Why does the depletion region in a reverse biased diode grow?

Question

From what I understand, at the pn-junction of a diode, electrons diffuse from the n-region into the p-region until the resulting electric field grows large enough to stop any further flow of electrons. So, roughly speaking, if $\mu_n$, $\mu_p$ are the respective chemical potentials (not including any electrostatic potentials) and $V_n$, $V_p$ are the electrostatic potentials due to the field, equilibrium is obtained when $\mu_n - eV_n = \mu_p - eV_p$.

One might then reason that if there was an external electric field acting on the charges, equilibrium would insead be obtained when $\mu_n - eV_n - eU_n = \mu_p - eV_p - eU_p$ where $U_n, U_p$ denote the potentials dude to the external field. This would seem to imply that if a voltage source is connected to the diode with the positive terminal connected to the n-region and the negative to the p-region the depletion region would shrink since the voltage difference $|V_p - V_n|$ required for equilibrium is smaller.

This reasoning is clearly fallacious somehow since the depletion region actually grows in the situation described above. Why is it fallacious?

Answer 1

Electrons form the conduction band from N-region migrate to the P-region due to chemical potential, until an equilibrium is reached as in your first equation.

When a battery is placed in the circuit, with the positive side to the N-region, some electrons also migrate from its conduction zone through the battery and are placed in the valence zone of the P-side until an equilibrium is reached.

The process is similar, and the only difference is presence of the chemical potential of the battery instead of the chemical potential of the junction PN.

So, the chemical potential of the battery has to be added in the second equation, and explains why the total electric potential is increased in this reverse polarization.