pn junction 10¹⁸/10¹⁶: potential, width, field

Data used: $n_i = 1.45\times10^{10}\ \text{cm}^{-3}$, $kT/e = 0.02585$ V at 300 K, $\varepsilon = \varepsilon_0\varepsilon_r = 8.85\times10^{-14}\times11.9 = 1.053\times10^{-12}\ \text{F cm}^{-1}$, $e = 1.6\times10^{-19}$ C. $N_a = 10^{18}\ \text{cm}^{-3}$ (p side), $N_d = 10^{16}\ \text{cm}^{-3}$ (n side) — a $p^+n$ junction.

i) Built-in potential

$$\begin{aligned} V_0 &= \frac{kT}{e}\ln\frac{N_aN_d}{n_i^2} = 0.02585\ln\frac{10^{18}\times10^{16}}{(1.45\times10^{10})^2}\\ &= 0.02585\ln(4.756\times10^{13}) = 0.02585\times31.49\\ &= 0.814\ \text{V} \end{aligned}$$

ii) Width of depletion layer

$$\begin{aligned} W_0 &= \sqrt{\frac{2\varepsilon(N_a+N_d)V_0}{eN_aN_d}}\\ &= \sqrt{\frac{2\times1.053\times10^{-12}\times1.01\times10^{18}\times0.814}{1.6\times10^{-19}\times10^{34}}}\\ &= 3.29\times10^{-5}\ \text{cm} = 0.329\ \mu\text{m} \end{aligned}$$

Split between the sides:

$$W_n = W_0\frac{N_a}{N_a+N_d} = 3.26\times10^{-5}\ \text{cm},\qquad W_p = W_0\frac{N_d}{N_a+N_d} = 3.26\times10^{-7}\ \text{cm}$$

Almost all (99%) of the depletion layer is in the lightly doped n side.

iii) Maximum electric field

The field peaks at the metallurgical junction:

$$\begin{aligned} \mathcal{E}_0 &= \frac{eN_dW_n}{\varepsilon} = \frac{1.6\times10^{-19}\times10^{16}\times3.26\times10^{-5}}{1.053\times10^{-12}}\\ &= 4.95\times10^{4}\ \text{V cm}^{-1} \end{aligned}$$

Check: $\mathcal{E}_0 = 2V_0/W_0 = 2\times0.814/3.29\times10^{-5} = 4.95\times10^{4}$ V/cm. ✔

Answer: $V_0 \approx 0.814$ V, $W_0 \approx 0.329\ \mu$m, $\mathcal{E}_{max} \approx 4.95\times10^{4}$ V/cm ($4.95\times10^{6}$ V/m).

   E
   |   p+ |      n
   |      |\
   |      | \
   |      |  \
   |      |   \
   +------+----\---- x
        -Wp  0   Wn
   Emax at x = 0