1
\$\begingroup\$

enter image description here

For the 1st question, is this a Sallen-Key filter? What criteria must a circuit possess in order for it to classified as a Sallen-Key filter? I searched the internet but found no clear answer.

For finding the frequency: my attempt, see below, defines node voltages Va and Vb and uses nodal analysis to find the transfer function, treating vo = vb since the opamp will regulate the voltage at the inverting input the same as the non-inverting input.

After finding the transfer function, I differentiated with respect to f to find the value in terms of R and C.

enter image description here

$$\frac{v_i-v_a}{R}=\frac{v_a-v_b}{R}+\frac{v_a-v_b}{\frac{1}{jwc}}$$ $$\frac{v_a-v_b}{\frac{1}{jwc}}=\frac{v_b}{R||\frac{1}{jwc}}$$

After substituting Va in terms of Vb,A = Vb/Vi, I got the equation below:

$$gain=\frac{jwc}{3+w^2Rc-j(6wRc+wc)}$$

This is as far as I got, I am stuck on how I can get the above answer. Is the nodal analysis equation I derived correct?

\$\endgroup\$
1
  • \$\begingroup\$ Assuming that something is Sallen & Key if they described it in their paper, then yes as it's described by Sallen & Key in TR-50, 6 May 1954 on page 6. Or in their later paper in March of 1955 on page 77 of IRE Transactions. (Of course, anyone can make arbitrary distinctions about what is and isn't.) \$\endgroup\$
    – jonk
    Dec 2 '21 at 19:39
2
\$\begingroup\$

Nodal analysis is the way to go. You can work on s-domain to make the life easier.

According to your schematic, \$\mathrm{v_b=V_o}\$. Using the properties of a closed loop operational amplifier, you'll get

$$ \mathrm{ V_i+(1+s\ RC)V_o=(2+s\ RC)V_a } $$

where

$$ \mathrm{ V_a=V_o\frac{1+2s\ RC}{sRC} } $$

So the gain function can be written as

$$ \mathrm{ G=\frac{V_o}{V_i}=\frac{u-1}{u^2+2u-1} } $$

where $$ \mathrm{ u=1+s\ RC } $$

Before you ask, this \$u\$ comes from using the paralleled impedances R and C across \$\mathrm{v_b}\$ and GND.

After the substitution, you'll get $$ \mathrm{ G=\frac{s/(RC)}{s^2+\frac{4}{RC}s+\frac{2}{R^2C^2}} } $$

It's obvious that this is a transfer function of a second order system which has a transfer function of

$$ \mathrm{ G=A\frac{\omega_n^2}{s^2+2\zeta \omega_n+\omega_n^2} } $$

but with an extra zero. So, re-arranging the final equation so that it forms like the one above gave me

$$ \mathrm{ G=\frac{s}{2RC}\cdot \frac{\frac{2}{R^2C^2}}{s^2+\frac{4}{RC}s+\frac{2}{R^2C^2}} } $$

From here, you can go for either substituting \$s\$ with \$j\omega\$, or differentiating, or any other technique. But eventually, you'll find that the maximum gain occurs at \$\mathrm{\omega=\sqrt{2/(R^2C^2)}=\sqrt2/(RC)}\$.

\$\endgroup\$
1
  • \$\begingroup\$ Your 5th formula is the one. The last one has a gain larger by 1/(RC)^2. \$\endgroup\$ Dec 2 '21 at 16:26
1
\$\begingroup\$

Here are two answers to your questions:

1.) Sallen&Key topology:

When an RC network is combined with a (positive) finite-gain amplifier (gain of "one" or "two" or any other value <3) and if there is single-element signal feedback.

2.) Center frequency fo: For finding the center frequency (for maximum output) you must not differentiate - it is possible to find this frequency by visual inspection of the transfer function.

This is a bandpass function with an imaginary numerator ("s=jw"). Therefore, you must find the minimum of the denominator which makes the denominator also imaginary (to make the whole function real). Remember that at f=fo (maximum) the phase shift of the bandpass is zero.

Hence - set the real part of the denominator equal to zero.

\$\endgroup\$
-2
\$\begingroup\$

First, I will present a method that uses Mathematica to solve this problem. When I was studying this stuff I used the method all the time (without using Mathematica of course).

