Question

an abrupt $p-n$ junction has build -in potential of 0.6V. If the junction capacitance (C) at a reverse bias voltage of $V_{R1}$ is 10 pF and this junction capacitance will become 20 pF at a reverse voltage of 5 V. Then the value of VR1 is
(where, $V_{R1}and{V}_{R2}$ are magnitude of reverse voltages).

• A. 9.8 V
• B. 21.8 V
• C. 12.4 V
• D. 13.8 V

Answer 1

The correct option is B 21.8 V

$\begin{array}{cc}\text{For junction capacitance}{C}_j& \propto \frac{1}{\sqrt{V_{bi}+V_R}}\\ \text{where}{V}_{bi}& =\text{built-in potential}\\ V_R& =\text{reverse voltage}\\ \text{given,}{C}_{j1}=10pF;C_{j2}& =20pF\\ \therefore \frac{C_{j1}}{C_{j2}}& =\sqrt{\frac{V_{bi}+V_{R2}}{V_{bi}+V_{R1}}}\\ \frac{10}{20}& =\sqrt{\frac{0.6+V_{R2}}{0.6+V_{R1}}}\\ \frac{1}{4}& =\frac{0.6+V_{R2}}{0.6+V_{R1}}\\ 0.6+V_{R1}& =2.4+4V_{R2}\\ \text{Given,}{V}_{R2}=5V& \\ 0.6+V_{R1}& =2.4+(4\times 5)\\ \therefore V_{R1}& =21.8V\end{array}$