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Short note on Resistive Load
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CS stage with resistive load.
$-V_{in} \lt V_{Th}$ -> $M_1- OFF$

$\therefore\, V_{out}=V_{DD}$

As, $V_{in}$ increases to $V_{Th},\,\,M_1$ begins to turn ON.

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

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

Differentiating w.r.t $V_{in}$, we get,

$A_v=\frac{\partial V_{out}}{\partial V_{in}}=0-R_D(\frac{1}{2}\mu_n\,C_{ox}\,2\, \frac{W}{L}(V_{in}-V_{Th})\,(1)) $

$A_v=-gmR_D \hspace{3cm}$

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

$A_v$ can be increase => increase W/L or decrease $I_D$.

Source Degeneration :
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$A_v=\frac{\partial V_{out}}{\partial V_{in}}=-R_D(\frac{\partial I_{D}}{\partial V_{in}})$

let, $G_m=\frac{\partial I_{D}}{\partial V_{in}}$

$G_m=\frac{\partial I_{D}}{\partial V_{GS}}*\frac{\partial V_{GS}}{\partial V_{in}}$

$V_{GS}=V_{in}-I_D\,R_S$

$\frac{\partial V_{GS}}{\partial V_{in}}=1-R_S\,\frac{\partial V_{D}}{\partial V_{in}}$

$G_m=\frac{\partial I_{D}}{\partial V_{GS}}(1-R_S\frac{\partial I_{n}}{\partial V_{in}})$

$G_m=g_m(1-R_S\,g_m)$

$G_m=\frac{g_m}{1+R_s\,gm}$

$\therefore A_v=-G_m\,R_D$
$\hspace{1.3cm}\frac{-gm\, R_d}{1+R_S\,gm}$

As, $A_v$ decreases, $R_s$ increases.

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