Electrical – What the purpose of a diode in parallel with a power switch

In your first example, before the switch opens, there is a current flowing through L. When the switch opens the current wants to continue flowing in the same direction so this has to create a positive voltage on the right-hand side of L relative to the left hand side (which is fixed at +Vs).

This positive voltage causes current to continue to flow and this current finds the path of least resistance through diode D. It continues to flow until all the magnetic energy in the inductor is gone then current = 0 and the voltage on the right hand side of L is the same as the left hand side i.e. +Vs.

In the 2nd example with the H bridge, all the diodes are needed because it has to cater for scenarios when the motor is deactivated i.e. pin 3 and pin 6 are open circuit. This is the simplest way to explain it I think. The motor current may have previously been upwards or downwards so all four diodes are needed to catch the back emf when the motor is switched off.

Also, current flow back into the H-Bridge circuit does not appear very practical.

That's how they work.

The left circuit (A) is what he means.

The Zener diode limits the voltage across the inductor to its nominal value. The other diode is there just to block the Zener diode in the forward direction.

Concerning the decay time, this is better than just having the normal diode, because a normal diode just yields a voltage of ca. 0.7 V across the diode. For fast decay, we want the maximum tolerable voltage (e.g. as much as the switching transistor can handle) in order to have high power which means dissipating energy in less time.

More detailed explanation:
Let's consider the time after the switch opens (after the coil has been energized):
The current through the inductor drops suddenly which results in a reverse voltage across the inductor ("reverse" means that now the bottom terminal is more positive than the top terminal). That voltage can damage the transistor (acting as the switch) or create unwanted sparks etc.
In order to avoid or limit that voltage it can be clamped by a normal diode (circuit A without Z-diode). What actually happens is that the total energy, \$E = \frac{1}{2}LI^2\$, stored in the inductor must be dissipated. Energy is the integral of power over time \$E = \int P dt\$, i.e. it takes some time depending on the dissipating power and the time is shorter if the power is higher. The power on the other hand is the product of voltage squared over resistance (inductor and diode resistance) \$P = \frac{V^2}{R}\$, i.e. we have a higher dissipating power, i.e. higher energy decay rate, if the voltage is kept higher (circuit A with a Zener diode).

Here is an LTspice simulation of the situation

• without a Zener diode (left sub-circuit) and
• with a 10 V Zener diode (right sub-circuit)

The top graph shows the power dissipation in both inductors. The area under both power curves must be the same (= energy stored in the inductor).

The bottom graph shows the switching signal (red) and the voltages at the bottom terminals of the inductors.