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Confusion about op-amp statement

  1. Jan 26, 2008 #1
    In my text it says,

    "As a result of the negative feedback, it can be shown that the closed-loop gain is almost insensitive to the open-loop gain A of the op amp. For this reason, op amps are used in circuits with feedback paths."

    I am a bit confused on what it means by closed-loop gain is almost insensitive to the open-loop gain. Any thoughts?

  2. jcsd
  3. Jan 26, 2008 #2


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    I would think it means that when one has a feedback path, the open loop gain no longer affects how the input and output are related. With the feedback path, you are not subjected to the enormous open loop gain of the op amp, so the circuit essentially become insensitive to open loop gain.
  4. Jan 27, 2008 #3
    this is due to the fact that the input of the open-loop amplifier is very low, almost zero (this is the role of the high gain open-loop chain!). imagine this is not the case, in stationary conditions: then, the output will be so big to suppress the input signal, when reapplied to the input through the feedback path. this bring to instability, in disagree with assumption of stationariety. so, the open-loop amplifier practically doesn't affect total gain, which is fixed by the feedback net. all this is true in stationary conditions, when the feedback net is "more fast" than the amplifier in restoring a low-signal condition at the open-loop chain input.
  5. Jan 28, 2008 #4


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    there need to be some caveats: the open-loop gain of the op-amp, is the gain of the op-amp itself. and they design the op-amp so that this gain is very, very large. apply a microvolt to teeny-weeny + and - input terminals, and you'll get 100000 microvolts coming out (assuming you powered this up correctly).

    however, with negative feedback, if you have an inverting amp (a circuit with an op-amp and some resistors) with gain of -100 or a non-inverting amp of gain +100 or similar, if the op-amp gain (open-loop) inside the circuit could change from 100000 to 200000 and the gain of your circuit would increase from 100 to 100.0001 or something like that. you wouldn't know the difference.

    when you have astronomical gains (as high input resistance) in the op-amp that you can count on, then the characteristics of the circuit with negative feedback depends only on the other parts, not the op-amp.

    negative feedback is cool. unless you're a psychologist, then they turn it around. what we engineers call "negative feedback", they call "positive feedback" because they make a value judgement that stability is a "positive" thing. what we call "positive feedback", they call "negative" because of the same value judgement. psychologists don't know squat.
  6. Jan 28, 2008 #5


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    You guys are making this way too complicated.

    In the basic negative-feedback configuration, you have two gains: forward gain, a, and the feedback factor, f. The forward gain is what you would call the "open loop" gain.

    The transfer function of this topology is:

    [itex]\frac{ V_{out} }{ V_{in} } = \frac{ a } { 1 + af } = \frac{ 1 } { 1/a + f } [/itex]

    To see the effect of negative feedback, just assume f is some reasonable value, say, f = 0.5.

    Now, consider what happens when you change a, the open-loop gain. If a = 100, you get a closed-loop gain of:

    [itex]\frac{ V_{out} }{ V_{in} } = \frac{ 1 } { 1/100 + 0.5 } = 1.96078 [/itex]

    If a = 1000, you get a closed-loop gain of:

    [itex]\frac{ V_{out} }{ V_{in} } = \frac{ 1 } { 1/1000 + 0.5 } = 1.996007 [/itex]

    That's what they mean when they say the closed-loop gain is insensitive to the open-loop gain. In fact, that's why we use negative feedback in circuits: you want your circuit to deliver a precise gain for the customer, regardless of the myriad manufacturing errors that can easily change the open-loop gain by a factor of 10 from one part to another.

    - Warren
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