Concept:A reverse-biased \(\text{p-n}\) junction behaves like a parallel-plate capacitor: the p and n neutral regions act as conducting plates separated by the non-conducting depletion layer of width \(W\). As reverse voltage increases, \(W\) widens, reducing the transition capacitance.
Formula:$$C_T = \frac{\varepsilon A}{W} \quad \text{where} \quad W \propto \sqrt{V_0 + V_r}$$
$$C_T \propto \frac{1}{\sqrt{V_0 + V_r}}$$
Solution:- Increasing reverse voltage \(V_r\) widens the space charge depletion layer \(W\).
- Because capacitance is inversely proportional to plate separation distance (\(C_T = \varepsilon A / W\)), \(C_T\) decreases as \(V_r\) increases.
- (This voltage-dependent capacitance is the operating principle of varactor / varicap diodes).
Why other options are incorrect:- Option A: Capacitance decreases with plate separation (depletion width), so it cannot increase with reverse bias.
- Option C: Transition capacitance is strongly voltage-dependent.
- Option D: Capacitance decreases smoothly as \(1/\sqrt{V}\); it does not drop abruptly to zero.
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