Si cube with As then 10¹⁴ boron

The size of the cube is not stated; take the usual 1 cm³ cube (L = 1 cm, A = 1 cm²), so $R = \rho L/A$ is numerically equal to $\rho$. The boron dose is taken as $10^{14}$ atoms per cm³. Atomic concentration of Si = $5\times10^{22}\ \text{cm}^{-3}$.

Data used (Si, 300 K): $n_i = 1.45\times10^{10}\ \text{cm}^{-3}$, $kT = 8.617\times10^{-5}\times300 = 0.02585\ \text{eV}$, $\mu_e = 1350$, $\mu_h = 450\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$, $e = 1.6\times10^{-19}\ \text{C}$.

(a) Pure silicon

$$\begin{aligned} \sigma_i &= n_ie(\mu_e+\mu_h) = 1.45\times10^{10}\times1.6\times10^{-19}\times1800\\ &= 4.176\times10^{-6}\ \text{S cm}^{-1}\\ R_i &= 1/\sigma_i = 2.39\times10^{5}\ \Omega \end{aligned}$$

(b) 1 As in $10^9$ Si atoms

$$N_d = \frac{5\times10^{22}}{10^9} = 5\times10^{13}\ \text{cm}^{-3}, \quad n \approx N_d,\quad p = \frac{n_i^2}{N_d} = 4.2\times10^{6}\ \text{cm}^{-3}$$ $$\begin{aligned} \sigma &\approx eN_d\mu_e = 1.6\times10^{-19}\times5\times10^{13}\times1350\\ &= 1.08\times10^{-2}\ \text{S cm}^{-1}\\ R &= 92.6\ \Omega \end{aligned}$$

(c) Further doped with $10^{14}$ boron cm⁻³

Boron is an acceptor. Since $N_a > N_d$ the sample becomes p-type (compensated):

$$p = N_a - N_d = 10^{14} - 5\times10^{13} = 5\times10^{13}\ \text{cm}^{-3},\quad n = \frac{n_i^2}{p} = 4.2\times10^{6}\ \text{cm}^{-3}$$ $$\begin{aligned} \sigma &\approx ep\mu_h = 1.6\times10^{-19}\times5\times10^{13}\times450\\ &= 3.6\times10^{-3}\ \text{S cm}^{-1}\\ R &= 277.8\ \Omega \end{aligned}$$

Answer: $R_{pure} \approx 2.39\times10^5\ \Omega$, $R_{As} \approx 92.6\ \Omega$, $R_{As+B} \approx 277.8\ \Omega$. The resistance rises after boron doping because the carrier number is the same but holes have a lower mobility than electrons.