is--will there be a difference when the two coils are made to do work?
Have you been looking at the electric field induced EMF across your pickup coil,or have you been looking at the magnetically induced current through your pickup coil?
Have you confused your self between the two?
Have you decided that the higher EMF value across the pickup coil,is a result of a stronger magnetic field from the bifi coil to that of the single wound coil?.
It would seem that you (Mags) have accused me of doing the very same that you think i would accuse you off,as seen above in your own words.
It would seem that your coils can be identical (your own words),and yield different result's,where as mine should yield the same result,because they have the same resistive and inductive value's--WTF ::).
Well one thing is clear Mag's--your not interested in anyone's result's unless they are the same as yours.
Brad
@Tinman,
I mean it Brad I really believe you need to find help for your chronic psychosis. Read this comment:
When the Tesla serial bifilar solenoid is fully charged it goes into self resonance. The coil transforms into an LC tank that has a frequency determined by the ratio of the inductance of the coil in Henries and the capacitance between the windings in Farads. All one needs to do to sustain the oscillation is replace the power lost to resistance, which is minimal.
When the field begins to collapse in the bifilar inductor, the elastic held ferrite magnet piston is pulled away from the coil's ferrite core. The power of the magnet field is stored as an electrical charge between the windings. When the charge reaches it's maximum level, it returns to the inductor as a magnetic field and the overhead magnet piston is once again attracted back towards the coil. This work is free of charge, minus the small amount of loss we need to replace with the inductive kickback. There's no size limit on the serial bifilar or the magnet piston. All one needs to do is run the magnet piston up and down inside an output coil and voila, FREE POWER for the people forever and ever.
Quote from: synchro1 on March 07, 2018, 04:44:37 PM
These guy's kill me. I've been watching all these failed attempts to get the bifilar to work for five or six years now. They always make the same mistake. They test the bifilar dead. The bifilar coil has a quarter million times the capacitance as the single wire coil between the windings. How many times has everyone listened to me repeat myself.
The serial bifilar coil needs to be charged before it will work.
Tinman needs to shock charge his serial bifilar coil by shorting the coil electrodes across two hot battery leads, like we do when we lock a Leedskalnin PMH, causing a large spark to jump. This brings it's enormous capacitance to high potential. The coil begins to audibly ring and spontaneously generates a powerful magnetic field.
The only comedy show here syncro--is you.
QuoteMagluvin ran a series of tests on his bifilar, shock charging it with BEMF before he grabbed his video. Tinman just starts off cold with an inert coil and gets the same tedious no results all these other failed testers show. No difference! I get angry with them and grow insulting and Augustus Snodgrass rings off his there district fire alarm. Same bullshit routine.
You must have been drinking to much coolaid syncro.
I was sending 300 volt pulses to the bifi and single wound coil's--how much shock do you want?
How about you come and hold onto the output of my inductive kickback coil,and we'll see how much of a shock you get.
And how about you lay the fuck off TK,and stop being an ass wipe-->you know TK would wipe the floor with you,when it comes to anything electrical.
Everything you have posted is garbage--you dont even know the difference between inductive kickback and alternating current,and yet you think you have the right to rubbish people like TK,who are years ahead of you in everything in the EE world.
Brad
Quote from: tinman on March 07, 2018, 06:24:22 PM
snip....
I mean,if it is so much better,then why not use them in transformers,or maybe even winding configurations in electric motors.
snip...
I ask myself the same question. Especially since my last little motor build using air core bi-fi coils is so much more efficient than any pulse motor I've built in the last 30 years.
Cheers
@Magluvin,
Look Mags, this Tinman guy is using the "Fuck Off" vulgarity. I want you to have Stephan place him on moderated status.
Quote from: synchro1 on March 07, 2018, 07:30:32 PM
@Tinman,
I mean it Brad I really believe you need to find help for your chronic psychosis. Read this comment:
When the Tesla serial bifilar solenoid is fully charged it goes into self resonance. The coil transforms into an LC tank that has a frequency determined by the ratio of the inductance of the coil in Henries and the capacitance between the windings in Farads. All one needs to do to sustain the oscillation is replace the power lost to resistance, which is minimal.
