I am looking for a circuit that I expected to find everywhere but I don't. The "active" lowpass filter I often see in the web uses an RC filter with a buffer in series.


simulate this circuit – Schematic created using CircuitLab

The alternativ version I found quite often is, in my opinion, not a true one pole lowpass but a lowpass shelving filter. The capacitor in the feedback is paired with a parallel resistor.


simulate this circuit

Now my question is, why don't I see the second variant without the additional parallel resistor in the feedback path more often? Are there any major disadvantages to the RC - buffer implementation?

  • \$\begingroup\$ other than , only being 1st order -6dB/octave with unbalance inputs for offset voltage. Use basic configuration for a Sallen-Key second order (two-pole) low pass filter \$\endgroup\$ Aug 19 '18 at 15:07
  • \$\begingroup\$ Neither one of your circuits is "active filter". Also, notice that your second circuit cannot have a voltage gain lower than one. So it's not a good filter. \$\endgroup\$
    – G36
    Aug 19 '18 at 15:23
  • \$\begingroup\$ Second one is taken from here and labeled an amplifier circuit. Was wondering the same. electronics-tutorials.ws/filter/filter_5.html Why would the second one not be active? What are the criteria here? \$\endgroup\$ Aug 19 '18 at 15:26
  • \$\begingroup\$ It's only an active filter if the op-amp (amplifier) is used in a way that "rewrites" the impedances of resistors and capacitors by using feedback. Neither of those do and that site should not really be calling them active filters because the filtering is performed by passive components and not the amplifier. \$\endgroup\$
    – Andy aka
    Aug 19 '18 at 16:46
  • 1
    \$\begingroup\$ @Soldersmoke I just ran a simulation of the circuit and it creates some odd lowpass shelving filter. So I guess with the right values, you get something close to a lowpass but it's theoretically no true lowpass since the gain at inf Hz is not -inf, right? \$\endgroup\$ Sep 18 '19 at 12:33

If you are asking why you don't see the second circuit with no R2, that's because it has (with ideal components) infinite DC gain. At DC the impedance of C1 is infinite and the gain will be (one plus) the ratio of the feedback impedance to R3.

Also, in this non-inverting configuration, the minimum stop-band gain is unity.

In the inverting configuration, you do in fact sometimes see the circuit without the parallel resistor, when integrating low-frequency signals over a finite period of time. With a short enough period of time, the very high DC gain will not have time to dominate, assuming the DC component is sufficiently small.


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