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I'll preface by saying I'm aware of questions like these 1, 2, and 3, but none of them answer my specific question.

All of these questions say that the kickback from an inductor can be given by:

\$V=L\frac{dI}{dt}\$

However, I've seen other sources say things along the lines of "the inductor tries to keep the current flowing through it the same as it was before switching". The exact quote from the above source is:

this component now leaps into action and does everything in its power to maintain the status quo (keep the current flow going just as it was).

To me, keeping the current flowing exactly as it was means the current is the same. As in, if the steady state (i.e. after several time constants) current through the inductor was 1 A, then the inductor would try its hardest to keep the current at 1 A (at least at the instant after the switch switches). The inductor therefore acts as an instantaneous current source. If you had a resistive load connected in parallel with the inductor, the voltage over it would be \$V=IR\$ where \$I=1\text{ A}\$. This voltage will begin to drop though as the current is only 1 A for an instant.

To aid in understanding, I've drawn a circuit below. In this case, I would expect the voltage over the inductor/resistor after switching the current source off to be \$V = IR = 100 \text{ V}\$.

schematic

simulate this circuit – Schematic created using CircuitLab

But, as far as I'm aware, \$L\frac{dI}{dt} \neq IR\$. Since everyone says that \$V=L\frac{dI}{dt}\$, I'm inclined to think that it is correct, but I'm struggling to understand why \$V=IR\$ is wrong on this occasion. Surely the parallel resistance must come into play, otherwise flyback diodes would do nothing (as they would count as low resistance, therefore low kickback).

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    \$\begingroup\$ "..if the steady state ... current through the inductor was 5 mA, then the inductor would try its hardest to keep the current at 1 A". How did you jump from 5 mA to 1 A? \$\endgroup\$ – Transistor Nov 18 '20 at 5:40
  • \$\begingroup\$ @Transistor my bad. I changed the numbers but I guess I missed that one \$\endgroup\$ – JolonB Nov 18 '20 at 7:46
  • \$\begingroup\$ Mistake 1. LdIdt≠IR is wrong equation because you used I on both sides instead left side (in equation ) it should be current (function) through inductor and in right side it should be current (function) through resistor \$\endgroup\$ – user215805 Nov 18 '20 at 10:56
  • \$\begingroup\$ There will be always be some voltage and some di/dt for which Ldi/dt = IR. \$\endgroup\$ – user_1818839 Nov 18 '20 at 14:32
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    \$\begingroup\$ Do you have Flash player? If so, visit this page and play with the Flash movie at the bottom; it is about the inductor property discussed here. \$\endgroup\$ – Circuit fantasist Nov 22 '20 at 21:57
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To me, keeping the current flowing exactly as it was means the current is the same.

The inductor "tries" to keep the current constant, or "does everything in its power" to keep the current constant. That doesn't mean that it actually keeps the current constant.

Similarly, a resistor "resists" the flow of current. That doesn't mean the current through a resistor is always 0.

The inductor therefore acts as an instantaneous current source.

Correct. If you implement what SPICE calls a transient analysis, you will find that at each time step, the inductor is treated as a current source (representing this tendency to keep the current constant) in parallel with a resistor (representing the possibility, if the inductor is not shorted, of the current changing infinitesimally before the next time step).

Since everyone says that \$V=L\frac{dI}{dt}\$, I'm inclined to think that it is correct, but I'm struggling to understand why \$V=IR\$ is wrong on this occasion.

An ideal inductor doesn't have any parameter "\$R\$", so there's no way you could write the equation \$V=IR\$ to describe the ideal inductor.

\$V=IR\$ is an equation that describes the behavior of an ideal resistor.

\$V=L\frac{dI}{dt}\$ is an equation that describes the behavior of an ideal inductor.

In some particular circuit, \$V=IR\$ might well describe what is happening to a resistor that is connected in parallel with an inductor. But it can't describe what's happening to the inductor itself, because an inductor is not a resistor, and doesn't even have a parameter \$R\$ to determine its voltage or current.

I've drawn a circuit below. In this case, I would expect the voltage over the inductor/resistor after switching the current source off to be V=IR=100 V.

This is not correct. If \$V\$ is the voltage at the terminal where the resistor and inductor are connected, but not the one that is designated as ground, then for current to continue flowing top-to-bottom through the inductor, the resistor current must be flowing bottom to top. And thus the voltage must be -100 V, not +100 V.

Which is a good thing, because if it were +100 V, then the inductor current would be increasing rather than decreasing, and the whole circuit would be violating conservation of energy (since the resistor and inductor would both be absorbing power but nothing would be providing power).

