Voltage gain of differential pair.

Since, current source - increase resistance, we simplify differebtial pair in fig(a).

Using superposition theorem,
(1) Consider $V_{in_{1}}$ active and $V_{in_{2}}=0$, obtain $V_x\,\,and\,\,\,V_y$

• To obtain $V_x$, $M_1$ forms a CS stage, with $R_S$ equal to impedance looking into the source of $M_2$ i.e $\frac{1}{g_{m_{2}}}$

  $\therefore\,\,\,\frac{V_x}{V_{in_{1}}}=\frac{-R_D}{\frac{1}{g_{m_{1}}}+\frac{1}{g_{m_{2}}}}\hspace{2cm}$.....(1)
  [ Since,$R_S=\frac{1}{g_{m_{2}}}$ also $R_{D_{1}}=R_{D_{2}}=R_D$ ]

• To calculate $V_y$, use Thevenin's theoram,

$V_T=V_{in_{1}}$ and $R_T=\frac{1}{g_{m_{1}}}$
$\therefore $ circuit looks like CG stage.
$\therefore \,\,\,\frac{V_y}{V_{in_{1}}}=\frac{R_D}{\frac{1}{g_{m_{1}}}+\frac{1}{g_{m_{2}}}}\hspace{2cm}$.....(2)

$\therefore$ from (1) and (2),
$(V_x-V_y)|_{V_{in_{1}}}=V_{in_{1}}\frac{-2\,R_D}{\frac{1}{g_{m_{1}}}+\frac{1}{g_{m_{2}}}}\hspace{2cm}$.....(3)

if $g_{m_{1}}=g_{m_{2}}$ then,
$(V_x-V_y)|_{V_{in_{1}}}=-g_m\,R_D\,V_{in_{1}}\hspace{2cm}$.....(4)

(2) Consider $V_{in_{2}}$ active and $V_{in_{2}}=0$,
Since, circuit is symmetric, the effect of $V_{in_{2}}$ at X & Y is identical to that of $V_{in_{1}}$ with opposite polarity.

$\therefore\,\,\,(V_x-V_y)|_{V_{in_{2}}}=g_m\,R_D\,V_{in_{2}}\hspace{2cm}$.....(5)

(4)+(5), gives,

$V_{out}=(V_x-V_y)_{tot}=-g_m\,R_D(V_{in_{1}}-V_{in_{2}})$
$V_{out}=\frac{(V_x-V_y)_{tot}}{V_{in_{1}}-V_{in_{2}}}=-g_m\,R_D=A_v$

if $g_{m_{1}} \neq g_{m_{2}}$, then,
$A_v=\frac{-2\,R_D}{\frac{1}{g_{m_{1}}}+\frac{1}{g_{m_{2}}}}$