When the field begins to collapse in the bifilar inductor, the elastic held ferrite magnet piston is pulled away from the coil's ferrite core. The power of the magnet field is stored as electrical charge between the windings. When the charge reaches it's maximum level, it returns to the inductor as a magnetic field and the overhead magnet piston is once again attracted back towards the coil. This work is free of charge, minus the small amount of loss we need to replace with the inductive kickback. There's no size limit on the serial bifilar or the magnet piston. All one needs to do is run the magnet piston up and down inside an output coil and voila, FREE POWER for the people forever and ever.
As i said in my last post--you post garbage.
You have no idea of what you are reading--it is beyond you.
Learn the difference between the electric field,and the magnetic field.
Quote from: hoptoad on March 07, 2018, 07:36:31 PM
I ask myself the same question. Especially since my last little motor build using air core bi-fi coils is so much more efficient than any pulse motor I've built in the last 30 years.
Cheers
How did you determine the efficiency of your pulse motor hoptoad?.
Quote from: tinman on March 07, 2018, 07:38:50 PM
As i said in my last post--you post garbage.
You have no idea of what you are reading--it is beyond you.
Learn the difference between the electric field,and the magnetic field.
@Tinman,
I'll come down there and kick your ass you dick sucker.
Quote from: synchro1 on March 07, 2018, 07:36:35 PM
@Magluvin,
Look Mags, this Tinman guy is using the "Fuck Off" vulgarity. I want you to have Stephan place him on moderated status.
You need to be careful syncro,as it wont be me that will be getting moderated.
I believe you have been there a couple of times before--for undue harassment of other members--such as you are doing now to TK.
Mags has already had to edit some of your posts for that very reason,and although myself and Mags may not be seeing eye to eye on this subject,he will still remain on the right side of the tracks.
Quote from: synchro1 on March 07, 2018, 07:41:39 PM
@Tinman,
I'll come down there and kick your ass you dick sucker.
Ah,a physical threat toward me lol
Be my guest.
Address is 5 Zanadu Court,Gelorup,Western Australia.
Better bring some mates with you as well.
See you when you get here.
Quote from: tinman on March 07, 2018, 07:51:06 PM
Ah,a physical threat toward me lol
Be my guest.
Address is 5 Zanadu Court,Gelorup,Western Australia.
Better bring some mates with you as well.
See you when you get here.
@Tinman,
Listen dick head. 300 volts ain't gonna cut it. You need current to generate a magnetic field. The current surge from shorting across battery electrodes generates a powerful magnetic field and the collapse charges the bifilar coil. 300 volts can't generate any magnetism fool!
author=synchro1 link=topic=16203.msg517953#msg517953 date=1520472501]
So,Mags was using how many volts?--i was using 7.2
Mag's push button switch is different from a transistor being used as a switch how?
My power supply is different to a battery how?
Was Mags using a power supply or battery?
QuoteYou need current to generate a magnetic field.
Well duh ::)
And you need an electric field to produces an EMF across a pickup coil--is it sinking in yet?
Quote300 volts ain't gonna cut it.
What was the kickback voltage value in Mag's test?
QuoteListen dick head.The current surge from shorting across battery electrodes generates a powerful magnetic field and the collapse charges the bifilar coil.
Which is exactly what i was doing--> Dickhead,only i used a power supply like Mag's did.
Quote300 volts can't generate any magnetism fool!
Sooo,300 volts dropped across a coil with only 3.2 ohms resistance,and an inductive value of only 1.2mH,isnt going to produce any magnetic field ? ::)
So the power being dissipated from the 100 ohm resistor on the pickup coil-came from where?
Like i said--you are clueless.
My coil capacitance measures 3.33 uF, the inductance 78mH and the electrolytic capacitor 70uF. The resonant frequency of the bifilar coil is 312 Hertz and the resonant frequency of the LC tank 70 Hertz. This yields a co-efficient of 265 Cycles per minute. Look at the video again and tell me from counting the vibrations in the cork cocktail coaster if it's in that range or not.
author=synchro1 link=topic=16203.msg517953#msg517953 date=1520472501]
Quote@Tinman,
Listen dick head.
