The differential-mode voltage gain is proportional to the output resistance looking into the active load transistor. The voltage gain can be increased, therefore, if the output resistance can be increased. An increase in output resistance can be achieved by using, for example, a cascode active load. This configuration is shown in figure.

The output resistance Ro is given by

$R_0 = r_{04} + r_{06}(1+g_mr_{04}) \cong g_mr_{04}r_{06}$

The small-signal differential-mode voltage gain is then

$A_d = \frac{v_0}{v_d} = g_m(r_{02}||R_0)$

The differential-mode voltage gain can be further increased by incorporating a cascode configuration in the differential pair as well as in the active load.

One such example is shown in figure. Transistors M3 and M4 are the cascode transistors for the differential pair M1 and M2. The differential-mode voltage gain is now

$A_d = \dfrac{v_o}{v_d} = g_m(R_{o4}||R_{o6})$

where $R_{o4} \cong g_m r_{o2} r_{o4}$ and $R_{o6} \cong g_m r_{o6} r_{o8}$

The small-signal differential-mode voltage gain of this type of amplifier can be on the order of 10,000.