Math help to Sallen-Key transfer function

[olemariendal]

Hi,

I need help to determine the transfer function for a Sallen-Key low pass filter. I need help step by step for the mathematics steps for the transfer function.

The circuit is this:


The steps i already have and end up with, is this.

\[\begin{equation}
V_{out}\cdot \left(\dfrac{\left(G_3+sC_4\right)\left(G_1+sC_2+G_3\right)}{G_3}\cdot (k-G_3)\cdot (k-sC_2)\right)=G_1\cdot (V_{in})
\end{equation} \]

\[
\begin{equation}
V_{out}=\dfrac{G_1\cdot (V_{in})}{\left(\dfrac{\left(G_3+sC_4\right)\left(G_1+sC_2+G_3\right)}{G_3}\cdot (k-G_3)\cdot (k-sC_2)\right)}
\end{equation}
\]

And then is \[G_{1,3}=\frac{1}{R_{1,3}}\]

\[\begin{equation}
V_{out}=\dfrac{\dfrac{1}{R_1}\cdot (V_{in})}{\left(\dfrac{\left(\dfrac{1}{R_3}+sC_4\right)\left(\dfrac{1}{R_1}+sC_2+\dfrac{1}{R_3}\right)}{\dfrac{1}{R_3}}\cdot (k)-\dfrac{1}{R_3}\cdot (k-sC_2)\right)}
\end{equation}\]

I'm stock here, please help me to continue, without CAS tools.

I know that the solution should be:
\[\begin{equation}
\dfrac{V_{out}}{V_{in}}=\dfrac{k}{s^2\cdot (R_1)\cdot (R_2)\cdot (C_2)\cdot (C_4+s)\cdot \left(R1\cdot (C2+R_3)\cdot (C_4+R_1)\cdot (C_4-k)\cdot (R1)\cdot (C2)\right)+1}
\end{equation}\]

Thanks for your time.
Sorry about me English writing skills.

 

[andre_luis]

It is not clear the mean of parameter K on development above.
Anyway, you can find something on this topology here.

 

[LvW]

In your post, the last transfer function ("should be") is wrong.
Please check the units. You have "funny" expressions like the following:
(s+C4), (C2+R3), (C4-k),...

And in your Vout equation I see (k-sC2).
I suppose k=(1+Ra/Rb), correct?

Please, verify everything.

 

[olemariendal]

In your post, the last transfer function ("should be") is wrong.
Please check the units. You have "funny" expressions like the following:
(s+C4), (C2+R3), (C4-k),...

And in your Vout equation I see (k-sC2).
I suppose k=(1+Ra/Rb), correct?

Please, verify everything.

Yes, k is the gain. The funny expressions is coming because I got error when I copied the LaTeX text to the forum.

It should be like this:
\[\dfrac{V_{out}}{V_{in}}=\dfrac{k}{s^2* R_1* R_2* C_2* C_4+s* \left(R1* C2+R_3* C_4+R_1* C_4-k* R1* C2\right)+1}\]

I just change the \cdot to *

 

[LvW]

Here is my advice how to find the transfer function: Apply Kirchoff`s current law for each node (currents in=currents out) using only conductances Y.
For example, for the first node:
(Vin-Vx)Y1=(Vx-V+)Y3+(Vx-Vout)Y2

Together with a similar equation for the other node and the gain expression Vout=(1+Ra/Rb)V+ = kV+
you have three equations for the three unknown quantities Vx, V+ and Vout/Vin.
Hence, you can solve for Vout/Vin.

 

[olemariendal]

My problem is to reorganize and solve this equation

\[
V_{out}=\dfrac{\dfrac{1}{R_1}* V_{in}}{\left(\dfrac{\left(\dfrac{1}{R_3}+sC_4\right)\left(\dfrac{1}{R_1}+sC_2+\dfrac{1}{R_3}\right)}{\dfrac{1}{R_3}}* k-\dfrac{1}{R_3}* k-sC_2\right)}\]

to the solution
\[
\dfrac{V_{out}}{V_{in}}=\dfrac{k}{s^2* R_1* R_2* C_2* C_4+s* \left(R1* C2+R_3* C_4+R_1* C_4-k* R1* C2\right)+1}\]

 

[LvW]

Are you sure that your expression is correct?
Because k should appear in the numerator, you could multiply numerator and denominator with k - however, in this case we have k^2 in the denominator which is not correct.
Therefore my question above.
I think, already the first equation in your post#1 is wrong.

 

[olemariendal]

Are you sure that your expression is correct?
Because k should appear in the numerator, you could multiply numerator and denominator with k - however, in this case we have k^2 in the denominator which is not correct.
Therefore my question above.
I think, already the first equation in your post#1 is wrong.

