I'm going to buy a holding electromagnet and a strike plate to hold some things, and I want to design my circuit (arduino-controlled) not to fry like bacon. I'm aware that since a holding magnet is an inductor, I should use a flyback diode and possibly a capacitor to handle the back EMF for when current is interrupted. However, what happens if the holding magnet is physically forced away from the strike plate? Work is being done to overcome the magnetic force, so I figure the energy goes somewhere, but how does that momentary change manifest in the circuit? Do I see increased current through the coil? Decreased current? And for that matter, what happens in the circuit when the magnet meets and locks to the strike plate?

Basically, I'm trying to determine if I need to handle a forward EMF spike as well as a back EMF spike, and my research hasn't taught me enough about magnetic fields to figure it out on my own.


I'm currently using this circuit:


simulate this circuit – Schematic created using CircuitLab

L1 is the magnet; I don't know its inductance, but it has a series resistance of 20 Ohms. D1 is the zener that protects against overvoltage; R1 is there because the only zener I had was exactly 12V and I wanted some safety margin to avoid a short in case the power supply went high for some reason other than L1. D2 is the flyback; it protects against voltages less than -1V, which hopefully aren't enough to ruin the cap (a schottky would be better, but I don't have one lying around).

I operate this by turning the power supply on and off. In the future I'll put a Darlington between C1 and V1. It SEEMS to work and not damage anything even when I force the plates apart, so that's good, hopefully I'm not doing anything nasty to the power supply. I still need to look at this with a scope to make sure.

I did have the idea of putting my own inductor in series with L1. This would act to limit current changes caused by L1's inductance changing. Not sure if I'll do that.

  • \$\begingroup\$ Good question, to which I don't know the answer, but have you tried measuring what happens? \$\endgroup\$ Dec 5, 2014 at 6:03
  • \$\begingroup\$ Well, I believe I'd want a scope for that, and I've never used one. The one I'd be using would not be mine, and so I'd REALLY rather not damage it. How big are EMF kickbacks? I could start with a Megohm:Ohm voltage divider and work my way down, but I'm not even sure these voltages won't make the resistors fail short. I'm really out of my depth. I'd be happy to test and report back if I had some advice on a process. \$\endgroup\$
    – Ed Krohne
    Dec 5, 2014 at 9:50
  • \$\begingroup\$ Umm, English is not my native language, can you tell more detail about "holding magnet", or some links are just OK.:) \$\endgroup\$
    – diverger
    Dec 5, 2014 at 10:01
  • \$\begingroup\$ catalog.apwcompany.com/viewitems/electromagnets/…? Here it is. Pass current through it, and it turns into a magnet which can hold to a strike plate (also linked on that page). They can be used to pick up and release anything with a strike plate mounted on it. \$\endgroup\$
    – Ed Krohne
    Dec 5, 2014 at 10:13
  • \$\begingroup\$ Work is done when you pull two stationary magnets apart too: the energy is in the form of increased potential energy between the two magnets. I don't think it's any different if one or both parts are electromagnets. \$\endgroup\$ Dec 5, 2014 at 10:57

2 Answers 2


You may know the formula

$$ U_L(t)=\,L\frac{dI}{dt} $$ for the voltage over an inductor.

  • One consequence: if you stop the current flow through an inductor, e.g. by a switch, you get a high voltage peak, which may damage things.

However, this formula comes from the change of magnetic flux over time:


where L is considered constant over time. If not, you get

$$ U_L(t)=L\frac{dI}{dt}+I\frac{dL}{dt}$$

The problem is that you have no idea how the inductivity L changes over time. It will change non-linear over distance between coil and plate. Also, force on the plate increases when it's near the coil, so the speed does, leading to an even higher change of L.

Even if we assume a linearity over time, the solution of the equation is ugly.

I tried to write a simulation allowing to specify the behavior of L over time, but I have to think about the result, as I'm currently not sure if it makes sense. I'll let you know.

However, you should consider that at one point, the plate gets energy from your coil / circuit, and on the other point, it gives energy back. This can lead to voltage spikes, even in both directions, so I would not only use a flyback diode, but also a zener (with voltage above supply voltage).

I'd also suggest to measure it with a scope.


I was on a long tour now, but last Friday I had the chance to play in our lab for a short time.

We have several reels of enameled copper wire, the problem is to find one with both ends of the wire being accessible. I found only this one:

  • wire diameter: 0.22mm
  • wire resistance: 200 Ohm
  • solenoid diameter: 3cm
  • solenoid length: 3cm

I connected it to a constant voltage supply via a 2kOhm resistor and applied 50V to get at least a little current. There is the voltage over the coil when inserting and removing an iron screw:

enter image description here

The scope was set to AC coupling, so you don't see the ca. +5V base line.

