Question

A Zener diode has a contact potential of $1\text{V}$ in the absence of biasing. It undergoes Zener breakdown for an electric field of ${10}^6\text{V/m}$ at the depletion region. If the width of the depletion region is $2.5\mu \text{m}$, what should be the reverse biased potential for the zener breakdown to occur ?

• A. $3.5\text{V}$
• B. $2.5\text{V}$
• C. $1.5\text{V}$
• D. $0.5\text{V}$

Answer 1

The correct option is B $2.5\text{V}$

The breakdown field of the Zener diode $={10}^6\text{V/m}$

The width of the depletion region $=2.5\times {10}^{-6}\text{m}$

$\therefore V_{\text{breakdown}}=E\times d={10}^6\times 2.5\times {10}^{-6}=2.5\text{V}$.

Hence $\left(B\right)$ is the correct answer.