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, Nd, and Na can only be determined if the Fermi energy, Ef, is known. Typically, Ef must first be determined from the charge neutrality condition,

n + Na- = p + Nd+.

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, Nd+, and Na- in a spreadsheet. Then choose a temperature and calculate n, p, Nd+, Na- for every value of the Fermi energy between Ev and Ec. For one of these Ef values, the charge neutrality condition will be satisfied.

When n + Na- and , p + Nd+ are ploted as a function of EF, 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] 

Ef [eV]

Nc(300 K) =


Nv(300 K) =


Nd =


Na =


Ed =


Ea =


Eg =


T =


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

n-type Nd > Na:

n = Nd - Na
p = ni²/n
Ef = Ec - kBTln(Nc/(Nd - Na))

p-type Na > Nd:

p = Na - Nd
n = ni²/p
Ef = Ev + kBTln(Nv/(Na - Nd))

The plot is generated with the following code:

for (i=0; i<500; i++) {
  Ef = i*Eg/500;
  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)));