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) …