This is all me steps, I hope it can explain what i'm trying to do:

\[
V_x\] is:

\[
\begin{equation}
V_x\left(\dfrac{1}{R_1}+sC_2+\dfrac{1}{R_3}\right)-\dfrac{1}{R_1}* V_{in}-\dfrac{1}{R_3}* V_+-sC_2* V_{out}=0
\end{equation}
\]
\[V_+\] is:
\[
\begin{equation}
V_+\left(\dfrac{1}{R_3}+sC_4\right)-\dfrac{1}{R_3}* V_x=0
\end{equation}
\]

Isolate \[V_x\] in equation 2:

\[
\begin{align*}
V_+\left(\dfrac{1}{R_3}+sC_4\right)-\dfrac{1}{R_3}* V_x=0 \\
V_+\left(\dfrac{1}{R_3}+sC_4\right)=\dfrac{1}{R_3}* V_x \\
\dfrac{V_+\left(\dfrac{1}{R_3}+sC_4\right)}{\dfrac{1}{R_3}}=V_x
\end{align*}
\]

We know that \[V_-\] is:

\[
\begin{equation}
V_-=V_{out}* \frac{R_a}{R_a+R_b}=V_+
\end{equation}
\]

It is an ideal op amp, so that means:

\[
V_+=V_-\]



All
\[\dfrac{1}{R_{1,3}}=G_{1,3}\]

and
\[\dfrac{1}{C_{2,4}}=sC_{2,4}\]



We substitute equation 3 into equation 1.

\[ \begin{equation}
\dfrac{\left(G_3+sC_4\right)\left(G_1+sC_2+G_3\right)}{G_3}* V_+-G_1* V_{in}-G_3* V_+-sC_2=0
\end{equation}
\]

Isolating V_{in} and V_{out}:

\[\begin{equation}
\dfrac{\left(G_3+sC_4\right)\left(G_1+sC_2+G_3\right)}{G_3}* V_+-G_3* V_+=G_1* V_{in}+sC_2* V_{out}
\end{equation}
\]

Replacing \[V_+\] with \[\dfrac{R_a}{R_a+R_b}*V_{out}\]

\[\begin{equation}
\dfrac{\left(G_3+sC_4\right)\left(G_1+sC_2+G_3\right)}{G_3}* \frac{R_a}{R_a+R_b}* V_{out}-G_3* \frac{R_a}{R_a+R_b}* V_{out}=G_1* V_{in}+sC_2* V_{out}
\end{equation}
\]

Lets say that \[\dfrac{R_a}{R_a+R_b}=k\] and then we get:

\[
\begin{equation}
\dfrac{\left(G_3+sC_4\right)\left(G_1+sC_2+G_3\right)}{G_3}* k* V_{out}-G_3* k* V_{out}=G_1* V_{in}+sC_2* V_{out}
\end{equation}
\]

Then isolate all \[V_{out}\] on left side.

\[
\begin{equation}
\dfrac{\left(G_3+sC_4\right)\left(G_1+sC_2+G_3\right)}{G_3}* k* V_{out}-G_3* k* V_{out}-sC_2* V_{out}=G_1* V_{in}
\end{equation}
\]

Then can we move \[V_{out}\] out of a braces:

\[
\begin{equation}
V_{out}* \left(\dfrac{\left(G_3+sC_4\right)\left(G_1+sC_2+G_3\right)}{G_3}* k-G_3* k-sC_2\right)=G_1* V_{in}
\end{equation}
\]

Take V_{out} out, so only that is on left side:

\[
\begin{equation}
V_{out}=\dfrac{G_1* V_{in}}{\left(\dfrac{\left(G_3+sC_4\right)\left(G_1+sC_2+G_3\right)}{G_3}* k-G_3* k-sC_2\right)}
\end{equation}
\]

Again can we replace \[G_{1,3}\] with \[\dfrac{1}{R_{1,3}}\]:

\[
\begin{equation}
V_{out}=\dfrac{\dfrac{1}{R_1}* V_{in}}{\left(\dfrac{\left(\dfrac{1}{R_3}+sC_4\right)\left(\dfrac{1}{R_1}+sC_2+\dfrac{1}{R_3}\right)}{\dfrac{1}{R_3}}* k-\dfrac{1}{R_3}* k-sC_2\right)}
\end{equation}
\]

 

[D.A.(Tony)Stewart]

Less complicated method https://www.vyssotski.ch/BasicsOfInstrumentation/AnalysisOfTheSallen-KeyArchitecture.pdf

 

[LvW]

Less complicated method https://www.vyssotski.ch/BasicsOfInstrumentation/AnalysisOfTheSallen-KeyArchitecture.pdf

Yes - that´s what I already have proposed in my post#5