Well, we are trying to analyze the following opamp-circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$ \begin{cases} \text{I}_2=\text{I}_1+\text{I}_5\\ \\ \text{I}_2=\text{I}_3+\text{I}_4 \end{cases}\tag1 $$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1-\text{V}_2}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_2}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_2}{\text{R}_4}\\ \\ \text{I}_5=\frac{\text{V}_3-\text{V}_1}{\text{R}_5} \end{cases}\tag2 $$

Now, we can put equations \$(1)\$ into \$(2)\$ to get:

$$ \begin{cases} \frac{\text{V}_1-\text{V}_2}{\text{R}_2}=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\frac{\text{V}_3-\text{V}_1}{\text{R}_5}\\ \\ \frac{\text{V}_1-\text{V}_2}{\text{R}_2}=\frac{\text{V}_2}{\text{R}_3}+\frac{\text{V}_2}{\text{R}_4} \end{cases}\tag3 $$

Now, using an ideal opamp, we know that:

$$\text{V}_+=\text{V}_-=\text{V}_2=\text{V}_3:=\text{V}_x\tag4$$

So, we can rewrite \$(3)\$ as follows:

$$ \begin{cases} \frac{\text{V}_1-\text{V}_x}{\text{R}_2}=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\frac{\text{V}_x-\text{V}_1}{\text{R}_5}\\ \\ \frac{\text{V}_1-\text{V}_x}{\text{R}_2}=\frac{\text{V}_x}{\text{R}_3}+\frac{\text{V}_x}{\text{R}_4} \end{cases}\tag5 $$

Hence, we can solve for the transfer function:

$$\mathcal{H}:=\frac{\text{V}_x}{\text{V}_\text{i}}=\frac{\text{R}_3\text{R}_4\text{R}_5}{\left(\text{R}_1\left(\text{R}_2+\text{R}_5\right)+\text{R}_2\text{R}_5\right)\left(\text{R}_3+\text{R}_4\right)+\text{R}_3\text{R}_4\text{R}_5}\tag6$$

Where I used the following Mathematica-code:

In[1]:=Clear["Global`*"];
V2 = Vx;
V3 = Vx;
FullSimplify[
 Solve[{I2 == I1 + I5, I2 == I3 + I4, I1 == (Vi - V1)/R1, 
   I2 == (V1 - V2)/R2, I3 == V2/R3, I4 == V2/R4, 
   I5 == (V3 - V1)/R5}, {I1, I2, I3, I4, I5, V1, Vx}]]

Out[1]={{I1 -> ((R3 + R4) (R2 + R5) Vi)/(
   R3 R4 R5 + R2 (R3 + R4) R5 + R1 (R3 + R4) (R2 + R5)), 
  I2 -> ((R3 + R4) R5 Vi)/(
   R3 R4 R5 + R2 (R3 + R4) R5 + R1 (R3 + R4) (R2 + R5)), 
  I3 -> (R4 R5 Vi)/(
   R3 R4 R5 + R2 (R3 + R4) R5 + R1 (R3 + R4) (R2 + R5)), 
  I4 -> (R3 R5 Vi)/(
   R3 R4 R5 + R2 (R3 + R4) R5 + R1 (R3 + R4) (R2 + R5)), 
  I5 -> -((R2 (R3 + R4) Vi)/(
    R3 R4 R5 + R2 (R3 + R4) R5 + R1 (R3 + R4) (R2 + R5))), 
  V1 -> ((R3 R4 + R2 (R3 + R4)) R5 Vi)/(
   R3 R4 R5 + R2 (R3 + R4) R5 + R1 (R3 + R4) (R2 + R5)), 
  Vx -> (R3 R4 R5 Vi)/(
   R3 R4 R5 + R2 (R3 + R4) R5 + R1 (R3 + R4) (R2 + R5))}}

My equation was also confirmed using LTspice.


