0
\$\begingroup\$

What precisely thermal equilibrium implies in case of a semiconductor. One thing I know is that Fermi Level remains the same. But what other things does this implies, is there net current flow in case of a thermal equilibrium?

\$\endgroup\$
  • \$\begingroup\$ More context for your question is needed. In general devices are probably not in thermal equilibrium, but some sort of steady state. Currents are flowing with (perhaps) thermal gradients here and there. \$\endgroup\$ – George Herold Sep 17 '15 at 13:37
  • \$\begingroup\$ A lot. Probably more than fits into 30k characters. \$\endgroup\$ – PlasmaHH Sep 17 '15 at 15:19
2
\$\begingroup\$

It's a fairly broad question so I'll try to cover the key points.

The "thermal equilibrium" is a special case of the "steady state" classification. A semiconductor is said to be in steady state when the generation and recombination rates are equal and current is constant.

Note that this holds true even for optical generation and any other conceivable type of generation. This means that when a semiconductor is in the steady state, the number of electrons and holes does not change throughout the semiconductor.

The special thing about thermal equilibrium is that the generation is due solely to the surrounding heat. This means that all the generation in the semiconductor is thermal generation. In addition, there is no current in a thermal equilibrium semiconductor.

It is only in thermal equilibrium that we discuss Fermi levels. When a device is not in thermal equilibrium but is in steady state (with no current), we have what is called a "quasi-Fermi level" which is simply the Fermi level that would be computed if the non-equilibrium carrier concentration is taken into consideration instead.

It is worth noting though that an absence of band bending is not a necessary characteristic of thermal equilibrium. This can be seen in non-uniformly doped semiconductors where diffusion gradients lead to the creation of electric fields within the semiconductor and cause band bending. Despite this, since the diffusion and drift currents cancel out, the differentially doped semiconductor is still said to be at thermal equilibrium.

|improve this answer|||||
\$\endgroup\$

Not the answer you're looking for? Browse other questions tagged or ask your own question.