I've encountered the following circuit and I'm puzzled about the effective resistance that the opamp will see.

I know that R1 is in parallel with R2+R3, so this would be easily answered by R1*(R2+R3)/R1+(R2+R3). But there's a ground between R2 and R3 and I don't know if that will affect the total resistance the opamp will see.

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1 Answer 1


I know that R1 is in parallel with R2+R3

No it isn't; R2 and R3 both connect to ground.

I'm puzzled about the effective resistance that the opamp will see.

I'm assuming you mean the resistance/impedance present at the inverting input node.

Because it's a virtual ground (due to negative feedback), the op-amp's inverting input (if it had feelings or the ability to understand impedance) would "detect" an impedance of near-zero-ohms.

It's a virtual ground because it's an integrator with negative feedback from output to inverting input and, we can assume that the op-amp has a massive open-loop gain hence, the voltage at the inverting input will, under normal operational conditions, be at the same voltage as the non-inverting input (ground potential).

If in fact you meant "what's the effective input resistor for the integrator" then, the answer is 68 kΩ. R3 is nulled-out by the virtual ground and, for this circuit to work as intended, there will be a voltage source on the left hand node pointing off the page. This voltage source will have a very low output impedance (ideally zero) and therefore, any effect that R2 has is lost hence, R1 is the effective impedance between the unseen voltage source (left of page) and the inverting input.

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  • \$\begingroup\$ "R1 is the effective impedance ..." Shouldn't that be R1 || R2? R2 is grounded. R1 is virtually grounded. \$\endgroup\$
    – Transistor
    May 28, 2022 at 11:23
  • \$\begingroup\$ @Transistor R2 is shunted by the zero impedance of the voltage source that is needed to drive the circuit. There is little point in analysing it without a signal source. \$\endgroup\$
    – Andy aka
    May 28, 2022 at 11:26
  • \$\begingroup\$ If input impedance is defined as \$ \frac {V_{in}} {I_{in}} \$ then the current through both R1 and R2 must be taken into account - regardless of the source impedance. \$\endgroup\$
    – Transistor
    May 28, 2022 at 11:37
  • \$\begingroup\$ My answer explores two options of what the OP might mean namely (a) the impedance seen by the inverting input and (b) the effective input resistor for the integrator. If you want to explore a third then I encourage you to leave it as an answer. But, remember what the op said: I'm puzzled about the effective resistance that the opamp will see. @Transistor - input impedance of the circuit seems a bigger distance away from the op's quote than either of my (a) or (b). \$\endgroup\$
    – Andy aka
    May 28, 2022 at 11:40
  • \$\begingroup\$ Thank for orienting me, I see I was analysing it from a wrong perspective. My question referred to (b) mostly. The integrator forms a low pass filter whose cutoff frequency depends on C1 and the integrator's input resistor(s). If R2 and R3 have no influence, then the cutoff frequency is solely defined by 68K and 100n (23Hz). Correct? \$\endgroup\$
    – Domingo
    May 28, 2022 at 13:15

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