Carrier concentration vs. Fermi energy

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

the density of holes in the valence band is,

the density of ionized donors is,

and the density of ionized acceptors is,

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.

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)));
}