Question

The small signal capacitance of an abrupt $P^{+}n$ junction is $1nF/c{m}^2$ at zero bias. If the built-in bias voltage is 1 volt, the capacitance at a reverse voltage of 99 volts is

• A. $100$
• B. $0.1$
• C. $10$
• D. $0.01$

Answer 1

The correct option is B $0.1$

Ans : (b)

For abrupt p-n junction
$C_j\propto V^{-1/2}$
$C_j\propto \frac{1}{\sqrt{V}}$
$\frac{C_2}{C_1}=\sqrt{\frac{V_1}{V_2}}=\sqrt{\frac{1+0}{1+99}}=\sqrt{\frac{1}{100}}=\frac{1}{10}$
$C_2=\frac{C_1}{10}=\frac{1}{10}=0.1nF/c{m}^2$