Electronic – Zener diode – Zener voltage vs reverse voltage specification

It's just the basic operating principle of a diode. An ideal diode would allow current to flow in one direction but block current flow in the other direction. This is based on how it's made, with a p-type region, an n-type region, and a depletion zone in between. Like the bottom diode in this picture:

When you apply a some voltage, in your case 0.9V then the p-type holes and the n-type electrons move into the depletion region because they are repelled by their respective battery terminal. With enough voltage (0.9V in your case) the free electrons in the depletion region get moving and current begins to flow like this:

Now in the ideal case if you were to reverse that battery the opposite will happen and you'll get no current flowing:

In the real world though you can only apply so much reverse voltage or push before you hit the breakdown voltage and current begins to flow freely in the reverse direction. Zener diodes take advantage of this fact and are constructed to breakdown at lower voltages such as your 3.3V.

Sources:
You can read more about how zeners are made here Or see the article I got all the pictures from here

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.