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Analog Electronics - Engineering

Q1:

In figure the input and output Miller capacitances are

A 0.25 μF each

B 500 pF each

C 0.25 μF and 500 pF respectively

D 500 pF and 0.25 μF respectively

ANS:A - 0.25 μF each

Ci = cf(A-1).
Co = cf(A-1)/A.

If we can apply then correct answer is:

Ci = 0.25 micro farad.
Co = 50 pf.