What is the role of the feedback resistor in an op-amp circuit?

[niehaoma]

I just started learning about Op-Amps. Working through basic equations seems straightforward, however I was curious about the function of the feedback resistor. As the name implies, one might infer the current/voltage potential "feeds back" from the output back into the input (I know there are two voltage potentials at each end of the resistor, Vo and the potential from the input pin). However, when the equations are defined (nodal analysis), the current is defined as moving through the feedback resistor towards the output pin. If you take a basic setup, the first resistor (say R1) is in series with Rf (assuming Ideal op-amp, hence no current flows into the inputs due to Rin = infinity), basically forming a voltage divider. In an ideal op amp, the closed loop voltage gain (inverting) is basically the ratio of Rf/R1 = Vo/Vin.

I guess I am looking for a simple English explanation, instead of the mathematical relationship, regarding the effect the feedback resistor imparts on the op-amp circuit, relative to input and/or output.

Thanks. Let the schooling begin ;)

 

[jegues]

The feedback connection between the output and the inverting input terminal (i.e. Negative Feedback) forces the differential input voltage towards zero. (Since the op-amp has very high open loop gain)

In other words, this feedback is the reason why the voltage at the negative input terminal of the op amp is the same at the positive input terminal.

 

[carlgrace]

Another way to look at it (jegues is correct BTW) that may be helpful is to thing of Rf as developing the output voltage based on the input current created by R1.

Basically, the two input terminals of an op amp in negative feedback are at the same potential (to the extent the open-loop gain is infinite). Then, the non-inverting input is usually connected to ground. Therefore, the inverting input is ground as well. So, a current flows from the input to the inverting input. Now, the input has infinite input resistance, so the current has nowhere to go but through Rf. You can then use Ohm's law to calculate Vo.

This is a kind of intuitive way to calculate the output equation of an inverting op-amp amplifier.

 

[Averagesupernova]

...Then, the non-inverting input is usually connected to ground.

Only in inverting mode.

 

[carlgrace]

Only in inverting mode.

Thank you for the helpful reply. Please continue reading to this part of my earlier post:

This is a kind of intuitive way to calculate the output equation of an inverting op-amp amplifier.

The same intuitive explanation is valid for a non-inverting amplifier as well. It is slightly more complicated, though, because you have to solve KCL to determine the dc voltage at the op-amp inputs.

 

[Averagesupernova]

The same intuitive explanation is valid for a non-inverting amplifier as well. It is slightly more complicated, though, because you have to solve KCL to determine the dc voltage at the op-amp inputs.

That was kinda the point. Explain it in such a way as to cover both cases. Bottom line in negative feedback with op-amps is that the inverting input will try to track the non-inverting input. The output will do whatever it has to in order to get this to happen. The DC voltage on the inputs will be whatever the previous stage puts it at. Allowing the input to float DC-wise is poor practice in my opinion. There was a thread a couple of months ago about measuring the current in the neutral wire on mains power that allowed this to happen. The thread seems to have died. I PM'd the OP but got no reply.
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Anyway, sorry for the 'helpful reply'.

 

[carlgrace]

Only in inverting mode.

That was kinda the point. Explain it in such a way as to cover both cases. Bottom line in negative feedback with op-amps is that the inverting input will try to track the non-inverting input. The output will do whatever it has to in order to get this to happen. The DC voltage on the inputs will be whatever the previous stage puts it at. Allowing the input to float DC-wise is poor practice in my opinion. There was a thread a couple of months ago about measuring the current in the neutral wire on mains power that allowed this to happen. The thread seems to have died. I PM'd the OP but got no reply.
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Anyway, sorry for the 'helpful reply'.

No, worries. Thanks for the explanation. I agree with you now that i think about it. While what I said was correct, i suppose someone could get confused.

Thanks.

 

[Averagesupernova]

Oh you were definitely correct. Somehow I never felt I had a good grasp of op-amps until someone explained the way I did about the output doing 'what it has to' in order to get the inputs to track. Somehow this made it so clear.