Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 81 of 494
Under reverse-biased conditions, a semiconductor \( \text{P-N} \) junction exhibits a junction capacitance (\( C_j \)) because the depletion region acts as:
A
An inductor storing magnetic field energy
B
A dielectric insulating layer sandwiched between two conducting regions
C
A pure resistor with zero phase shift
D
A constant DC voltage source
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: A dielectric insulating layer sandwiched between two conducting regions
Concept:

Under reverse bias, the mobile-carrier-free depletion layer behaves as a dielectric medium separating the conductive P and N regions, forming a voltage-variable capacitor (varactor).

Formula:

$$C_j = \frac{\varepsilon A}{W} \quad (\text{where } W \propto \sqrt{V_{\text{reverse}}})$$

Solution:

  • The P and N bulk regions contain abundant mobile carriers and behave like parallel conducting plates.


  • The depletion region between them is depleted of free carriers and acts as a dielectric insulator of width \( W \).


  • This structure forms a junction transition capacitance \( C_j = \varepsilon A / W \).


Why other options are incorrect:

  • Option A: Inductive properties arise from magnetic flux linkage in coils, not electrostatic depletion layers.


  • Option C: The charge storage across the depletion boundary introduces a reactive capacitive impedance.


  • Option D: The junction stores charge electrostatically; it is not an active energy source.

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