Semi-conductors

Question

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 $$\bar n={N\over \exp\beta(\mu-E_f)+1}$$

where $\mu$ is the chemical potential.

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I can see that the Fermi Dirac statistics say that for one fermion the mean occupation number is $$\bar n = {1\over \exp\beta(E-\mu)+1}$$ 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 $N$ bound $e^-$ states. I am not quite so confident as to why $$\mu\to E_f$$ $$E\to \mu$$

Could someone please explain?

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Anybody? :(

Answer 1

The first equation is not correct, the Fermi-level is the chemical potential of the electrons.

In semiconductor the carrier density $n$ (units of $\text{m}^{-3}$) in the conduction band is the integral of the Fermi-Dirac function over the conduction band density of states $g_c(E)$ (units of $\text{m}^{-3}\text{J}^{-1}$),

$$n = \int_0^{\infty} g_c(E) f(E, E_f) dE$$

where $$ f(E, Ef) = \frac{1}{\exp\left(\frac{E - E_f}{k_BT}\right)}$$

In practice you can actually use the Boltzmann distribution provided $E_f << E_c$ where $E_c$ is the conduction band energy. This allows the so called effective density of state to be used which looks like slightly your original equation,

$$n \approx N_c \exp\left(-\frac{E - E_f}{k_BT}\right)$$