Unlike others Syncro (although i think your syncro's are shot,and you are nothing more than a crash box now),im happy to take your insults all day-every day,as it's nothing more than dribble.
So feel free to carry on,you have my full permission to waffle on as much as you like about me,say what ever you want about me,it wont bother me a bit,as i am enjoying reading your funny post's about how electric and magnetic fields work,and which way the inductive kickback current flows through an inductor.
I also love how you measure current flow with your magic DMM,that reads the same value on both AC and DC :D
Your inability to understand how an alternating current can be produced by a spinning ball magnet,is second to none.
The fact that you dont understand that your reed switch will be closed twice per revolution of the spinning ball magnet is also comical ;D
So carry on syncro--your doing a marvelous job there. ::)
@Tinman,
Just shut you shithead.
The R.P.M. of the Reed switch spinner needs to match the resonance of the bifilar to pump the magnet piston. If the ball speeds up to fast the piston oscillation stops. The resonant frequency of the bifilar coil alone is over 18 thousand cycles per minute, so the addition of the 70uF capacitor is needed to bring the frequency co-efficent down to 265 cycles per minute where the resonant magnetic magnetic field can grab ahold of the magnet piston.
Quote from: synchro1 on March 07, 2018, 09:04:56 PM
@Tinman,
Just shut you shithead.
The R.P.M. of the Reed switch spinner needs to match the resonance of the bifilar to pump the magnet piston. If the ball speeds up to fast the piston oscillation stops. The resonant frequency of the bifilar coil alone is over 18 thousand cycles per minute, so the addition of the 70uF capacitor is needed to bring the frequency co-efficent down to 265 cycles per minute where the resonant magnetic magnetic field can grab ahold of the magnet piston.
Let me guess--you have not tried this with a monofilar coil,nor have you any energy in to energy out calculations to show-->of course you will be using that magic DMM of yours to make the calculations ::)
You have a spinning ball magnet,that closes a reed switch(twice each revolution of the ball magnet),which powers up the magical bifilar coil,which in turn provides the energy required to pump your magnetic piston that is suspended from a string,which dose what?
Keep going syncro-love reading your post's.
Dont forget to keep those insults up ;)
Quote from: tinman on March 07, 2018, 07:40:41 PM
How did you determine the efficiency of your pulse motor hoptoad?.
Power draw versus torque. Admittedly, I'm NOT using a bi-filar in the classical sense that you are all talking about here.
Cheers
Quote from: hoptoad on March 08, 2018, 01:08:52 AM
Power draw versus torque. Admittedly,
Cheers
Well it is actually refreshing to see some one take the torque into consideration,as most only go for P/in verses charge battery voltage :D
QuoteI'm NOT using a bi-filar in the classical sense that you are all talking about here.
Care to share your design?
Brad
author=synchro1 link=topic=16203.msg517958#msg517958 date=1520474696]
Quote@Tinman,
Just shut you shithead.
Quote from: tinman on March 08, 2018, 01:38:46 AM
Well it is actually refreshing to see some one take the torque into consideration,as most only go for P/in verses charge battery voltage :D
Care to share your design?
Brad
As you are well aware, I've shared much information on this site, freely and without hesitation. But that's been long after I have had the opportunity to fully explore and understand a given path looking for a given prerequisite.
At the moment I am playing with old ideas in a new way, and when I think I more fully understand the operation and limitations of the method I'm playing with, I'll directly share.
However, here's a little hint. It closely relates to an experiment carried out in 2011 and detailed on my old blog in 2012.
Page 13 Fig 2 and Fig 4 give a starting point. Some elements need changing or discarding.
https://adamsinfo.000webhostapp.com (https://adamsinfo.000webhostapp.com)
Going over old work with a fresh eye provided by another person has presented an interesting torque result.
Not overunity, but rather, what I have been particularly looking for. An electric motor that has raw grunt under load over a broad rpm range.
Being time poor myself due to many commitments, I would not hold your breathe waiting on me to reveal a simple circuit (after fully exploring its parameters myself) when I know you are capable of deducing it yourself, in the meantime. The circuits shown are not exactly what I am using now, but close enough for a starting point. The changes needed are minor and few, but elemental to function for the purpose of running torque.
