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I'm having trouble understanding the simple voltage regulator that can be built using a zener diode (from section 2.04 in the Art of Electronics). I know that it would be better to use amplifiers, et cetera, but I'm just trying to understand how this circuit works.

diagram of resistor-diode power supply using zener/avalanche diode

I don't really understand how the circuit works, but I am guessing that when a load is applied to the output, it drains current from the source (Vin) and thus causes the voltage to drop? How does the zener diode help to maintain the voltage and thus make this circuit act as a regulator?

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2 Answers 2

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Look at the Zener diode curve. You will see that the device breaks down at the Zener voltage when reverse-biased, and conducts. That property will fix the output voltage at the breakdown voltage, over a range of output currents, when used with a resistor, with relatively small voltage changes. It will also stabilise the output against changes in the input voltage.

Strictly speaking, Zener diodes are low-voltage devices (up to about 5V6). Higher-voltage ones have a different mode of operation and are called avalanche diodes. Both types are commonly referred to as Zeners, though.

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  • \$\begingroup\$ Thanks, for some reason I wasn't see the diode as being reverse-biased, my mistake. So when V < Vz(breakdown), there is no regulation, but when V == Vz(breakdown), the zener diode then holds the output voltage at Vz(breakdown)? \$\endgroup\$
    – Dr. Watson
    Jan 21, 2011 at 19:35
  • \$\begingroup\$ Yes, similar to a forward-biased ideal diode with Vz voltage source at the cathode. \$\endgroup\$
    – tyblu
    Jan 21, 2011 at 21:00
  • \$\begingroup\$ @Leon Heller I didn't understand the "relatively small voltage changes" part......should the input voltage be always be increased after the current for zener breakdown has been reached? \$\endgroup\$
    – Hydrous Caperilla
    Feb 17, 2018 at 14:15
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    \$\begingroup\$ Zener breakdown is caused by voltage, not current. \$\endgroup\$
    – Leon Heller
    Feb 17, 2018 at 14:45
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    \$\begingroup\$ You don't have to. It's just a (not very good) stabiliser. \$\endgroup\$
    – Leon Heller
    Feb 17, 2018 at 15:22
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Let's assume your input voltage is 10V, the zener is a 5V type and the resistor is 100\$\Omega\$. The zener will set \$V_{OUT}\$ to 5V, so that the current through the resistor will be \$I = \frac{10V - 5V}{100\Omega } = 50mA\$. When you put a load on the output, say 500\$\Omega\$ the zener will still keep the output at 5V, so R will still have 50mA through it, but part of that current will go through the load: \$I_L = \frac{5V}{500\Omega} = 10mA\$, so that there's only 40mA through the zener. For lower load resistances the current though the load will increase, taking away current from the zener diode, until it has too little current left to regulate \$V_{OUT}\$ properly. The 50mA from the example wasn't chosen at random; it's often the current at which the zener voltage is specified (especially for older zeners). More modern zeners can work at currents far below 1mA. If \$R_L\$ is less than 100\$\Omega\$ the resistor divider \$R/R_L\$ will pull \$V_{OUT}\$ below the zener voltage, the zener won't play anymore and \$V_{OUT}\$ will drop below 5V.
So far for load regulation.

Line regulation tells how a regulator reacts to variations in input voltage. Let's take our example with the 500\$\Omega\$ load, and decrease the input voltage to 9V. The output voltage is still kept at 5V by the zener, so the current through the load remains 10mA, but the current though R will be \$I = \frac{9V - 5V}{100\Omega } = 40mA\$, and therefore the current through the zener 30mA. Again the input voltage can decrease to the point where the zener has too little current left to operate properly. The upper limit of the input voltage is determined by the maximum current allowed through zener and R.

This type of voltage regulation is very simple, but not very good. Line regulation is poor, which means that the output voltage will still vary a bit when the input voltage increases/decreases. Same with load regulation: output voltage will vary with varying loads. And compared to the maximum load there's a rather big loss in the zener, so it's not very efficient. A small integrated regulator like an LM78Lxx is always a better choice.

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  • \$\begingroup\$ I would expect that a Zener regulator may be advantageous in cases where there's never much load (e.g. 250uA max) and it's okay if the output voltage varies a lot, but where the output voltage must not be allowed to get above a certain level. For example, a couple of decent 100K resistors in series with rectified AC120 line voltage will pass an average current of a little more than a quarter milliamp but should have no trouble withstanding 170 volts (each would see a peak power well below 0.1 watts). The 78Lxx regulators can't withstand anything near that on their inputs. \$\endgroup\$
    – supercat
    Nov 19, 2012 at 17:11

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