Concept:Forward bias lowers the electrostatic potential barrier across the junction, shifting the N-side energy bands upward relative to the P-side bands by \( q V_F \) and splitting the equilibrium Fermi level into two quasi-Fermi levels: \( E_{Fn} - E_{Fp} = q V_F \).
Formula:$$q V_{\text{barrier,net}} = q (V_0 - V_F) \quad \text{and} \quad E_{Fn} - E_{Fp} = q V_F$$
Solution:- Applying a forward voltage \( V_F \) lowers the potential barrier from \( q V_0 \) to \( q(V_0 - V_F) \).
- In the energy band diagram, this appears as an upward shift of the N-side bands relative to the P-side bands.
- The separation between the electron and hole quasi-Fermi levels equals the applied energy: \( E_{Fn} - E_{Fp} = q V_F \).
Why other options are incorrect:- Option A: Band edges move closer in energy under forward bias; they separate further under reverse bias.
- Option B: A single flat Fermi level exists only at thermal equilibrium (zero external bias).
- Option C: The material bandgap \( E_g \) is an intrinsic property and does not drop to zero under normal bias.
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