The p-n Junction — Question 1

Question 1

Equilibrium carrier ratios establish the built-in junction potential

A nondegenerate abrupt homojunction has fully ionized dopants $N_A=10^{23}\,\mathrm{m^{-3}}$ and $N_D=10^{22}\,\mathrm{m^{-3}}$, with $n_i=10^{16}\,\mathrm{m^{-3}}$ and $V_T=.0250\,\mathrm V$. In neutral regions, $n_n\simeq N_D$ and $n_p\simeq n_i^2/N_A$.

1. Use $n\propto e^{\phi/V_T}$ to derive the built-in potential $V_{bi}=\phi_n-\phi_p$.

2. Evaluate $V_{bi}$ and the increase if both doping densities are multiplied by ten.

Question 1 - Solution

Equilibrium carrier ratios establish the built-in junction potential

(a) Method and calculation

$$\frac{n_n}{n_p}=\frac{N_AN_D}{n_i^2}=e^{(\phi_n-\phi_p)/V_T}.$$$$\boxed{V_{bi}=V_T\ln(N_AN_D/n_i^2)}.$$

(b) Interpretation and further calculation

$$\boxed{V_{bi}=.0250\ln10^{13}=.7483\,\mathrm V}.$$$$\boxed{\Delta V_{bi}=V_T\ln100=.1151\,\mathrm V}.$$

Physical check

The equilibrium electrochemical potential is uniform. The internal built-in potential is not a usable open-circuit battery voltage at identical external contacts.