Question
A Zener diode has a contact potential of 0.8 V in the absence of biasing. It undergoes Zener breakdown for an electric field of 106 Vm–1 at the depletion region of p-n junction. If the width of the depletion region is 2.4 μm, what should be the reverse biased potential for the Zener breakdown to occur?
Solution
Here,
Breakdown electric field of the Zener diode, E = 106Vm–1.
Width of the depletion region, d =2.4 = 2.4 x 10–6m
∴ Reverse biased potential, Vbreakdown = E x d
= 106 x 2.4 x 10–6
= 2.4 V.
