pn junction 10¹⁷/10¹⁶: potential, widths, field

Formulas (abrupt junction, depletion approximation):

$$V_0 = \frac{kT}{e}\ln\frac{N_aN_d}{n_i^2}, \quad W = \sqrt{\frac{2\varepsilon(N_a+N_d)V_0}{eN_aN_d}}, \quad W_n = \frac{N_a W}{N_a+N_d}, \quad W_p = \frac{N_d W}{N_a+N_d}$$

Constants: $kT/e = 0.02586$ V at 300 K, $e = 1.602\times10^{-19}$ C, $\varepsilon = \varepsilon_r\varepsilon_0 = 11.9 \times 8.854\times10^{-14} = 1.054\times10^{-12}$ F/cm (silicon assumed).

Given: $N_d = 10^{17}$ cm$^{-3}$ (n-side), $N_a = 10^{16}$ cm$^{-3}$ (p-side). $n_i$ and $T$ are not given; assume silicon at 300 K with $n_i = 1.45\times10^{10}$ cm$^{-3}$.

(i) Built-in potential

$$\begin{aligned} \frac{N_aN_d}{n_i^2} &= \frac{10^{16}\times10^{17}}{(1.45\times10^{10})^2} = 4.76\times10^{12} \\ V_0 &= 0.02586 \times \ln(4.76\times10^{12}) = 0.02586 \times 29.19 \\ &= 0.755\ \text{V} \end{aligned}$$

(ii) Depletion widths

$$\begin{aligned} W_0 &= \sqrt{\frac{2 \times 1.054\times10^{-12} \times (10^{16}+10^{17}) \times 0.755}{1.602\times10^{-19} \times 10^{16} \times 10^{17}}} \\ &= 3.30\times10^{-5}\ \text{cm} = 0.330\ \mu\text{m} \\ W_n &= \frac{N_a}{N_a+N_d}W_0 = \frac{10^{16}}{1.1\times10^{17}} \times 0.3305 = 0.0300\ \mu\text{m} \\ W_p &= \frac{N_d}{N_a+N_d}W_0 = \frac{10^{17}}{1.1\times10^{17}} \times 0.3305 = 0.300\ \mu\text{m} \end{aligned}$$

The depletion region lies mostly in the lightly doped p-side.

(iii) Field at the metallurgical junction

$$\begin{aligned} |E_0| &= \frac{eN_dW_n}{\varepsilon} = \frac{1.602\times10^{-19} \times 10^{17} \times 3.005\times10^{-6}}{1.054\times10^{-12}} \\ &= 4.57\times10^{4}\ \text{V/cm} = 4.57\ \text{MV/m} \end{aligned}$$

It points from the n-side to the p-side ($E_0 = -4.57\times10^4$ V/cm if $x$ is measured from p to n). Check: $\frac{1}{2}|E_0|W_0 = 0.5 \times 4.57\times10^4 \times 3.305\times10^{-5} = 0.755$ V $= V_0$.

Answer: $V_0 = 0.755$ V; $W_0 = 0.330$ $\mu$m, $W_n = 0.030$ $\mu$m, $W_p = 0.300$ $\mu$m; $|E_0| = 4.57\times10^4$ V/cm.