# Calculating voltage spike of an inductor

Suppose there is a circuit like this where the inductor has no resistance

In this circuit, if Switch 1 is initially closed, then there is

10/100 = 0.1 A

of current flowing through the inductor.

Then, if Switch 1 is instantly opened and Switch 2 is instantly closed, the inductor discharges through the resistor.

However, I have read that since the current through the inductor cannot instantly change, that must mean the instantaneous current flowing through the inductor after Switch 2 is closed must be the same as before.

Am I correct when I calculate the instantaneous voltage across the resistor the moment after Switch 2 is closed as

V = I * R

V = 0.1 * 10000 = 1000V?

Also, does this mean that the instantaneous voltage has nothing to do with the inductance of the inductor?

• In practice the properties of the inductor like leakage inductance is going to have a significant impact on what sort of voltage you will get. – squarewav Apr 1 '16 at 8:49

For the circuit you have drawn, yes, the voltage spike has nothing to do with the value of the inductor. It is given entirely by the value of the current in the inductor, and the resistive load R2 across it.

What does that mean if R2 is a higher value, or even absent? In theory, with what you have drawn, if R2 was open circuit, then the voltage spike would be infinite. As you can guess, that doesn't happen in real life.

In practice, there are two things omitted from your drawing.

a) the stray capacitance across the inductor, and due to any wires from the inductor terminal to ground
b) any breakdown mechanism for your switch

capacitance

As the switch opens, the current will start to charge the stray capacitance, which will limit the rate of rise of the voltage. For large value inductors, with many turns in close proximity, this capacitance can be surprisingly large.

Sometimes an external capacitor is added to the inductor deliberately to reduce the rate of voltage rise.

No matter whether your switch is mechanical one with opening contacts, or a semi-conductor one like a MOSFET, it will not support an infinite voltage.

the switch

Mechanical switches are especially poor at breaking the current flow, as at the first break, the contact separation is very small, and an arc needs very little voltage to form. This arc will keep the current flowing, and damage the contacts. It is responsible for switch and relay failure, unless controlled.

In the old-style contact breaker car ignition system, the 'points' that connected and disconnected the coil to the battery could be subject to excess erosion from arcing. Often, the first sign that your 'condenser' (capacitor) had failed would be excessive wear at the points. The capacitor, fitted across the points, slows the rate of voltage rise so that the points are sufficiently far apart before the voltage gets high enough to create an arc.

The specifications for a MOSFET will typically give a breakdown voltage figure. Good ones will also give an energy they can withstand when the breakdown voltage is exceeded. As long as the stored energy in the inductor is less than that figure, a MOSFET can switch current off to a coil, limit the open circuit voltage to its breakdown voltage figure, and survive.

Short answer: Yes and Yes

When you close the switch 2, the current that flow in the circuit is given by this equation:

$$i = Io.e^{-t/\tau} \\ \tau = \frac{L}{R}$$

with initial conditions given by: $$i = 0.1A~~ when~~t = 0~~ and \\ i = 0~~ when~~ t = \inf$$ which deduce that equation to: $$i = 0.1*e^{-200,000*t}$$ as you can see, (tau) T = 5us, which is very small. In practice, that means that after 5us, the voltage at resistor will decay 63% of original voltage, i.e., 37% of 1000 = 370 V. After 3*5us = 15us, the voltage will decay 95% of original voltage, i.e. 5% * 1000 = 50V. And, after 5*5us = 25us, only 1% will remains on the resistor, i.e. only 10 volts.