# Semiconductors - Fermi energy and function

As I understand it, the Fermi function $$f(E) = \frac{1}{1+e^{(E-E_f)/(kT)}}$$ gives the probability of finding an electron at a particular energy in a material.

It also has the following property : $$f(E_f+E_0) = 1 - f(E_f-E_0)$$ which means that the probability of finding an electron at an energy $E_0$ above $E_f$ (Fermi energy) is equal to the probability of not finding it an energy $E_0$ below $E_f$.

It is then logical to expect the Fermi energy to be between the valence and conduction band in a metal. For semiconductors however, I have some difficulties to reason about the Fermi energy. Here are my questions :

1. In a non-doped semiconductor, $E_f$ is also midway between the valence and conduction band (in the bandgap). However, if this "Fermi theory" still applies, wouldn't there be a non-zero probability of finding an electron immediately just below and above $E_f$ contradicting with the fact that no electron can be in the bandgap?
2. In a doped semiconductor (let's say n-type) at $0$ $K$, $E_f$ is still situated below but closer to the conduction band. As the temperature increases, $E_f$ returns midway between the valence and conduction band. Why is that?

P.S. : I am not very comfortable with quantum mechanics so I would appreciate if your answers could contain the least possible of it.

• It is logical to expect the Fermi energy of a metal to be right inside the conduction band. Aug 22 '17 at 13:06

1) The Fermi function tells you the probability that a state with energy $E$ is filled. However, it does not guarantee that such a state exists! You are correct that no states exist within the band gap (assuming a perfect crystal).

So, there is no contradiction: the Fermi function only applies if a state exists, and states don't exist in the gap.

2) The Fermi energy is defined at zero temperature, so you can't talk about the Fermi energy changing as a function of temperature. You're probably interested in is the chemical potential $\mu$, which is temperature dependent and as is equal to $E_f$ at zero temperature. If you're interested in non-zero temperatures, you need to replace $E_f$ with $\mu$ in your definition of the Fermi function.*

$E_f$ also doesn't need to be below the conduction band. You can dope a semiconductor hightly enough that the Fermi energy is in the conduction band ("degenerate doping").

I'm not sure why $\mu$ would go to the middle of the gap as the temperature is increased. I'm not sure it's true. Could you provide some context or a source?

EDIT: I see why the chemical potential would go to the midpoint of the gap, at least for a semiconductor that's not degenerately doped. An explanation is here. I can try to flesh out the explanation if the page is confusing.

*(That said, even at non-zero temperature, people often work with $E_f$ instead of $\mu$: when you convert $E_f$ to a temperature -- the Fermi temperature -- you get temperatures like 10,000 K for many materials, and compared to that, room temperature might as well be absolute zero. For that reason, $E_f$ and $\mu$ sometimes get used interchangeably, even tho they shouldn't.)

1) The reason the Fermi Energy is midway(only true if the valence and conduction band have same parabolic dispersion relation) is the following : assume you have N carriers in the valence band at 0K and now you consider T=5K. if a carrier gets into the conduction band thanks to thermal generation, it should generate a hole in the valence band. Note that it is note because there is a non negligible proabailty of finding a carrier in the band gap that there will be a carrier there. Indeed, the number of carriers in an energy level $\varepsilon$ are given by : $n(\varepsilon) = f(\varepsilon)g(\varepsilon)$ where g is the density of states. Since g=0 in the gap there can not be any charges although there is a non zero probability of finding a carrier at that energy at that temperature.