ENERGY AMPLIFICATION

[Magluvin]

@dave45


a collapsing field doesn't make the current reverse.


1) Initiially an inductor opposes a change (resistance to acceleration)
2) then an inductor mtaintains a same current (body in motion remains in motion)
3) then when the current starts to diminish, the coil collapse wants to keep the current going in the same direction


back EMF is the force against the initial current increase... (1)
forward EMF is the force that keeps the current going (3)

field collapse doesn't not mean current in reverse

If we switch on a power source to a coil, the mag field builds. When we disconnect the source, does the field collapse cause negative or positive current in reference to the source current?

I say forward, initially.    If we use a diode and cap to capture the collapse, like in a basic switching power supply.  If we place the diode and cap in a series loop, and apply a source across the coil then disconnect it, the field collapse will charge the cap.  Right? Say the source is 10v, and we get 20v into the cap, just as an example. Not impossible. Right? Now, reverse the diode.    Use a non polarized cap for these experiments, or reverse the electrolyitic cap accordingly with diode bias direction.

Now when we reversed the diode(and cap if electrolytic) on the coil, using the same source polarity,  we still get greater than source voltage into the cap.    Why is that? Granted, when we configure the diode and cap across the coil to capture forward collapse current, the source is not getting direct connection to the cap when across the coil. So no source current enters the cap because the diode is not biased in that polarity. So the cap has no charge till the source is disconnected. 

But, when we reverse the diode, the cap and diode are biased in line with the source, and the cap charges to 10v while the coil is taking on source current. But when the source is disconnected, the cap ends up with more voltage than the source.    How can that be??

If the coil collapse current is the same direction as when the source was connected, then how did the cap get charged, even higher than the source voltage, with an actual 'reverse emf' The diode is a one way street, so no glitches here.

I had found this while working on a pulse motor.  I was using a cap and diode to capture from my drive coils. It was working fine. Checking the voltage on the cap and it was well above source voltage.  But when I had my meter on the cap continuously for some testings, I noticed that my cap was filled to source voltage right when I connected my source.    Well that cant be right. The rotor was in a locked position but the reed was closed.  Being that I knew that the diode should not be in the bias on position with the switch, I reversed the diode and I still got higher than source in the cap.

My only explanation could be that the coils collapse, if it does not get loaded while collapsing, goes into a reverse collapse by bouncing off of the coils own tiny self capacitance, thus being able to actually go into full reversal, due to LC oscillation in the coil. But upon reversal, the diode is now biased and the collapse can now enter the cap.


It works on Falstad sim also.   The first circuit running when you start the sim, just change out the upright wire on the left of the circuit with a diode, and then reverse the diode. Notice what happens to the cap each time on the scopes below on the sim.

Weird isnt it?

Mags

[a.king21]

d3Xor


OK then.


Here is a simple question.
An ne2 bulb has a strike voltage of 80 to 100 volts.


Is it possible to light an ne2 bulb using
1 a single coil (2 ohms up)  (no secondary)
2 a 1.5v volt AA battery?

[d3x0r]

If we switch on a power source to a coil, the mag field builds. When we disconnect the source, does the field collapse cause negative or positive current in reference to the source current?



I say forward, initially.    If we use a diode and cap to capture the collapse, like in a basic switching power supply.  If we place the diode and cap in a series loop, and apply a source across the coil then disconnect it, the field collapse will charge the cap.  Right? Say the source is 10v, and we get 20v into the cap, just as an example. Not impossible. Right? Now, reverse the diode.    Use a non polarized cap for these experiments, or reverse the electrolyitic cap accordingly with diode bias direction.

Now when we reversed the diode(and cap if electrolytic) on the coil, using the same source polarity,  we still get greater than source voltage into the cap.    Why is that? Granted, when we configure the diode and cap across the coil to capture forward collapse current, the source is not getting direct connection to the cap when across the coil. So no source current enters the cap because the diode is not biased in that polarity. So the cap has no charge till the source is disconnected. 

But, when we reverse the diode, the cap and diode are biased in line with the source, and the cap charges to 10v while the coil is taking on source current. But when the source is disconnected, the cap ends up with more voltage than the source.    How can that be??

If the coil collapse current is the same direction as when the source was connected, then how did the cap get charged, even higher than the source voltage, with an actual 'reverse emf' The diode is a one way street, so no glitches here.

Because, you previuosly were going from a electron surplus to an electron deficit ... when you remove the ability to have a supluss, you genereate a deficit (a higher positive)... which is a continuing forward motion... it's similar to momentum or inertia...

a spring attached at the end of a body that is in motion will continue to extend the string, imparting more potential... excessive to the source; because inductance is the mass... the more the inductance the heavier the board, for a certain speed (current) you will get more or less stress on the spring


if your cap was already partially charged; you would have less of a gain.  The diode allows the current to start, until it doesn;t, at which point the current wants to keep going forward... but you started pushing the board (got it up to speed), then stopped... which then the spring(capacitor) is tensed by the continuing current.


