question_answer

How is it that the reverse current in a zener diode starts increasing suddenly at a relatively low breakdown voltage of \[\mathbf{5}\text{ }\mathbf{V}\] or so?

Answer:

                    In a zener diode, both p-and n-sides are heavily doped with acceptor and donor impurities respectively. Due to this depletion layer formed is very thin\[(<{{10}^{-6}}m)\]. Even a small reverse bias voltage of \[\text{5 V}\] sets up a very high electric field of\[~\text{5 }\times \text{ 1}{{0}^{\text{6}}}\text{V}{{\text{m}}^{\text{-1}}}.\]This field is strong enough to pull valence electrons from the host atoms on the p-side which are accelerated to n-side.                     These electrons give rise to a large reverse current or breakdown current. The emission of electrons from the host atoms due to high electric field is known as internal field emission or field ionisation. The breakdown of the diode due to internal field emission is called zener breakdown.