Concept:Reverse bias raises the electrostatic potential barrier, shifting the N-side energy bands downward relative to the P-side bands and increasing the total barrier height to \( q(V_0 + V_R) \).
Formula:$$q V_{\text{barrier,net}} = q (V_0 + V_R) \quad \text{and} \quad E_{Fp} - E_{Fn} = q V_R$$
Solution:- Applying a reverse voltage \( V_R \) adds to the built-in potential barrier, increasing the total barrier height to \( q(V_0 + V_R) \).
- The N-side bands shift downward relative to the P-side bands, preventing majority carrier diffusion across the junction.
Why other options are incorrect:- Option A: The barrier height increases under reverse bias; it decreases under forward bias.
- Option C: The N-side bands shift downward under reverse bias, not upward.
- Option D: Applied bias does not change the intrinsic bandgap energy \( E_g \).
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