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2nd order opamp low pass filter " BW = 100 Hz

trying to find the transfer function for a 2nd order opamp low pass filter, can i get a help to find my transfer function ?

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can i get a help to find my transfer function ?

Nice Mr Okawa's website shows the formula for a 2nd order low-pass Sallen Key filter: -

enter image description here

You can even plug in the component values and get a bode plot. Good luck.

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  • \$\begingroup\$ thank you for your answer. But in my case there is a third capacitor and a voltage at 5 volts. what would you recommend me to do in this case ? \$\endgroup\$ – Omar Kharoub Sep 26 '20 at 14:44
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    \$\begingroup\$ 3rd cap and 5V are not relevant \$\endgroup\$ – Stefan Wyss Sep 26 '20 at 14:48
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    \$\begingroup\$ That 5V is the power supply and the capacitor is a filter and local storage for providing current quickly when needed. \$\endgroup\$ – Justme Sep 26 '20 at 15:04
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You can apply the fast analytical circuit techniques or FACTs to determine this transfer function almost by inspection. This is actually an example of the book I wrote on the subject. The principle lies in finding the time constants involving each of the energy-storing elements, the two capacitors in your circuit. First, you start for \$s=0\$ and open the caps. You have a simple voltage follower whose gain is 1: enter image description here

Then, determine the resistance \$R\$ driving each capacitor to determine the time constant \$\tau=RC\$ when the stimulus source is zeroed (\$V_{in}\$ is replaced by a short circuit). You determine a resistance across terminals by installing a test generator \$I_T\$ which develops a voltage \$V_T\$: the resistance you want is simply: \$R=\frac{V_T}{I_T}\$. Here we go for \$\tau_1\$ and \$\tau_2\$:

enter image description here

In the left side, you see that the current source generates a voltage \$V_1\$ which biases the (+) pin and is thus duplicated on the output and the (-) input: there is 0 V across the current source (considering an infinite open-loop gain for the op-amp) and thus \$\tau_1 = 0\cdot C_1=0\;\mathrm s\$ and since the current source in the right-side drawing only biases the resistances in series, you have: \$\tau_2=(R_1+R_2)C_2\$. We can already form the first-order coefficient of our denominator: \$b_1=(R_1+R_2)C_2\$.

For the second-order time constant, we set \$C_2\$ in its high-frequency state and determine the resistance offered by \$C_1\$'s terminals.

enter image description here

In this mode, because the voltage at the (+) pin is 0 V, so is the right-side of the current source. Therefore, the voltage \$V_T\$ depends on the two paralleled resistors: \$\tau_{21}=C_1(R_1||R_2)\$. And that's it! We have our denominator formed as: \$D(s)=1+sb_1+s^2b_2=1+s(\tau_1+\tau_2)+s^2(\tau_2\tau_{21})=1+\frac{s}{Q\omega_0}+(\frac{s}{\omega_0})^2\$. There are no zeroes in this transfer function and a quick Mathcad sheet plots the response for you:

enter image description here

The FACTs get you straight to the answer by inspection, without writing a single line of algebra. Furthermore, the transfer function is directly expressed in a low-entropy form where meaningful parameters such as a quality factor and a resonant frequency show up naturally.

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