# How do Zener diodes maintain the potential across their terminals?

My physics book has a topic about Zener diodes being used as voltage regulators in the reverse bias.

Well, I'm curious to know how does a Zener diode maintain the potential across its terminals after it has undergone avalanche breakdown? Does it start conducting in full offering almost zero resistance? If so, how can there be a potential gradient across it?

Is the principle that for high current change, there is a minimal and negligible change in potential across the Zener? But doesn't it behave as a pure conductor in avalanche breakdown? If so, how is it possible for there to be a drop in potential? After all it allows large amounts of current through it. Finally, can you keep answers somewhat simple?

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en.wikipedia.org/wiki/Zener_effect Note that zener effect and avalanche are totally different! –  Georg Apr 16 '13 at 12:38
There's a difference between resistance (pure V/I) and dynamic resistance, deltaV/deltaI . What you're seeing after the zener voltage is reached is a low dynamic resistance. –  Carl Witthoft Nov 12 '13 at 16:20

In avalanche breakdown the zener diode does not behave like a pure conductor. It behaves like a "something that consumes N volts" followed by a perfect conductor. An intuitive way to think of it is: it costs you N volts worth of energy to keep the diode in breakdown. If you apply less than N volts breakdown stops and it barely conducts at all (it becomes a very good resistor.)

The way avalanche breakdown works is: there are some charge carriers (e.g. electrons) that are being accelerated by the voltage. When the electron hits a bond between two other atoms if the energy is low enough it just bounces off. But if the voltage is large enough then a loose electron will get accelerated (by the voltage) so that it will hit with sufficient energy to break a molecular bond and release another electron. Now there are two electrons being accelerated to fast enough speeds to break bonds. The instant you reduce the voltage below the breakdown limit, the electrons are no longer accelerated enough to break any more bonds, so the free electrons "settle back" into the bonds that are missing electrons and the current stops almost immediately. All that energy from the acceleration is released as heat.

So in a voltage regulator circuit like this:

Kirchoff's voltage law says that the voltage around any closed loop is 0. So you get +10 volts from the input, and you know you are going to drop -6 volts across the diode. Thus there must be 4 volts across the $40\Omega$ resistor and 6 volts across the $60\Omega$ resistor. So you can figure out the currents across the resistors. Now Kirchoff's current law says that the current going through the diode is the current through the $40\Omega$ resistor minus the current through the $60\Omega$ resistor.

For an input voltage >8.4 Volts (8.4 = 6.0 * 140/100) there will be 6 Volts across the load. Any remaining current gets shunted across the diode (which is now in breakdown.) At an input voltage <8.4 Volts there will be <6 Volts across the diode so there will be almost no current across the diode. The current through the resistors will be (approximately) the input voltage divided by $140\Omega$.

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