The depletion layer of a reverse-biased $ \text{P-N} $ junction acts like the dielectric of a capacitor. This junction · Electronics MDCAT Past Paper MCQ

Physics Electronics PMDC Conceptual Practice

PMDC Verified Question 360 of 494

The depletion layer of a reverse-biased $ \text{P-N} $ junction acts like the dielectric of a capacitor. This junction capacitance ($ C_j $) varies with reverse voltage ($ V_R $) according to which behavior?

Capacitance increases linearly with increasing reverse voltage

Capacitance remains constant regardless of the applied reverse voltage

Capacitance decreases as reverse voltage increases because the depletion width widens

Capacitance drops instantaneously to zero as soon as any reverse voltage is applied

Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:

Option C: Capacitance decreases as reverse voltage increases because the depletion width widens

Concept:

A reverse-biased $ \text{P-N} $ junction functions as a parallel-plate capacitor (transition or space-charge capacitance), where the neutral P and N regions serve as conducting plates separated by the insulating depletion layer.

Formula:

$$C_j = \frac{\varepsilon A}{W} \quad \text{where} \quad W \propto \sqrt{V_B + V_R} \implies C_j \propto \frac{1}{\sqrt{V_B + V_R}}$$

Solution:

Increasing the reverse bias voltage $ V_R $ widens the depletion width $ W $.

Because capacitance is inversely proportional to plate separation ($ C = \frac{\varepsilon A}{W} $), junction capacitance $ C_j $ decreases as $ V_R $ increases.

This voltage-variable capacitance property is utilized in varactor (varicap) diodes.

Why other options are incorrect:

Option A: Capacitance decreases rather than increasing because the plate separation (depletion width) expands.
Option B: Capacitance varies with applied voltage; it is not constant.
Option D: The capacitance remains non-zero for all reverse voltages below breakdown.

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