The p-n Junction — Question 4

Question 4

The depletion width follows from electrostatics and charge balance

For an abrupt junction use $N_Ax_p=N_Dx_n$, $W=x_p+x_n$, and $V_j=e(N_Ax_p^2+N_Dx_n^2)/(2\varepsilon)$.

1. Derive W in terms of $V_j$, $N_A$, and $N_D$.

2. Evaluate W for $N_A=N_D=10^{22}\,\mathrm{m^{-3}}$, $V_j=.800\,\mathrm V$, $\varepsilon=1.00\times10^{-10}\,\mathrm{F/m}$, and $e=1.60\times10^{-19}\,\mathrm C$.

Question 4 - Solution

The depletion width follows from electrostatics and charge balance

(a) Method and calculation

$$x_p=\frac{N_DW}{N_A+N_D},\quad x_n=\frac{N_AW}{N_A+N_D}.$$$$\boxed{W=\sqrt{\frac{2\varepsilon V_j}{e}\left(\frac1{N_A}+\frac1{N_D}\right)}}.$$

(b) Interpretation and further calculation

$$W^2=\frac{2(10^{-10})(.800)}{1.60\times10^{-19}}\frac{2}{10^{22}}=2.00\times10^{-13}\,\mathrm{m^2}.$$$$\boxed{W=.4472\,\mu\mathrm m,\quad x_p=x_n=.2236\,\mu\mathrm m}.$$

Physical check

The dimensions under the square root are length squared. Larger doping reduces the distance needed to support the specified potential.