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Junction Capacitance

Consider an a...

Question

Consider an abrupt p-junction.Let $V_{b i}$ be the built-in potential of this junction and $V_{R}$ be the applied reverse bias. If the junction capacitance $(C_{j})$ is $1 p F$ for $V_{b i} + V_{R} = 1 V$ , then for $V_{b i} + V_{R} = 4 V, C_{j}$ will be

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Solution

Ans : (d)

$C_{j} \propto V_{j}^{- 1 / 2}$
$\frac{C_{j 2}}{C_{j 1}} = \sqrt{\frac{V_{1}}{V_{2}}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$
$C_{j 2} = \frac{C_{j 1}}{2} = \frac{1 p F}{2} = 0.5 p F$

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