Why is $V_a=-V_b$ in this op-amp?

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

$$\V_a = V_b\$$ because the op-amp has ideally infinite open loop gain and of course, if $$\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.

• Vb is measured with inverse polarity (+ is on ground), that's where the sign gets reversed. Commented Jul 5, 2021 at 15:42
• 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 $$V_a = -V_b$$ Commented Jul 5, 2021 at 15:45
• @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. Commented 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 .

• What is "feedback gain"? I rather think that Vdiff=Vout/Aol with Aol=open loop gain.
– LvW
Commented 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.

• You've ignored the marked polarity with which V_b is measured. Commented Jul 5, 2021 at 15:52