What are the requirements for describing charge carriers (e.g. electrons) in a semiconducting material by - Fermi distribution - Boltzmann distribution

When do we apply the one or the other? If the explanation to this question is in Ashcroft/Mermin, reference to the relevant chapter would be appreciated.

I don't agree with both commentators so far.

Edit Consider Eq. 21 in Sze, PHysics of Semiconductor Devices:

$$n = N_C exp\left(-\frac{E_C-E_F}{kT}\right)$$

or Eq. 2.16 in Pierret, Semiconductor device fundamentals.

  • $\begingroup$ Can you please clarify your question a bit? Are you asking about limits of application or something else? In case this may be important, Fermi (or Bose, if you are dealing with bosons - not in this case, of course) distribution is applicable when quantum effects become important, e.g. at very low temperature. $\endgroup$
    – Eugene B
    Mar 19, 2013 at 11:50

2 Answers 2


Electrons are fermions (as are holes), therefore you should always use the fermi distribution, never boltzmann.

It is OK to use the boltzmann distribution as an approximation to the fermi distribution. (Makes the math simpler sometimes.) When is it a good approximation? When any given conduction-band state has probability << 1 of having an electron in it. (That's for n-type.)

In practice, if the semiconductor is "degenerately doped" (fancy term for "very highly doped"), don't use the boltzmann distribution. If you're using the semiconductor as a laser, don't use the boltzmann distribution. In most other circumstances, the boltzmann distribution is a pretty close approximation to the true fermi distribution.

  • $\begingroup$ Alright. Can you quantify this somehow? $\endgroup$
    – TMOTTM
    Mar 25, 2013 at 14:56
  • 1
    $\begingroup$ For n-type, the criterion is "When any given conduction-band state has probability << 1 of having an electron in it." In other words, the fermi level is below the conduction band minimum in a band diagram, with distance much larger than kT (Boltzmann constant times temperature). How much larger than kT? Well, the larger it is, the more accurate the Boltzmann approximation is! There is no quantitative criterion that I can tell you: It depends on what you're calculating and how accurate you want the answer to be. $\endgroup$ Mar 25, 2013 at 17:40
  • $\begingroup$ Thanks, that's a good answer. Can you point me out to some literature where your answer is maybe elaborated on? $\endgroup$
    – TMOTTM
    Mar 25, 2013 at 22:27

I would suggest that it is always better to use Fermi distribution when dealing with semiconductors. This is clear even from the wikipedia page. Anyway, I don't think I've even met semiconductors described via Boltzmann distribution. Moreover, if you find a system that would actually be well-described by Boltzmann distribution, you still can apply Fermi distribution to it, as Boltzmann distribution can be treated as a classical limit of Fermi or Bose distribution, so you shouldn't go wrong anyway. Similarly, you can apply special relativity to classical mechanics problems.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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