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

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 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? 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.

• Oh, considering the switch as an infinite resistance was just the insight which was needed, thanks! And therefore the voltage spike in the plot mentioned in the original question is the inductance 'trying' to drive current through an infinite resistance (to oppose the sudden variation). Shouldn't the peak voltage still be of the same sign as the continuous regime voltage in the inductance, in that case? Is the sign change in the plot above (original question) therefore due to being measured at the battery, or some sort of LCR oscillation ? – Dugas Jul 1 at 14:28
• No. In an inductor, it's the current that cannot change direction quickly. The voltage can change arbitrarily -- it has one polarity when the current is increasing, and the opposite polarity when the current is decreasing. This is directly related to the sign of the derivative. – Dave Tweed Jul 1 at 14:43

Current limit is defined by V/DCR of coil and is not saturated. Then L/R has slow rise time and fast flyback as R opens.

• but instead of high side drive, it is low side drive with engine ground and +Ve ignition voltage to >35kV or until detonation under 8 or 9 atmospheres of compression with fuel and current limited by carbon resistor wire ~50k.

• ringing is a sign of LC resonance and damping shape factors determine condition of actual ignition analyzed by Engine Tech. On scope. Sync’d to index pulse for PLL controlled sweep rate with variable RPM loading ignition detonation/ preignition and fuel burning.

• Plug like any arc gap looks like an SCR or actual a DIAC but much higher threshold with activation energy releasing electrons above BDV with negative incremental resistance to go from a gas insulator to a conductor until holding current is extinguished with fuel burn.

X yes it is basic EE theory with inductors and flyback effects amp-seconds input = output minus losses in cable to reduce EMI on AM radios.

But we do no integrate to infinity, your engine warranty isn’t that long. ;)

During the off-to-on transition, the inductor voltage is determined by the battery voltage. That causes the current to ramp up at a specific rate.

During the on-to-off transition, the battery voltage is irrelevant: it’s not even connected. The voltage can rise higher can higher until it drives the spark.