I can solve this circuit with nodal analysis, however, why is $$V_a=-V_b$$ in this ideal op-amp? Can I find this relationship from Kirchhoff's voltage law?


simulate this circuit – Schematic created using CircuitLab


3 Answers 3


\$V_a = V_b\$ because the op-amp has ideally infinite open loop gain and of course, if \$V_a\$ didn't pretty much equal \$V_b\$ then the output would be slammed against the power rails. Negative feedback ensures that \$V_a\$ pretty much equals \$V_b\$.

I'm unsure why you think that \$V_a = -V_b\$ because that would be wrong. Maybe you have messed a sign up somewhere.

  • \$\begingroup\$ Vb is measured with inverse polarity (+ is on ground), that's where the sign gets reversed. \$\endgroup\$
    – Ben Voigt
    Jul 5, 2021 at 15:42
  • 2
    \$\begingroup\$ That is, the op-amp provides that $$(V^+ - GND) = (V^- - GND)$$ but that's beside the point. The definitions $$V_a = (V^- - GND)$$ and $$V_b = (GND - V^-)$$ lead directly to $$V_a = -V_b$$ \$\endgroup\$
    – Ben Voigt
    Jul 5, 2021 at 15:45
  • \$\begingroup\$ @BenVoigt well that's open to interpretation. The + and - symbols don't particularly seem attached to Vb and I read them as random scrawls. Apart from anything else, it makes no logical or common sense to regard Vb as the inverted polarity of Va. Thanks for the downvote. \$\endgroup\$
    – Andy aka
    Jul 6, 2021 at 7:15

In order to compare voltages, they must share the same reference for One of the two +/- probe paths.

Yours does not. Its inverted. So Vin- = Vin+

Since ground by definition is defined by your choice of 0V, even if floating, that is the reference you should choose.

That statement Va=-Vb is just to make you think and understand this point.

But in reality, Vdiff = Vout / feedback gain (typ) 1e5 for BJT’s .

  • \$\begingroup\$ What is "feedback gain"? I rather think that Vdiff=Vout/Aol with Aol=open loop gain. \$\endgroup\$
    – LvW
    May 23, 2022 at 9:43

Why is Va=−Vb in this op-amp?

It isn't.

IF all input and output voltages are within the device's normal operating parameters AND IF there are no over-currents or current limiting AND IF the device is a theoretically perfect opamp

THEN Va = Vb, not -Vb

In a classic inverting amplifier topology, R1 and R2 form a simple voltage divider between V1 and Vo (Vin and Vout). Negative feedback drives the R1-R2 noce at the inverting input to be equal to the non-inverting input.

  • \$\begingroup\$ You've ignored the marked polarity with which V_b is measured. \$\endgroup\$
    – Ben Voigt
    Jul 5, 2021 at 15:52

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