Amplitude of EMF/BEMF when a coil is de-energised 1

[Lakey]

Good Morning,

I’ve read somewhere that “ theoretically, the peak voltage that is generated when you disconnect a coil form its supply, without suppression of any kind, may be infinite”.

If it is theoretical then, I guess, there must be a recognised formula that can be applied. I done the usual searches etc, but I have been unsuccessful finding any information which relates to this particular issue.

My question is then – is the above statement true, if so, what is the formula that one would apply?

My understanding is that although the peak voltage from a de-energised coil is very high it is impossible to predict, which is largely due to the coil’s (inter-winding) capacitance.

I appreciate your thoughts.

Regards.

 

[Skogsgurra]

Absolutely right, Lakey. But, given the data, you should be able to predict it.

The "formula" is simply that current in an inductive circuit continues to flow when interrupted. So, if you have a winding with 1 H inductance and 2 A current, your current will still be 2 A when you open the switch. It is then up to the switch and the capacitance in the circuit to handle this current.

Let's say that the total capacitance (interwinding, switch and wiring) is 1 nanofarad. Your voltage over the switch will then rise initially with 2000 000 000 V/s or 2 kV/microsecond (dv/dt = i/C). If your breaker opens fast enough (withstand voltage increases faster than 2kV/us) and nothing else breaks down, you will have a rather high voltage very quickly.

The highest voltage you will have is when all the energy in the coil (W=I^2xL/2) has been transferred to the capacitance. The energy is 2 joules. This corresponds to a voltage across 1 nF being equal to (using W = U^2xC/2) a little more than 63 000 volts.
You will most likely get a breakdown in your switch or in your coil or just about anywhere the isolation is weakest.

The ignition coil used in cars (older models) used this principle to produce voltage for the spark plugs. The "points" opens rather slowly and the withstand voltage didn't build up fast enough to prevent losing energy through sparking across them. That's why there's always a capacitor parallel to the contacts. The capacitor prevents sparking and lets the voltage build up to 10 - 20 kV before it sparks in the right place (inside the motor).

Gunnar Englund

http://www.gke.org

 

[Lakey]

Skogs,

Excellent explanation – a star for you.

At the nearest opportunity I will conduct an experiment where the switch opening times versus peak voltage can be monitored.

Thanks & regards,

 

[waross]

Hi skogsgurra;
I learned something today. Thank you.
I was taught a different method to calculate/estimate the voltage. The current continues to flow in the resistance of the circuit. The higher the resistance, the higher the voltage. The resistance through the arc at the opening contacts, although dificult to calculate, will limit the voltage. A resistance in the discharge circuit of whatever value will provide a means to estimate the back emf.
From your explanation I see that the voltage will not approach infinity but will approach the limit described by your explanation. This value will be reduced by any resistance in the circuit and by any energy lost in arcing as the contacts open. Note; The resistance of the coil should be included in the total resistance. This is consistant with the voltage across a freewheeling diode dropping from applied voltage to zero when the circuit is opened.
Perhaps my method will predominate with slow opening switches and measurable circuit resistance.
I can see that your method will predominate with very fast switching and very high circuit resistance. The insulation resistance of the coil will have a bearing also.
And we shouldn't forget that the value of theoretical voltage is so high by either method that the actual limit will be the voltage at which the insulation fails.
Respectfully

 

[Skogsgurra]

Yes, waross. The equivalent (arc) resistance method is simpler. And absolutely applicable if you have a parallel resistance, for example like you have across some field windings. But since Lakey already seemed to master that part of the theory, I took him a little bit further. That's actually what he was asking for.

Gunnar Englund

http://www.gke.org

 

[VEBill]

Measuring the transient voltage will probably change it (depending on the ratio of your instrument input Z to the baseline strays). Even 10Mohm//1pF is a lot less than a really, really good and clean open circuit.

 

[Skogsgurra]

You are a real engineer, VE1BLL.

I had a professor once. That was in the days when a 10 MOhm//1pF probe was unheard of. When I protested and said something like what you just said, his answer was: "Why measure? Just calculate". But then again, he was not an engineer...

Gunnar Englund

http://www.gke.org