Concept:The built-in electric field points from the positive donor ions (N-side) to the negative acceptor ions (P-side). Because electric potential decreases in the direction of field lines, the N-side is at a higher potential.
Formula:$$V(x) = -\int E(x) dx \implies V_N - V_P = V_B > 0$$
Solution:- The N-side of the depletion layer contains positively charged donor ions (\( N_D^+ \)).
- The P-side contains negatively charged acceptor ions (\( N_A^- \)).
- Because electric field lines point from positive to negative charges (N to P), the electric potential is higher in the N-region than in the P-region by an amount equal to \( V_B \).
Why other options are incorrect:- Option A: The P-side contains negative space charge and is at a lower potential.
- Option B: The built-in potential across the depletion layer is non-zero at equilibrium.
- Option D: The space-charge distribution is monotonic across the step junction rather than oscillatory.
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