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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