Subject: CMOS VLSI Design

Topic: Single Stage Amplifier

Difficulty: Medium

-> For $V_{in} \geq B_b-V_{Th}, M_1$ is OFF and $V_{out}=V_{DD}$

$\therefore I_D =\frac{1}{2}\mu_n\,C_{ox} \frac{W}{L}(V_b-V_{in}-V_{Th})^2$

$V_{out}=V_{DD}-R_D\,I_D$

$\therefore V_{out}=V_{DD}-R_D \Big[ \frac{1}{2}\mu_n\,C_{ox} \frac{W}{L}(V_b-V_{in}-V_{Th})^2 \Big]$

Differentiating on both sides w.r.t $V_{in}$, we get,
$\frac{\partial V_{out}}{\partial V_{in}}=-\mu_n\,C_{ox} \frac{W}{L}(V_b-V_{in}-V_{Th})\,(-1-\frac{\partial V_{Th}}{\partial V_{in}})\,R_D$

$\hspace{1cm}$ But, $\frac{\partial V_{Th}}{\partial V_{in}}=\frac{\partial V_{Th}}{\partial V_{SB}}=\eta\,\hspace{2cm}......\eta=\frac{gm_b}{gm}$

$\therefore \,\,\frac{\partial V_{out}}{\partial V_{in}}=A_V=\mu_n\,C_{ox} \frac{W}{L}(V_b-V_{in}-V_{Th})\,(1+\eta)\,R_D$

$\therefore \, …