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 ${{10}^{6}}V{{m}^{-1}}$ at the depletion region of p-n junction. If the width of the depletion region is $2.5\mu m$, 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 1

Hint: Reverse biased potential or breakdown voltage of a diode is the minimum voltage required so that the diode breakdowns and conducts electricity in reverse biased. Use the formula for potential difference across a length in a constant electric field.

Formula used:
$\Delta V=Ed$

Complete step by step answer:
When a diode is in reverse biased condition, it does not conduct any electricity. However, if the voltage (potential difference) across the diode exceeds a certain value then the diode breaks down and it conducts electricity.It is given that the Zener undergoes breakdown when the electric field of ${{10}^{6}}V{{m}^{-1}}$ is present at the depletion region of the p-n junction of the diode.

It is said that the width of the depletion region is equal to $2.5\mu m$.We know that the potential difference or voltage across a length d in a uniform electric field of magnitude E is given as $\Delta V=Ed$ … (i).
In this case, $E={{10}^{6}}V{{m}^{-1}}$ and $d=2.5\mu m=2.5\times {{10}^{-6}}m$.
Substitute the values of E and d in equation (i).
$\Delta V={{10}^{6}}\times 2.5\times {{10}^{-6}}\\
\therefore \Delta V=2.5V$
This means that the breakdown voltage or the reverse biased potential of the Zener diode is 2.5V.

Hence, the correct option is B.

Note:The potential difference across a zener diode remains constant for a wide range of current, when it is in reverse biased condition. Therefore, a zener diode is a special type of diode that is used as a voltage regulator.