# Transfer function for Sallen-Key Low-pass Filter

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} } \$$

• I assume i1 = i2 + i3. Sep 29, 2019 at 19:50
• Also if you want to find voltages, you generally express all currents in terms of voltages and you end up with an equation of voltage variables and have no current variables (node voltage method). Sep 29, 2019 at 19:51
• $i_2$ is the current through R2, but you are making it dependent on $X_{C1}$. Why? Sep 29, 2019 at 19:54
• Have a look here (electronics.stackexchange.com/questions/439745/…) and see how the fast analytical circuits techniques or FACTs can get you the transfer function in a record time without using nodal analysis. Sep 29, 2019 at 20:44

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:

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