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When the current flowing in an inductor is interrupted the voltage will rise until (typically) there is a flashover or insulation breakdown in the switch. Assuming extremely good insulation what would limit the voltage rise in practice? I can't really believe it would head into the megavolt region.

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    \$\begingroup\$ self capacitance \$\endgroup\$ – Marko Buršič May 28 at 8:55
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    \$\begingroup\$ Lets have some numbers; 10 milliHenry and 10picoFarad (perhaps those old auto ignition systems) and 1 amp in the inductor. 0.5 * L * I ^2= 0.5 * C * V^2, and we solve for V = I * sqrt(L/C); thus V = 1 * sqrt(10 milli/10pico) or about 30,000 volts. \$\endgroup\$ – analogsystemsrf May 29 at 2:38
  • \$\begingroup\$ Mainly the breakdown voltage of whatever is breaking the current - air gap in a switch (circuit breaker), avalanche voltage in a transistor. I think these would be much more limiting than any (tiny) capacitance in the inductance. \$\endgroup\$ – Reversed Engineer May 29 at 11:51
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TL;DR: It also depends on the actual setup.

It can be useful to see the problem from an energy point of view. Since the energy stored in the inductor is \$E={1 \over 2} L I^2\$, when you "instantly" bring the current to zero that energy has to go somewhere. It cannot vanish.

If you model the current interruption as an extremely rapid increase of resistance at the point of the circuit where the switching element is placed, what you get is that the voltage across the inductor begins to raise rapidly. When does it stop raising? It depends on the inductor setup and its surrounding environment.

Marko Bursic has already mentioned self capacitance in his answer.

You say "assuming extremely good insulation", but "extremely good" is still "not perfect". There is resistance and capacitance toward anything surrounding the inductor. You can get an increase of leakage currents due to voltage increase or displacement current into nearby object due to any tiny parasitic capacitance, whose "reactance" becomes low when the rise time is small.

For example If you have a regular, mechanical switch, you may get an arc between the contacts that dissipates the energy into heat, ionization of the air and generation of EM waves.

At high enough voltages, electrons can be extracted from the metal and an electric arc can be formed in vacuum, too (there are also switches that operate using a vacuum arc).

BTW, unless you do want to get high voltage spikes, that's why you put overvoltage protection devices or snubber circuits in parallel to inductors whose current may be interrupted. So, in practice it is you (the circuit designer) that wants to limit the voltage rise.

In a sense, controlled generation of inductor overvoltage spikes is what is done in some step-up DC-DC converters: you "interrupt" the current in an inductor in order to get an increase in voltage, then you "dump" the increased voltage into an output capacitor to "store" it for the load. Of course the switching controller is key to useful operation.

On a related note: it can come as a surprise, but vacuum has not "infinite resistance", i.e. it doesn't impede the flow of current at all! It simply has no free charges to support currents. Once a charge is extracted from a nearby object things change dramatically. This is explained in detail in this article by Charles Chandler. Excerpts (emphasis mine):

The answer is that the vacuum doesn't impede the electrons at all. In a perfect vacuum (except for the test charge), the charged particle's behavior can be calculated by just three factors: 1) its mass, which gives it inertial forces, 2) its electric charge, which makes it responsive to an electric field, and 3) the electric field acting on the charged particle. Then the acceleration of the particle is just the equilibrium between the inertial and electric forces.

[...]

This has been mistaken to be a measure of resistance, which seems to become infinite at low pressures, but this is not correct. At higher pressures, the breakdown voltage increases steadily with pressure (off the right side of Figure 2), and so does the resistance. So there is a direct relationship between breakdown voltage and resistance in that range. If we erroneously take that as a hard and fast rule, and we observe the breakdown voltage shooting up at very low pressures, we conclude that the resistance must be increasing at very low pressures, asymptotically approaching infinite resistance at a pressure characteristic for that gas. But here we have to remember that Paschen was studying breakdown voltages, and made no mention of resistance. If we make direct measurements of the resistance, we find that it varies directly with the pressure, and continues straight down to nothing at no pressure, without the sudden deviation at the threshold discovered by Paschen. This can be confirmed by putting an ammeter into the circuit, and finding the resistance by the volts divided by the amps.

So an inductor placed in an "ideal" vacuum could have its terminal voltage increase enormously, but once the vacuum breaks down, because even a small quantity of electrons is extracted from nearby objects (e.g. the terminals of the inductor itself, for example), a large current may be produced in the vacuum (hence an arc).

That's enough for the concept: the rest boils down of how good are the insulators surrounding the inductor and how big is the extraction potential of electrons from the materials on which the inductor voltage spike is applied.