-- edit
so maybe that's not right about the cap. 
But; intially pushing the board, does happen with/through the spring, it's not a rigid motion... the coil mass will push back on the cap and make a lower potential than the voltage supply  (back EMF, the first motion)... it will compress the spring slightly.... but somehow they are also tied together because the current flow inducer encompasses the system... pushing on both the spring and the board...


actually... the spring has to always be in a tensed state... oh no it doesn't, first effect would be ground satisfying the cap, just because 1) it's resistive path is lower... less 1/T... 2) unless the coil's ... I guess this is inductive reactance... inductance/mass is lower than some amount... the capacitor will give first... so at the top end of the coil the potential should go to +V or G depending on which you're looking at.  so really there could be a force back on the cap to make it a negative potential (I'm still working in the framework of the ground-switched ignition coil; but multiply by -1 everything, if on the other side)  ... continuing is still continuing, and relative directions will be self consistent

[d3x0r]

d3Xor


OK then.


Here is a simple question.
An ne2 bulb has a strike voltage of 80 to 100 volts.


Is it possible to light an ne2 bulb using
1 a single coil (2 ohms up)  (no secondary)
2 a 1.5v volt AA battery?

Yes.  Without reversing the current.
maybe not; a AA batter is current limited.  The coil inductance would have to be very large, which, without very large conductors to reduce the resistance per length; which increases capacitance per length... and requires more current.


So No.  there is a limit to the power of an AA battery...
maybe... given enough time; but with too much time, self inductance will cancel the current also...


Yes, if I can replace the battery with a super cap charged to 1.2V with the same (or less) Joules stored.... because the current isn't as limited.


Excessive current on the AA will create a notable Heat reaction, which will accelerate the current it can produce, which produces more heat, and that will be another limiting point.

[Magluvin]

Because, you previuosly were going from a electron surplus to an electron deficit ... when you remove the ability to have a supluss, you genereate a deficit (a higher positive)... which is a continuing forward motion... it's similar to momentum or inertia...

a spring attached at the end of a body that is in motion will continue to extend the string, imparting more potential... excessive to the source; because inductance is the mass... the more the inductance the heavier the board, for a certain speed (current) you will get more or less stress on the spring


if your cap was already partially charged; you would have less of a gain.  The diode allows the current to start, until it doesn;t, at which point the current wants to keep going forward... but you started pushing the board (got it up to speed), then stopped... which then the spring(capacitor) is tensed by the continuing current.


-- edit
so maybe that's not right about the cap. 
But; intially pushing the board, does happen with/through the spring, it's not a rigid motion... the coil mass will push back on the cap and make a lower potential than the voltage supply  (back EMF, the first motion)... it will compress the spring slightly.... but somehow they are also tied together because the current flow inducer encompasses the system... pushing on both the spring and the board...


actually... the spring has to always be in a tensed state... oh no it doesn't, first effect would be ground satisfying the cap, just because 1) it's resistive path is lower... less 1/T... 2) unless the coil's ... I guess this is inductive reactance... inductance/mass is lower than some amount... the capacitor will give first... so at the top end of the coil the potential should go to +V or G depending on which you're looking at.  so really there could be a force back on the cap to make it a negative potential (I'm still working in the framework of the ground-switched ignition coil; but multiply by -1 everything, if on the other side)  ... continuing is still continuing, and relative directions will be self consistent

Not sure if you understand what I was getting at. Im not claiming any gain with it.  Its just valuable info that shows other avenues to possibly explore, and may have a particular good use.

In the falstad sim, the transformers have some capacitance programmed in. They can oscillate.

Infact, if someone tries the circuit suggestion I gave on the sim, with the diode reversed, if you watch the coil on the scope, it swings to kilovolts due to no load while going into full magnetic and electrical full reversal, compared to a normal collapse current as we know it.

Below is pics of each diode direction functioning. The scope shots are the inductor left, capacitor right.

The first pic is the diode set up to capture forward collapse currents from the coil. Like using a diode as a snubber that directs the collapse currents back into the coil and it dies there. Here we have a cap in series with the diode, and the capacitor is charged only by the coil collapse.

In the second shot, I exchange the diodes.  We see the cap being charged(through the 100ohm resistor) and when the switch is released, the cap continues to charge further by the coil collapse current, which must have had a reversal in order to do so.

Notice the high voltage the coil reaches during reversal, where it has no load to discharge into.

Mags