What Happens to an Inductor When the Circuit Is Broken?

[goc9000]

Hi all !

I have a question regarding the behaviour of an inductor upon breaking the circuit it belongs to. Now I don't know that much about electronics (I've just started formal training in this area), so I might have gotten some things wrong. If this is the case, I would appreciate corrections :)

As far as I've understood, an inductor can be thought of as a sort of a "magnetic capacitor", in that it picks up energy while being powered up, stores it, up to a certain limit, in the form of a magnetic field, and returns the energy to the circuit as the magnetic field collapses. The inductor transfers energy to and from the field in such a way so as to resist changes in current flow.

Now this sounds pretty simple, but I'm having trouble imagining what will actually happen in a very simple situation. Let's assume that we have a current source connected to an inductor. It can be safely assumed that the inductor's field won't affect other parts of the circuit. Now when the 'juice' is turned on, the inductor will gradually charge its magnetic field until it is 'saturated' (AFAIK, the greater the intensity of the current, the more energy it can store). Now what happens if we simply break the circuit (via a switch) ? The "global" current intensity through the circuit will drop to 0 extremely sharply, which implies that the inductor must release all of its energy (the expression indicating the maximum amount of energy it can store becomes 0). The inductor will try to prevent the loss of current flow by releasing energy from the stored field, but how can it generate current if the circuit is broken ? Will it generate a current locally and cause electrons/positive charges to "pile up" on either sides of the gap in the circuit ?

Many sources mention that in a real-life case, that the inductor would deal with this by temporarily forcing a closed circuit, in that an electric arc would occur between the ends of the "gap" in the circuit (i.e. through the switch). But what's really going on ? I'm actually interested in what happens if the resistance of the gap is way too high even for electric arcs to occur.

Now if my hypothesis is true, I guess that the final state would be unstable. Would the electrons then surge back through the inductor in the opposite direction, thereby creating a sort of an oscillator until all energy is dissipated through heat ? Would the charges somehow get distributed evenly across the conductor, so that there is no current and therefore no magnetic effect to worry about ?

Thank you all for your attention.

 

[pervect]

If you interrupt the current to an inductor, it will indeed generate a very high voltage in an effort to keep the current flowing.

In many cases, when you try and switch the current off suddenly in a large inductor, the switch will indeed fail, and allow some current to pass, as you have already mentioned.

If the switch does not fail, the energy stored in the inductor goes into stray capacitance. (There is always some stray capacitance to ground in any circuit). So the voltage will not become infinite - it will be limited by the stray capacitance. Since the exact value of the stray capacitance depends strongly on the geometry of the inductor and it's immediate surroundings, it's hard to calculate the details - generally it will be "high" and the exact value will depend on a lot of factors. (For instance, standing near the circuit might will influence the stray capacitance and hence the voltage it generates.)

Electronic transistors that have to switch inductive loads have to have some sort of safety device to keep the voltage from getting too high and ruining the transistor. A normally reversed biased diode from the inductor to the positive supply is one solution (a diode clamp), but it's not always a good one if it is desired to turn the inductor off quickly (V=Ldi/dt, and the voltage from a reversed biased clamp diode is low, meaning di/dt is low).

Various possibilities for protecting the transistor exist, a few examples are adding an resistor/capacitor "snubber circuit" to store the extra energy in a capacitor, or putting in a zener diode with a series resistor to ground to bleed off the energy.