Say we had a voltage source connected to a Low Pass RC filter and that this filter was connected to the negative node of an Operational Amplifier, so we would have a filter followed by an inverting amplifier topology.

enter image description here

Please keep in mind that I understand this circuit may be incorrect, I just built it as I had it pictured in my mind as the simplest way to illustrate my question.

In this case, how can I define the cutoff frequency of the filter, as well as the phase shift that the signal experiences? I know that the Low Pass filter delays the signal in 45º, and that the inverting OpAmp inverts the signal in 180º. Then, would the phase shift be -45º-180º=-225º?

  • \$\begingroup\$ Either you talk about delay, or pahse delay, or phase shift; they're different. You say "the filter delays 45<sup>o</sup>", which is wrong (and also suggests insecurity over the basics). The phase shift (which is what you say, ultimately) is not fixed, it varies across the spectrum, so you can only talk about a specific phase shift at a specific frequency. The cutoff also depends on the transfer function, usually considered where the magnitude drops by -3dB; not set in stone, though, as there can be filters with ripple and/or high Q that need special care. \$\endgroup\$ Commented May 8, 2018 at 11:16
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    \$\begingroup\$ The circuit is incorrect - it won't act as a filter because the capacitor is at the virtual ground of the input. If you want to know about the phase shift of a simple RC low pass filter then ask that and don't clutter your question with ideas that are wrong and thus make it hard to answer. \$\endgroup\$
    – Andy aka
    Commented May 8, 2018 at 11:20
  • \$\begingroup\$ Also, how do you know that the filter adds 45 deg phase shift? \$\endgroup\$ Commented May 8, 2018 at 11:20
  • \$\begingroup\$ I am sorry, I just assumed to be working at the cutoff frequency so that for a low pass filter, the phase shift is 45º. I also apologize for mixing up the expressions. I study in French, so it can be a bit difficult to translate everything into English. \$\endgroup\$
    – Bee
    Commented May 8, 2018 at 21:20

1 Answer 1


To be clear and to protect against future edits, here is the circuit you show that this answer pertains to:

There is no low pass filter here. That is because the negative input of the opamp stays at 0 V during normal operation. The voltage across C2 doesn't change, so there is no current thru it, so there is no difference if it were removed.

With C2 removed, this is simple block with a gain of -1. It does have frequency limits over which it works, but those are not well defined. One limit is a result of the gain-bandwidth product of the opamp. However, that is usually only specified as a minimum value, with little guidance how high it might be. There are other low pass filters due to parasitic capacitance, but again, those are hard to quantify.

If you want a predictable low pass filter with a single dominant pole, then you need to add your own deliberate filter component, like a inductor or capacitor, that dominates stray and parasitic values.

  • \$\begingroup\$ ......or connect the R1-C2 combination to the non-inverting opamp input terminal. In this case, you have a simple first-order lowpass followed by a unity-gain amplifier (which provides a low-resistive voltage output). \$\endgroup\$
    – LvW
    Commented May 8, 2018 at 12:30
  • \$\begingroup\$ @LvW: There are lots of ways to "fix" this. If we're just talking about the theoretical output voltage, no amplifier is needed at all, just R1 and C2. \$\endgroup\$ Commented May 8, 2018 at 12:42
  • \$\begingroup\$ Yes - no doubt about it. But in the title line the questioner has explicitedly mentioned "...filter + amplifier". That was the background of my comment. \$\endgroup\$
    – LvW
    Commented May 8, 2018 at 12:53

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