# Solenoid Back EMF

Discussion in 'General Electronics Discussion' started by Arouse1973, Jan 16, 2014.

5,164
1,081
Dec 18, 2013
Hi Guys

What is the best formula for working out the back EMF of a solenoid. Now I have the results but wanted to calculate it for fun to see how close I can get to real life. This is a sealed unit so I can't see the coil.

The coil is 24.6 Ohms and has an inductance of 20.6mH. The voltage rating is 12Volts and the coil draws about 500mA. The only thing I can't do is measure the capacitance, although I don't see much of an inrush current when it powers up (measuring VD across 0.33R).

I have 0V connected all the time and tab the power on and then release to produce the EMF spike, my scope is connected across this clip to 0V, probe is x10 12pF. I see negative spike first and then a series of +/- ringing.

The formula I am using is -L*dI/dt and for the current V/R*(1-e^-tr/L) should I be using V*e^-tr/L?. Also if I use time of say slightly greater than 0 I get a massive voltage as you would expect. The coil has capacitance I can't measure, is this going to be series or parallel?

What do I use for the time part of these equations? Do I have to work out the L/R for the capacitance and series resistance of the coil.
Thanks

2. ### Laplace

1,252
184
Apr 4, 2010
What is the circuit being used to produce the EMF spike? You show the current as an exponential decay through the winding resistance as though the terminals of the solenoid coil were shorted together. But that is a method to avoid producing any EMF spike. So how exactly is the current through the solenoid interrupted? Maximum back EMF will be produced by a step function going from full current to zero current instantly so the dI/dt is infinite. Is the current interrupted by the opening of a mechanical switch? If so, is there a visible electrical arc generated across the switch contacts at the moment of opening? That arc would mean that dI/dt is somewhat less than infinite. There is also the distributed inter-winding capacitance associated with the inductance of the coil so the solenoid will be self-resonant at some frequency which leads to the observed ringing. The back EMF is dependent on how fast the current is turned off, so how fast is your circuit able to turn off the current?

5,164
1,081
Dec 18, 2013
Hi Laplace. I was working out the correct protection diode for this circuit and had a look at the spike out of interest with the diode disconnected. I was after some accurate formula to be able to calculate what I saw, for a bit of fun. I don't know what else to add to the standard formula to get close.

4. ### KrisBlueNZSadly passed away in 2015

8,393
1,270
Nov 28, 2011
The diode's voltage rating needs to be comfortably higher than the voltage applied to the coil, because when the coil is energised, the diode is reverse-biased and has the applied voltage across it.

The diode's current rating needs to be comfortably higher than the current flowing in the coil when the driver transistor turns off. When that happens, the inductor tries to maintain the current at the same value; this causes the voltage to flip to the opposite polarity, but the current is initially the same, before it decays to zero.

So in this case, the diode needs to comfortably withstand 12V and 0.5A so a 1N4001 (50V, 1A) will be fine, unless the switching frequency is high, in which case you need a faster diode.

Edit: Sorry, I think you already knew all of that. I read that you were working out which diode to use, and answered that question, but I don't think you were asking that question.

5,164
1,081
Dec 18, 2013
Yes Kris
What I was after for a bit of fun was a formula forworking out the voltage spike from a real world inductor which takes into account the capacitance and resistance of the coil.

6. ### KrisBlueNZSadly passed away in 2015

8,393
1,270
Nov 28, 2011
Right. I don't know how to calculate that :-(

7. ### duke37

5,364
769
Jan 9, 2011
The energy in an inductance is I * I * L /2
The energy in a capacitor is V * V * C/2
If the solenoid is abruptly switched off, the energy is transferred from the inductance to the capacitance. Thus the theoretical voltage can be calculated.
In practise the turn off will not be abrupt, especially if the voltage is high, and a lot of energy may be lost in the switch.

There is of course a limit to the insulation of the solenoid.

