So i read about the Zener diode voltage regulator circuit in the The Art of Electronics book, it made no sense to me, then I read about it in this article which was helpful but I still don't get it:


simulate this circuit – Schematic created using CircuitLab

How is it reglating the voltage exactly?! and why is the diode "reverse biased" (I guess it means connected in reverse) which doesn't conduct current unless a very high reverse voltage is applied?

If for example the input voltage is 12V, what would be the output voltage and current? and what would be the voltage and current values through the diode?


When a zener is connected in reverse, it will only conduct when a voltage higher than its zener voltage is applied. If you have a 5V zener diode, and you apply 9 volts backwards across it, it will conduct. However, its voltage drop (that is, the voltage dropped across it) will be its zener voltage (in this case, 5V). Since Vout is connected ACROSS the diode, then that means the voltage on the output will be the same as the voltage drop, or 5V. Thus, the output is "regulated" to 5V even though 9 volts are applied.

  • \$\begingroup\$ When you say 5V zener diode do you mean breakdown voltage? if so doesn't zener diods have like 100+ breakdown voltages like 1N4148 diodes i recently bought? \$\endgroup\$ – razzak Jun 21 '16 at 18:19
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    \$\begingroup\$ Zener diodes have a special construction allowing them to recover after their reverse breakdown voltage has been exceeded. They are not damaged like 1N4148 diodes are. The reverse voltages are also lower than the reverse voltages in standard silicon diodes like the '4148. For Zener diodes, their reverse breakdown voltage is called their "zener voltage". Once again, what sets Zener diodes apart from standard diodes is that they can recover after their reverse breakdown voltage is exceeded. Standard diodes would be permanently damaged. \$\endgroup\$ – DerStrom8 Jun 21 '16 at 18:22
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    \$\begingroup\$ Zener diodes are constructed to have a known low reverse breakdown voltage. They are available with a wide selection of reverse breakdown voltages. \$\endgroup\$ – Peter Bennett Jun 21 '16 at 18:24
  • \$\begingroup\$ Thanks for adding that @PeterBennett , I forgot to mention that. \$\endgroup\$ – DerStrom8 Jun 21 '16 at 18:25
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    \$\begingroup\$ +1 for everything you said. One thing I would add on per the OP question is that the "output current" would be dependent on the load applied. The voltage through the diode is dependent on its spec'ed zener voltage. The current through the diode is dependent on its internal physics, I think 20-30 mA is a good aim to get the spec'ed zener voltage. But again, check your individual datasheet. \$\endgroup\$ – Jim Jun 21 '16 at 18:26

It is easiest to understand just by looking at the following curve: enter image description here

The vertical axis is the current, you can see that in reverse bias "breakdown" region, that no matter how much current you flow (unless you are flowing a small amount) that the reverse voltage changes very little (see the breakdown voltage label on the graph).

That means that this circuit has very low compliance, i.e it doesn't move (in voltage terms) very much with changes in current. This is the characteristic of a nice regulated voltage supply.

Now obviously there are limits to the amount of current and the amount of regulation, but as a first order system it works well.

  • \$\begingroup\$ How would i find out the current and voltage regulation limits? \$\endgroup\$ – razzak Jun 21 '16 at 18:44
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    \$\begingroup\$ @razzak the data-sheet for the device will have all those parameters. \$\endgroup\$ – placeholder Jun 21 '16 at 18:44

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