Electronic – Is back emf inductance in a ignition coil circuit symmetric for on/off cases

back-emfinductance

I was trying to understand how an ignition coil circuit works practically, and came upon this website explaining the behavior of an ignition coil when subjected to changes in voltage:
http://www.learnabout-electronics.org/ac_theory/inductors.php

Especially the following paragraph:

This voltage will however now be much larger than the original supply voltage; this is because the amplitude of a voltage induced into a conductor is proportional to (among other factors) the rate of change of the magnetic field. At switch on, because there were two opposing voltages changing, the supply increasing and the back e.m.f. decreasing, the rate of change was slowed down. However at switch off there is no supply voltage so the magnetic field collapses extremely quickly causing a very rapid rate of change and therefore producing a very large voltage pulse.

What bothers me is that according to my rusty EE knowledge, the equations governing inductance
$v_r = L \frac{di}{dt}$

should lead to symmetric behavior when a switch is closed and opened.
In fact, according to my understanding, the voltage spike shown in the following graph should be in the other direction, as it would oppose the decrease in current. Am I understanding inductors incorrectly, or is there in practice much more to the topic which simple EE theory does not cover?

Image taken from http://www.learnabout-electronics.org/ac_theory/inductors.php

Best Answer

No, it isn't symmetric in that sense.

In a circuit that consists primarily of inductance and resistance, there's a time constant associated with its transient response1 that works out to

$$\tau = \frac{L}{R}$$

In the igntiion circuit, you basically have a battery, a switch and the ignition coil in series. \$R\$ represents the total resistance of all three. Two of them have constant resistance, but the switch varies between very low and very high resistance.

When the switch is closed, the total \$R\$ is low (dominated by the resistance of the battery and the coil), which means that \$\tau\$ is relatively large — the current changes more slowly. When the switch is open, the total \$R\$ is extremely large (dominated by the switch), so \$\tau\$ is very small and the current changes very rapidly.

Therefore, since the voltage is determined by

$$V(t) = L\frac{dI(t)}{dt}$$

when \$\frac{dI}{dt}\$ is large, so is the voltage.


1 The transient response is of the form

$$I(t) = I_{\infty}+(I_0 -I_{\infty})e^{-\frac{t}{\tau}}$$

At t = 0, I(t) = I0 and at t = ∞, I(t) = I.