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

enter image description here

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:

enter image description here

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?

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    \$\begingroup\$ Possible duplicate of Are op-amp inputs interchangeable? \$\endgroup\$ – KingDuken Apr 17 '19 at 2:13
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    \$\begingroup\$ 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. \$\endgroup\$ – brhans Apr 17 '19 at 2:14
  • \$\begingroup\$ Read the answers to KingDuken's duplicate link. \$\endgroup\$ – brhans Apr 17 '19 at 2:16
  • \$\begingroup\$ Your equation V+=V- must not be used. This is valid only if the device operates in the linear region of its transfer characteristics. And this is possible for negative feedback only! \$\endgroup\$ – LvW Apr 17 '19 at 7:44

Your second circuit will not work as an amplifier because it has positive feedback. It's actually a comparator with hysteresis.

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  • \$\begingroup\$ The input stage of an op-amp is symetrical. However, the second stage is not! That's where the negative and positive inputs come from. \$\endgroup\$ – Janka Apr 17 '19 at 2:47
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    \$\begingroup\$ axsvl77 - the two input voltages are only supposed to be equal if there is proper negative feedback; in an ideal op amp, the fact that the input voltages are equal is a consequence of the fact that the amplifier has very high gain and the feedback is negative; in case the feedback is positive the amplifier will simply saturate to one of the rails and the non-inverting input voltage in your case will be defined by the voltage divider between Vout and Vin. \$\endgroup\$ – joribama Apr 17 '19 at 2:54
  • \$\begingroup\$ The key here is to keep in mind that the input pins are high impedance and have no ability to actually drive the input voltages. Instead, it's the output that drives the voltage and the presence of proper feedback makes the input voltages to be the same. Think of it as an equilibrium condition. \$\endgroup\$ – joribama Apr 17 '19 at 3:05
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    \$\begingroup\$ @joribama Yes, good answer. I want to warn you that some academic produced material showing that there is a metastable point for the positive-feedback circuit where the relationship between input voltage and output voltage is the same as for the negative-feedback case. Some have mistakenly interpreted this to mean that the positive-feedback circuit has gain when it is, as you say, nonlinear. Keep fighting the good fight \$\endgroup\$ – Elliot Alderson Apr 17 '19 at 11:53

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