Suppose I have a n-type semiconductor whose fermi-level lies (say) 0.2 eV below the conduction band. Why would this level change if I changed the doping by making the donor concentration (say) 4 times the original value?

And what does it even mean? Does this mean that chemical potential also changes ? (I guess it, probably, does but what it means?)

Also could we estimate how would it change in some approximate way?

  • $\begingroup$ The impurities can shift $E_c$ and $E_v$ therefore shifting $E_F$ by its definition. I think it probably has to do with the fact that electrons are fermions and can't occupy the same level, aka Pauli Exclusion. $\endgroup$
    – M Barbosa
    Jun 12, 2016 at 12:57
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    $\begingroup$ @MBarbosa shifting of $E_C$ and $E_V$ happens, but the main point is the increased carrier concentration in the conduction band due to the donor states. As the donor states are close to the conduction band, some of them will be ionized. As a result the CB carrier concentration is higher as one would expect just from the band gap and the temperature. Therefore the band gap has to be shifted by $E_F$ to describe the electron distribution. Maybe this documents helps to clear up things: uwyo.edu/cpac/_files/docs/kasia_lectures/… $\endgroup$
    – user_na
    Jun 12, 2016 at 14:35

1 Answer 1


Fermi level is an energy in which the electron distribution probability is 1/2. Due to Pauli Exclusion, the electrons will pile up so the Fermi level for electrons will move up.

Chemical potential is the same with Fermi level.

To calculate the value, you have to integral the Fermi-Dirac distribution function times the density of states. For a rough estimate, you can compare the doping density with the effective density of states. If they are comparable, then the Fermi level is close to the conduction band edge.

  • $\begingroup$ Strictly speaking, the Fermi energy (or Fermi level) of a solid is the chemical potential at zero-temperature, i.e. it is the energy of the most energized electron in the solid when said solid is in its ground-state. See for example, Ashcroft and Mermin's book Solid State Physics. $\endgroup$
    – lcortesh
    Mar 1, 2018 at 1:03

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