Q96 / 120 · VITEEE 2008 · Physics

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

  • A
    $3.5 \mathrm{~V}$
  • B
    $2.5 \mathrm{~V}$
  • Correct answer
    C
    $1.5 \mathrm{~V}$
  • D
    $0.5 \mathrm{~V}$

Solution

Here, electricfield, $E=10^{6} \mathrm{~V} / \mathrm{m}$ width of depletion region, $\mathrm{d}=2.5 \mu \mathrm{m}=2.5 \times 10^{-6} \mathrm{~m}$ $\therefore$ Potential required for breakdown $(\mathrm{V})=\mathrm{Ed}=\left(10^{6} \times 2.5 \times 10^{-6}\right) \mathrm{V}=2.5 \mathrm{~V}$ Contact Potential $=1.0 \mathrm{~V}$ $\therefore$ Reverse biased potential for zener breakdown $=(2.5-1.0) \mathrm{V}=1.5 \mathrm{~V}$

Hence, the correct answer is option: 'C'.

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