Doped Semiconductors — Question 2

Question 2

Compensation changes the majority type through the net dopant density

A fully ionized sample has $N_D=2.00\times10^{22}\,\mathrm{m^{-3}}$, $N_A=5.00\times10^{22}\,\mathrm{m^{-3}}$, and $n_i=10^{16}\,\mathrm{m^{-3}}$. It is in nondegenerate thermal equilibrium.

1. Find the majority and minority carrier densities to appropriate accuracy.

2. If $\mu_p=.0400\,\mathrm{m^2/(V\,s)}$ and $e=1.60\times10^{-19}\,\mathrm C$, estimate conductivity and state why the total dopant count is not the carrier count.

Question 2 - Solution

Compensation changes the majority type through the net dopant density

(a) Method and calculation

$$p-n=N_A-N_D=3.00\times10^{22}\,\mathrm{m^{-3}}\equiv\Delta N.$$$$\boxed{p\simeq\Delta N=3.00\times10^{22},\quad n=n_i^2/p=3.333\times10^9\,\mathrm{m^{-3}}}.$$

(b) Interpretation and further calculation

$$\boxed{\sigma\simeq ep\mu_p=192\,\mathrm{S/m}}.$$$$\boxed{p\simeq N_A-N_D\ne N_A+N_D}\quad(n\ll p).$$

Physical check

Donors compensate acceptors, leaving the difference to set the majority population. Both kinds of ionized impurities can still contribute to scattering.