Ethan,
I think you're looking at a signal application and trying to apply it to a power supply application. Reverse-biased diodes are used for signal clamping all the time. On opamp input stages, you'll often see reverse-biased diodes going from the input line to the + and - power supply rails. This ensures that the input signal never swings greater than the rail voltage + 0.6V. It helps to protect the input device from damage.
You also see this on power amp outputs, to shunt "flyback" from reactive loads to the supply rails, to avoid breakdown of the output transistors.
There are some instances where you see regular silicon diodes used reverse-biased in a power supply. The most common example is across a voltage regulator IC. This will shunt any overvoltages on the supply line of 0.6V or more greater than the unregulated input voltage across the IC rather than through the IC, avoiding output stage breakdown. Also, a second reverse-biased diode is sometimes seen between the regulated supply and ground for reverse-polarity protection.
In a power supply, a zener diode, gas arrestor tube or whatever used as a brute force clamp to short out overvoltages is often called a "crowbar." Since a crowbar's whole function is to short the supply if it rises above a certain voltage, it's strictly a safety device and should not be used for "regulation" in the usual sense. And it should be preceded by a fuse or fusible link in series with the supply line, since the whole idea of shorting out the supply is to interrupt it by causing a fuse to blow! (If the overvoltage condition is more than momentary, that is).
Zeners are also used for regulation. This type of regulation is known as "shunt regulation" and while it's not as good as an active regulator circuit using a series pass element, it's "good enuff" in some cases, especially in low power circuits that draw a constant current and don't need very tight regulation. When used as a regulator, as opposed to a crowbar, a zener needs a series resistor to form an "inverted-L" attenuator, with the resistor acting as the upper leg of the attenuator and the zener as the bottom leg. The output voltage is taken from the junction of the two.
Zeners need a certain amount of current (a few mA, I don't recall the exact "typical" figure) passing through them at all times to maintain their zener voltage. The series resistor is calaculated thus:
R = (Vin - desired Vout) / (Iload + Izener)
OK, I just got up off my ass and checked one of my books. It gives 5mA as a "typical" value of zener current. So let's say we want to take 15VDC unregulated and spit out 12VDC regulated. Our circuit draws 20mA at 12VDC.
R = (15 - 12) / (.02 + .005) = 3 / .025 = 120 ohms.
Under normal operation, the resistor will dissipate (15 - 12)sq / 120 = .075W. If the supply were shorted, the resistor would have to dissipate 15sq/120=1.8watts. So we'd probably use a 5W resistor if we DIDN'T want it to blow. If we DID want it to blow, we'd use, say, a 1W resistor (flameproof, of course!).
But wait, what happens if no load is connected? The zener will draw whatever current is necessary to maintain about 12V across its terminals, in other words, the whole 25mA will pass through the zener. Therefore, the zener should have a rating of
P = .025 * 12 = 0.3W. So we'd use a 500mW zener.
Zeners are noisy. The zener in our hypothetical power supply should be bypassed by a filter cap, and ideally kept away from the signal circuits.
This has been a fun mental exercise. But in the real world, I'd probably reach for an LM317 :green: