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Op amp circuits are designed to achieve a specific gain regardless of the differences between individual op amps. One very common circuit has a gain of -R2/R1. Here's a (corrected) schematic:

inverting op amp schematic

Another common configuration has a gain of R2/R1+1 and is non-inverting:

non-inverting schematic

What I can't see is why on earth anyone would use the inverting one, except for the odd case where you actually want inversion. The non-inverting one has high input impedance without an extra input stage, and almost the same gain. Is there any advantage to the first example?

Also, since the first example does not have high input impedance, it can take significant current to drive. So, often a source follower is placed before the amplifier. For the second configuration, is there any reason why a source follower would ever be necessary?

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    \$\begingroup\$ In the first case (inverting), since it is a single-supply op-amp, wouldn't one need to have a "virtual ground" of VCC/2 feeding into the + input, otherwise the output would try to go negative? (I'm a digital guy still learning to use op-amps, so I could easily be wrong about this). \$\endgroup\$ – tcrosley Aug 7 '12 at 1:10
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    \$\begingroup\$ You've answered your own question. Sometimes you want an amplifier with low input impedance. For example, if you're amplifying current. \$\endgroup\$ – user33657 Dec 4 '13 at 19:44
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The inverting configuration is capable of gains less that 1, and can be used as a mixer. Here is a good primer.

http://chrisgammell.com/2008/08/02/how-does-an-op-amp-work-part-1/

I don't know exactly why (anyone feel free to chime in), but the fact that negative feedback is holding the negative input terminal at 0v means that node is a proper place to sum currents, making the mixer circuit viable (although inverting). Op amps are also cheap and come in packages with more than one, so you can usually just invert something again if it's "upside down"

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One factor not yet mentioned is that some op amps work best when the common-mode input voltage is kept within a narrow range. It's very difficult to design an op amp in which the same circuitry handles common-mode voltages near both rails. Typically, an op amp will either not work correctly when the inputs are too near near one of the rails, or else it will have one set of input circuitry for use when voltages are near one rail, another set for when voltages are near the other, and circuitry to automatically switch between them. If the two input circuits are not perfectly matched, switching between then may disturb the output. Keeping the common mode voltage at a fixed value eliminates this problem.

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    \$\begingroup\$ Having a varying common mode voltage creates distortion, through mechanisms like channel length modulation/Early effect in the current mirror feeding the input differential pair. With large source resistances this distortion can be quite significant. \$\endgroup\$ – Bitrex Aug 7 '12 at 2:56
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    \$\begingroup\$ In his first book, Jim Williams formulates Williams's Rule: "a little known tenet of precision op amp circuits... always invert (except when you can't)." \$\endgroup\$ – markrages Jun 29 '13 at 5:04
  • \$\begingroup\$ @markrages: I like that rule. It's common to use op amps in cases where one wants to minimize input currents, and for those a non-inverting configuration is usually the way to go, but if one's input is going to have known finite resistance characteristics, that's a good sign one should use an inverting config. \$\endgroup\$ – supercat Jun 29 '13 at 15:25
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    \$\begingroup\$ @markrages: Curiously, the same principle appears in some other fields as well. For example, standard black and white film can be processed to produce a positive or negative, and both positive and negative color films are available. A direct positive can be better than a print made from a negative, but a print made from a negative is better than one made from a positive. \$\endgroup\$ – supercat Jun 29 '13 at 17:39
  • \$\begingroup\$ Although there are several cases where inverting is not ideal. For example, if your source impedance is high, you would have to use potentially high value resistors in which case noise could be higher than necessary. Even if the source impedance isn't terribly high, generally you want to minimize series resistances. Particularly for keeping noise low in audio circuits. If you study a really good low-noise audio circuit, you will usually find very low or even no series resistance in the signal path. \$\endgroup\$ – squarewav Dec 25 '14 at 20:32
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In any case, inverting is not a problem. We can get a positive signal just by changing the wiring. Furthermore, I think using several amp stages is pretty common, and an even number of inverting amps make a bigger non-inverting one.

Wikipedia gives some disadvantages for the non-inverting configuration: http://en.wikipedia.org/wiki/Operational_amplifier_applications#Non-inverting_amplifier

I don't think that placing a buffer in the input of the second configuration provides any advantage.

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    \$\begingroup\$ Using positive feedback will usually make the output "run away" to the positive rail and then stay there, basically latching the first positive output voltage event until powered down. This is of no use when trying to amplify some varying input signal. \$\endgroup\$ – JimmyB Aug 6 '12 at 21:50
  • \$\begingroup\$ Could you include disadvantages stated in Wikipedia in your answer? \$\endgroup\$ – Dmitry Grigoryev Jan 24 '18 at 14:57
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Really, nowadays the humble inverting amplifier has almost no advantages over the non-inverting amplifier (excluding the absence of a common-mode error and, of course, the inversion). But in the past, when there were no differential amplifiers, this was the only way to make an amplifier with negative feedback.

The generalized inverting configuration with various elements E1 and E2 (resistors, capacitors, inductors, diodes, transistors, sensors, etc.) connected in the place of R1 and R2, is extremely useful. There the op-amp removes the undesired voltage drop across E2 by an equivalent output voltage thus providing ideal load conditions (short connection) for E1... the op-amp acts as an element with negative impedance neutralizing the positive impedance of E2. See more about this technique in my wikibooks story of Voltage compensation.

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protected by Community Dec 4 '13 at 21:52

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