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OP1 has a finite input resistance, but an infinite open loop gain (other parameters are also ideal). The other two op amps are ideal as well. Can I still assume that there is a virtual ground between the positive and negative terminals of OP1 and the input resistance (Rin in the schematic) is actually R1?

It's a pretty basic question but I want to know if my assumption is correct.

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  • \$\begingroup\$ how finite is "finite"? Because, even older Opamps like the LM358 have at most input bias currents in the nA. Compare that with your R1, and you see how much "finite" is a relative term. \$\endgroup\$ Dec 4, 2021 at 18:52
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    \$\begingroup\$ @MarcusMüller, finite's an absolute term, though - it means quantifiable, limited in size. The ratio between R1 and Rinmop1 may be huge, may make it irrelevant in practice, but it's not infinite. Anyway, the OP's exploring to understand as much as possible, beyond what's actually used - great to see and encourage :-) \$\endgroup\$
    – TonyM
    Dec 4, 2021 at 19:01
  • \$\begingroup\$ @TonyM absolutely! \$\endgroup\$ Dec 4, 2021 at 19:05
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    \$\begingroup\$ Sorry if my comment sounded dismissive! I probably should have said: "Hey, I know this is one imperfection you start taking into account. But look at the current that actually flows into the inputs of non-ideal real-world opamps. When you compare that to the currents that flow e.g. through R1, you'll notice the error would really be negligible. There's other imperfections you might care more about!" I was to abrasive maybe, there, my apologies. \$\endgroup\$ Dec 4, 2021 at 19:09
  • \$\begingroup\$ We have virtual ground when the inverting input of the opamp is directly connected to the output. \$\endgroup\$
    – Miss Mulan
    Dec 4, 2021 at 22:33

2 Answers 2

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Can I still assume that there is a virtual ground between the positive and negative terminals of OP1 ...

With negative feedback as shown the output will adjust until OP1's inverting input is very close to that of the non-inverting input. Since the non-inverting input is physically grounded you can assume that the inverting input is virtually grounded.

... and the input resistance (Rin in the schematic) is actually R1?

Correct.

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Assuming that arrangement is stable and balanced, it should be a virtual ground.. the finite input resistance won't matter because of the infinite gain.

So DC input resistance is just R1 to ground.

Chances are a real op-amp will oscillate if R3 is too low, and that means all bets are off.

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  • \$\begingroup\$ Right ... I add behavior very "weird" ... as I have simulated ... \$\endgroup\$
    – Antonio51
    Dec 4, 2021 at 20:26
  • \$\begingroup\$ @Antonio51 As R3-> \$\infty\$ there negligible phase-lagged feedback so it will be stable for any reasonable op-amp but once R3 gets into the same order of magnitude as the other resistors it's going to feel like oscillating I think. \$\endgroup\$ Dec 4, 2021 at 20:29

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