In applications requiring a large voltage gain in a single stage, the relationship $A_v = -g_m.R_d$ suggests that we increase the load impedance of the CS stage. With a resistor or diode connected load, however increasing the load resistance limits the output voltage swing.

A more practical approach is to replace the load with a current source as shown in the figure.

Both transistors operate in saturation. Since the total impedance seen at the output node is equal to $r_{o1} || r_{o2}$, the gain is $A_v = g_m( r_{o1} || r_{o2} )$.

Let us consider $M_1$ and $M_2$ separately. If $L_1$ is scaled by a factor 'a' > 1, then $W_1$ may need to be scaled proportionally as well. This is because, for a given drain current, $V_{gs1} - V_{th1}\propto \frac {1}{\sqrt{(\frac{W}{L})_1}}$

i.e if $W_1$ is not scaled, the overdrive voltage increases, limiting the output voltage swing. Also, since $g_{m1}\propto\sqrt{(\frac{W}{L})_1}$ scaling up only $L_1$ lowers $g_{m1}$.

In applications where these issues are unimportant, $W_1$ can remain constant while $L_1$ increases. Thus, the intrinsic gain of the transistor can be written as

$g_{m1}r_{01} = \sqrt{2(\frac{W}{L})_1 \mu_nC_{ox}I_D \frac{1}{\lambda I_D}}$

indicating that the gain increases with L because $\lambda$ depends more strongly on $L$ than $g_m$ does. Also, note that $g_m.r_o$ decreases as Id increases. Increasing $L_2$ while keeping $W_2$ constant increases $r_{o2}$ and hence the voltage gain, but at the cost of higher $|V_{DS2}|$ required to maintain $M_2$ in saturation.