Concept:Reverse bias applies an external voltage that reinforces the built-in potential barrier, increasing the total junction potential to \( V_0 + V_R \) and widening the depletion layer.
Formula:$$W = \sqrt{\frac{2 \epsilon_s (V_0 + V_R)}{q} \left( \frac{1}{N_A} + \frac{1}{N_D} \right)} \propto \sqrt{V_0 + V_R}$$
Solution:- The applied reverse voltage \( V_R \) pulls majority carriers away from the junction.
- This exposes more fixed donor and acceptor ions, widening the depletion layer \( W \) proportionally to \( \sqrt{V_0 + V_R} \).
Why other options are incorrect:- Option B: Depletion width narrows toward zero under forward bias, not reverse bias.
- Option C: The depletion width depends directly on the applied reverse voltage \( V_R \).
- Option D: Physical layer widths are always positive quantities.
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