How to derive the transfer function of this op-amp?

Question

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.

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

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?

Answer 1

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

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

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)}\$.

Answer 2

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.