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Question for stability of negative feedback OP-AMP

  1. Jul 14, 2015 #1

    I've studied the golden rules of the feedback OP-AMP and applying this to voltage follower shows that voltage gain (Vout/Vin) is 1.

    Thus, Vout should eventually follows Vin when Vin suddenly changed. I've tried to follow this process for clear feeling by drawing pictures as shown in the attached image.

    Since AV is very large for typical OP-AMP, It looks Vout becomes divergent! (or becomes saturation value.) This divergence is even more clear when feedback equation is approximates to (V+ - V-)AVn where n is number of looping. (You can get this approximation by looking last shown Vout in the image.)

    What was I wrong? Why does feedback make the system crazy?

    Attached Files:

  2. jcsd
  3. Jul 14, 2015 #2


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    I am afraid, you are mixing open-loop gain Aol with closed-loop gain Acl=1.
    Hence, the third figure in your sequence is wrong:
    Here is the correct sequence assuming Acl=1(because of full negative feedback):
    1.)Vin=3V, Vout=3V
    2.) Sudden jump from Vin=3V to Vin=5V: In the very first moment we still have Vout=3V (time constant of the device),
    3.) Hence Vdiff=5-3=+2V (this is the diff. voltage directly between the input nodes)
    4.) Therefore, Vout tends to increase (direction to pos. supply) and passes a value of slightly below 5V, which causes Vdiff~some µV
    5.) This is a new equilibrium because the tiny Vdiff (µV range) multiplied with the (large) open-loop gain "holds" the value of Vout at a value very close to 5V (perhaps 4.99995, depending on the open-loop gain Aol).
  4. Jul 14, 2015 #3
    Thanks for replying! But I think I need a little more help for the detailed process.

    Could you tell me more about details of 4th process in your reply that how Vout suddenly gets the value slightly below 5V at 1st forward feeding? I thought it should be 2Aol in volt since open loop gain Aol is intrinsic property of OP-AMP so Vout = VdiffAol always holds from inputs to output regardless of feedback loop.
  5. Jul 14, 2015 #4


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    Yes - that`s correct. Let me try to explain in more detail.
    ....4.) Because Vdiff is heavily enlarged (by 2 V) the opamp "feels" that the input is overdriven and the output tends to go into positive saturation. However, it does not "jump" to the supply rail (because of internal time constants) but there will be a finite rise time.
    And - as I have mentioned - on the way to the pos. supply voltage the output crosses a voltage which again (as bevor in the 3V case) exactly fulfills the condition mentioned by you: Vout = Vdiff*Aol . At this point (in this moment) the rising is stopped (equilibrium) because a further increase of the feedback signal would again decrease the diff voltage Vdiff.
    Example: Vout=5.1 V. That means: Feedback voltage (at the inv. terminal) is also 5.1V and Vdiff=5-5.1=-0.1 V.
    Hence, the opamp sees a negative diff. voltage and tends to go back to negative output voltages (and will stop again at an output which fulfills Vout=Vdiff*Aol).

    There is only one stable condition (equilibrium) caused by negative feedback: In my example:
    Vin=5V, Vout=+4.99995V, Vdiff=0.00005V.
    From this, we can calculate back:
  6. Jul 14, 2015 #5
    Yes thanks!

    So what my picture does not show is finite response time OP-AMP. My picture implicitly assumed that response time of OP-AMP is a brink of instance, which is impossible in real world! Thus as on the way of Vout increase triggered by sudden input voltage change, It finds equilibrium point where OP-AMP gain equation Vout = AOLVdiff holds and stay there. It looks like OP-AMP finds comfort "home" during journey at which it intended to go somewhere at first time.
  7. Jul 14, 2015 #6


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    Yes - that is the secret of stabilization caused by negative feedback. And this is only ONE benefit of negative feedback. There are some more.
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