 Solution

At equilibrium, majority Electrons undergo diffusion from n to p, while majority electron holes diffuse from p to n. Their departure exposes positive donors on the n side and negative acceptors on the p side. The resulting electric field points from n to p and opposes further diffusion. Equilibrium is reached when the chemical potential is constant across the junction. The built-in potential is this electrostatic adjustment, not a discontinuity of the material's band gap.

Since electronic band energies shift by $-eV(x)$, both the conduction band and valence band undergo band bending down from p to n by $e{V}_b$ at zero applied bias. Their separation remains the same gap. In the nondegenerate thermal equilibrium limit the built-in voltage can also be expressed using neutral-edge densities as $e{V}_b=k_{B}T\log (n_n/n_p)=k_{B}T\log (n_n{p}_p/n_i^2)$.

Take the n-contact potential and its electrochemical potential as reference. Applying $V_{\mathrm{ext}}$ to the p side changes the total depletion-layer drop toThus $V_{\mathrm{ext}}<0$ is reverse bias: it increases the barrier and widens the depletion region. The bands have horizontal tangents at the neutral edges in this approximation and a smooth cubic bend within the linearly graded layer.

A biased junction is not in global thermal equilibrium. Its contact electron electrochemical potentials obeyUnder reverse bias the p-contact level is higher by $e|V_{\mathrm{ext}}|$. A single horizontal equilibrium chemical potential cannot describe this state; electron and hole quasi-Fermi levels describe the nonequilibrium populations. The sketch marks the contact levels. Detailed spatial quasi-Fermi profiles require transport and recombination information beyond the electrostatic doping profile.