# Back EMFs

Discussion in 'Electronic Basics' started by Mark Taylor, Dec 23, 2004.

1. ### Mark TaylorGuest

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

2. ### Andrew HolmeGuest

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?

3. ### Larry BrasfieldGuest

Hi, one.
If not limited by the switching speed of the interrupting
device, shunt impedance across the coil will do the limiting.
That impedance consists of stray winding capacitance,
external capacitance, and (for a non-air coil) losses due
to eddy current flow in the core material.
I cannot imagine why ESR matters in normal cases. It
would be ununsual for ESR to determine the interrupted
current level and the external circuit will dominate once
the interruption occurs.
Yes. See above.
Better that than non-resistance wire. But wire made from
materials intentionally made resistive? Why bother?

4. ### John FieldsGuest

---
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.

5. ### Mark TaylorGuest

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!

6. ### John PopelishGuest

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?

8. ### Mark TaylorGuest

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.

9. ### John PopelishGuest

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.