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Junction Capacitance

The built-in ...

Question

The built-in potential of an abrupt p-n junction is $0.75 V$ .If its junction capacitance $(C_{j})$ at a reverse bias ( $V_{R})$ of $1.25 V$ is $5 p F$ , the value of $C_{j} (i n p f)$ when $V_{R} = 7.25 V$ is
2.5

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Solution

The correct option is A 2.5
$C_{j} \propto \frac{1}{\sqrt{V_{b i} + V_{R}}}$
$\frac{C_{2 j}}{C_{i j}} = \sqrt{\frac{V_{b i} + V_{R_{1}}}{V_{b i} + V_{R_{2}}}}$
$C_{2 j} = C_{1 j} \sqrt{\frac{2}{8}} = \frac{C_{1 j}}{2} = 2.5 p F$
So, answer is 2.5.

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Q. Consider an abrupt p-junction.Let

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