The p-n Junction — Question 6

Question 6

Differentiating depletion charge yields junction capacitance

For a fixed abrupt junction, $Q=AeN_AN_DW/(N_A+N_D)$ is the magnitude of charge on either depletion side, and $W^2=(2\varepsilon/e)(1/N_A+1/N_D)V_j$. Define $C_j=dQ/dV_j$.

1. Derive the small-signal depletion capacitance.

2. For $A=1.00\,\mathrm{mm^2}$, $\varepsilon=10^{-10}\,\mathrm{F/m}$, and $W=.500\,\mu\mathrm m$, evaluate $C_j$ and the change when reverse bias quadruples $V_j$.

Question 6 - Solution

Differentiating depletion charge yields junction capacitance

(a) Method and calculation

$$2W\frac{dW}{dV_j}=\frac{2\varepsilon}{e}\frac{N_A+N_D}{N_AN_D}.$$$$\boxed{C_j=A\frac{eN_AN_D}{N_A+N_D}\frac{dW}{dV_j}=\varepsilon A/W}.$$

(b) Interpretation and further calculation

$$\boxed{C_j=10^{-10}(10^{-6})/(.500\times10^{-6})=200\,\mathrm{pF}}.$$$$V_j\to4V_j\Rightarrow W\to2W\Rightarrow\boxed{C_j\to100\,\mathrm{pF}}.$$

Physical check

This is a differential capacitance. Using Q divided by Vj would give twice the correct small-signal value because depletion width changes with voltage.