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Question

Physics Question on Electric Field

A Zener diode has a contact potential of 1V1\, V in the absence of biasing. It undergoes Zener breakdown for an electric field of 106V/m10^{6} \,V / m at the depletion region of pnp-n junction. If the width of the depletion region is 2.5μm2.5 \,\mu\, m, what should be the reverse biased potential for the Zener breakdown to occur?

A

3.5 V

B

2.5 V

C

1.5 V

D

0.5 V

Answer

2.5 V

Explanation

Solution

Reverse biased potential for the zener breakdown Vr=EdV_{r} =E d =106×2.5×106=10^{6} \times 2.5 \times 10^{-6} =2.5=2.5 volt