The method I am currently using does NOT increase maximum breaking torque (at nearly zero rpm), instead it acts like a turbo charger, increasing the running torque at higher rpm.
Cheers and don't forget we all have the capacity to K.I.S.S. :)
Quote from: hoptoad on March 08, 2018, 02:06:40 AM
As you are well aware, I've shared much information on this site, freely and without hesitation. But that's been long after I have had the opportunity to fully explore and understand a given path looking for a given prerequisite.
At the moment I am playing with old ideas in a new way, and when I think I more fully understand the operation and limitations of the method I'm playing with, I'll directly share.
However, here's a little hint. It closely relates to an experiment carried out in 2011 and detailed on my old blog in 2012.
Page 13 Fig 2 and Fig 4 give a starting point. Some elements need changing or discarding.
https://adamsinfo.000webhostapp.com (https://adamsinfo.000webhostapp.com)
Going over old work with a fresh eye provided by another person has presented an interesting torque result.
Not overunity, but rather, what I have been particularly looking for. An electric motor that has raw grunt under load over a broad rpm range.
Being time poor myself due to many commitments, I would not hold your breathe waiting on me to reveal a simple circuit when I know you are capable of deducing it yourself, in the meantime. The circuits shown are not exactly what I am using now, but close enough for a starting point. The changes needed are minor and few, but elemental to function for the purpose of running torque.
The method I am currently using does NOT increase maximum breaking torque (at nearly zero rpm), instead it acts like a turbo charger, increasing the running torque at higher rpm.
Cheers and don't forget we all have the capacity to K.I.S.S. :)
Thanks Hoptoad.
I can see why the hall sensor would get fried in the emitter follower circuit (circuit A),and not in circuit B.
I will take a closer look tonight,but AM i am half way through my own pulse motor build,to see how my new coil design go's.
Brad
Quote from: tinman on March 07, 2018, 09:52:05 PM
Let me guess--you have not tried this with a monofilar coil,nor have you any energy in to energy out calculations to show-->of course you will be using that magic DMM of yours to make the calculations ::)
You have a spinning ball magnet,that closes a reed switch(twice each revolution of the ball magnet),which powers up the magical bifilar coil,which in turn provides the energy required to pump your magnetic piston that is suspended from a string,which dose what?
Keep going syncro-love reading your post's.
Dont forget to keep those insults up ;)
@Tinman,
How can anyone pay serious attention to you when you persistently continue to misspell "does" as "dose", after having been corrected repeatedly by me in the past?
Quote from: synchro1 on March 08, 2018, 08:05:58 AM
@Tinman,
How can anyone pay serious attention to you when you persistently continue to misspell "does" as "dose", after having been corrected repeatedly by me in the past?
Here syncro--this is so you.
https://www.sciencealert.com/people-who-pick-up-grammar-mistakes-jerks-scientists-find
Quote: Scientists have found that people who constantly get bothered by grammatical errors online have "less agreeable" personalities than those who just let them slide.
Seems science even knows what type of person you are syncro--they nailed it.
Quote from: synchro1 on March 08, 2018, 08:05:58 AM
@Tinman,
How can anyone pay serious attention to you when you persistently continue to misspell "does" as "dose", after having been corrected repeatedly by me in the past?
And while we're at it,put this in ya pipe and smoke it ;D
https://www.youtube.com/watch?v=IxLstEtzf_M
Quote from: tinman on March 07, 2018, 08:47:49 PM
author=synchro1 link=topic=16203.msg517953#msg517953 date=1520472501]
So,Mags was using how many volts?--i was using 7.2
Mag's push button switch is different from a transistor being used as a switch how?
My power supply is different to a battery how?
Was Mags using a power supply or battery?
Well duh ::)
And you need an electric field to produces an EMF across a pickup coil--is it sinking in yet?
What was the kickback voltage value in Mag's test?
Which is exactly what i was doing--> Dickhead,only i used a power supply like Mag's did.
Sooo,300 volts dropped across a coil with only 3.2 ohms resistance,and an inductive value of only 1.2mH,isnt going to produce any magnetic field ? ::)
So the power being dissipated from the 100 ohm resistor on the pickup coil-came from where?
Like i said--you are clueless.
Im going to work and will get back to reply to other stuff here after.. But this i have to address here now...
