What keeps the kickback voltage from reaching an infinite voltage?

Question

We know that the voltage over an inductor is defined by the formula:

\$V = L * \frac {di}{dt} \$

So in the case where the current flow is suddenly interrupted (like when a mechanical contact is opened), voltage spikes occur in real life.

However, this is not always the case: we don't see arcs happen in small inductive loads. (By small inductive loads I mean a toy car motor, for example.) However, the formula says that the \$ \frac{di}{dt} \$ term should approach infinity when mechanical contacts are opened, therefore the \$L\$ term (which should be small in small inductive loads) shouldn't have a significant effect. Simply, we should be able to see sparks any time we open any inductive load - independent of the inductance.

What are the practical factors that stop the voltage from reaching infinity? Does the current flow actually decrease slower, or is the formula perhaps insufficient for such a "discontinuity"?

Answer 1

A real inductor looks like this (shown below is an inductor with 4 coils) there is a small amount (usually in the pF-fF range) of capacitance between each coil. Each piece of wire also has some resistance associate with it.

Because each coil in an inductor has resistance (or each section of wire if you consider one coil) this impedes the current and reduces the voltage. The small amount of capacitance will also store some of the voltage and prevent an instantaneous change in voltage.

These all soak up energy that prevents the Electro Motive Force (EMF) that has been stored around an inductor from generating an infinite voltage. An inductor can actually be simplified into a circuit such as the one to the left below.

^{simulate this circuit – Schematic created using CircuitLab}

A superconducting coil would be able to generate much more massive voltages because of much lower losses due to parasitics.

Answer 2

Any energy storage system (an inductor) has non-zero size.

Anything of non-zero size has non-zero electric fields, or capacitance. Device junctions are usually a large source of parasitic capacitance. Flyback systems use a diode to transfer energy into a load capacitor.

At peak voltage excursion, all the inductive energy has (1) been dissipated as heat (2) been radiated as EM field (3) been stored in the electric-field of the intentional and the parasitic capacitances.

Answer 3

The series resistance matters a great deal with the "kickback" voltage due to the series capacitance of the "switch" when opened. This forms a classic series RLC resonant circuit which has properties of voltage gain by impedance ratio of

\$Q=\dfrac{|X_C|}{R} = \dfrac{|X_L|}{R}=\dfrac{\omega _0 L}{R}\$ at resonant frequency \$\omega _0= \dfrac{1}{\sqrt{LC}}\$

For the situation of kickback voltage peak, it can be proven that \$|V_p| = Q * V_{dc}\$ for Quality Factor, Q (above) and loop supply voltage Vdc at some resonant frequency.

When de-energizing a circuit with a contact switch as t goes to 0, V/L=dI/dt, V does not go to infinity due to this parasitic capacitance.

Example

^{simulate this circuit – Schematic created using CircuitLab}

e.g. Consider a series circuit, Vdc=1V, L=1uH, R=1 Ohms , Idc= 1A. What is the switch voltage kickback, when just opened, if Csw = 1pF?

1V , 100V, 1kV, 1e6 V or infinite?

Now consider the same for a FET switch with 1nF output capacitance with RdsOn << 1% of R=1. What is dV?

p.s. if you learnt something, then comment your answer.

The intuitive answer is that the switch goes from a conductor to a tiny stray capacitor which limits the slew rate of the voltage and as does the inductor limit the slew rate of the current and at their resonant frequency the voltage gain, Q at ω0 is inversely proportional to R, so bigger series R dampens the voltage.

Answer \$ V_p= I_{dc} \sqrt{\dfrac{L}{C}} \$ = 1A * √(1uH/1pF)= 1kV

Misc

It can be proven the open circuit impedance like a transmission line "characteristic impedance" \$ Zo= \sqrt{\dfrac{L}{C}} \$

We see the voltage kickback looks like Ohm's Law. \$ V_p = I_{dc}*Z_0\$ The peak voltage Vp, generated from interrupting an inductive current, \$I_{dc}\$.

Answer 4

Just consider a simple example of 100 uH and 1 amp flowing. When the contact in series with the inductor opens, there may be 5 pF of parasitic capacitance left across the inductor and that 1 amp will create a high kick-back voltage but how much?

$$I = C\dfrac{dV}{dt}$$

So potentially (no pun intended) the voltage across the 5 pF capacitor could rise at a rate of 200 kV/microsecond. Given that its starting voltage is potentially neglible in comparison, within a few micro seconds a pretty big voltage could develop. However this is mitigated by the lack of energy stored in the inductor: -

$$W = \dfrac{L\cdot I^2}{2}$$

Or 5 micro joules. All this energy will cyclically transfer to the capacitor and we can equate the capacitor energy formula to 5 uJ to give us the maximum voltage: -

$$W = \dfrac{C\cdot V^2}{2}$$

This produces a peak capacitor voltage of 1414 volts.