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schematic

simulate this circuit – Schematic created using CircuitLab

So, here we have a typical op-amp with negative feedback. The non-inverting input is at virtual-ground, and if the input is higher or lower than ground, the voltage is inverted, and fed back into the non-inverting input through the voltage divider until the non-inverting input is equal to the inverting input.

But why do we need the voltage divider? The gain of the amplifier is essentially unlimited, so we're not really helping "cut-down" the output voltage to apply to our inputs incrementally. In fact, it seems that the voltage divider is actually dentrimental to the op-amps task.

While it doesn't really help us tame the gain of the output, it does effectively (if slightly) reduce the input voltage. Thus, our inverting input has to be very specific, and equal to the dampened non-inverting terminal's input.

For example, let's say that I want to make sure that the non-inverting input is at 5V. If I use a 1KΩ resistor at R2, use a 100Ω resistor at R1, and the input signal is 3V the input signal after R1's dampening will be around 2.75V.

What are the resistors for?

Allenph
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  • This is going to take a minute. Please [join me in chat.](http://chat.stackexchange.com/rooms/23828/room-for-sean-boddy-and-allenph) –  May 16 '15 at 06:11
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    Some corrections: The non-inv. input is NOT at "virtual" ground. Indeed, it is really grounded. Moreover, a portion of the output is fed back to the inverting input. This portion is determined by the voltage divider. To understand how this feedback reduces the gain and stabilizes the dc operating point you should try to become familiar with the concept of negative feedback. – LvW May 16 '15 at 08:50
  • For more information about "how negative feedback works" see here:http://electronics.stackexchange.com/questions/170117/understading-negative-feedback – LvW May 16 '15 at 08:57
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    Too many incorrect statements to begin to sensibly answer this. – Andy aka May 16 '15 at 09:44
  • There is no voltage divider here. The non-inverting input *is* grounded, and not virtually. The output voltage is inverted regardless of the inverting input's polarity. Nothing is fed back to the non-inverting input. Too many errors to deal with in a comment or an answer. It's just nonsense from start to finish. I suggest you re-read your textbooks. – user207421 May 16 '15 at 11:20
  • Why all the downvotes? It is not like he asked some homework questions, he has a lot of incorrect statements, but his question is not a illegitimate one. – Rick van Loo Nov 08 '15 at 18:44

3 Answers3

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There are a couple things that need to go with looking at how opamps work with negative feedback.

  1. Very high input impedance, so essentially no current flow into the terminals.
  2. The opamp will do all that it can do ensure that the inverting and non inverting terminals are at the same voltage. (V+ - V- = 0)

Consider the following.

schematic

simulate this circuit – Schematic created using CircuitLab

This does not have any feedback. We have 1V on the non inverting terminal and the inverting terminal is grounded. The opamp, will do everything it can to make sure that the terminals are at the same potential. So the opamp is going to increase its voltage in the hopes that the negative terminal will balance out, so that the difference between the two terminals is zero. But, because there is no feedback, the opamp's "efforts" are in vain and the output just increases until you reach the limit of the opamp. It's trying, but nothing it does can bring the terminal voltages to be the same.

schematic

simulate this circuit

In this case, the opamp has feedback. So as the output voltage changes, the voltage at negative terminal changes. When a balance is found, the opamp maintains that voltage. If the input changes, the opamp will then find what the new balance is. All because it wants to make sure that the voltage at the terminals is 0V. (V+ - V- = 0)

This is all great, but what if want gain ? Right now, the output will change so that it matches the input, and for a buffer, that's great but its not useful if you have a low voltage signal that you want to boost up to something more readable.

And this is where feedback resistors come into play.

schematic

simulate this circuit

Remember the rules ? Opamp will do everything it can to make sure the inverting and non inverting terminals are the same.

Let's see what happens now.

We have a 1V signal. If the opamp outputs 1V, then the voltage seen at the inverting terminal is only 0.5V. But the opamp is not "happy" at this because it want's the difference to be 0. So it increases the voltage. When the opamp increases the voltage to 2V, then the inverting terminal sees 1V. The opamp is happy. So we fed in 1V, and we got 2V outs, this is a gain of 2.

So feedback resistors are used to attenuate the voltage at at the inverting terminal from the output to trick the output terminal to produce a larger voltage.

efox29
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    Without negative feedback the op amp's function is to amplify the differential input voltage by its own gain. *With* negative feedback that acts to equalize the input voltages. To state that this still happens without negative feedback is simply nonsense. – user207421 May 16 '15 at 11:17
  • The opamp can do nothing else than to amplify the diff. voltage between both opamp terminals - of course, also in case of negative feedback. However, in case of feedback this differential voltage (microvolt range) is determined by TWO voltages of opposite polarity (signal input and opamp output). This leads to the mentioned small diff. voltage which, normally, is neglected (set to zero; principle of virtual ground). – LvW May 16 '15 at 13:40
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To answer your question in a single sentence: in this circuit configuration the resistors set the gain of the circuit. (Vout = A x Vin). Where A = -R2/R1

It's as simple as that.

Dean
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In opamp circuits, you are trying to design a system which ultimately is insensitive to the actual opamp gain. This can be done if the opamp gain is very high ('infinite'). With sufficiently high gain, then overall characteristics of the system can be defined with 'simple' components -- resistors and capacitors. These are relatively precise, and therefore accurate performance can be achieved with inaccurate (unknown high gain) opamps, but with accurate passive (resistors, capacitors) components.

In your circuit, the goal is not very high gain, but precise controlled gain. The high gain opamp, and precise resistor ratio allow this to be achieved easily.

jp314
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