# Why are the inputs of an ideal op-amp “inverting input” and “non-inverting input”?

Here is the first ideal op-amp circuit, called an "Inverting Amplifier", that many students will encounter:

The gain here is $$\G=-\frac{R_F}{R_{IN} }\$$. Thus, with a negative gain, $$\V_{OUT}\$$ is inverted with respect to $$\V_{IN}\$$. Also, since $$\V_{IN}\$$ goes into the inverting input, this all makes sense.

Now, if we flip this all around like this:

For this circuit, $$\V_{IN}\$$ goes into the non-inverting input. However, the gain still has a negative sign: $$\G=-\frac{R_F}{R_{IN} }\$$, and $$\V_{OUT}\$$ is still inverted.

So why is this called the inverting input?

Solving the lower circuit, incorrectly:

$$I_{IN} = I_F$$ $$\frac{V_{IN}-V_+}{R_{IN}} = \frac{V_{+}-V_{OUT}}{R_{F}}$$

$$\textrm{If } V_{+} = V_{-} \textrm{ , as is true by definition for an ideal op-amp, and } V_{-} = 0, \textrm{ then } V_{+} = 0 \textrm{ thus }$$

$$\frac{V_{IN}}{R_{IN}} = \frac{-V_{OUT}}{R_{F}}$$

$$\frac{V_{OUT}}{V_{IN}} = -\frac{R_{F}}{R_{IN}}$$

What's wrong with this circuit analysis?

• Possible duplicate of Are op-amp inputs interchangeable? – user103380 Apr 17 '19 at 2:13
• Your 2nd circuit has positive feedback. Your attempt at a gain formula for that circuit is completely incorrect, since the circuit won't behave as an amplifier. – brhans Apr 17 '19 at 2:14