pn junction built-in potential and diffusion coefficients

Data used: Si at 300 K, $n_i = 1.45\times10^{10}\ \text{cm}^{-3}$, $kT/e = 0.02585$ V, $\mu_e = 1350$, $\mu_h = 450\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$ (standard values; doped-Si mobilities are somewhat lower). $N_d = 10^{17}$, $N_a = 10^{16}\ \text{cm}^{-3}$.

Built-in potential

$$\begin{aligned} V_0 &= \frac{kT}{e}\ln\frac{N_aN_d}{n_i^2}\\ &= 0.02585\ln\frac{10^{16}\times10^{17}}{(1.45\times10^{10})^2}\\ &= 0.02585\ln(4.756\times10^{12})\\ &= 0.02585\times29.19 = 0.755\ \text{V} \end{aligned}$$

Diffusion coefficients (Einstein relation)

$$D = \mu\frac{kT}{e}$$

Diffusion across a junction is by minority carriers: electrons in the p part and holes in the n part.

p part (electrons diffusing):

$$D_e = 1350\times0.02585 = 34.9\ \text{cm}^2\text{s}^{-1}$$

n part (holes diffusing):

$$D_h = 450\times0.02585 = 11.6\ \text{cm}^2\text{s}^{-1}$$

Answer: $V_0 \approx 0.755$ V; $D_e \approx 34.9\ \text{cm}^2\text{s}^{-1}$ (p side) and $D_h \approx 11.6\ \text{cm}^2\text{s}^{-1}$ (n side).

The ratio $D/\mu = kT/e = 0.02585$ V is the same for both carriers at a given temperature; if mobilities read from a doping-dependent graph are used, D scales in the same proportion.