# 3 band audio equalizer circuit transfer function

I am trying to find the transfer function of part of a 3 band audio equalizer circuit that amplifies low frequencies but I am confused about the topology. Here is the topology: I was under the impression that the circuit could be reduced down to your normal inverting amplifier where the transfer function would be -(Z_f)/(Z_S) but it looks like part of "Z_s" is also a part of "Z_f" in the picture above and this is confusing me.

Update: I used the node between c1 and c2 and the node between the negative terminal of the op amp and RF to find the transfer function. I got the following: from my understanding, this is a typical form for a transfer function for a bandpass filter. How can you use this transfer function to create a bandpass filter of a particular center frequency? no information is given about bandwidth or cutoff frequencies. However, for the particular equalizer I am making I want a center frequency of 500 Hz and all component values are given in the above diagram. using those component values I should be able to come up with a particular bandwidth and reuse that bandwidth but shift the center frequency to 500Hz.

• Show the bass, mid and treble full circuit please. Jun 19 '20 at 8:33
• You have found hopefully the right transfer function formula. It has the well known 2nd order form and its coefficients A,B and C can be presented with functional parameters such as midband gain and frequency and the Q-factor or bandwidth. You'll find these from sourcebooks. Infinitely many component combinations can produce the same functional parameter values. You can search for ex. with Excel Solver or some math software a working component value combination. Then find with a circuit simulator is it right and is it acceptable with some component value tolerances. See also filter handbooks. Jun 22 '20 at 17:53

This cannot be reduced to inverting amp with gain -(Zf)/(Zs). The RC circuit has three ports. But virtual ground simplifies the case a little. One of the ports is grounded. You should calculate with complex variable phasors which output voltage pulls the inverting input of the opamp to GND. The result will be an equation between Uout and Uin.

If you write the usual node voltage equation (=KCL) for the node where R1, R2, C1 and C2 meet you have the essential law for the circuit. You must eliminate the voltage V2 of that node. It can be done by setting the current through Rf same as through C1. Then you can replace V2 with -Uout/(sC1Rf) and have only Uin and Uout in the equation.

What you have shown us is nothing more than a Multiple Feedback Band-Pass Filter see it yourself: simulate this circuit – Schematic created using CircuitLab

http://sim.okawa-denshi.jp/en/OPtazyuBakeisan.htm

http://eet.etec.wwu.edu/hardyc/project/docs/Filter%20Design/Multiple%20Feedback%20Bandpass.pdf

But can you show us the full circuit please?