Concept:The built-in potential barrier represents the energy difference between the majority carrier levels on either side of the junction at equilibrium, which depends logarithmically on doping levels.
Formula:$$V_B = \frac{k_B T}{e} \ln\left( \frac{N_A N_D}{n_i^2} \right) = V_T \ln\left( \frac{N_A N_D}{n_i^2} \right)$$
Solution:- Increasing \( N_A \) shifts the P-side Fermi level closer to the valence band.
- Increasing \( N_D \) shifts the N-side Fermi level closer to the conduction band.
- Aligning the Fermi levels at equilibrium creates a larger built-in potential barrier \( V_B \), which grows logarithmically with \( N_A N_D \).
Why other options are incorrect:- Option B: The relationship is logarithmic rather than linear.
- Option C: Doping concentrations directly determine the majority carrier concentrations and the resulting barrier height.
- Option D: An inverse relationship is mathematically inconsistent with the semiconductor equilibrium equation.
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