Carrier concentration vs. Fermi energy

For a doped semiconductor, the density of electrons in the conduction band is,

$$n = N_c(T)\exp\left(\frac{E_F-E_c}{k_BT}\right).$$

The density of holes in the valence band is,

$$p = N_v(T)\exp\left(\frac{E_v-E_F}{k_BT}\right).$$

The density of ionized donors is,

$$N_D^+=\frac{N_D}{1+2\exp\left(\frac{E_F-E_D}{k_BT}\right)}.$$

and the density of ionized acceptors is,

$$N_A^-=\frac{N_A}{1+4\exp\left(\frac{E_A-E_F}{k_BT}\right)}.$$

The factor of 4 is valid in the formula for the acceptors if the semiconductor has a light hole and a heavy hole band as Si and Ge do.

The four quantities $n$, $p$, $N_D$, and $N_A$ can only be determined if the Fermi energy, $E_F$, is known. Typically, $E_F$ must first be determined from the charge neutrality condition,

$$n+N_A^- = p+N_D^+.$$

The Fermi energy can be found by solving the charge neutrality condition numerically. One way to do this is to program the formulas for n, p, N_d⁺, and N_a^- in a spreadsheet. Then choose a temperature and calculate n, p, N_d⁺, N_a^- for every value of the Fermi energy between E_v and E_c. For one of these E_f values, the charge neutrality condition will be satisfied.

When n + N_a^- and , p + N_d⁺ are ploted as a function of E_F, the Fermi energy is where the two lines cross. A new plot like the one below can be generated by pressing the 'Replot' button.

log [1/cm³]
	E_f [eV]

Normally it is not necessary to determine E_f numerically and the following approximation is sufficient.

n-type N_d > N_a:

n = N_d - N_a
p = n_i²/n
E_f = E_c - k_BTln(N_c/(N_d - N_a))

p-type N_a > N_d:

p = N_a - N_d
n = n_i²/p
E_f = E_v + k_BTln(N_v/(N_a - N_d))

The plot is generated with the following code:

for (i=0; i<500; i++) {
  Ef = i*Eg/500;
  n=Nc*pow(T/300,1.5)*exp(1.6022E-19*(Ef-Eg)/(1.38E-23*T));
  p=Nv*pow(T/300,1.5)*exp(1.6022E-19*(-Ef)/(1.38E-23*T));
  Namin = Na/(1+4*exp(1.6022E-19*(Ea-Ef)/(1.38E-23*T)));
  Ndplus = Nd/(1+2*exp(1.6022E-19*(Ef-Ed)/(1.38E-23*T)));
}