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Why does negative feedback stabilize an operational amplifier?

  1. Apr 17, 2012 #1
    I'm studying for an exam and i'm struggling to understand op amps, here's my problem:
    I understand that for an op amp to have a controllable gain, the difference between the input terminals must be very very close to zero (and you can assume it is zero). And to do this you use negative feedback. For example a non inverting op amp http://upload.wikimedia.org/wikipedia/commons/6/66/Operational_amplifier_noninverting.svg
    If the voltage divider sends half the voltage from the output to the inverting terminal, then the output will need to be twice as big as Vin in order for the input terminals to be the same. So the amp will have a gain of 2.
    A lot of website explain this by saying the op amp "Wants" the inputs to be the same, so it adjusts the output so that they are.
    What I don't understand is why the op amp "Wants" the inputs to be the same. Obviously it doesn't actually "Want" anything, but how does the circuit stabilize and manage to produce the desired output?
  2. jcsd
  3. Apr 17, 2012 #2
    First I want to clarify about the word "stabilize". We use this word mainly for stability of the op-amp where it does not burst into oscillation. What you talked about, we usually don't refer as stabilization.

    Having said that. The definition of op-amp is it has very high internal gain. It has a +ve and -ve input and an output. The amp amplifier any signal between the +ve and -ve input. Say the open loop gain of the amp is 10000, and you put 1mV between the two input with the +ve side to the +ve input. The output will be 1mV X 10000=+10V.

    There are at least two obvious problem with this:
    1) The open loop gain of the op-amp is not exact between each device even when it is from the same wafer. So you get different gain. One can be 10000, one can be 13000, another one can be 8000. So you can't count on getting a predictable gain.
    2)The gain also temperature dependent. Say you have a gain of 10000, if you raise 20 deg, the gain can drop down to 8000. So again, you can't count on it.

    The cure is by using negative feedback to lower the gain but stabilize the gain value so regardless of different op-amp, at different temperature, you get the same gain.

    Op-amp don't want to have the input to be the same. It is the way negative feedback will make the op-amp behave this way. Like you said, when the output of the amp go through a resistor network and take the divided by 2 point and connect to the -ve input.

    So if you put 1V to the +ve, the op-amp having high internal gain:
    1) At t=0 before op-amp even respond, you will have +1V at +ve input, and 0V at -ve input. That is 1V across the inputs. The output starts to pull the output high. The divided by 2 of the output voltage will rise to half of the value.
    2) Say the output swing to +1V, the -ve input will be +0.5V. You still have 0.5V across the inputs. The output will therefore keep going up.
    3) Say the output swing to +1.8V, the -ve input will be +0.9V. You still have 0.1V across the inputs. The output will therefore keep going up.
    4) Say the output swing to +1.99V, the -ve input will be +0.0.995V. You still have 0.005V across the inputs. Now you are reaching to point you have very little voltage across the input and the output will start to slow down in rising.

    This will keep going until the difference voltage at the inputs give the required output, which is very close to 2V.

    By looking at this steps, you can see how the negative feedback works.

    Why do we do all these? because by negative feedback, even the open loop gain, change only affect the last little bit. Say even if the gain varies from 5000 to 10000, it will take only the difference between 0.4mV for 5000 to 0.2mV for 1000 at the input to produce 2V. If you take into consideration of divided by 2. The output error will be (2X0.2mV)-(2X0.4mV)= -0.4mV!!!! This is 0.4mV out of 2V and the output error if the open loop gain of the op-amp change by 50%!!!! This is stabilization.

    Go through it step by step and see how the op-amp use negative feedback to improve accuracy.
    Last edited: Apr 17, 2012
  4. Apr 17, 2012 #3
    You're thinking about it the wrong way. Let's use an analogy.

    Take a large, semi-circular bowl like a wok, and drop a small marble into it. What happens? The balls settles into the bottom. What happens when you push the ball somewhere else? It rolls back to the center.

    How does that happen? Because of basic physics, the ball will roll to where there is a local minimum of gravitation potential energy. Any other spot on the wok is unstable.

    Same for an op amp. Because of the ludicrously massive amount of gain that a normal op amp has, any significant difference between the (+) and (-) inputs will saturate the output. Because of this, the only way to have a stable signal without saturation is to find some way to keep both inputs extremely close to each other in value.

    Negative feedback is used to have the level of one of the inputs be affected by the level of the output so that both inputs are always very close to the same level.
  5. Apr 18, 2012 #4
  6. Apr 18, 2012 #5
    Thanks yungman, that's explained it perfectly. The reason I was confused was because I assumed that the output instantly changes. I thought that when the inputs see a difference, say 1V and 0V, the amps massive gain (say 1000) amplifies the difference and the output will instantly be 1000V. This confused me because if half of this was fed back to the inverting terminal, the inputs will be 1V and 50V, so the difference will be 49, and the output will become 4900, half of this will be fed into the inverting terminal again so the inputs will be 1V and 2450V, and so on... This seemed like positive feedback to me, witch is why I was confused.
    I now see that the key is that the output doesn't instantly change, it gradually builds up, making the difference between the terminals smaller each time, and that the difference tends towards 0, so the overall gain caps off at a certain value.
    Please tell me if this is correct or if I'm still getting things confused :/
  7. Apr 18, 2012 #6
    Yes, this is what we call closing the loop. Also, the accuracy of the op-amp circuit drop as frequency goes up because the open loop gain drops and more error voltage needed to give the output.
  8. Apr 18, 2012 #7


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    Some amazing stuff the boys wrote up above. Definitely beyond me.

    But if you are looking for a simple way to look at IDEAL op amps....like the ones you see in college or state license exams...

    Theres really just 4 simple rules in my opinion.
    No current enters op amp.
    And current in equals current out where the feedback meets the input (Simple KCL)
    And yes there is -RF/RA and 1+RF/RA....but those two formulas are just derived from the four things above.

    Those simple rules can answer all of the IDEAL op amp circuit problems for finding gain, transfer functions and other frequency response things.

    That's the simple route. If you want to go beyond......re-read the other posts above.
    Last edited: Apr 18, 2012
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