Power Gain in a Small Signal Mosfet Circuit
Well, we are looking at the following circuit:







simulate this circuit &ndash; Schematic created using CircuitLab

Using KCL, we can write:

$$
\begin{cases}
\text{I}_{\text{R}_1}=\text{I}_{\text{C}_1}+\text{I}_{\text{C}_2}\\
\\
\text{I}_{\text{C}_2}=\text{I}_2+\text{I}_{\text{R}_2}\\
\\
\text{I}_{\text{R}_2}+\text{I}_2=\text{I}_3\\
\\
\text{I}_3+\text{I}_{\text{C}_1}=\text{I}_1
\end{cases}\tag1
$$

Using KVL, we can write:

$$
\begin{cases}
\text{I}_{\text{R}_1}=\text{I}_1=\frac{\text{V}_0-\text{V}_1}{\text{R}_1}\\
\\
\text{I}_{\text{R}_2}=\frac{\text{V}_2-\text{V}_3}{\text{R}_2}=\frac{\text{V}_2-0}{\text{R}_2}=\frac{\text{V}_2}{\text{R}_2}\\
\\
\text{I}_{\text{C}_1}=\frac{\text{V}_1-\text{V}_4}{\left(\frac{1}{\text{sC}_1}\right)}=\frac{\text{V}_1-0}{\left(\frac{1}{\text{sC}_1}\right)}=\text{V}_1\cdot\text{sC}_1\\
\\
\text{I}_{\text{C}_2}=\frac{\text{V}_1-\text{V}_2}{\left(\frac{1}{\text{sC}_2}\right)}=\left(\text{V}_1-\text{V}_2\right)\cdot\text{sC}_2
\end{cases}\tag2
$$

Substituting \$(2)\$ into \$(1)\$, gives:

$$
\begin{cases}
\frac{\text{V}_0-\text{V}_1}{\text{R}_1}=\text{V}_1\cdot\text{sC}_1+\left(\text{V}_1-\text{V}_2\right)\cdot\text{sC}_2\\
\\
\left(\text{V}_1-\text{V}_2\right)\cdot\text{sC}_2=\text{I}_2+\frac{\text{V}_2}{\text{R}_2}\\
\\
\frac{\text{V}_2}{\text{R}_2}+\text{I}_2=\text{I}_3\\
\\
\text{I}_3+\text{V}_1\cdot\text{sC}_1=\text{I}_1
\end{cases}\tag3
$$