When we want to apply the derivation from above to your circuit we need to use Laplace transform (I will use lower case function names for the functions that are in the (complex) s-domain, so \$\text{y}\left(\text{s}\right)\$ is the Laplace transform of the function \$\text{Y}\left(t\right)\$):

  • $$\text{R}_2=\frac{1}{\text{sC}_1}\tag7$$`
  • $$\text{R}_4=\frac{1}{\text{sC}_2}\tag8$$

So, we can rewrite the transfer function as:

$$\mathscr{H}\left(\text{s}\right)=\frac{\text{R}_3\cdot\frac{1}{\text{sC}_2}\cdot\text{R}_5}{\left(\text{R}_1\left(\frac{1}{\text{sC}_1}+\text{R}_5\right)+\frac{1}{\text{sC}_1}\cdot\text{R}_5\right)\left(\text{R}_3+\frac{1}{\text{sC}_2}\right)+\text{R}_3\cdot\frac{1}{\text{sC}_2}\cdot\text{R}_5}=$$ $$\frac{\text{C}_1\text{R}_3\text{R}_5\text{s}}{\text{C}_1\text{C}_2\text{R}_1\text{R}_3\text{R}_5\text{s}^2+\left(\text{C}_1\text{R}_5\left(\text{R}_1+\text{R}_3\right)+\text{C}_2\text{R}_3\left(\text{R}_1+\text{R}_5\right)\right)\text{s}+\text{R}_1+\text{R}_5}\tag9$$

Now, when working with sinusoidal signals we can use \$\text{s}:=\text{j}\omega\$ (where \$\text{j}^2=-1\$ and \$\omega=2\pi\text{f}\$ with \$\text{f}\$ is the frequency of the input signal in Hertz). So, we get:

$$\underline{\mathscr{H}}\left(\text{j}\omega\right)=\frac{\text{C}_1\text{R}_3\text{R}_5\text{j}\omega}{\text{C}_1\text{C}_2\text{R}_1\text{R}_3\text{R}_5\left(\text{j}\omega\right)^2+\left(\text{C}_1\text{R}_5\left(\text{R}_1+\text{R}_3\right)+\text{C}_2\text{R}_3\left(\text{R}_1+\text{R}_5\right)\right)\text{j}\omega+\text{R}_1+\text{R}_5}=$$ $$\frac{\text{C}_1\text{R}_3\text{R}_5\omega\text{j}}{\text{R}_1+\text{R}_5-\text{C}_1\text{C}_2\text{R}_1\text{R}_3\text{R}_5\omega^2+\left(\text{C}_1\text{R}_5\left(\text{R}_1+\text{R}_3\right)+\text{C}_2\text{R}_3\left(\text{R}_1+\text{R}_5\right)\right)\omega\text{j}}\tag{10}$$

So, the absolute value if given by:

$$\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\frac{\text{C}_1\text{R}_3\text{R}_5\omega}{\sqrt{\left(\text{R}_1+\text{R}_5-\text{C}_1\text{C}_2\text{R}_1\text{R}_3\text{R}_5\omega^2\right)^2+\left(\left(\text{C}_1\text{R}_5\left(\text{R}_1+\text{R}_3\right)+\text{C}_2\text{R}_3\left(\text{R}_1+\text{R}_5\right)\right)\omega\right)^2}}\tag{11}$$


Now, when:

  • $$\text{R}:=\text{R}_1=\text{R}_3=\text{R}_5\tag{12}$$
  • $$\text{C}:=\text{C}_1=\text{C}_2\tag{13}$$

We can simplify \$(11)\$:

$$\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\frac{\text{CR}\omega}{\sqrt{4+12\left(\text{CR}\omega\right)^2+\left(\text{CR}\omega\right)^4}}\tag{14}$$

Now, we can find:

$$\frac{\partial\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|}{\partial\omega}=0\space\Longleftrightarrow\space\text{CR}\left(4-\left(\text{CR}\omega\right)^4\right)=0\space\Longrightarrow\space\omega=\frac{\sqrt{2}}{\text{CR}}\tag{15}$$

And we get:

$$\left|\underline{\mathscr{H}}\left(\text{j}\cdot\frac{\sqrt{2}}{\text{CR}}\right)\right|=\frac{1}{4}\tag{16}$$

\$\endgroup\$
1
  • \$\begingroup\$ Thanks so so much for the seriously well-motivated downvotes, I really really appreciate it. \$\endgroup\$ Dec 3 '21 at 19:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.