Math help to Sallen-Key transfer function

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.

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

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.

LvW said:

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.

Click to expand...

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 *

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.

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

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.

LvW said:

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.

Click to expand...

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 \[\dfrac{1}{R_{1,3}}\]:

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

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

SunnySkyguy said:

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

Click to expand...

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