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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?

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Well, one simple answer is that the number of electrons and holes are equal in equilibrium. –  PhHEP Mar 21 '13 at 19:46
    
This is fine for intrinsic semiconductor. but what about doped ones, they have also single fermi level during equilibrium? –  user22206 Mar 22 '13 at 4:59
    
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. –  PhHEP Mar 22 '13 at 6:02
    
Possible duplicate: physics.stackexchange.com/questions/68162/… –  Steve B Dec 29 '13 at 2:24
    
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. –  Nanite Jan 28 at 8:27
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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.

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