Electron concentration in an extrinsic semiconductor

In an extrinsic semiconductor the electron concentration is found from two conditions: charge neutrality and the mass action law. The result is valid for n-type, p-type and compensated material.

Step 1: Carrier concentrations in terms of $E_F$

For a non-degenerate semiconductor,

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

Step 2: Mass action law

Multiplying,

$$np = N_cN_v\exp\left(-\frac{E_g}{kT}\right) = n_i^2$$

This product does not depend on $E_F$, so it holds for doped material as well.

Step 3: Charge neutrality

With $N_d$ donors and $N_a$ acceptors, all ionised at room temperature:

$$n + N_a = p + N_d$$

Step 4: Solve for n

Put $p = n_i^2/n$:

$$n + N_a = \frac{n_i^2}{n} + N_d \;\Rightarrow\; n^2 - (N_d - N_a)n - n_i^2 = 0$$

Taking the positive root:

$$n = \frac{N_d-N_a}{2} + \sqrt{\left(\frac{N_d-N_a}{2}\right)^2 + n_i^2},\qquad p = \frac{n_i^2}{n}$$

Special cases

Fermi level form

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

$$n = n_i\exp\left(\frac{E_F-E_{Fi}}{kT}\right) \;\Rightarrow\; E_F - E_{Fi} = kT\ln\frac{n}{n_i}$$

Example: Si, $N_d = 10^{16}$, $N_a = 0$: $n = 10^{16}$ cm⁻³, $p = 2.1\times10^{4}$ cm⁻³, $E_F - E_{Fi} = 0.348$ eV.

At low temperature (freeze-out) the donors are not fully ionised and $n \approx \sqrt{N_cN_d/2}\,e^{-\Delta E/2kT}$, where $\Delta E = E_c - E_d$; the formula above applies to the normal extrinsic range.