Edit: I think maybe I am confusing terms (it has been a few years since my last electronics class). I am trying to get the expression for the gain of this active filter amplifier circuit in terms of R1 and R2

So, I am looking at the following circuit from my textbook:

enter image description here

And I want to know how to calculate \$A_V\$, so I get the transfer function: $$\frac{V_{out}}{V_{in}}= \frac{-R_2}{R_1 + \frac{1}{sC}}$$ Now for a DC frequency of 0 Hz to get the DC gain, the capacitor acts as an open circuit with infinite impedance, so then \$\frac{1}{sC}\$ tends to infinity, and I feel like that would mean that the DC gain is 0 just by looking at the transfer function! But, as the schematic suggests, it \$A_V= \frac{-R_2}{R_1}\$, implying that when the frequency is 0, then \$\frac{1}{sC}\$ tends to zero. What am I missing?

Thank you.

  • \$\begingroup\$ As no DC can flow in R1 and C1, delete them and see what you are left with. Hint - disconnect the non-inverting terminal from the common rail and vary it at DC. \$\endgroup\$ Feb 15 at 14:37
  • \$\begingroup\$ Please, can you explain why do you think that "the schematic suggests Av=-R2/R1" ? \$\endgroup\$
    – LvW
    Feb 15 at 14:38
  • \$\begingroup\$ @LvW the schematic has the gain and the cutoff frequency in it, in a shaded box \$\endgroup\$
    – Bee
    Feb 15 at 14:48
  • 1
    \$\begingroup\$ Consider that the expression \$A_V= \frac{-R_2}{R_1}\$ isn't valid for all frequencies. If we assume that the capacitor is a short for the frequencies of interest, then \$A_V= \frac{-R_2}{R_1}\$ is correct. For much lower frequencies (like DC = 0 Hz) the expression is incorrect. The mentioning of "High pass filter" should tell you already that this circuit isn't going to pass DC. \$\endgroup\$ Feb 15 at 14:54
  • \$\begingroup\$ You may inject input offset but not DC gain \$\endgroup\$ Feb 15 at 15:47

The DC gain of that circuit is zero. Once the capacitor impedance drops significantly below R1, the gain becomes -R2/R1. That only happens when the input frequency is significantly greater than \$\frac{1}{2\pi f R_1 C}\$ as semi-alluded-to in the picture. Not all web resources are good at explaining everything.

I feel like that would mean that the DC gain is 0V

Gain is not defined in terms of units because it is volts per volt.

  • 1
    \$\begingroup\$ I would say it is 1 as it becomes a follower (admittedly driven from 0V). \$\endgroup\$ Feb 15 at 14:40
  • \$\begingroup\$ @PeterSmith I don't understand what you are saying. \$\endgroup\$
    – Andy aka
    Feb 15 at 14:42
  • \$\begingroup\$ @PeterSmith What is the impedance of capacitor C1 at the input? What will that mean for any DC voltage at the input? Yes, the follower has a DC gain of 1 but there's an AC coupling capacitor as well. \$\endgroup\$ Feb 15 at 14:47
  • \$\begingroup\$ I edited the gain units, that was a typo and I understand it was incorrect. I don't understand why the capacitor impedance would drop? As far as I know, low frequencies going through a capacitor means the impedance would rise? \$\endgroup\$
    – Bee
    Feb 15 at 14:47
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    \$\begingroup\$ Bee - Correct. Peter Smith - disagree. The overall gain of the (((circuit))) is not 1 (or -1), because the output is not an image of the input. The output is 0 V no matter what the DC voltage is at the left side of C1 (Bee - reference designators - YES!). This is a gain of 0, not 0 dB. The output is 0 times any (DC) input. \$\endgroup\$
    – AnalogKid
    Feb 15 at 15:05

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