The p-n Junction — Question 2

Question 2

Depletion charge neutrality fixes how the junction width is divided

In the depletion approximation, an abrupt junction has charge density $-eN_A$ over $-x_p<x<0$ and $+eN_D$ over $0<x<x_n$. Fields vanish outside. Let $N_A=4N_D$ and total width $W=.500\,\mu\mathrm m$.

1. Use Gauss law to find $x_p$ and $x_n$.

2. For $N_D=10^{22}\,\mathrm{m^{-3}}$ and $e=1.60\times10^{-19}\,\mathrm C$, find the charge magnitude per area on each side.

Question 2 - Solution

Depletion charge neutrality fixes how the junction width is divided

(a) Method and calculation

$$\int_{-x_p}^{x_n}\rho\,dx=0\Rightarrow N_Ax_p=N_Dx_n,\quad x_p+x_n=W.$$$$\boxed{x_p=W/5=.100\,\mu\mathrm m,\quad x_n=4W/5=.400\,\mu\mathrm m}.$$

(b) Interpretation and further calculation

$$Q_n/A=eN_Dx_n,\quad Q_p/A=-eN_Ax_p.$$$$\boxed{Q_n/A=+6.40\times10^{-4},\quad Q_p/A=-6.40\times10^{-4}\,\mathrm{C/m^2}}.$$

Physical check

Depletion extends farther into the more lightly doped material. Equal total charge magnitudes do not imply equal widths or equal charge densities.