GATE 2018 EC Q.18: solution and answer

GATE 2018 · ECQ.18 · 1 mark · NAT

A p-n step junction diode with a contact potential of $0.65 V$ has a depletion width of $1 μ m$ at equilibrium. The forward voltage (in volts, correct to two decimal places) at which this width reduces to $0.6 μ m$ is ______.

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Answer and solution

ANSWER

The forward voltage is $V \approx 0.42 V$ .

Official answer: 0.40 to 0.43Source: Official answer key, GATE 2018 EC (IIT Guwahati)

For a p-n step junction, the depletion width satisfies $W \propto V_{bi} - V$ , where $V$ is the forward bias voltage (taken positive, since forward bias lowers the barrier potential and thus shrinks the depletion width).
At equilibrium ( $V = 0$ ): $W_{0} = k V_{bi} = 1 μ m$ , with $V_{bi} = 0.65 V$ .
At forward bias $V$ : $W = k V_{bi} - V = 0.6 μ m$ .
Dividing the two relations eliminates the constant $k$ : $(\frac{W}{W _{0}})^{2} = \frac{V _{bi} - V}{V _{bi}} \Rightarrow (\frac{0.6}{1})^{2} = \frac{0.65 - V}{0.65}$
So $0.36 = \frac{0.65 - V}{0.65} \Rightarrow 0.65 - V = 0.36 \times 0.65 = 0.234 V$ .
Therefore $V = 0.65 - 0.234 = 0.416 V \approx 0.42 V$ .

KEY IDEA

For a p-n step junction, the depletion width scales with the square root of the effective barrier potential: $W \propto V_{bi} - V$ , where $V$ is the applied forward bias (taken positive, since forward bias reduces the barrier and hence the width). Equating the ratio of widths at two bias points eliminates the proportionality constant and the junction's physical parameters entirely.

WATCH OUT FOR

Treating the forward-bias case with the wrong sign, i.e. writing $W \propto V_{bi} + V$ instead of $V_{bi} - V$ — forward bias lowers the built-in barrier, so the depletion width shrinks, not grows, with increasing forward voltage.

From the GATE 2018 EC paper. Every question of this paper · Download the full paper · All 6 P-N junction questions, with notes

GATE 2018 EC Q.18: solution and answer

Answer and solution

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