Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 412 of 494
How does increasing the doping concentration of donor and acceptor impurities affect the width (\( W \)) of the depletion layer in a \( \text{P-N} \) junction?
A
The depletion layer width increases because more total ions are created
B
The depletion layer width decreases because a higher spatial charge density satisfies the built-in potential over a shorter distance
C
The depletion layer width remains unchanged because width depends only on dielectric permittivity
D
The depletion layer immediately collapses to zero thickness under all biasing conditions
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: The depletion layer width decreases because a higher spatial charge density satisfies the built-in potential over a shorter distance
Concept:

Depletion layer width is inversely related to dopant concentration because higher ionic charge densities satisfy the required potential difference across a narrower physical distance.

Formula:

$$W = \sqrt{\frac{2 \varepsilon V_B}{e} \left(\frac{1}{N_A} + \frac{1}{N_D}\right)}$$

Solution:

  • Higher doping levels (\( N_A, N_D \)) place more dopant atoms near the junction.


  • As carriers diffuse and expose these ions, the necessary space-charge required to support \( V_B \) is established over a shorter physical distance.


  • Therefore, higher doping concentrations result in a narrower depletion layer width \( W \).


Why other options are incorrect:

  • Option A: A higher density of ions means fewer lattice layers must be depleted to balance the barrier potential.
  • Option C: Depletion width depends directly on dopant densities \( N_A \) and \( N_D \).
  • Option D: A finite depletion width is maintained at equilibrium for any non-infinite doping concentration.

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