Resistance of Si cube doped with As and B

Resistance of a cube with side $L = 1$ cm: $R = \rho L/A = \rho \times \dfrac{1\ \text{cm}}{1\ \text{cm}^2}$, so $R$ in $\Omega$ equals $\rho$ in $\Omega$ cm.

Data: $n_i = 1.45\times10^{10}$ cm$^{-3}$, $N_{Si} = 5\times10^{22}$ cm$^{-3}$, $e = 1.602\times10^{-19}$ C. Assumed room-temperature mobilities for Si: $\mu_e = 1350$, $\mu_h = 450$ cm$^2$V$^{-1}$s$^{-1}$ (at such low doping, impurity scattering is negligible).

(a) Pure silicon

$$\begin{aligned} \sigma &= e n_i(\mu_e + \mu_h) \\ &= 1.602\times10^{-19} \times 1.45\times10^{10} \times 1800 \\ &= 4.18\times10^{-6}\ \Omega^{-1}\text{cm}^{-1} \\ R &= \frac{1}{\sigma} = 2.39\times10^{5}\ \Omega \end{aligned}$$

(b) Doped with 1 As per $10^9$ Si atoms

Arsenic is a donor:

$$\begin{aligned} N_d &= \frac{5\times10^{22}}{10^9} = 5\times10^{13}\ \text{cm}^{-3} \gg n_i \\ n &\approx N_d = 5\times10^{13}\ \text{cm}^{-3} \\ p &= \frac{n_i^2}{n} = \frac{(1.45\times10^{10})^2}{5\times10^{13}} = 4.2\times10^{6}\ \text{cm}^{-3} \ (\text{negligible}) \\ \sigma &= e n\mu_e = 1.602\times10^{-19} \times 5\times10^{13} \times 1350 \\ &= 1.081\times10^{-2}\ \Omega^{-1}\text{cm}^{-1} \\ R &= \frac{1}{\sigma} = 92.5\ \Omega \end{aligned}$$

(c) Doped with 1 B per billion Si atoms

Boron is an acceptor: $N_a = 5\times10^{13}$ cm$^{-3}$.

If boron is added together with the arsenic (both present): $N_a = N_d$, so the acceptors capture all the electrons given by the donors (full compensation). The net doping $N_d - N_a = 0$, so the crystal behaves as intrinsic: $n = p = n_i$.

$$R \approx 2.39\times10^{5}\ \Omega$$

If the crystal is doped with boron alone (p-type): $p \approx N_a = 5\times10^{13}$ cm$^{-3}$.

$$\begin{aligned} \sigma &= e p\mu_h = 1.602\times10^{-19} \times 5\times10^{13} \times 450 \\ &= 3.60\times10^{-3}\ \Omega^{-1}\text{cm}^{-1} \\ R &= \frac{1}{\sigma} = 277\ \Omega \end{aligned}$$

Answer: Pure Si: $R = 2.39\times10^{5}$ $\Omega$. With 1 As per $10^9$: $R = 92.5$ $\Omega$. With 1 As and 1 B per $10^9$: compensated, so $R$ returns to about $2.39\times10^{5}$ $\Omega$. (With 1 B per $10^9$ alone: $R \approx 277$ $\Omega$.)