B Fermi Energy Dependence on Free Charge Carrier Concentration

  The number of allowed k-values per unit volume of k-space in a piece of material with volume V is $V/8\pi^3$. The number of energy levels in the Fermi sphere, considering two spin values for each k-value, is
$$\begin{gather}N=2\frac{4\pi k_F^3}{3}\frac{V}{8\pi^3}. \end{gather}$$
Thus if there are N electrons in a volume V the concentration is n = N/V, which leads to
$$\begin{gather}n=\frac{k_F^3}{3\pi^2}, \quad\text{with}\quad E(k)=\frac{\hbar^2 k... ...*} \quad\rightarrow\quad E(n)=\frac{\hbar^2}{2m^*}(3\pi^2n)^{2/3}. \end{gather}$$
In the case of semiconductors the band gap has to be taken into account. Using the  Boltzmann approximation to the Fermi distribution the carrier concentrations in the conduction and valence band are
$$\begin{gather}n_c = N_c e^{-\frac{E_c - E_F}{k_BT}}, \qquad p_v = P_ve^{-\frac{E_F-E_v}{k_BT}}. \end{gather}$$
Noting that for the intrinsic case, E_F = E_F,i, n_c=p_v=n_i the concentrations may be written as
$$\begin{gather}n_c=n_ie^{\frac{E_F-E_{F,i}}{k_BT}}, \qquad p_v=n_ie^{-\frac{E_F-E_{F,i}}{k_BT}}. \end{gather}$$
And with the net carrier concentration n=n_c-p_v one derives
$$\begin{gather}\frac{n_{\text{net}}}{n_i} = 2\sinh\frac{E_F - E_{F,i}}{k_B T}. \end{gather}$$