Semiconductors — Question 2

Question 2

Unequal band densities shift the intrinsic Fermi level from midgap

Use the nondegenerate population formulas $n=N_ce^{-(E_c-E_F)/k_BT}$ and $p=N_ve^{-(E_F-E_v)/k_BT}$. An intrinsic sample has n=p and $N_c/N_v=4$.

1. Derive the intrinsic Fermi energy $E_i$ relative to midgap.

2. At $k_BT=.0250\,\mathrm{eV}$, calculate the shift and explain its direction.

Question 2 - Solution

Unequal band densities shift the intrinsic Fermi level from midgap

(a) Method and calculation

$$\ln(N_c/N_v)=\frac{E_c+E_v-2E_i}{k_BT}.$$$$\boxed{E_i=\frac{E_c+E_v}{2}+\frac{k_BT}{2}\ln(N_v/N_c)}.$$

(b) Interpretation and further calculation

$$\boxed{E_i-E_{\mathrm{mid}}=-(.0250/2)\ln4=-.0173\,\mathrm{eV}}.$$$$\frac{n}{p}=\frac{N_c}{N_v}e^{2(E_F-E_{\mathrm{mid}})/k_BT}=1\quad\text{at }E_F=E_i.$$

Physical check

A larger conduction-band state count requires the Fermi level to move downward to maintain equal electron and hole populations.