Topic: Analog Circuits
In Miller's theorem, if a capacitor C is connected between the input and output of an amplifier with gain A, the equivalent input capacitance is :
- A. C
- B. $(1+A)C$
- C. $(1-A)C$
- D. $C/(1+A)$
Answer: C. $(1-A)C$
Explanation: According to Miller's theorem, an impedance $Z$ connected between the input and output nodes of an amplifier with voltage gain $A$ can be represented as an equivalent input impedance $Z_{in} = \frac{Z}{1-A}$. For a capacitor, $Z = \frac{1}{sC}$. Thus, $Z_{in} = \frac{1}{sC(1-A)}$, which corresponds to an equivalent input capacitance of $C_{in} = C(1-A)$.