Question

Consider the circuit shown in the figure below:


Both the transistors $M_1$ and $M_2$ have output resistance $r_{01}$ and $r_{02}$ and the transconductance of both the transistors are $g_{m1}$ and $g_{m2}$ respectively. Assume that the biasing condition is neglected that will keep the transistors in saturation region. Then the value of small voltage gain $\left|\frac{V_0}{V_{in}}\right|$ is equal to

• A. $\frac{r_{01}||r_{02}}{\frac{1}{g_{m1}}+r_{01}||r_{02}}$
• B. $\frac{r_{01}||r_{02}}{\frac{1}{g_{m1}}+r_{01}}$
• C. $\frac{r_{02}}{\frac{1}{g_{m1}}+r_{01}||r_{02}}$
• D. $\frac{r_{01}}{\frac{1}{g_{m1}}+r_{01}||r_{02}}$

Answer 1

The correct option is A $\frac{r_{01}||r_{02}}{\frac{1}{g_{m1}}+r_{01}||r_{02}}$

Drawing small-signal equivalent model, we will get,


Thus, the circuit can be reduced to


$Therefore{V}_0=\left(g_{m1}{V}_{\pi 1}\right)\times r_{01}||r_{02}$

$Now,V_{in}=V_0+V_{\pi 1}$
$V_{\pi 1}=V_{in}-V_0$
Substituting it in the above equation, we get,
$V_0=\left(g_{m1}\right)\left(r_{01}||r_{02}\right)(V_{in}-V_0)$

$V_0=\left[\frac{g{m}_1\left(r_{01}||r_{02}\right)}{1+g_{m1}\left(r_{01}||r_{02}\right)}\right]V_{in}$

$Thus,\frac{V_0}{V_{in}}=\frac{g_{m1}{r}_{01}||r_{02}}{1+g_{m1}\left(r_{01}||r_{02}\right)}=\frac{r_{01}||r_{02}}{\frac{1}{g_{m1}}+\left(r_{01}||r_{02}\right)}$