Semiconductors — Question 2

Question 2

Donor density enters an exact carrier-balance quadratic

At equilibrium, a nondegenerate semiconductor with fully ionized donors and no acceptors obeys $np=n_i^2$ and $n-p=N_D$. Let $N_D=3n_i$.

1. Find exact n and p in units of ni.

2. Compare n with the approximation $n\simeq N_D$, and state the condition under which that approximation improves.

Question 2 - Solution

Donor density enters an exact carrier-balance quadratic

(a) Method and calculation

$$n(n-N_D)=n_i^2\Rightarrow n=\frac{N_D+\sqrt{N_D^2+4n_i^2}}2.$$$$n/n_i=(3+\sqrt{13})/2=3.30278,\quad p/n_i=(\sqrt{13}-3)/2=.302776.$$

(b) Interpretation and further calculation

$$\frac{n-N_D}{n}=\frac{.302776}{3.30278}=.0916731.$$$$N_D\gg n_i:\quad n=N_D+n_i^2/N_D+O(n_i^4/N_D^3).$$

Physical check

Charge neutrality includes immobile positive donor ions; the mobile electrons and holes need not have equal densities.