I have design an active filter with an online tool. The circuit looks as follows: enter image description here

It should be a low pass filter of 3rd order with a cut-off frequency at 50kHz.

Unfortunately, it attenuatues DC components and some lower frequencies as well as seen in the transfer function below:

enter image description here

In case of DC input, the whole voltage drops at the potential shown below as a red dot.

enter image description here

I cannot explain this behaviour. Does anyone know what might cause such a behaviour?


PS: For the measurement setup I used a signal generator for the input signal and I measured the transfer function with a Sampled ACV measurement function of the DMM HP3458.

  • \$\begingroup\$ I guess the 2nd order filter is not correct. Give some links, references where you have found it. \$\endgroup\$ Commented Sep 2, 2019 at 10:29
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    \$\begingroup\$ Split your problem into smaller parts. I'd measure at the output of the left opamp, what transfer do you get? If it is still not flat at lower freqs. then the problem is on the left of the circuit. If the response is flat then the right side is the issue. The circuit you show should not have the shown behavior so maybe something outside that circuit is causing the issue. What supply voltages are you using? Is there supply decoupling present? \$\endgroup\$ Commented Sep 2, 2019 at 10:53
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    \$\begingroup\$ What is your power supply config? You need a split supply. Either that or create a virtual ground at your single supply midpoint. Without this your physical circuit won't work properly. \$\endgroup\$ Commented Sep 2, 2019 at 12:31
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    \$\begingroup\$ Theer's something else going on, nothing to do with the filter as described. Test the sig gen directly into the DMM for a start, or find an oscilloscope. \$\endgroup\$
    – user16324
    Commented Sep 2, 2019 at 19:50
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    \$\begingroup\$ " What do you mean by split supply?" Oh boy. There's your problem, or at least your first problem. You need two supplies, call them A and B. Let's say that both of them are 5 volts. Connect the - of A to ground, and the + of B to ground. Now connect A+ to the V+ of the 823, and B- to V-. With your meter referenced to ground, you should read +5 on V+ and -5 on V-. An op amp cannot drive an output voltage higher than its + supply or lower than its - supply, so you can't expect a usable result with a single supply,. The only puzzle is why it works at all at higher frequencies. \$\endgroup\$ Commented Mar 8, 2021 at 15:01

2 Answers 2

  1. Your DMM HP3458 setup is incorrect for a low frequency signal. Perhaps the number of readings is sufficient if using the 100ksps rate in order to capture the entire wave.

  2. Your filter is a bit weird.
    R1A * C2A = 50 kHz ok
    R1B * C1B = 13 kHz
    R2B * C2B = 4 MHz

Update :
The Red Dot is supposed to be null DC as the non-inverting input is reference is 0V this means the DC gain is unity by virtue of R18=R58. So the conclusion is measurement setup error, again.

  • \$\begingroup\$ The 3458 AC function is good down to 1 Hz. Or at least that's what the data sheet says. \$\endgroup\$ Commented Mar 8, 2021 at 14:30

The whole point of an operational amplifier is to keep 0V between inverting and non-inverting inputs. As there is 0V on non-inverting input, the voltage at red dot will be kept 0V as per the feedback.

  • \$\begingroup\$ I think his only applies for DC. For AC, the red dot can theoretically be anything. The red dot node =/= inverting input node of the opamp. The inverting input of the opamp will be kept 0V as per the feedback. \$\endgroup\$
    – Huisman
    Commented Sep 2, 2019 at 13:37
  • \$\begingroup\$ The red dot is not directly connected to ground or the any of the op amp inputs. I don't understand your argument. \$\endgroup\$
    – Hilmar
    Commented Sep 2, 2019 at 13:45
  • \$\begingroup\$ This is not "the whole point of an operational amplifier". The virtual short between inverting and non-inverting inputs is only seen with negative feedback, it is not inherent in all operational amplifier circuits. \$\endgroup\$ Commented Mar 8, 2021 at 16:20
  • \$\begingroup\$ Whenever making a frequency response of a circuit, you should feed one frequency to see if the circuit is clipping. Your circuit has no "virtual ground" so it is rectifying the signal. \$\endgroup\$
    – Audioguru
    Commented Mar 8, 2021 at 16:52

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