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Explain common mode response of differential pair with necessary derivations.
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Common Mode Characteristics

(a) Circuit is symmetric but tail current source with finite resistance:

i) As $V_{in,cm}$ increases, $V_D$ increases, thereby increases drain current of $M_1\,\,and \,\,M_2$ and decreases both $V_x,\,\,V_y$.
ii) Due to symmetry, $V_x=V_y$, so they can be shorted together. iii) Now, $M_1\,,\,M_2$ are parallel. Therefore, circuit can be simplified as per fig (c)

iv) $-(M_1+M_2)$ has twice the width (i.e twice bias current). Therefore twice the transconductance ($g_m$).
$\therefore \,\,$ CM gain of the circuit will be,

$A_{v,cm}=\frac{V_{out}}{V_{in,cm}}$

$A_{v,cm}=\frac{-R_D/2}{1/2g_m+R_{SS}}$

Note: In a symmetric circuit, CM variation disturb the bias points, altering the small-signal gain and limiting the output voltage swings.

(b) Effect of mismatches in $R_D$:

i) let, $R_{D1}=R_D$ and $R_{D2}=R_D+ \Delta R_D$ small mismatch and circuit is symmetrical otherwise
ii) assume $M_1\,\,,\,M_2$ identical.
iii) As, $V_{in,cm} increases, I_{D1} \,\,and\,\,I_{D2}$ increases by $\Big[ \frac{g_m}{1+2\,g_m\,R_{SS}} \Big] \Delta\,V_{in,cm}$ iv) But, $V_x$ and $V_y$ change by different amount.
$\Delta V_x=-\Delta\,V_{in,cm}\,\frac{g_m}{1+2\,g_m\,R_{SS}}(R_D)$
$\Delta V_y=-\Delta\,V_{in,cm}\,\frac{g_m}{1+2\,g_m\,R_{SS}}(R_D+\Delta\,R_D)$

$\Big[ I_D=\frac{1}{\frac{1}{g_m}+2R_{SS}}V_{in}=\frac{g_m}{1+2\,g_m\,R_{SS}}V_{in} \Big]$

$\therefore$ Change in differential output voltage can be calculated by subtracting $\Delta V_y$ from $\Delta V_x$.

(c) Mismatch in transistor $M_1$ and $M_2$.

$A_{v,cm}=\frac{V_x-V_y}{V_{in,cm}}$ $V_x=I_{D1}\,R_D=g_{m_{1}}\,V_{GS}\,R_D$
$\hspace{0.3cm}=-g_{m_{1}}(V_{in,cm}-V_P)R_D \hspace{2cm}$....(1)

$V_y=I_{D2}\,R_D$
$\hspace{0.3cm}=-g_{m_{2}}(V_{in,cm}-V_P)R_D \hspace{2cm}$....(2)

$V_p=R_{SS}(I_{D1}+I_{D2})=R_{SS}(g_{m_{1}}V_{GS1}+g_{m_{2}}V_{GS2})$
$\hspace{0.3cm}=R_{SS}(g_{m_{1}}(V_{in,cm}-V_P)+g_{m_{2}}(V_{in,cm}-V_P))$

$V_p=\frac{(g_{m_{1}}+g_{m_{2}})R_{SS}}{(g_{m_{1}}+g_{m_{2}})R_{SS}+1}V_{in,cm}\hspace{3cm}$......(3)

Substituting (3) in (1) and (2), we get,
$V_x=\frac{-g_{m_{1}}}{(g_{m_{1}}+g_{m_{2}})R_{SS}+1}\,R_D\,V_{in,cm}$

$V_y=\frac{-g_{m_{2}}}{(g_{m_{1}}+g_{m_{2}})R_{SS}+1}\,R_D\,V_{in,cm}$

$\therefore\,\,V_x-V_y=\frac{-(g_{m_{1}}-g_{m_{2}})}{(g_{m_{1}}+g_{m_{2}})R_{SS}+1}\,R_D\,V_{in,cm}$

$A_{v,cm}=\frac{V_x-V_y}{V_{in,cm}}=\frac{-(g_{m_{1}}-g_{m_{2}})}{(g_{m_{1}}+g_{m_{2}})R_{SS}+1}R_D$

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