Question

The built-in potential of an abrupt p-n junction is 0.75 V. If its junction capacitance (C_J) at a reverse bias (V_R) of 1.25 V is 5 pF, the value of C_J (in pF) when V_R = 7.25 V is _________

Answer 1

Concept:

 junction capacitance

\({{\rm{C}}_{\rm{j}}} \propto \frac{1}{{\sqrt {1 - \frac{{{{\rm{V}}_{\rm{a}}}}}{{{{\rm{V}}_{{\rm{bi}}}}}}} }}\)

Where V_a is applied bias, Thus,

\(\frac{{{{\rm{C}}_{{\rm{j}}1}}}}{{{{\rm{C}}_{{\rm{j}}2}}}} = \frac{{\sqrt {1 - \frac{{{{\rm{V}}_{{\rm{a}}2}}}}{{{{\rm{V}}_{{\rm{bi}}}}}}} }}{{\sqrt {1 - \frac{{{{\rm{V}}_{{\rm{a}}1}}}}{{{{\rm{V}}_{{\rm{bi}}}}}}} }}\)

Application:

We have \({{\rm{V}}_{{\rm{bi}}}} = 0.75{\rm{V}},{\rm{\;}}{{\rm{C}}_{{\rm{j}}1}} = 5{\rm{pF\;and\;}}{{\rm{V}}_{{\rm{a}}1}} = - 1.25{\rm{V}}\)

Thus,

\(\begin{array}{l} \frac{5}{{{{\rm{C}}_{{\rm{j}}2}}}} = \frac{{\sqrt {1 - \frac{{\left( { - 7.25} \right)}}{{0.75}}} }}{{\sqrt {1 - \frac{{\left( { - 1.25} \right)}}{{0.75}}} }}\\ {{\rm{C}}_{{\rm{j}}2}} = 5 \times \frac{{\sqrt {1 + \frac{5}{3}} }}{{\sqrt {1 + \frac{{29}}{3}} }}\\ \Rightarrow {{\rm{C}}_{{\rm{j}}2}} = 5 \times \frac{{1.633}}{{3.266}} = 2.5{\rm{pF}} \end{array}\)