Transfer function for Sallen-Key Low-pass Filter
Well, we have the following circuit:







simulate this circuit &ndash; Schematic created using CircuitLab

Using KCL, we can write:

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

Using KVL, we can write:

$$
\begin{cases}
\text{I}_{\text{R}_1}=\frac{\text{V}_\text{in}-\text{V}_1}{\text{R}_1}\\
\\
\text{I}_{\text{R}_2}=\frac{\text{V}_1-\text{V}_+}{\text{R}_2}\\
\\
\text{I}_{\text{C}_1}=\frac{\text{V}_+}{\left(\frac{1}{\text{sC}_1}\right)}=\text{V}_+\cdot\text{sC}_1\\
\\
\text{I}_{\text{C}_2}=\frac{\text{V}_--\text{V}_1}{\left(\frac{1}{\text{sC}_2}\right)}=\left(\text{V}_--\text{V}_1\right)\cdot\text{sC}_2=\left(\text{V}_\text{out}-\text{V}_1\right)\cdot\text{sC}_2
\end{cases}\tag2
$$

Substituting \$(2)\$ into \$(1)\$, we get:

$$
\begin{cases}
\frac{\text{V}_\text{in}-\text{V}_1}{\text{R}_1}+\left(\text{V}_\text{out}-\text{V}_1\right)\cdot\text{sC}_2=\frac{\text{V}_1-\text{V}_+}{\text{R}_2}\\
\\
\frac{\text{V}_1-\text{V}_+}{\text{R}_2}=\text{V}_+\cdot\text{sC}_1
\end{cases}\tag3
$$

Now, in the ideal opamp circuit, we know that \$\text{V}_+=\text{V}_-=\text{V}_\text{out}\$. So we get:

$$
\begin{cases}
\frac{\text{V}_\text{in}-\text{V}_1}{\text{R}_1}+\left(\text{V}_\text{out}-\text{V}_1\right)\cdot\text{sC}_2=\frac{\text{V}_1-\text{V}_\text{out}}{\text{R}_2}\\
\\
\frac{\text{V}_1-\text{V}_\text{out}}{\text{R}_2}=\text{V}_\text{out}\cdot\text{sC}_1
\end{cases}\tag4
$$

So, we get:

$$\mathcal{H}\left(\text{s}\right):=\frac{\text{V}_\text{out}\left(\text{s}\right)}{\text{V}_\text{in}\left(\text{s}\right)}=\frac{1}{\text{R}_1\text{R}_2\text{C}_1\text{C}_2\text{s}^2+\text{C}_1\left(\text{R}_1+\text{R}_2\right)\text{s}+1}\tag5$$