# 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

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 $140\Omega$.