CD Stage:

• Known as source follower {$A_v \approx 1$}

• CS Stage -> to achieve high $A_v$ with limited power supply the load impedance must be as large as possible.

• If load impedance is low, then a buffer must be placed after amplifier so as to drive the load with negligible loss of signal.

• Hence source follower can operate as BUFFER.

• When $V_{in} \lt V_{Th}, \,\,M_1 - off \,\,\,\therefore \,NO\,\,Current \,\,\,\,\therefore\, V_{out}=0$

• As $V_{in}$ crosses $V_{Th}, \,\,M_1-ON$ in saturation.

  As $V_{in}$ increases, $I_D$ increases, $V_{out}$ increases. $V_{out}$ follows $V_{in}$ with difference equal to $V_{GS}$.

$\hspace{1cm} V_{out}=I_D\,R_S$
$\hspace{1.8cm}=\frac{1}{2}\mu_n\,C_{ox} \frac{W}{L}(V_{GS}-V_{Th})^2 \,R_S$
$\hspace{1cm} V_{out}=\frac{1}{2}\mu_n\,C_{ox} \frac{W}{L}(V_{in}-V_{out}-V_{Th})^2 \,R_S \hspace{2cm}$

$...(V_{GS}=V_{in}-V_{out})$

• To Calculate $A_V$, differential w.r.t $V_{in}$,

  $\frac{\partial V_{out}}{\partial V_{in}}=\frac{1}{2}\mu_n\,C_{ox}\,2\, \frac{W}{L}(V_{in}-V_{out}-V_{Th})\,(1-\frac{\partial V_{out}}{\partial V_{in}}-\frac{\partial V_{Th}}{\partial V_{in}})\,R_S$

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

  $\therefore\,\,A_V=\frac{1}{2}\mu_n\,C_{ox}\,2\, \frac{W}{L}(V_{in}-V_{out}-V_{Th})\,(1-A_V-\frac{\partial V_{Th}}{\partial V_{in}})\,R_S$

$\therefore\,\,A_V=\frac{gm\,R_S}{1+(gm+gm_b)R_S}\hspace{2cm}$

$.....\Big[ gm=\mu_n\,C_{ox} \frac{W}{L}(V_{GS}-V_{Th}) = \mu_n\,C_{ox} \frac{W}{L}(V_{in}-V_{out}-V_{Th})\Big]$

-> Small signal Equivalent.

-> Expression for output resistance $R_O$.

$R_O=\frac{V_{out}}{I_{out}}|_{V_{in}=0}$

At X,
$I_x-gmV_x-gm_b\,V_x=0 \hspace{1cm}$ ......(KCL at X)

$\frac{V_x}{I_x}=R_O=\frac{1}{gm+gm_b}$