I'm trying to know the fr equation for the following op amp configuration: enter image description here I have some capacitors values for specific fr frequencies:

fr = 34Hz -> C = 33nF

fr = 50Hz -> C = 22nF

fr = 160Hz -> C = 6.8nF

fr = 330Hz -> C = 3.3nF

fr = 1.1kHz -> C = 1nF

fr = 2.3kHz -> C = 470pF

fr = 6kHz -> C = 180pF

fr = 16kHz -> C = 33nF

Capacitors are not exact values, they are commercial. I know these values are correct because I saw the frequency response in an oscilloscope.

I was searching on web and found this: enter image description here

This is the page where I found it: https://www.electronics-tutorials.ws/filter/filter_7.html The circuits are very similar. The difference is R9 resistor which add an offset to the input signal. This offset is to use just one power source instead of two. The page give me an equation to calculate the fr frequency. enter image description here

But when I replace the values in the equation, it doesn't give a close values as the practical capacitance.

fr = 15.25Hz -> C = 33nF

fr = 22.87Hz -> C = 22nF

fr = 74.01Hz -> C = 6.8nF

If I multiply the equation by 2, it give me a closer value.

fr = 30.5Hz -> C = 33nF

fr = 45.74Hz -> C = 22nF

fr = 148.02Hz -> C = 6.8nF

If it is the solution, I would like to understand why.

If someone could explain me the first op amp configuration, I would be very grateful.

  • \$\begingroup\$ from schematics is not clear how the offset is added and where is your gnd connection. expand the sch. with valuable details. \$\endgroup\$ Aug 9, 2019 at 5:56
  • \$\begingroup\$ I put the original circuit. The offset comes from a voltage divider in series with a voltage follower op amp, so any other impedance affects the positive op amp pin. \$\endgroup\$ Aug 9, 2019 at 14:36
  • \$\begingroup\$ if you look at your schematics and tutorial, you see that is much different. the tutorial does not have a resistor from r1 to the non-inverting input of the opamp. \$\endgroup\$ Aug 9, 2019 at 15:06
  • \$\begingroup\$ It has. It's 100k \$\endgroup\$ Aug 9, 2019 at 15:31

2 Answers 2


If someone could explain me the first op amp configuration, I would be very grateful.

Well, the shown circuit is a second-order bandpass in "multifeedback configuration".

You are asking for explanation of the circuit? One general remark: There are two different configurations, which both work: With or without a resistor at the common capacitive node to ground. However, in both cases, the design equations are different.
There are several approaches to explain the working principle:

1.) With a feedback loop R2-C1 the opamp works as an integrator (lowpass function) and the feedback loop C2-R1 is also a lowpass - however, because it constitutes the feedback loop it works - together with the opamp - as a highpass. Hence, the lowpass-highpass combination can work as a bandpass.

2.) With the opamp output grounded, we have the classical passive R1-C2-R2-C1 bandpass with a maximum Q-value of 0.5 only. However, if we lift the ground and connect this node to the output of an opamp the feedback effect enhances the midband region and, thus, increases the Q value to much larger values.

3.) The complete feedback network is a "bridged -T" network: R1 is grounded and both capacitors form the "T", which is bridged by R2. It is a well known fact from system theory that a bridged-T can produce complex zeros. If such a "bridged -T" is used as a feedback circuit, the complex zeros turn into complex poles. This is needed for high-Q bandpass circuits.

4.) This last circut interpretation can be used to explain the role of the 27k resistor in the first circuit. From the feedback point of view, both resistors (100k and 27 k) are connected to ground and, thus, are in parallel. Hence, the role of these two resistors can be taken over by one single resistor R1 only (second circuit) which has the value of the parallel combination of both.

Any further questions?

  • \$\begingroup\$ Wao... Thanks, my friend. Your explanation was perfect. Thank you very much! \$\endgroup\$ Aug 10, 2019 at 0:39

You are correct to use that equation. Use the parallel combination of 100k and 27k as R1 in your calculations.

  • \$\begingroup\$ That's true. Thanks, my friend. Do you know why? Why do I have to consider this resistors in parallel? That's because both signals, input and offset, form a single signal? I would like to understand that circuit. \$\endgroup\$ Aug 9, 2019 at 14:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.