Inductive kickback

Question

I know this is a very basic concept, and has been described in many places, but I still can't 100% understand inductive kickback.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

If the switch in the circuit above is initially closed, and then immediately opened, we will have an instantaneous drop in current. Using the following equation describing voltage across an inductor: $$V = L\frac{dI}{dt},$$ we would get the voltage at point A to be a high positive number, to induce a negative voltage across the inductor. Since inductors are suppose to resist changes in current, how would a very high voltage at point A maintain the flow of current? Wouldn't it result in current in the opposite direction?

score 2 · Accepted Answer · edited Oct 02 '18 at 01:49

Once you close the switch, the inductor will face a really high $$\frac{Di}{Dt}$$, and from that the inductor will try to maintain the current flowing to the same direction yielding a high voltage(which can be understood from the inductive voltage formula you provided). This kickback can be understood from Lenz-Law since the change in current causes a change in magnetic field and as the inductor faces a change in the opposite direction, this field will force a current in the same direction it was flowing in the beginning, which is entering point A.

So answering your question: No, the current will still flow to the same direction.

Once the inductor tries to maintain this current flowing, the switch is now going to behave as a high resistance path, and for the current to flow there should be a really high voltage produced across the switch, which causes the famous arc, a phenomenon when the voltage is too big to the point of breaking that little gap insulation, causing dielectric rupture.

Got it. Having a high voltage across the switch makes sense. But what prevents current from flowing in the reverse direction? Since there is a voltage drop in that direction as well? Also, let's say we add a kickback diode, that will clamp point A at about V+ + 0.6V. What's forcing the current to specifically flow up through the diode and down the inductor, rather than up through the inductor? — user198450, Oct 02 '18 at 01:29
@user198450, the current starts out flowing down the inductor when the switch is closed. It can't change instantly because \$\frac{dI}{dt}=V/L\$ and \$V\$ is finite. So it keeps flowing downward when the switch opens, either by creating an arc across the switch or by flowing up through a diode that you added to the circuit. — The Photon, Oct 02 '18 at 01:36

Andy aka · Answer 2 · 2018-10-02T07:49:44.700

If the switch in the circuit above is initially closed, and then immediately opened, we will have an instantaneous drop in current

No, you might have a high rate of change of current but it won't be instantaneous due to there always being a practical current flow through the coils parasitic capacitance (causing a ringing waveform) and the voltage breakdown of the switch as it opens will produce a spark that continues the current.

how would a very high voltage at point A maintain the flow of current? > Wouldn't it result in current in the opposite direction?

The coil turns into a generator and forces current in the same direction it was previously travelling (albeit decaying quite rapidly). This means that point 'A' produces a big positive voltage so that current is forced in the same magnitude and direction through the external circuitry connected around the coil.

Inductive kickback

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