The p-n Junction — Question 3

Question 3

Poisson integration gives the triangular electric field

An abrupt junction has depletion charges $-eN_A$ for $-x_p<x<0$ and $+eN_D$ for $0<x<x_n$, uniform permittivity $\varepsilon$, and zero field at both outer depletion edges.

1. Integrate $dE/dx=\rho/\varepsilon$ to find the signed field on each side.

2. Find the potential rise $V_j=\phi(x_n)-\phi(-x_p)$ in terms of peak field magnitude and W, then in terms of doping and widths.

Question 3 - Solution

Poisson integration gives the triangular electric field

(a) Method and calculation

$$\boxed{E_p(x)=-\frac{eN_A}{\varepsilon}(x+x_p),\quad E_n(x)=\frac{eN_D}{\varepsilon}(x-x_n)}.$$$$\boxed{E_{\max}=|E(0)|=eN_Ax_p/\varepsilon=eN_Dx_n/\varepsilon}.$$

(b) Interpretation and further calculation

$$V_j=-\int_{-x_p}^{x_n}E\,dx=\boxed{\tfrac12E_{\max}(x_p+x_n)=\tfrac12E_{\max}W}.$$$$\boxed{V_j=\frac{e}{2\varepsilon}(N_Ax_p^2+N_Dx_n^2)}.$$

Physical check

The field points from the positive donor ions on the n side toward negative acceptor ions on the p side. Its negative area gives a positive potential rise toward n.