My power supply is different to a battery how?
Was Mags using a power supply or battery?"
lol. Maybe someone that did watch my vids can inform you as to everything about my tests. Clearly you did not watch them if you cannot remember what sort of supply I used.
Maybe you can give the nice folks here a clear explanation as to why you think your single wire coil did better than the bifi coil. Id love to hear that explanation, it might be helpful to us little folk ;)
"So,Mags was using how many volts?--i was using 7.2"
How many volts did I use Brad? .5v? 3v? 10v? 20v? Maybe someone that actually watched my vids will go back and watch them again and keep you updated on all the details of what I had shown..
Ya know something, You kept asking, how many times, "what is the difference between a bifi coil and a normal coil with a cap across it?" I show what that difference is and you post a vid claiming Mags this and Mags that. Mags switching is bad so his vid shows nothing worth looking at. Mags didnt show measurements of his coils so they must not be the same. So now you do this vid and 'claim' the the single wire coil takes on inductive kickback pulses better than a bifilar coil.. Well good for you. Now you can explain why it is that you have gotten that result? Because my rebuttal vid is coming, and I 'did' watch your vid, and I 'do' remember what went on in your vid. Maybe you should show an exact diagram of your circuit as I did. Just so others can try to replicate your results. ;)
Mags
Quote from: Erfinder on March 08, 2018, 10:09:04 AM
If the highlighted is a reference to you know who.... please say so. Other free thinking tinkerers may begin to take the individual serious and possibly benefit as you say have.....
Regards
I omitted your name mainly because I know some individuals here would possibly automatically discard anything I might have written after it. On another note, thanks for the crumbs that led me to the cake. One day hopefully, you'll lead me to the cake topping, if I don't find it myself in the meantime.
Cheers Erfinder .... keep on keepin on.
author=Magluvin link=topic=16203.msg517982#msg517982 date=1520519125]
QuoteMy power supply is different to a battery how?
Was Mags using a power supply or battery?"
lol. Maybe someone that did watch my vids can inform you as to everything about my tests. Clearly you did not watch them if you cannot remember what sort of supply I used.
Guess you missed the boat on that one to mag's.
I was asking syncro,as he said i needed to use a battery.
I watched your videos-both of them.
QuoteMaybe you can give the nice folks here a clear explanation as to why you think your single wire coil did better than the bifi coil. Id love to hear that explanation, it might be helpful to us little folk ;)
I can do that,and i can show it.
Will write up the explanation tonight,and we'll see if it sinks in.
Quote"So,Mags was using how many volts?--i was using 7.2"
How many volts did I use Brad? .5v? 3v? 10v? 20v? Maybe someone that actually watched my vids will go back and watch them again and keep you updated on all the details of what I had shown..
Once again,i was asking syncro,as he says that 300 volt pulses are not high enough,so i needed a higher drive voltage-->from a battery
As i said ,i watched your video's.
QuoteMags didnt show measurements of his coils so they must not be the same.
You will show us all where i said that?
QuoteYa know something, You kept asking, how many times, "what is the difference between a bifi coil and a normal coil with a cap across it?" I show what that difference is
And im still asking.
When you find out what your actually looking at,as i have,then you might understand as to why i am asking you what the difference is.
You are making the mistake in thinking that your bifi coil is absorbing more of the inductive kickback because you see a higher EMF value across the sniffer coil.
Quoteand you post a vid claiming Mags this and Mags that. Mags switching is bad so his vid shows nothing worth looking at.
And when you get a clean switching setup,with an ongoing pulse train,then maybe you can look as close as i did.
QuoteSo now you do this vid and 'claim' the the single wire coil takes on inductive kickback pulses better than a bifilar coil..
And it does.
QuoteNow you can explain why it is that you have gotten that result?
I can-and more.
I can tell you why a single wound coil will be more efficient as well.
QuoteBecause my rebuttal vid is coming, and I 'did' watch your vid, and I 'do' remember what went on in your vid. Maybe you should show an exact diagram of your circuit as I did. Just so others can try to replicate your results
Sure,i can post a circuit--it's very simple.
Looking forward to your vid.
