I came across this circuit in another post and started looking at the op amp filter and how to apply traditional circuit analysis ( using 1/jwc for capacitors) and couldn't derive the transfer function. Image of Circuit

Question: How would we derive the transfer function for the filter topology? Ignore the HP Filter on the V+ terminal and ignore the components beyond (and including) the zener diode. Use the Generic names, C1, R1, etc.

assume Vin = V+ and we want to find Vo = output of OpAmp.

  • 1
    \$\begingroup\$ Note to the reader: it's a photodetector with AC coupling, followed by an op-amp with a bandpass filter, then a peak detector. D5 is a Schottky diode, not a zener diode. \$\endgroup\$
    – Jason S
    Apr 25, 2012 at 20:13
  • \$\begingroup\$ you mean DC decoupling, AKA High Pass Filter. \$\endgroup\$
    – CyberMen
    May 8, 2012 at 12:41
  • \$\begingroup\$ @JasonS Wouldn't it make more sense to not have R31 there if you are doing DC decoupling and rely on the high impedance input of the OpAmp? $$\frac{s}{s+\frac{1}{RC}}$$ if R is huge the gain of the system is MORE flat across all frequencies except 0 (aka, DC component) where it is Zero. \$\endgroup\$
    – CyberMen
    May 8, 2012 at 12:45
  • \$\begingroup\$ absolutely not! Taking R31 out leaves no defined mechanism to regulate the DC average voltage, and what will happen is the average voltage will drift up or down depending on the opamp's input leakage current, until the op-amp's protection diodes kick in and you risk introducing nonlinear clipping. You pick R31 high enough so that the high-pass filter lets through the frequencies of interest. \$\endgroup\$
    – Jason S
    May 8, 2012 at 12:58

2 Answers 2


While formulating my answer to that question, I analyzed that circuit in some detail. It looks like a standard second-order bandpass filter, but used in a non-inverting configuration. Since a non-inverting amplifier cannot have a gain less than 1, I was intrigued to know what it's response should actually be.

The form of the transfer function is:

\$\dfrac{V_o}{V_{in}} = \dfrac{\mathrm s^2+a\mathrm s+\omega_0^2}{\mathrm s^2+b\mathrm s+\omega_0^2}\$

You can do some inspection by mentally removing or shorting the capacitors from which it is apparent that the LF & HF gains will be 1 as the equation predicts.

OK, here goes:

To simplify things a bit, we can guess that the ratio of R17 to R18 is important, so lets call it k (401.6). So if we replace R18 with just R, we can replace R17 with kR. Also since C1 & C5 are the same we can just call them C. Also, putting s=j\$\omega\$ is cleaner (and we get a Laplace transform).

Calling the voltage at the R18, C5 C1 junction Vx and summing the currents into that node we get :-

\$\dfrac{0-V_x}{R}+\dfrac{V_{in}-V_x}{\dfrac{1}{\mathrm sC}}+\dfrac{V_{out}-V_x}{\dfrac{1}{\mathrm sC}}= 0\$

\$V_x.(\dfrac{1}{R}+2\mathrm sC)=(V_{in}+V_o).\mathrm sC\$

\$V_x=\dfrac{(V_{in}+V_o).\mathrm sC}{\dfrac{1}{R}+2\mathrm sC}\$

Now the voltage at the inverting input of U1 is Vin (if the circuit is stable!) and summing the current at this node we get :-

\$\dfrac{V_x-V_{in}}{\dfrac{1}{\mathrm sC}}+\dfrac{V_o-V_{in}}{kR} = 0\$

So :- \$V_o=V_{in}.(1+\mathrm skRC)-V_x\mathrm skRC\$

Substituting for Vx, we get :-

\$\dfrac{V_o}{V_{in}}=\dfrac{1+\mathrm skRC-\dfrac{\mathrm s^2kR^2C^2}{1+2\mathrm sRC}}{1+\dfrac{\mathrm s^2kR^2C^2}{1+2\mathrm sRC}}\$

And :- \$\dfrac{V_o}{V_{in}}=\dfrac{\mathrm s^2+\mathrm s.\dfrac{2+k}{kRC}+\dfrac{1}{kR^2C^2}}{\mathrm s^2+\mathrm s.\dfrac{2}{kRC}+\dfrac{1}{kR^2C^2}}\$

(The plot for this exactly matches Telaclavo's graph.)

Now we can see that the natural frequency is given by :-

\$\omega_0 = \dfrac{1}{RC\sqrt k}\$ (ie \$f_0\$=14.5kHz)

... and that the maximum gain when \$\mathrm s^2+\omega_0^2=0\$ is given by :-


As for the time domain, since we have a Laplace transform, we can just take it's inverse to get the impulse response. In traditional textbook style I will simply say that this is left as an exercise for the student (ie too damn hard :)


Equivalent circuit:

Equivalent circuit

Apply KCL to the two nodes where I defined Vx and Vi. Solve for Vo in those two simultaneous equations. Make VGND=0 for AC response. See details here.

Results: frequency response of H(s)=Vo(s)/Vi(s) is

Frequency response

The peak is at 14.5 kHz, and there, the gain is 202.

  • 1
    \$\begingroup\$ I'll give you the check mark if you show your proof step by step to derive the transfer function. \$\endgroup\$
    – CyberMen
    Apr 25, 2012 at 21:01
  • 4
    \$\begingroup\$ @CyberMen Then change the title of the question, and ask for help as to how to solve a system of equations. \$\endgroup\$
    – Telaclavo
    Apr 25, 2012 at 21:09

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