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Consider two diodes, A is step graded, B is linear graded. Find the ratio of the capacitance of A to B, when the applied voltage in reverse bias is 64V.(a) 0.2(b) 2(c) 0.5(d) 5The question was asked during an interview.I need to ask this question from Transition Capacitance in chapter Diode Circuit of Analog Circuits

Answer»

Correct OPTION is (d) 5

The EXPLANATION: In A, CA ∝ \(\frac{1}{\sqrt{V}}\)

CB ∝ \(\frac{1}{\sqrt[3]{V}}\)

\(\frac{C_A}{C_B} = \frac{4}{8} = \frac{1}{2}\) = 0.5.


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