The p-n Junction — Question 7

Question 7

A capacitance-voltage slope measures the lightly doped side

A one-sided abrupt $p^+n$ junction has $N_A\gg N_D$, area A and permittivity $\varepsilon$. Use $C_j=\varepsilon A/W$ and $W^2=2\varepsilon(V_{bi}+V_R)/(eN_D)$ for reverse voltage $V_R\ge0$.

1. Derive the slope of $1/C_j^2$ versus $V_R$ and its voltage-axis intercept.

2. Given slope $S=1.25\times10^{19}\,\mathrm{F^{-2}V^{-1}}$, $A=10^{-6}\,\mathrm{m^2}$, $\varepsilon=10^{-10}\,\mathrm{F/m}$, and $e=1.60\times10^{-19}\,\mathrm C$, find $N_D$.

Question 7 - Solution

A capacitance-voltage slope measures the lightly doped side

(a) Method and calculation

$$\boxed{\frac1{C_j^2}=\frac{2(V_{bi}+V_R)}{e\varepsilon A^2N_D}}.$$$$\boxed{S=2/(e\varepsilon A^2N_D),\quad V_{R,\mathrm{intercept}}=-V_{bi}}.$$

(b) Interpretation and further calculation

$$N_D=\frac{2}{(1.60\times10^{-19})(10^{-10})(10^{-6})^2(1.25\times10^{19})}.$$$$\boxed{N_D=10^{22}\,\mathrm{m^{-3}}}.$$

Physical check

The intercept is an extrapolation of the depletion model, where its formal zero-width limit would predict divergent capacitance. The area must be squared in the doping extraction.