Thermodynamics – Why Use Boltzmann Distribution Instead of Fermi-Dirac in Transistors?

Question

Considering electrons are fermions I would think it is better to use Fermi-Dirac distribution when discussing the physics of the electrons and holes in a transistor. However, I have been told Boltzmann distribution is what should be used. Can Boltzmann distribution apply to fermions?

Answer 1

Short version: it's an approximation, but it's a good one that's accurate for normal transistors and makes a lot of math easier.

In many cases, you're only interested in the "tail" of the distribution, in which case Fermi-Dirac and Maxwell-Boltzmann statistics produce approximately the same results. For Fermi-Dirac

$$n_{FD} = \frac{1}{e^{\left(\epsilon - \mu\right)/k_BT} + 1},$$

and for Maxwell-Boltzmann

$$n_{MB} = \frac{1}{e^{\left(\epsilon - \mu\right)/k_BT}}.$$

The two converge in the limit where $e^{\left(\epsilon - \mu\right)/k_BT}$ is large (equivalently, if $\left(\epsilon - \mu\right)/k_BT$ is medium sized). Put another way, they converge in the limit where $n$ is small (i.e. you're dealing with the tail of the distribution). This is generally the case in transistors. E.g., if you have a typical n-type semiconductor, that means there are electrons in the conduction band, but the number of electrons in the conduction band is small compared to the number of available states (even close to the band edge). In other words, $n$ is small, and Maxwell-Boltzmann statistics makes for a good approximation. This all falls apart if your conduction band has a lot of electrons in it (e.g. you've degenerately doped it), but that's not the case in common transistors.

EDIT: if you want a visual of the "tails", see the figure here. The red curve is the Fermi-Dirac distribution, and it has a low value in the conduction band because $\mu$ falls in the bandgap for normal doping.