Quote from: Erfinder on March 08, 2018, 10:47:14 AM
you're better than this...
QuoteThe ideas you are trying to communicate in this video are lost through your say nothing approach,
ATM,i am trying not to communicate to much of my idea,and so my say nothing approach is about as much as i want to say.
Quotescope displaying a blast from the past multiple pulse per pass scope shot, and constant referencing to voltage as it relates to events leading up to the production of the what you call oscillations.... this latter reference can lead one to speculate that your switch is not a BJT.....
It could,but who will give it a second thought ?. ;D
Mags
You may want to read this.
https://physics.stackexchange.com/questions/99060/capacitors-why-is-the-energy-not-stored-in-a-magnetic-field
Quote: while charging a capacitor there will be a magnetic field present due to the change in the electric field. And of course BB contains energy as pointed out. However: As the capacitor charges, the magnetic field does not remain static. This results in electromagnetic waves which radiate energy away. The energy put into the magnetic field during charging is lost in the sense that it cannot be feed back to the circuit by the capacitor.
In the limit of a fully charged capacitor, there is no displacement current maintaining a magnetic field and all the energy is stored in the electric field.
Quote from: tinman on March 08, 2018, 05:59:48 PM
author=Magluvin link=topic=16203.msg517982#msg517982 date=1520519125]
Guess you missed the boat on that one to mag's.
I was asking syncro,as he said i needed to use a battery.
I watched your videos-both of them.
I can do that,and i can show it.
Will write up the explanation tonight,and we'll see if it sinks in.
Once again,i was asking syncro,as he says that 300 volt pulses are not high enough,so i needed a higher drive voltage-->from a battery
As i said ,i watched your video's.
You will show us all where i said that?
And im still asking.
When you find out what your actually looking at,as i have,then you might understand as to why i am asking you what the difference is.
You are making the mistake in thinking that your bifi coil is absorbing more of the inductive kickback because you see a higher EMF value across the sniffer coil.
And when you get a clean switching setup,with an ongoing pulse train,then maybe you can look as close as i did.
And it does.
I can-and more.
I can tell you why a single wound coil will be more efficient as well.
Sure,i can post a circuit--it's very simple.
Looking forward to your vid.
"Guess you missed the boat on that one to mag's.
I was asking syncro,as he said i needed to use a battery.
I watched your videos-both of them."
Ok
"I can do that,and i can show it.
Will write up the explanation tonight,and we'll see if it sinks in."
Ok
"Once again,i was asking syncro,as he says that 300 volt pulses are not high enough,so i needed a higher drive voltage-->from a battery
As i said ,i watched your video's."
Ok
"You will show us all where i said that?"
It felt that you implied it in the vid as you were referring to everything else you felt was going wrong in my vid.
"And im still asking.
When you find out what your actually looking at,as i have,then you might understand as to why i am asking you what the difference is.
You are making the mistake in thinking that your bifi coil is absorbing more of the inductive kickback because you see a higher EMF value across the sniffer coil."
We will see.
"And when you get a clean switching setup,with an ongoing pulse train,then maybe you can look as close as i did."
it only seems to be you that insists I didnt have any consistent switching happening. We will see.
"And it does."
We will see
"I can-and more.
I can tell you why a single wound coil will be more efficient as well."
We will see
"Sure,i can post a circuit--it's very simple.
Looking forward to your vid."
Ok
Mags
Quote from: tinman on March 08, 2018, 06:07:51 PM
ATM,i am trying not to communicate to much of my idea,and so my say nothing approach is about as much as i want to say.
Why is that? Why show it at all? I mean you come down on Erfinder for not completely divulging what he talks about, and he has shown vids of some of his stuff. So maybe that all can end now, being you are using the same approach to your project.
Mags
Quote from: tinman on March 08, 2018, 06:13:19 PM
Mags
You may want to read this.
https://physics.stackexchange.com/questions/99060/capacitors-why-is-the-energy-not-stored-in-a-magnetic-field (https://physics.stackexchange.com/questions/99060/capacitors-why-is-the-energy-not-stored-in-a-magnetic-field)
Quote: while charging a capacitor there will be a magnetic field present due to the change in the electric field. And of course BB contains energy as pointed out. However: As the capacitor charges, the magnetic field does not remain static. This results in electromagnetic waves which radiate energy away. The energy put into the magnetic field during charging is lost in the sense that it cannot be feed back to the circuit by the capacitor.