Some interesting data on this point can be found in this lengthy slide collection from NASA site (it is a set of presentations by several authors); the first part is: High Voltage Engineering Techniques For Space Applications by Steven Battel

In slide #29 (page 25 of the PDF) you get (emphasis mine):

Intrinsic Dielectric Strength Limits

  • The breakdown of Gasses at STP is dependent on type.
    • Air : ~3 kV/mm (75 V/mil)
    • He : ~0.37 kV/mm (9.3 V/mil)
    • SF 6 : ~9 kV/mm (222 V/mil)
  • High Vacuum breakdown is surface and configuration dependent but is in the range of 20 to 40 kV/mm (500 to 1000 V/mil)
  • Ignoring surface effects, Liquids and Solids bulk properties are generally similar in the same range of 15 to 20 kV/mm (375 V/mil to 500 V/mil).

So here you go: even in vacuum the best you can get is a dielectric strength of 40kV/mm. Can this lead to MV voltages? It depends on the setup, but probably something will break down before that happens.

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self capacitance

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ You may want to refer to winding resistance too so as to damp any resultant ring. \$\endgroup\$ – Warren Hill May 28 at 11:07
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If you have a non-physical pure inductor, as in a thought experiment, or in a SPICE simulator, then nothing limits the rise in voltage. Why stop at megavolts, you could go to zettavolts and beyond.

However, a physical inductor has dimensions, and that means capacitance. The capacitance makes the rate of voltage rise finite, and ultimately limits the peak voltage if nothing else does.

Establish a current in the inductor, then open the switch. The current starts flowing into the stray capacitance, and charging it up, at an initial rate of \$\frac{I_{ind}}{C_{stray}}\$ volts per second. With a few pFs stray capacitance, and 1 amp flowing, that's going to be pretty quick.

If something breaks down, the insulation, air between slowly opening switch contacts, then the current will divert to that path, and the voltage will stop rising. If nothing breaks down, then the current will continue to flow into the stray capacitance, and the voltage will rise such the energy stored in it, \$0.5CV^2\$, is (ideally) equal to the original energy stored in the inductor, \$0.5LI^2\$. That's the first quarter cycle of an LC resonance. After that, the circuit will continue to ring, losing energy through dissipation and EM radiation.

If the stray capacitance is enhanced, for instance by the 'condensor' fitted across the points in an old-style car ignition, then the rate of voltage rise will be slower. In this case, the voltage rise rate is slowed to allow the mechanical points to open to an adequate gap before being stressed with the several hundred volts needed for the coil to break down the spark plug.

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  • \$\begingroup\$ If nothing breaks down, wouldn't it end up resonating in this LC tank? \$\endgroup\$ – Huisman May 28 at 9:59
  • \$\begingroup\$ @Huisman absolutely, I was wondering whether to add that to my answer. \$\endgroup\$ – Neil_UK May 28 at 10:08
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    \$\begingroup\$ @Huisman (shameless plug): you can observe the resonance phenomenon in a LTspice simulation of a Zener snubber circuit in this answer of mine. \$\endgroup\$ – Lorenzo Donati supports Monica May 28 at 10:29
  • \$\begingroup\$ What does Quantum Electrodynamics say about this? \$\endgroup\$ – David Tonhofer May 28 at 23:17
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    \$\begingroup\$ @DavidTonhofer I'm not a physicist, but I have some doubts QED can be usefully applied here. These are not quantum phenomena (well, extracting an electron from a metal is, but that's not the key point here), the lumped element model of an inductor is all Maxwell's equations at work, albeit in a very simplified way (quasi-static fields hypotesis, etc. etc.).... \$\endgroup\$ – Lorenzo Donati supports Monica May 29 at 16:21
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Mainly the breakdown voltage of whatever is breaking the current - the air-gap in the mechanical switch (circuit breaker) as it opens (~3 kV/mm for dry air), forward voltage of any flyback diode (if there is one in the circuit), or breakdown or avalanche voltage of a transistor turning off.

If it's a mechanical switch, remember that it's a race between the air-gap as it opens, and the energy in the inductor dissipating. By the time the gap between the switch contacts is only a millimetre wide, there will only be a 3000 Volt spark across it even if there is still energy left in the inductor to support that, as it's magnetic field collapses.

These may be much more limiting than any (tiny) capacitance in the inductance.

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Voltage across inductor will rise to v =L.di/dt

di/dt is determined by how fast the transistor switch switches off and cuts the current off.

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  • \$\begingroup\$ Down markers - You may not like it but that's the way it is! You obviously don't know your Horowitz & Hill! \$\endgroup\$ – James May 30 at 14:06
  • \$\begingroup\$ I don't think downvoters disagree with the truth you are saying. The key point is that you are not trying to answer the question of the OP. Please review the help center on how to answer a question successfully. \$\endgroup\$ – Lorenzo Donati supports Monica May 30 at 15:32

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