Topic: Analog Circuits
Following is the small signal high frequency equivalent circuit of a common source amplifier. $V_o/V_i$ will be of the form ($K, z_1, a_0, a_1, a_2, a_3$ are constants containing circuit elements)
- A. $K(s-z_1)/(a_0+a_1s+a_2s^2)$
- B. $K \cdot s/(a_0+a_1s+a_2s^2+a_3s^3)$
- C. $K(s-z_1)/(a_0+a_1s)$
- D. $K/(a_0+a_1s)$
Answer: A. $K(s-z_1)/(a_0+a_1s+a_2s^2)$
Explanation: The high-frequency equivalent circuit of a Common Source amplifier includes the gate-drain capacitance ($C_{gd}$) creating a Right Half Plane (RHP) zero. It also has two poles caused by the input and output node capacitances. Thus, the transfer function has one zero and two poles, taking the form $\frac{K(s-z_1)}{a_0+a_1s+a_2s^2}$.