Ohm's law across a semiconductor junction

Question

Given a semiconductor junction between a p-type and an n-type semiconductor, the free charges redistribute themselves until an equilibrium is attained i.e. some of the free electrons (holes) will move from the n-type (p-type) to the p-type (n-type) semiconductor.
In equilibrium, where there is no applied field, the macroscopic charge density is zero, yet there is still a potential difference across the junction (albeit smoothed out to be continuous). My question is why this is not a violation of Ohm's law
$$
textbf{J}=sigmatextbf{E}
$$
where, in equilibrium, we can write $textbf{E}=-nablavarphineq 0$, where $varphi$ is the electrostatic potential (or the voltage). Thanks in advance for any help.

The Photon · Answer

My question is why this is not a violation of Ohm's law
$$
textbf{J}=sigmatextbf{E}
$$

In the semiconductor junction, the electric field that forms near the junction sweeps nearly all free carriers away from the junction, forming a depletion region.
In this region there are effectively no free carriers, therfore (if you want to view this in terms of Ohm's Law), the conductivity $sigma$ is very low, and so no current flows despite the presence of a nonzero electric field.
In general, though, even when you apply an external field to the junction and current does flow, it doesn't obey Ohm's law. That's because Ohm's law is a law about metallic conductors, which has been extended to certain other material systems that happen to behave similarly. It simply isn't expected to apply to non-homogeneous configurations like the semiconductor PN junction.

my2cts · Answer

Indeed, Ohm's law does not apply to junctions. Junction make up non-Ohmic devices. See https://en.m.wikipedia.org/wiki/Electrical_resistance_and_conductance#Static_and_differential_resistance.

Ohm's law across a semiconductor junction

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