Intrinsic Fermi level of pure semiconductor

Intrinsic Fermi level

The intrinsic Fermi level $E_{Fi}$ is the Fermi level of a pure (undoped) semiconductor, where $n = p = n_i$. It is the energy at which the probability of occupation is 1/2. Because each electron excited into the conduction band leaves one hole in the valence band, $E_{Fi}$ lies almost exactly at the middle of the band gap, shifted slightly towards the band with the smaller density of states. It serves as the reference level from which the Fermi level of doped material is measured ($E_F > E_{Fi}$ for n-type, $E_F < E_{Fi}$ for p-type).

Derivation (given $n_i$)

In any non-degenerate semiconductor

$$n = N_c\exp\left[-\frac{E_c-E_F}{kT}\right],\qquad p = N_v\exp\left[-\frac{E_F-E_v}{kT}\right]$$

For intrinsic material $n = n_i$ and $E_F = E_{Fi}$:

$$n_i = N_c\exp\left[-\frac{E_c-E_{Fi}}{kT}\right] \;\Rightarrow\; E_{Fi} = E_c - kT\ln\frac{N_c}{n_i}$$

Similarly from $p = n_i$:

$$E_{Fi} = E_v + kT\ln\frac{N_v}{n_i}$$

Adding the two forms and dividing by 2:

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

Doped material in terms of $n_i$

Dividing $n = N_ce^{-(E_c-E_F)/kT}$ by $n_i = N_ce^{-(E_c-E_{Fi})/kT}$:

$$n = n_i\exp\left(\frac{E_F-E_{Fi}}{kT}\right),\qquad p = n_i\exp\left(\frac{E_{Fi}-E_F}{kT}\right)$$

Example (Si, 300 K): $N_c = 2.8\times10^{19}$, $n_i = 1.45\times10^{10}\ \text{cm}^{-3}$: $E_c - E_{Fi} = 0.02585\ln(1.93\times10^9) = 0.553$ eV, i.e. about half of $E_g = 1.1$ eV.