In the limit of a fully charged capacitor, there is no displacement current maintaining a magnetic field and all the energy is stored in the electric field.
Well lets say we have a 10uf cap. At 0v, we have a balance between the 2 plates, no charge. Now we charge the cap to 10v. If we could count the electrons taken from the pos plate and count the electrons given to the neg plate, would you agree that every time we discharge the cap and recharge it to 10v, that the electron counts would be the same? If not, then why?
If so, then the reason that you are presenting for the possible loss you show in your bifi coil can also be directly related to the 50% loss in energy when we do a cap to cap experiment? What you are presenting as the loss in your bifi coil in that article has also been said here(by Poynt himself) to be the loss in the cap to cap, once I had proven that resistance and heat are not the loss in the cap to cap experiment as previously claimed here before I had presented my proof. But that still does not account for the electron count values for the voltages measured in the caps. Those numbers will be the same anyway you look at it. Even ideal caps the count will be relative to the voltage which firstly proves the resistance as a loss a falsehood.
So in the cap to cap experiment, would you agree that there is the possibility of this loss you claim in your bifi as the issue? If so, when we do the cap to cap deal, and we lose 50% of the energy total in the process, does that same loss you are presenting have anything to do with the loss in the cap to cap deal? If no, then why is it that your bifi has this particular loss you claim and the cap to cap does not? If you say yes, please explain why then once we do the cap to cap deal our electron count will still be relative to the voltage in each cap. If we add a 100 ohm resistance between the 2 caps and let them level out to 5v each, or if we just connect them directly, would there be this added loss you claim to have with the bifi capacitance, one way with the 100 ohm or a 1meg ohm resistor or if we direct connect the 2 caps?
I mean if 'you disagree' that the 'supposed' cap to cap magnetic field losses when the 2 caps are direct connected, is the same as the explanation for your bifi loss, then I am at a loss.
From what I got from that page you had shown is they(by the way, are they talking about Poynt at our?) are saying a magnetic field is created during the charging of the cap radiates some energy away from the cap as magnetic radiation. Is this what you are claiming as the loss you have in your bifi coil?
Mags
So brads test coils see a loaded secondary pickup coil. This will reduce the inductance of the test coil, but it will have an affect. How much of an affect? I dunno. Would have to see what the inductance increase is with the core and then what the inductance of the test coil is when the pickup coil is loaded. So why not rectify the pickup coil so the test coils could peak out as I have shown, and then the pickup coil only reacts to the collapsing field of the test coil. Now the loaded pickup wont affect the initial upswing of the test coils.
Will do that after I try to replicate Brads results, as it will be only the addition of a diode in series with the loaded pickup coil.
Mags
Hey Mags
Im at work ATM,and just had a quick read of your post's.
I will give detailed explanations tonight,when im on my computor.
Remember-there is a big difference between stored energy,and stored charge-such as in caps.
You loose no charge in a cap to cap transfer,but you loose energy,due to displacement currents. The same applies to a coil,be it single or bifi. As the bifi coil has more winding capacitance,then you loose more of the energy that is put into the coil.
I have mannaged to replicate your test results with my two coils,by removing the load on the sniffer coil-what does that tell you.
Brad
Check this one out:
https://youtu.be/3ykm-blvC-U (https://youtu.be/3ykm-blvC-U)
I'm getting better (Attraction-Neutralization) off the kickback from my "500 Newton Electro-Magnet Coil" then with the series bifilar coil. This setup generates a huge amount of current in the "Electro Magnet Coil" from the attraction stroke as I've demonstrated in my videos. All it would take is one commutator Reed switch for output off the "Electro-Magnet Coil".
I think I can place the output Reed switch on the other side of the Neo magnet spinner.
I'll upload another video of this soon. I believe this combination will prove to be way Overunity.
Member Tinhead has started a new enlightened thread on the bifilar and single wire coils first by banning me from posting, then deleting Erfinder's initial comment so only he and Thamsankooklala can post. Then he proceeds to lay in some completely worthless malarky.