# How does a Zener Diode work as a voltage stabilizer?

I mean I do have this diagram though and all the stuff about how it works, but I read it and not all of it makes sense. In my notes it says that when the input voltage increases beyond a certain limit the breakdown occurs, I get that line. But I can't understand the next line which says ""

How does the breakdown increase the potential drop across the resistor ? Shouldn't the potential drop be decreased as the voltage across the zener diode is also high and that due to the DC supply is also high ? And how can the voltage across the Zener diode be constant ? Didn't the supplied voltage just increase beyond the breakdown voltage, that's how the breakdown happens right ? When the voltage across the diode reaches a value higher than the zener diode the covalent bonds are broken and the electrons and holes flow to the opposite sides, so the current increases ? This is very confusing, any books that I can refer to study about semi-conductors

• Would Electrical Engineering be a better home for this question? Mar 6, 2021 at 9:01
• @Qmechanic, this is based on what I read from my high school Physics Textbook Mar 6, 2021 at 9:07

First a zener diode is nothing but a diode, just a heavily doped one at that (Let's get to that later).

When you apply a voltage across a diode, you get an I-V characteristic like this

What we need to focus on is the reverse bias part.

Suppose you apply a reverse biasing voltage. An electron crossing the depletion region (either in diffusion current or drift current) experiences a force towards the +ve side and a hole (assuming as an independent particle itself) towards the -ve side. In net effect this widens the depletion region, increasing the barrier voltage. This reduces the diffusion current( holes moving towards electrons and vice versa) to almost null but slightly increases the drift current (the hole-electron pair generated naturally by the semiconductor in the depletion region) which is still very low because of the small rate of generation of the hole electron pair.

But if you keep increasing the voltage, then the electrons due to high speed collide with the electrons in the covalent bonds, breaking them, hence releasing a lot of electron-hole pairs suddenly.This is termed as avalanche- the sharp spike you see to the left.

Is this useful? On a first glimpse no, as the extremely high energy of electrons results in the diode melting.

However then comes zener diode. It is heavily doped, which causes the depletion region to be really small initially. As above , when you apply a reverse biasing voltage, it breaks down, but at a much lower voltage. So the electrons are not extremely energetic preventing damage to the diode.

The voltage supplied is used up as energy to break the covalent bonds. Also ideally it has no resistance. So it cannot consume more energy than this and hence you cannot increase the voltage across this beyond the breakdown voltage. I am refering to voltage across the zener diode, not across the source.

If the voltage across the source is low, such that you cannot have enough voltage across the zener diode to cause breakdown, then it has negligible current through it, which in practice is just like an open circuit.

However when sufficient voltage is given it has a constant voltage across it, and the rest of the voltage supplied by the source is dropped in the other resistors etc.

Now if we connect a load resistor across the zener diode in reverse breakdown, by kirchoff's voltage law it MUST have the same voltage dropped across it as the diode, which is a constant because as discussed above, for ideal zener diodes you just can't increase the voltage beyond that.

A major misconception in your question is that covalent bonds break when "Voltage across it is higher than the zener breakdown voltage" - It happens at the zener breakdown voltage

• "If the voltage across the source is low, such that you cannot have enough voltage across the zener diode to cause breakdown, then it has negligible current through it, which in practice is just like an open circuit." How does the diode act as a voltage stabilizer for the load in that situation ? Mar 6, 2021 at 9:17
• If the voltage across the diode is low too then doesn't it mean that the voltage across the load will be low but when the voltage is enough to cause a breakdown the voltage is constant and a value higher than the condition when the voltage is low Mar 6, 2021 at 9:19
• @KoustubhJain Zener diode acts a voltage regulator, meaning it prevents voltage from exceeding a value. If V is the breakdown voltage of the zener diode, the voltage across the load ranges from 0 to V. It doesn't exceed V. Suppose an LED gets damaged beyond 2 V. You could probably use a 1.5V zener across it so that you can protect the LED Mar 6, 2021 at 9:53
• @KoustubhJain Also if you want the math when the voltage across the zener is less than breakdown, you can consider it as an infinite resistor and evaluate. All the current flows through the load and none through the zener. Also the voltage across it then is $I_{load}R_{load}$ Mar 6, 2021 at 9:55
• If I connect a voltage source of voltage higher than breakdown voltage,how can the voltage across the zenner diode be equal to the breakdown voltage which is not equal to the battery voltage?? Oct 23, 2023 at 3:06

[quote] when the input voltage increases beyond a certain limit the breakdown occurs, I get that line. But I can't understand the next line which says "https://i.stack.imgur.com/EaRVY.png" [/quote]

Note that the voltage "across" the zener diode must increase above the breakdown voltage. If there was no dropping resistor, then that would be your " input voltage". But since the dropping resistor and the load resistor is fixed, the input voltage would have to be higher for the voltage across the zener (or load) to cause the breakdown. At the zener breakdown voltage, the zener starts conducting ( it was acting like an open circuit all this while). But the voltage drop across the dropping resistor doesn't drop or rise magically. There will be 'sufficient increase in the voltage drop' only when you increase the input voltage further, because the rest will be maintained constant across the load. Hence why I think that quoted statement is not talking about the instant the breakdown happens but after it.