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Can somebody please explain to me why the output here is the negative of the "input"? I really can't wrap my head around it...enter image description here

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    \$\begingroup\$ Luck, and an unrealistic simulation. Add a microvolt of noise... \$\endgroup\$ Jan 22 '16 at 11:24
  • \$\begingroup\$ I suggest you read the pdf which you can download called "Op amps for everyone" by Ron Mancini. It is a free paper by Texas Instruments. It explains almost every aspect of opamps and how to use them. \$\endgroup\$ Jan 22 '16 at 11:32
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Try using negative feedback and not positive feedback. The circuit you have shown is nonesense unless you are trying to make a comparator with rather a lot of hysteresis.

When using a simulator it might theoretically "settle" on what seems to be an improbable scenario - If the output is -1V and the input is +1V then the voltage at the non-inverting input is 0V and this exactly matches the 0V at the inverting input.

Ergo, what seems to be a stable situation.

For a real circuit this just would not happen - a tiny bit of noise or drift would polarize the situation and the output would be hard against the positive rail.

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  • \$\begingroup\$ I know it's nonsense, I encountered this circuit while preparing for an exam, and I thought it was weird, so maybe the professor accidentaly switched the + and the - terminals of the op-amp \$\endgroup\$
    – Calin
    Jan 22 '16 at 11:30
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    \$\begingroup\$ Simulators can do strange things. \$\endgroup\$
    – Andy aka
    Jan 22 '16 at 11:31
  • \$\begingroup\$ @Andy aka, see my comment to Rmano`s answer. \$\endgroup\$
    – LvW
    Jan 22 '16 at 14:48
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If the simulator is using a linear model to solve the problem (for example, using the AC simulation mode) it can get easily fooled. The circuit will be basically this:

Equivalent

and the simulator will basically write down

$$ v_o = A_{vol} \left( v_o\frac{R}{R_1+R} - v_i\right)$$

...and find a solution, independently from the fact that the system is unstable.

(this is a typical pitfall in using simulators --- happens to my students almost always when simulating Bode plots. Very instructive, on the other hand).

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  • \$\begingroup\$ Is it a pitfall of the simulator? I don`t think so. The scheme is equivalent to a mechanical analogon: A small ball riding upon a larger one. Theoretically and under ideal (static) conditions (no external disturbance) this system is stable. The shown circuit seems to be stable under ideal static condidtions, which are the background of DC and AC analyses. The real behaviour can be observed for a TRAN analysis only with supply voltages switched-on at t=0 (and only THIS are real conditiones). Hence, it was the user who was wrong in his interpretation - and NOT the simulator. \$\endgroup\$
    – LvW
    Jan 22 '16 at 14:46
  • \$\begingroup\$ Yes, that was I meant, @LvW - it's a user pitfall. The simulator is right, and if you draw the Bode plots and you know your theory, you'll see the positive poles. \$\endgroup\$
    – Rmano
    Jan 22 '16 at 22:52
  • \$\begingroup\$ Yes - it is very important to know the capabilities, assumptions and limitations of the various analyses (DC, AC, TRAN,...). Only in this case, we can avoid misinterpretations. \$\endgroup\$
    – LvW
    Jan 23 '16 at 9:19

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