Concept:Applying a reverse bias pulls majority carriers away from the junction, exposing additional fixed ions and widening the space-charge region.
Formula:$$W = \sqrt{\frac{2 \varepsilon (V_B + V_R)}{e} \left(\frac{1}{N_A} + \frac{1}{N_D}\right)} \implies W \propto \sqrt{V_B + V_R}$$
Solution:- The total potential across the reverse-biased junction is \( V_{\text{total}} = V_B + V_R \).
- According to Poisson's equation for a step junction, depletion width \( W \) is proportional to the square root of the total reverse potential: \( W \propto \sqrt{V_B + V_R} \).
- Therefore, increasing \( V_R \) increases the width of the depletion layer.
Why other options are incorrect:- Option A: Depletion width increases with reverse voltage rather than decreasing.
- Option B: The depletion width depends directly on applied voltage; it is not constant.
- Option C: Depletion thinning occurs under forward bias, not reverse bias.
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