Fermi level at middle of band gap

In an intrinsic (pure) semiconductor the number of electrons in the conduction band equals the number of holes in the valence band. Equating their expressions shows that the Fermi level lies at (or very near) the centre of the gap.

Carrier concentrations

Using the density of states and the Fermi–Dirac function (Boltzmann approximation):

$$n = N_c\exp\left[-\frac{E_c-E_F}{kT}\right],\qquad N_c = 2\left(\frac{2\pi m_e^*kT}{h^2}\right)^{3/2}$$ $$p = N_v\exp\left[-\frac{E_F-E_v}{kT}\right],\qquad N_v = 2\left(\frac{2\pi m_h^*kT}{h^2}\right)^{3/2}$$

Equate n and p (intrinsic, $E_F = E_{Fi}$)

$$N_c\exp\left[-\frac{E_c-E_{Fi}}{kT}\right] = N_v\exp\left[-\frac{E_{Fi}-E_v}{kT}\right]$$

Take logarithms:

$$\begin{aligned} \frac{2E_{Fi}-E_c-E_v}{kT} &= \ln\frac{N_v}{N_c}\\ E_{Fi} &= \frac{E_c+E_v}{2} + \frac{kT}{2}\ln\frac{N_v}{N_c} \end{aligned}$$

Since $N_v/N_c = (m_h^*/m_e^*)^{3/2}$:

$$E_{Fi} = E_v + \frac{E_g}{2} + \frac{3}{4}kT\ln\frac{m_h^*}{m_e^*}$$

Why it is near the middle for Si

• The first term $E_v + E_g/2$ is exactly mid-gap.
• The second term is a small correction. For Si, $m_e^* = 1.08m_e$, $m_h^* = 0.56m_e$, $kT = 0.0259$ eV at 300 K:

$$\frac{3}{4}(0.0259)\ln\frac{0.56}{1.08} = -0.0127\ \text{eV}$$

This is only about 1% of $E_g = 1.1$ eV. So $E_{Fi}$ lies about 0.013 eV below the exact middle, i.e. practically at mid-gap. If $m_e^* = m_h^*$ (or at T = 0 K), it is exactly at the middle.

 Ec -------------------------
            0.55 eV
 mid ......................... Eg/2
 EFi - - - - - - (~0.013 eV below mid)
            0.55 eV
 Ev -------------------------

Physically, every electron excited to the CB leaves one hole in the VB; with similar densities of states on both sides, the level of 50% occupancy must sit midway between them.