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?
 A: 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 $140\Omega$.
A: Under normal operating conditions, Zener diodes do not undergo an avalanche breakdown. An avalanche breakdown is (usually) disastrous, irreversible and generally undesirable.
The Zener diode operates based on quantum tunnelling. To describe it simply, a high reverse bias voltage would increase the potential energy of the valence band electrons on the P side (since it would be connected to the negative terminal). Think of it as the electrons on the P side being strongly repelled.
At a certain point, the potential energy of these P side valence band electrons becomes so high that they actually possess a greater energy than some of the unoccupied energy states in the conduction band on the N side. In other words, it becomes energetically favourable for them to be in the conduction band on the N side.
Given a sufficiently narrow depletion region, electrons from the valence band on the P side can spontaneously move to the conduction band on the N side via  quantum tunnelling. This constitutes a current. If it helps in visualisation, you can see this as holes being injected into the P side valence band and electrons being injected into the N side conduction band.
The reason the voltage drop is maintained is that the potential energy of the P side valence band electrons must be maintained above the potential energy of the N side conduction band electrons in order for the flow of current across the junction to remain energetically favourable.
To put it as succinctly as possible, the process involves creating a junction voltage so high that electrons on the P side break free from their covalent bonds and move to the N side, creating a current. The voltage remains because it is required to maintain a constant flow of current by making it energetically favourable for electrons to move from the P side to the N side.
NOTE: The potential energies I talk about are with regards to electrons, voltages and potentials in conventional circuits are described with regards to positive charges. That's why I can describe the potential energy of electrons near the negative/ground terminal of a battery as higher than those near the positive terminal
A: The use of so-called Zener diodes for voltage "stabilization" is purely due to the experimental fact that their I-V characteristics have a very sharp (almost vertical) increase in reverse current upon reaching a critical reverse voltage ("breakdown voltage", Zener voltage Vz). Therefore, in a series connection with a resistor, the voltage drop over the Zener diode remains almost constant, close to Vz, even when the current through the series connection varies strongly. The physical mechanism for this steep rise in reverse current can be the band-to-band tunneling through a narrow depletion zone in highly doped pn-junctions (Zener effect) or the avalanche multiplication of charge carriers by impact ionization due to high fields in the depletion zone. The latter leads to the usual reverse breakdown of a pn-diode when the current is not limited by a resistor. This effect can be used for voltage stabilization or generation of a reference voltage in circuits. 
