Built-in potential from 1 Ω cm resistivity

Assumption: both the p-side and the n-side have bulk resistivity $\rho = 1\ \Omega$ cm, room temperature $T = 300$ K, so $kT/e = 0.02586$ V. In each extrinsic region the majority carriers alone decide the conductivity.

Step 1: Hole mobility

$$\mu_h = \frac{\mu_e}{3.1} = \frac{1400}{3.1} = 451.6\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$$

Step 2: Donor concentration on the n-side

$\sigma = 1/\rho = N_d e \mu_e$, so

$$\begin{aligned} N_d &= \frac{1}{\rho\, e\, \mu_e} = \frac{1}{1 \times 1.602\times10^{-19} \times 1400} \\ &= 4.46\times10^{15}\ \text{cm}^{-3} \end{aligned}$$

Step 3: Acceptor concentration on the p-side

$$\begin{aligned} N_a &= \frac{1}{\rho\, e\, \mu_h} = \frac{1}{1 \times 1.602\times10^{-19} \times 451.6} \\ &= 1.38\times10^{16}\ \text{cm}^{-3} \end{aligned}$$

Step 4: Built-in potential

$$\begin{aligned} V_o &= \frac{kT}{e}\ln\left(\frac{N_a N_d}{n_i^2}\right) \\ &= 0.02586 \ln\left(\frac{1.38\times10^{16} \times 4.46\times10^{15}}{(1.05\times10^{10})^2}\right) \\ &= 0.02586 \ln(5.59\times10^{11}) = 0.02586 \times 27.05 \\ &= 0.70\ \text{V} \end{aligned}$$

Answer: $N_d = 4.46\times10^{15}$ cm⁻³, $N_a = 1.38\times10^{16}$ cm⁻³, built-in potential $V_o \approx 0.70$ V.

The p-side needs about 3.1 times more dopant than the n-side for the same resistivity because holes are 3.1 times less mobile than electrons.