Concept:Higher doping concentrations provide a much greater density of donor and acceptor atoms per unit volume. Consequently, a smaller spatial volume of crystal near the junction is needed to uncover enough fixed ions to establish the equilibrium barrier potential, resulting in a narrower depletion layer.
Formula:$$W = \sqrt{\frac{2\varepsilon (V_0 - V)}{q} \left(\frac{1}{N_A} + \frac{1}{N_D}\right)}$$
Solution:- As acceptor density \(N_A\) and donor density \(N_D\) increase, the term \(\left(\frac{1}{N_A} + \frac{1}{N_D}\right)\) decreases.
- Therefore, the width \(W\) of the space charge region is inversely related to doping concentration.
- Heavily doped diodes have very thin depletion layers (e.g., Zener diodes), while lightly doped diodes have wide depletion layers.
Why other options are incorrect:- Option B: Depletion width increases when doping is light, not when it is heavy.
- Option C: Depletion width is directly governed by dopant density.
- Option D: A depletion layer always forms in an unbiased \(\text{p-n}\) junction to balance carrier diffusion.
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