Why do the Fermi level for electrons and holes coincide in equilibrium condition and why do they separate as quasi-Fermi levels in non equilibrium situations?

  • $\begingroup$ Well, one simple answer is that the number of electrons and holes are equal in equilibrium. $\endgroup$ – PhHEP Mar 21 '13 at 19:46
  • $\begingroup$ This is fine for intrinsic semiconductor. but what about doped ones, they have also single fermi level during equilibrium? $\endgroup$ – user22206 Mar 22 '13 at 4:59
  • $\begingroup$ Ah yes! Didn't think about that. I think what I was getting at was that the number of free electrons coming from the host atoms will be equal to the number of holes on the host atoms in equilibrium. But I guess that's not of much help to you since we don't distinguish electrons and holes from other electrons and holes respectively no matter where they come from! Sorry. $\endgroup$ – PhHEP Mar 22 '13 at 6:02
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/questions/68162/… $\endgroup$ – Steve Byrnes Dec 29 '13 at 2:24
  • $\begingroup$ A subtle point, it's more precise to talk about the quasi-Fermi-level of conduction band and quasi-Fermi-level of valence band. Holes don't really have a Fermi level, at least not in the way you're thinking about. $\endgroup$ – Nanite Jan 28 '14 at 8:27

A steady-state equilibrium, different from the thermodynamic equilibrium, can be triggered by external stimulus such as light-shining, which provoke the photoionization and the generation of electron-hole pairs, or current flowing, which injects electrons (or holes) to the system. Under these conditions, the concentrations of electron and holes are no longer governed harmoniously by the mass-action law and the Fermi-Boltzmann thermodynamic equilibrium, but are forced by the external conditions, and pulled-off from their reciprocal equilibrium. By this, the need to separate the two distinct quasi-fermi levels, one for electrons, one for holes, accounting for their out-of-thermodynamic-equilibrium concentrations.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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