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    \$\begingroup\$ @JolonB, since the inductor and resistor are in parallel (after the source is turned off), the voltage across them is the same. So if you wrote a mesh equation for the loop they form, you'd find that \$IR=L\frac{dI}{dt}\$. But this is because of the way you connected the inductor and resistor to each other, not an inherent behavior of either component on its own. \$\endgroup\$ – The Photon Nov 18 '20 at 2:46
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    \$\begingroup\$ Note that \$V\$ will actually be negative, and current will be flowing though the resistor from bottom to top. \$\endgroup\$ – The Photon Nov 18 '20 at 2:47
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    \$\begingroup\$ @JolonB, by choosing a different resistor. \$\endgroup\$ – The Photon Nov 18 '20 at 3:41
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    \$\begingroup\$ @JolonB, In the instant after the current source shuts off, the inductor current doesn't change. This means that current begins circulating in the inductor-resistor loop. And the voltage across the inductor is equal to the voltage across the resistor (nothing stops the voltage changing instantaneously). Then the current will drop with time constant L/R. If the supply shuts off quickly enough, the decay of the current will depend only on the inductor and resistor values, not on how quickly the supply changes from the intitial value to zero. \$\endgroup\$ – The Photon Nov 18 '20 at 19:59
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    \$\begingroup\$ @ThePhoton The cleared up a lot for me, but that last sentence confused me a bit. When I switch it off quickly, all the current flows through the resistor the instant after switching. If I switched it off slowly, how could I find the voltage over the resistor/inductor? I'm assuming it'll still be \$V=IR\$, but would \$I\$ be smaller in this case? \$\endgroup\$ – JolonB Nov 18 '20 at 20:20
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Without a component across the inductor, at transistor switch off, the voltage across the inductor will rise to be V = Ldi/dt.

With a resistor in place across the inductor, at transistor switch off, the voltage across the inductor (and resistor) will rise to be V = iR.

The resistor would usually be sized to make iR less than Ldi/dt without the resistor. That is to say a resistor would be put in place to limit the back EMF voltage. So typically iR is not equal to what Ldi/dt would be if there was no parallel resistor.

The resistor is being used to limit the back EMF so that it is below the VCE(max) of the driving transistor.

A resistor is used rather than the more usual parallel diode approach because the resistor technique speeds the decay of the inductor current.

di/dt = V/L and so rate of change of current, di/dt is proportional to the voltage across the inductor.

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  • \$\begingroup\$ The inductor always enforces the rule that V = Ldi/dt. It doesn't decide that sometimes it will ignore the rule. It ALWAYS enforces that rule. Whenever two components are in parallel (meaning they have the same voltage) the voltage will have to satisfy both of them. So V=IR AND V = Ldi/dt. \$\endgroup\$ – mkeith Nov 18 '20 at 16:47
  • \$\begingroup\$ @mkeith I totally agree, I've edited to clarify. \$\endgroup\$ – James Nov 18 '20 at 17:36
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You are right about everything as far as I can see. If you have 100 Ohms in parallel with an inductor, and there is 1A flowing through the inductor, then you open the switch that was supplying the current, then you will have 100V across the resistor and inductor (briefly) because the only path available to the inductor current is to loop through the resistor and inductor. The current circulating in this loop will then decay exponentially with a time constant of R/L. As the current decays, the voltage will also decay.

If there is nothing in parallel with the inductor, then you may end up with very high voltage when you open the switch. An arc may even form on the switch.

Whenever you switch something inductive, you want to provide some place for the current to go. Often a freewheel diode will be used for this (you can search for that).

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Consider the mechanical analog of flowing water. If you have a large mass of water, equivalent to a large current, but at a low speed, and you put an obstacle in it's way (e.g., a steep gradient or open switch), the water will rise to a much higher level, equivalent to an increase in voltage.

This abrupt change with time is \$V=L\frac{dI}{dt}\$, where dI is the change in current in "interval" dT. Of course, the time subdivision is infinitesimal in calculus, which is required to better understand this phenomena.

Abrupt shelving can cause a tsunami barely above sea-level to reach 500 meters, and the inductive "kick" of a doorbell running from 6 VDC can produce surge voltage of hundreds of volts. Put a neon lamp, e.g. NE-2, which requires ~70 V to light, across the contacts of a doorbell or buzzer, and it can light with input of just a few volts DC.

To further investigate how inductors work, and the use of a commutating diode, see Lekule Blog for a nice lab exercise!

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