So in playing around with some frequency response stuff I made this circuit: enter image description here

And I found that the cutoff frequency value is set by the Feedback Resistor and the capacitor. Take note that the feedback resistor and the feedback capacitor are in parallel.

Here however is a different RC circuit:

enter image description here

Here I found the exact opposite. When I would try to alter the cutoff frequency by changing the parallel Resistor it had no frequency response effect. Yet I found that changing the series Resistor was what set the cutoff frequency.

So I see this as a contradiction. In case A the frequency response is at the mercy of the resistor in parallel with the capacitor. In case B the frequency response is at the mercy of the resistor in series with the capacitor.

(I should mention as a side note that I was nowhere near the GBP of the OpAmp so that wasn't affecting my bandwidth).

I look at capacitors from the perspective of that they cannot change the voltage across them instantaneously.

Can someone make heads or tails of this? I would really appreciate it. I would love to have gain an intuition for how these RC circuits behave.

Thank you kindly.

  • 3
    \$\begingroup\$ In the second case, it's set by the parallel combination of both resistors. \$\endgroup\$ Mar 21, 2023 at 0:26
  • \$\begingroup\$ You need to look at the impedance of the capacitor at high and low frequencies in the circuits you posted. BTW, don't complicate the circuit in the second figure (get rid of R4) and fill in the function of V4 (I'm guessing you want an AC source). The second figure can be reduced to a single R & C (Thevenin's theorem). \$\endgroup\$
    – qrk
    Mar 21, 2023 at 0:34
  • \$\begingroup\$ To explain your results with the second circuit you should say what frequency source you considered, what value of C, and what range of values for R. \$\endgroup\$
    – The Photon
    Mar 21, 2023 at 0:51

4 Answers 4


It is correct that the principle topology of both circuits is comparable (neglecting the minus sign at the moment). However, there is a huge difference if we consider the values of the components and their influence on the frequency response. But in this case, it is most important that both circuits have the same frequency response!

This is because of the well-known Miller effect. Remember, the name of the opamp circuit - it resembles the classical Miller-integrator (DC-stabilized with a parallel resistor).

Explanation: Let us start with the most right (active) realization. It can be shown that the most left (passive) circuit is a kind of "equivalent" circuit diagram.

Both circuits (assuming an ideal opamp with a gain of Aol=999 only) will have the same filter response at Vout1. This is due to the Miller effect which increases the influence of the impedances by a factor of 10³, that means: It seems that these impedances are reduced by a factor of 1000.

Note the values for R3 and C2 (in comparison to R2 and C1). In particular, we can see that the parallel combination of both resistors R1||R2 resp. R1||R3 (determining the resulting time constant) is, of course, dominated by the lowest resistor value:

R2 in the first (passive) case and R1 in the second (active) case.

This explains why the resistors play such a different role in both circuits.

Fazit/summary: Both circuits have the same frequency response at the node "out1" - however, the active realization has two benefits: (1) Parts values are much more convenient and (2) we have a 2nd output providing amplitudes that are 999 times larger than the passive circuit.


simulate this circuit – Schematic created using CircuitLab


In case 1, input voltage is converted into current (V3/R2) and pushed into the R1/C1 parallel combination. The current is hence a constant across all frequencies till the amplifier's bandwidth. That is why R2 does not appear in the cut-off frequency. Only R1/C1 affect cut-off frequency.

In case 2, the current pumped into the R4/C2 is not constant across all frequencies because the voltage across R3 is not constant because capacitor voltage changes across frequencies. That is why the series resistor starts affecting cut-off frequency. However, your conclusion that the parallel resistor does not affect the cut-off frequency is incorrect. It will impact. You might have used very high R4 compared to R3. That is why you may not be seeing the impact.

  • \$\begingroup\$ Thank you that is an interesting point that the voltage in case 1 in the inverting terminal will always be gnd therefore the current through "R2" is unaffected by frequency variation. Whereas the voltage on top of the cap in case 2 is dependent on the capacitor reactance at that frequency therefore the current through "R3" is a function of the frequency of the input signal. I didn't see that before. People are telling me to refresh on my thevenin theorem so perhaps I need to look at that also but this type of intuitive feel was exactly what i was hoping for. Thanks again. \$\endgroup\$
    – Toasty
    Mar 21, 2023 at 16:50

In the first circuit, the key is to consider the current through the feedback path. As the frequency goes up, the capacitor's reactance decreases, and more feedback current can be delivered to the summing node (the inverting input of the op-amp). This increases the negative feedback and thus decreases the gain of the overall circuit.

In the second circuit, assuming R3 and R4 have equal value (as your diagram indicates by using the same symbol 'R' for both of their values), then you can use the Thevenin theorem to show that this is equivalent to a source with value V/2 feeding the capacitor through a resistance of value R/2. Changing the value of R absolutely will affect the transfer function of the circuit (taking the voltage across the capacitor as the output), provided you adjust it over a range comparable to the reactance of the capacitor (adjust it from, for example, \$0.1 X_C\$ to \$10 X_C\$) at the operating frequency.


Well, the transfer functions of your circuits are given by:

$$\mathscr{H}_{\text{circuit }1}\left(\text{s}\right):=\frac{\text{V}_\text{o}\left(\text{s}\right)}{\text{V}_\text{i}\left(\text{s}\right)}=-\frac{\frac{1}{\text{sC}_1}\space\text{||}\space\text{R}_1}{\text{R}_2}=-\frac{1}{\text{R}_2}\cdot\frac{\text{R}_1}{1+\text{C}_1\text{R}_1\text{s}}\tag1$$ $$\mathscr{H}_{\text{circuit }2}\left(\text{s}\right):=\frac{\text{V}_\text{o}\left(\text{s}\right)}{\text{V}_\text{i}\left(\text{s}\right)}=\frac{\frac{1}{\text{sC}_2}\space\text{||}\space\text{R}_4}{\text{R}_3+\left(\frac{1}{\text{sC}_2}\space\text{||}\space\text{R}_4\right)}=\frac{\text{R}_4}{\text{R}_3\left(1+\text{C}_2\text{R}_4\text{s}\right)+\text{R}_4}\tag2$$

The gain of the circuits is given by:

$$\left|\underline{\mathscr{H}}_{\text{circuit }1}\left(\text{j}\omega\right)\right|=\frac{1}{\text{R}_2}\cdot\frac{\text{R}_1}{\sqrt{1+\left(\text{C}_1\text{R}_1\omega\right)^2}}\tag3$$ $$\left|\underline{\mathscr{H}}_{\text{circuit }2}\left(\text{j}\omega\right)\right|=\frac{\text{R}_4}{\sqrt{\left(\text{R}_3+\text{R}_4\right)^2+\left(\text{C}_2\text{R}_3\text{R}_4\omega\right)^2}}\tag4$$

Solving for the cut-off frequency, we get:

$$\omega_{0\space|\space\text{circuit }1}=\frac{1}{\text{C}_1\text{R}_1}\tag5$$ $$\omega_{0\space|\space\text{circuit }2}=\frac{\text{R}_3+\text{R}_4}{\text{C}_2\text{R}_3\text{R}_4}=\frac{1}{\text{C}_2\text{R}_3}\cdot\left(1+\frac{\text{R}_3}{\text{R}_4}\right)\tag6$$


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