schematic of the circuit used

I did each filter alone and they work fine, but when they are made together to form a bandpass filter with the same components I used they don't work.

The output on the oscilloscope is a very weak signal that is similar to the output signal when the circuit is not fed with electricity (I guess it's negligible environment noise) while the input is a sine function from the function generator. Am I missing a middle stage or something? I tried all frequencies from 1Hz to 3MHz and nothing changes, even changing the amplitude of the input has no effects.

  • 2
    \$\begingroup\$ Please add an image of the oscilloscope waveform, a picture of your setup, etc. \$\endgroup\$
    – Null
    Commented Oct 27, 2021 at 19:24
  • \$\begingroup\$ I think you meant to list the values as R2 and R3, rather than R4 and R3, but I don't know which is which. \$\endgroup\$ Commented Oct 27, 2021 at 19:43
  • \$\begingroup\$ is fc1 the lowpass or the highpass stage? Not clear from your R2/R4 error. If 50Hz is the lowpass, you've got a band reject filter. \$\endgroup\$
    – Neil_UK
    Commented Oct 27, 2021 at 19:51
  • \$\begingroup\$ Make sure the low pass cut-off frequency is higher in frequency than the high pass cut-off frequency. If they are switched there will be no output. \$\endgroup\$
    – user173271
    Commented Oct 27, 2021 at 19:56
  • \$\begingroup\$ So your circuit: 1) rejects anything above 49 Hz and then 2) pass through anything above 497 Hz. So which of these frequencies can pass: 5 Hz, 200 Hz, 1 kHz? \$\endgroup\$ Commented Oct 27, 2021 at 20:06

2 Answers 2


It seems you've switched the corners you intended to implement.

What you're looking to build is a bandpass filter between 49 - 497 Hz. In order to accomplish this, first you need a high pass filter at 49 Hz, which will eliminate all low frequencies. Then the output of that stage needs to be filtered with a low pass filter with a 497 Hz corner.

So, fix the corner frequencies and you should be in good shape.

  • \$\begingroup\$ does it matter which circuit is first? or did I switch between high pass and low pass cut-off frequencies? \$\endgroup\$ Commented Oct 27, 2021 at 20:20
  • \$\begingroup\$ You're right - the fact that the corners are switched is the only problem. I corrected my answer. Does it make sense now that the low pass corner needs to be higher than the high pass corner? \$\endgroup\$
    – Kile
    Commented Oct 27, 2021 at 20:26
  • \$\begingroup\$ so why it didn't work as a reject band filter at least? all the frequencies that I tried were attenuated. \$\endgroup\$ Commented Oct 27, 2021 at 20:36
  • \$\begingroup\$ Band reject requires a different topology. You essentially have to bring the input signal to both filters and sum the outputs. Think of these two stages as the mathematical product of two functions. Once you filter out everything above 49 Hz it's gone. Then filtering that with a high pass filter eliminates all the signal that's left. \$\endgroup\$
    – Kile
    Commented Oct 27, 2021 at 20:44
  • \$\begingroup\$ I just noticed that my mistake R4 is actually R2 in the schematic, my high pass filter is actually the one with cut-off frequency equals 49Hz. I'm so sorry if that made confusion. in short, way the circuit is: filtering everything above 497Hz then filtering anything less than 49 Hz \$\endgroup\$ Commented Oct 27, 2021 at 20:52

Supplemental info.

As you now know, the sequence of breakpoints makes a difference in attenuation , but not the sequence of cascading the same stages. These are different filters.

See the difference is 40 dB in this case and a slight shift in the 0 deg phase.

enter image description here

  • \$\begingroup\$ but at least both ways of cascading are working normally except for the difference in performance. however, I got nothing at the output, just a very tiny noisy sine signal with max amplitude of less than 5 mV. \$\endgroup\$ Commented Oct 27, 2021 at 21:35
  • \$\begingroup\$ -40 dB or 1% ought to allow you to at least find which stage was lossy . You can always check Vin + to ensure it matches Vin-. \$\endgroup\$ Commented Oct 27, 2021 at 21:50
  • \$\begingroup\$ sorry for being a slow learner, can you elaborate more? I don't think I understand what you mean. \$\endgroup\$ Commented Nov 1, 2021 at 9:51
  • \$\begingroup\$ When using non-inverting input the feedback signal matches that Vin+ to make a null error input. Then check each stage.. Using an FM< sig gen , use x output to sweep FM input \$\endgroup\$ Commented Nov 1, 2021 at 12:34

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