Back EMFs

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ESR in this case would, I believe, be the resistance of the coil and
since I = E/R,  the output voltage of the coil would be:

                            E
                       L d --- 
              L dI          R  
         E = ------ = ----------
               dt        dt 


So, the larger R gets, the smaller I would get and the smaller E would
get.

I can\'t imagine why he was told that ESR had to lie in a specific
range though; ISTM that for the biggest spark the best of all possible
worlds would be zero coil resistance.  But, ESR is a term usually used
with capacitors, so maybe there\'s more going on than meets the eye.

>> He also said large
>> inductance isn\'t everything, either. Is this true?
>
>Yes.  See above.
>
>> Should all
>> ignition-type coils be wound from resistance wire?
>
>Better that than non-resistance wire.

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Look at your L.  A 72 turn close-wound air cored coil with about a 2
inch diameter won\'t have a lot of inductance, so you won\'t have a lot
of energy stored up to start with unless you\'re pumping a _lot_ of
current through it.

How many picofarads do you have to put across your simulated inductor to get the peak voltage to match the real one? What are you using as the switch to interrupt the current?

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John Popelish

Hi group,

If I interrupt the current supply to a large coil, what determines the level of back emf I can get? I'd imagine it was solely the inductance of the coil (bigger the better) and the load resistance the field collapses across, but some guy's told me the ESR of the coil is absolutely critical to getting maximum voltage and a little bit too much or too little will give a pretty lousy spark. He also said large inductance isn't everything, either. Is this true? Should all ignition-type coils be wound from resistance wire?

Thanks

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Fat, sugar, salt, beer: the four essentials for a healthy diet.

EMF is the product of inductance and rate of change of current:

V = L di / dt

The greater the current - and the faster you interrupt it - the greater the back e.m.f. Series resistance will reduce the voltage at the coil terminals; think of it as a resistor in series with a perfect voltage source. You want the lowest resistance possible. Why would you wind it from resistance wire? Is that to limit the pre- interruption current?? Why not use a current limtied supply e.g. an external series resistor?

Thanks. This all helps. I think maybe Larry was onto something when he spoke about inter-winding capacitance (I think that's what he meant). I've just made up an air core single layer solenoid of 70 turns at

52mm diameter, close wound and perhaps the proximity of the adjacent turns is creating a capacitive loss-path that damps the back EMF. Having checked it in Spice, it looks like that might be the answer; but I can't believe so much potential voltage can be lost in close-spaced turns. 9kV swiftly becomes < 100V!

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Fat, sugar, salt, beer: the four essentials for a healthy diet.

Keep in mind the stored energy for capacitors and inductors. If all the energy in the inductance gets transferred to the capacitance, the energy cannot increase.

For inductors, energy=I^2*L/2 For capacitors, energy=V^2*C/2 where the energy is in joules, the inductance in henries, the capacitance in farads, the current in amperes and the voltage in volts.

This energy equivalence allows you to calculate the best case (no losses, upper limit) of inductive energy converted to capacitive energy.

Switch arching, winding resistive, capacitive dielectric losses, etc. consume some of the energy during the transfer, so you never see the full transfer.

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John Popelish

Thanks, John. I can only go by what my buddy Steve told me, so I guess that would be 180pF., and as for a switch, I'm just using a crock-clip against a battery terminal.