I'm learning this MFB active LPF.
My questions regarding this as follows:
- How to derive transfer function for this configuration?
- Is there concept of virtual ground exist? If yes can you explain with derivation?
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\$\begingroup\$ I'm sure there are multiple derivations on the site, if you'd just do a search. Recently, I wrote this, which isn't the exact same case but close enough to show you how. But I'm sure a search will find more than a few other cases already solved for you. And yes, there is a virtual ground at the (-) node. \$\endgroup\$– periblepsisCommented Oct 30, 2023 at 5:52
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\$\begingroup\$ You have several answers and their effort and time deserves your time in response. Let them know if they have helped and where you may need a clarification, if so. \$\endgroup\$– periblepsisCommented Oct 30, 2023 at 22:12
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3 Answers
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The filter is very common and is studied in all technical schools and universities in electronics courses. The analysis (by inspection) begins by considering a very general active circuit of great practical use. We proceed as in the figure, in which there is also a numerical example:
The concept of virtual mass is fundamental when it comes to the voltage difference between the non-inverting input and the inverting one which is proved to be zero, so the voltages of the two inputs are at the same potential and in this case at ground.
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Is there concept of virtual ground exist? If yes can you explain with derivation?
Yes, an obvious one (by observation): -
How to derive transfer function for this configuration?
The inner loop is an integrator formed by \$C_2\$ and \$R_2\$. It has a transfer function of \$\dfrac{-1}{sC_2 R_2}\$.
And, because of the virtual ground we can say that the input impedance into the integrator is \$R_2\$ thus allowing us to redraw like this: -
Can you take it from here? Hint: Millman's theorem cuts-out a lot of lines of derivation.
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A method with s-domain impedances, generic node voltage analysis and virtual ground. No clever shortcuts are used:
Use voltage variable Va for the voltage of the joint of R1, R2, R3 and C1
Use voltage variable Vb for the voltage of the inverting input of the opamp.
Consider the output voltage like it was a voltage source Vo.
Write KCL equations for Va and Vb. Assume there's no current in the opamp input.
Assume virtual ground by writing Vb=0
Eliminate variable Va. You still have one equation which contains variables Vi and Vo. Solve Vo/Vi ; that's the transfer function.