It is clearly visible that there are spikes in both directions . When inserting the screw, the coils also sucks it in and consumes electric energy. When pulling out the screw, I invest energy into the system, and the coil propagates it to electric energy, resulting in the negative spike. It is also interesting that there is some kind of relaxation effect with inversed polarity after the spikes.

I have to mention that this setup is not comparable to your holding magnet. My coil is not really a magnet, as I notice no force on ferromagnetic material. My coil also is just an air coil, and as the hole in the reel has less than 1cm diameter, the screw as less, too. So I did not fill the full volume of the coil with material. (BTW: As it is hard to hit that hole with that screw, I could not push the screw in so fast, and so, the first peak is smaller than the second)

Your holding magnet is stronger by several orders, and so is the inductance. There is a yaw completed to a full yaw by the plate, so the effect of the plate will also be much larger than for my setup.

So, I'm sure that you will get really big spikes in both directions, which may damage your circuit, if it does not handle them.

  • 1
    \$\begingroup\$ I like it (+1) and it's what I was going to answer. One thing you could do is to measure the inductance of the activated coil both with and without the iron plate in place. you'd then have a number for delta L. Then the unknown is how fast the plate is removed. That's a bit of a guesstimate... but one could try some different numbers.. 1ms maybe? \$\endgroup\$ Dec 5, 2014 at 13:36
  • \$\begingroup\$ I'm thrilled that I inspired an investigation. I actually got into a conversation with the president of the company that sold me the electromagnets (APW company) and he told me that the magnet was a resistor and no kick would be measurable. I'm not about to argue with him, but that sounds tough to imagine. I still haven't managed to gain access to a scope, but I do measure a small, brief negative voltage when I move a weak magnet away from the electromagnet. \$\endgroup\$
    – Ed Krohne
    Jan 9, 2015 at 0:41

First thought is to think of this like an electric guitar pickup; A permanent magnet produces a constant field and when the strings move this field is slightly modulated and the result is that a small signal appears across the terminals of the coil. Would it matter if a constant current generator were connected to the coil and this generated the same static magnetic field?

No, I don't think there is a difference - the compliance of the current source would still allow the same signal to be produced across the terminals of the coil when strings move.

So, in the question, there is a DC electromagnet pulling on a magnetizable plate. There is a force of attraction and as the plate gets closer both the force increases as does the localized flux density. Looking at this in terms of a fixed magnet with a coil wrapped around it, the plate moving towards the coil/magnet will cause the local magnetic flux to increase and this will produce an emf pulse in one direction in the coil. As the plate moves away, the flux density decreases and this causes an emf pulse in the other direction.

The emf is a pulse because it is only generated while flux is being changed. Faraday's law of induction!

Going back to the electromagnet scenario (rather than physical magnet and coil), the effect of this "internal-to-the-coil" emf is seen at the terminals if the power feed is a current source just as a guitar pick-up works. However, because the electromagnet is voltage fed, the voltage pulse forces a current into or out-of the source depending on which way the plate moves.

Given that there is the normal DC current of the electromagnet, this pulse of current (limited by the self inductance and resistance of the coil) will cause a momentary increase/decrease in that current. This will be seen along the power rails to the coil.

So, the coil is energized and just sat there minding its own business. Then, the plate comes along and rapidly moves to the coil due to magnetic forces. This cause a modulation in the current taken by the coil BUT, importantly no voltage spike because the coil is energized with a voltage source via a transistor or a switch.

If you pull the plate away then there is another current pulse but for the reasons above there won't be a voltage spike.

Next, you open circuit the coil and immediately your flyback diode catches the back-emf - will the plate becoming detatched at this point make things any worse - no!

Does a relay need a special form of coil protection other than a flyback diode - no!

  • \$\begingroup\$ Well, relays aren't usually physically forced open, so even if there's a forward voltage spike when you do that, you wouldn't have to guard a relay against it. If I understand correctly, you're claiming that current is modulated but there's no voltage spike because the voltage source is strong enough to handle it. Wouldn't that depend on the power supply and how big the modulation is? \$\endgroup\$
    – Ed Krohne
    Dec 6, 2014 at 12:21
  • \$\begingroup\$ @EdKrohne if you are concerned then also put a diode from coil to ground. \$\endgroup\$
    – Andy aka
    Dec 6, 2014 at 13:28

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