A Zener diode has a contact potential of 1 V in the absence of biasing. It undergoes Zener breakdown for an electric field of 106 V / m at the depletion region of p-n junction. If the width of the depletion region is 2.5 μ m, what should be the reverse biased potential for the Zener breakdown to occur?

Question

A Zener diode has a contact potential of 1 V in the absence of biasing. It undergoes Zener breakdown for an electric field of 106 V / m at the depletion region of p-n junction. If the width of the depletion region is 2.5 μ 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

Tardigrade Admin · Accepted Answer

Reverse biased potential for the zener breakdown Vr =E d =106 × 2.5 × 10-6 =2.5 volt

3.5 V

2.5 V

1.5 V

0.5 V

Reverse biased potential for the zener breakdown Vr​=Ed =106×2.5×10−6 =2.5 volt

Reverse biased potential for the zener breakdown
$V_{r} = E d$
$= 1 0^{6} \times 2.5 \times 1 0^{- 6}$
$= 2.5$ volt