My highschool textbook states that Zener diodes are a special type of diode which is made of highly doped p and n junctions, and which can survive reverse bias - unlike normal diodes, which get damaged due to the excess heat produced in that mode. How can heavy doping help Zener diodes avoid damage in reverse bias?

Is it that it somehow gets rid of the excess heat produced in the diode? How does it do it?


2 Answers 2


The key difference between a Zener diode and a normal diode is that the Zener diode has a low breakdown voltage - typically in the few volts range. The breakdown voltage is low because the heavy doping means the depletion layer is very thin, and even at a low voltage the field strength over this thin depletion layer is very high. With a conventional diode the depletion layer is thicker and therefore a higher voltage is needed to achieve breakdown. However the breakdown phenomenon in a normal diode is basically the same as in a Zener diode.

Both diodes will be damaged if excessive current flows when the breakdown voltage is exceeded, and you need some resistance in series with the diode to limit the current.

  • $\begingroup$ Then, what are the benefits of having a low depletion layer? $\endgroup$ Nov 6, 2013 at 16:29

So,thin depletion layer means more internal electric field so less external voltage will be required for the breakdown. It also accounts for the less barrier region .


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