Compensated semiconductor concentrations and conductivity

Phosphorus is a donor and boron an acceptor. Since $N_a > N_d$, the sample is converted to p-type (compensated).

Given: $N_d = 10^{16}$, $N_a = 10^{17}\ \text{cm}^{-3}$. Data used: $n_i = 1.45\times10^{10}\ \text{cm}^{-3}$ (Si, 300 K), $e = 1.6\times10^{-19}$ C.

Carrier concentrations

$$\begin{aligned} p &= N_a - N_d = 10^{17} - 10^{16} = 9\times10^{16}\ \text{cm}^{-3}\\ n &= \frac{n_i^2}{p} = \frac{(1.45\times10^{10})^2}{9\times10^{16}} = 2.34\times10^{3}\ \text{cm}^{-3} \end{aligned}$$

Mobility

Mobility is limited by scattering from all ionised impurities: $N_a + N_d = 1.1\times10^{17}\ \text{cm}^{-3}$. From the graph (electron curve for $10^{17}$ at 300 K) $\mu_e \approx 800\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$. Holes have about 0.4 of that mobility in Si; the standard value for $\sim10^{17}\ \text{cm}^{-3}$ doping is $\mu_h \approx 330\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$ (assumed, not on the graph).

Conductivity

The electron term is negligible ($n \ll p$):

$$\begin{aligned} \sigma &= e(n\mu_e + p\mu_h) \approx ep\mu_h\\ &= 1.6\times10^{-19}\times9\times10^{16}\times330\\ &= 4.75\ \text{S cm}^{-1}\\ \rho &= 1/\sigma = 0.210\ \Omega\,\text{cm} \end{aligned}$$

Answer: $p = 9\times10^{16}\ \text{cm}^{-3}$, $n \approx 2.3\times10^{3}\ \text{cm}^{-3}$, $\sigma \approx 4.75$ S cm⁻¹ ($\rho \approx 0.21\ \Omega$ cm).

(If the lightly-doped value $\mu_h = 450$ is used, $\sigma = 6.48$ S cm⁻¹. The conductivity depends on the hole mobility chosen, but in either case the sample is p-type and conduction is by holes.)