8. ### KrisBlueNZSadly passed away in 2015

8,393
1,270
Nov 28, 2011
Right, so assuming perfect switching and no losses anywhere, the first voltage peak is when all the magnetic energy from the inductor has been transferred to the capacitor, right?

L = 0.0206H
I = 0.5A

Calculating energy in the inductor:
E = I^2 L / 2
= 0.25 * 0.0206 / 2
= 2.575 mJ

Calculating that amount of energy in the electric field in the capacitor:
E = V^2 C / 2
rearranges to V = sqrt(2 E / C)

Assuming 100 pF across the coil...
V = sqrt(5.15e-3 / 100e-12)
= sqrt(51500000)
= 7176V

That would be assuming no breakdown anywhere.

With 10 nF across the coil, the peak voltage would be 718V.

Sounds plausible.

9. ### duke37

5,364
769
Jan 9, 2011
I agree with the numbers but it is only necessary to equate the two energies, giving
V = I * SQR(L/C)

A car ignition has a 'condenser' across the switch which allows the points to open before the voltage has risen to a very high level. The coil gives about 300V on the primary and has a secondary with turns ratio of about 20:1 giving an output of 6kV.

I am not familiar with modern ignitions but I think they use a FET switch, no capacitor and a zener to limit the voltage. This would give a pulse with a much sharper leading edge.

10. ### KrisBlueNZSadly passed away in 2015

8,393
1,270
Nov 28, 2011
Thanks duke37

5,164
1,081
Dec 18, 2013
Close Kris, your adout double what you actually get in real life. Any more ideas?

12. ### duke37

5,364
769
Jan 9, 2011
If the switching is not instataeous then there will be energy lost in the switch and the voltage will be lower than calculated. If the capacitor is faulty in a car ignition, then there will be a large spark at the contacts and it is doubtful if there would be enough volts to get a spark at the plug.

Breaking a current is difficult because of the high voltage which can be generated. Opening a mechanical contact will draw an arc which can be maintained as the contacts open and the volts will be limited. Arc welders run at under 100V.
Mechanical power contactors have methods of controlling the arc, perhaps using magnetic fields or air blast.

5,164
1,081
Dec 18, 2013
Sorry you have confused me. Its got nothing to do with cars and there is no swich.

14. ### duke37

5,364
769
Jan 9, 2011
Quote
I have 0V connected all the time and tab the power on and then release to produce the EMF spike, my scope is connected across this clip to 0V, probe is x10 12pF. I see negative spike first and then a series of +/- ringing.
Quote

It seems to me that you have a switch of one sort or another, not specified.
The faster the switch, the less energy loss in the switch and the higher the voltage.

I talked about car ignitions because this is a well poven technology for using the high voltage generated when an inductance is turned off. It needs a good switch and an FET seems to be preferred these days.

A capacitor discharge ignition avoids the turn off switch problem by connecting a charged capacitor across the coil. The switch here is often an SCR.

15. ### KrisBlueNZSadly passed away in 2015

8,393
1,270
Nov 28, 2011
How are you interrupting the current to the coil?

5,164
1,081
Dec 18, 2013
By hand with a wire. I am not using a mechanical switch.

Thanks

17. ### KrisBlueNZSadly passed away in 2015

8,393
1,270
Nov 28, 2011
How quickly can you move the wire away from the contact point? Quickly enough to avoid arcing? Can you capture the event on a scope?

We're talking hundreds of volts here... are you sure your scope probe isn't arcing internally?

5,164
1,081
Dec 18, 2013
Yeah could be Kris I don't know. I think I am going to go with your approach, seems the most plausible. As you say there must be a loss somewhere. If I come up with a better solution I will let you know.
Cheers

19. ### duke37

5,364
769
Jan 9, 2011
I can think of two places which are lossy.

1. The contacts will always spark when opened. The rate of opening will control the time that the arc will last.
2. If the solenoid has an iron core, then there will be eddy current losses.

Being within a factor of two for the voltage seems to be reasonable to me.