Transfer function for Sallen-Key Low-pass Filter

Question

How can I find the transfer function H(s) of the following circuit by using the nodal analysis?

My thoughts woulb be:

the equation for the node \$ i_1 = i_1 + i_3 \$

\$ i_1 = \frac{U_e - U_1}{R_1} \$

\$ i_2 = \frac{U_1-U_2}{X_{c1}}\$

\$ i_3 = \frac{U_1 - U_a}{X_{c2}} \$

I expressed \$ U_2 \$ using the voltage divider

\$ U_2 = U_1\cdot\frac{X_{c1}}{X_{c1}R_2} \$

My idea was to express \$U_1\$ and \$U_1\$ set it all in the initial node equation \$ i_1 = i_1 + i_3 \$ and from that i might be able to derive the transfer function \$ \frac{U_a}{U_e} \$. But I don't know how to express \$ U_1 \$.

Accroding to some websites this is what the transfer function looks like for that typs of filters:

\$ \frac{U_{out}}{U_{in}} = \frac{\frac{1}{R_1C_1R_2C_2}}{s^2 + s(\frac{1}{R_2C_2} + \frac{1}{R_1C_2}) + \frac{1}{R_1C_1R_2C_2} } \$

Answer 1

Well, we have the following circuit:

^{simulate this circuit – 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:

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