Doped Semiconductors — Question 1

Question 1

Charge neutrality gives the exact carrier density near the intrinsic crossover

A nondegenerate semiconductor has fully ionized donors $N_D$ and no acceptors. Thermal equilibrium requires $n=p+N_D$ and $np=n_i^2$.

1. Solve exactly for both carrier densities.

2. For $N_D=3n_i$, evaluate n and p in units of $n_i$ and quantify the error of using $n=N_D$.

Question 1 - Solution

Charge neutrality gives the exact carrier density near the intrinsic crossover

(a) Method and calculation

$$n^2-N_Dn-n_i^2=0\Rightarrow\boxed{n=\tfrac12(N_D+\sqrt{N_D^2+4n_i^2})}.$$$$\boxed{p=n_i^2/n=\tfrac12(\sqrt{N_D^2+4n_i^2}-N_D)}.$$

(b) Interpretation and further calculation

$$\boxed{n/n_i=(3+\sqrt{13})/2=3.303,\quad p/n_i=.3028}.$$$$\boxed{(n-N_D)/n=.09167=9.17\%}.$$

Physical check

The positive quadratic root is required. At large donor density it approaches ND; at zero donor density both species approach ni.