Question 1.6: Objective: Calculate the junction capacitance of a pn juncti...

Microelectronics Circuit Analysis and Design

Objective: Calculate the junction capacitance of a pn junction. Consider a silicon pn junction at T = 300 K, with doping concentrations of $N_{a} = 10^{16} cm^{−3}$ and $N_{d} = 10^{15} cm^{−3}$ . Assume that $n_{i} = 1.5 × 10^{10} cm^{−3}$ and let $C_{jo}$ = 0.5 pF. Calculate the junction capacitance at $V_{R} = 1 V$ and $V_{R} = 5 V$

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The built-in potential is determined by

V_{bi} = V_{T} \ln \left(\frac{N_{a} N_{d}}{n_{i} ^{2}} \right) = (0.026) \ln \left[\frac{(10^{16})(10^{15})}{(1.5   \times   10^{10})^{2}} \right] = 0.637  V

The junction capacitance for $V_{R} = 1 V$ is then found to be

C_{j} = C_{jo} \left(1 + \frac{V_{R} }{V_{bi} } \right) ^{-1/2} = (0.5) \left(1 + \frac{1}{0.637 } \right) ^{-1/2} = 0.312  pF

For $V_{R} = 5 V$

C_{j} = (0.5) \left(1 + \frac{5}{0.637 } \right) ^{-1/2} = 0.168  pF

Comment: The magnitude of the junction capacitance is usually at or below the picofarad range, and it decreases as the reverse-bias voltage increases

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