Consider the circuit in this netlist. Assume all transistors are operating in saturation, \$V_{Tn} = │V_{Tp}│=V_T\$, \$\lambda ≠ 0\$ but \$\gamma = 0\$ (neglect body effect). Neglect all parasitic capacitances and consider only the two capacitances \$C_{L1}\$ and \$C_{L2}\$.
Find the transfer function expressed in first stage and the poles of the circuit.
Answer:
To find the transfer function \$ \frac{V_{\text{out}}}{V_{\text{in}}} \$ of the two-stage CMOS amplifier and identify the DC voltage gain and poles, we proceed step by step:
1. Small-Signal Modeling of the Circuit
First Stage:
Capacitors:
- \$ C_{L1} \$: Connected from node \$ D1 \$ to ground.
- \$ C_{L2} \$: Connected from \$ V_{\text{out}} \$ to ground.
2. Deriving the Transfer Function
First Stage Analysis:
\$V_s\$ is the source of \$M_{n1}\$ \$\Rightarrow V_s = V_x\$
$$ i_{d_{n1}} = -g_{m_{n1}} v_s + \frac{v_{D1} - v_s}{r_{o_{n1}}} $$
$$ i_{d_{p1}} = \frac{v_{D1}}{r_{o_{p1}}} $$
KCL at Node \$ D1 \$:
$$ -g_{m_{n1}} v_s + \frac{v_{D1} - v_s}{r_{o_{n1}}} + \frac{v_{D1}}{r_{o_{p1}}} + s C_{L1} v_{D1} = 0 $$
KCL at Node \$ V_x \$:
$$ \frac{v_{\text{in}} - v_s}{R_S} = g_{m_{n1}} v_s - \frac{v_{D1} - v_s}{r_{o_{n1}}} $$
However, some of classmates think KCl at node D1
$$ -g_{m_{n1}} v_s + \frac{v_{D1} - v_s}{r_{o_{n1}}} + \frac{v_{D1}}{r_{o_{p1}}} - s C_{L1} v_{D1} = 0 $$
