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Suppose there is a semiconductor with Fermi energy

  1. Mar 28, 2013 #1
    Suppose there is a semiconductor with Fermi energy $E_f$ and that there are $N$ bound electron states.

    I'd like to know why the mean number of excited electrons takes the form [tex]\bar n(T)={N\over \exp\beta(\mu-E_f)+1}[/tex]

    where [itex]\mu[/itex] is the chemical potential.

    I can see that the Fermi Dirac statistics say that for one fermion the mean occupation number is [tex]\bar n = {1\over \exp\beta(E-\mu)+1}[/tex]
    I am not sure however as to how for the semi-conductor it should assume the above form. The N is clearly due to the [itex]N[/itex] bound [itex]e^-[/itex] states. I am not quite so confident as to why [tex]\mu\to E_f, E\to \mu[/tex]

    Could someone please explain?

    Many thanks.
  2. jcsd
  3. Mar 30, 2013 #2
    i think by E, you mean the top of the valence band?
    Every electron that is not bound is excited (or makes a "hole" in the valence band and an electron in the conduction band).
    There are N bound states, so the number of excited electron is N* (1- fermi_dirac_dist(E))
    mathematically, this is the the same as changing the palces of Ef and E, try it.
    By the way, using E as the top of the valence band is an approximation which doesn't work if, for example, the semiconductor is very heavily p-type.
    Last edited: Mar 30, 2013
  4. Apr 2, 2013 #3
    Thank you, Giga_Man!
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