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Op-amp transient state

  1. Feb 3, 2016 #1
    sys35.gif

    sys36.gif





    Hello, I have two questions about feedback.

    1.

    If we look at the equation above, it says

    G(Vin-BVout)=Vout
    but this equation is actually,


    G(Vin-BVout1)=Vout2
    whereas Vout1 is Vout for first iteration and Vout2 is Vout for 2nd iteration.

    Therefore, they must have different values.
    How can we get the transfer function then?



    2.


    sys39.gif

    if we assume the very first iteration of the loop, V- will be greater than V+ as Vout>>>>>>>Vin.

    At this point, V+-V- <0, therefore, it will generate a negative feedback opamp response.

    and the next iteration vice versa.

    How can we prove that if we proceed this iteration over and over, we will get the resultR2/(R1+R2)
     
  2. jcsd
  3. Feb 3, 2016 #2

    Hesch

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    Why do you mention "iterations"?
    An analog system does not "iterate" the output voltage.
    Vout/Vin = G / ( 1 + Gβ ) is the correct expression. ( Mason's rule ).
    An iteration would imply a time-delay. There is no such time-delay in your upper figure.
    If there is a time-delay e.g. in the feedback, you must analyze a system, with the delay built in.
     
  4. Feb 3, 2016 #3
    I'd like to add;

    Q1. Your use of Vout1 and Vout2 would imply there was more than 1 output, where in this circuit there is only 1 output.
    What you have attempted to label as Vout1 would always resolve to an open loop gain, as B would be 0 for that jiffy or 2.

    Q2. You conclusion relies on your faulty conclusion for Q1. As the feedback loop is only connected to 1 of the +/- inputs, the gain shift will always be in the same direction.
     
  5. Feb 4, 2016 #4

    LvW

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    Kidsasd987, I know what you mean - and I will try to give you some information.
    However, I am afraid - it will be a rather long answer. Nevertheless....

    To answer your question it helps to analyze what may happen after switch-on the power supplies +/- Vs=+/-10V. More than that, you have to apply negative feedback (resistive netork between output node and inverting input).
    (The described timely sequence may be somewhat „formalistic“ - however, it helps to understand the feedback concept).

    Example
    : Non-inverting gain stage with desired gain of "+2". That means: Feedback factor k=0.5 using a voltage divider with two equal resistors
    Open-loop gain (assumption): Aol=1E4.
    1.) t=0: Apply at an input voltage Vin=1V. The opamp is not yet working in its linear range (feedback not yet active due to time constants within the circuit) and the output will immediately jump to Vs=+10V.
    2.) t>0: The voltage at the inverting terminal will rise to 0.5Vs=5V>Vin=1V. Hence, the voltage at this inverting terminal dominates (is larger) and the output voltage will change in the direction to minus 10V.
    3.) However, on its way to -10V the ouput voltage is crossing a positive value which produces at the inverting terminal a feedback voltage of +0.99980004V .
    4.) At this very moment (assuming an open-loop gain Aol=1E4) , the opamp is in its linear amplification region because the diff. voltage is Vdiff=Vin-0.99980004=1-0.99980004=0.00019996V.
    As a result, the output voltage is Vout=Vdiff*Aol=0.00019996*1E4=1.9996001V.
    5.) This is a stable equilibrium because: the classical feedback formula for a finite value of Aol also gives the output voltage Vout=Vin*[1E4/(1+0.5E4)]=1.9996001 V.
    6.) That means: We have an equilibrium because the output voltage has a value which exactly meets the condition Vout=Vdiff*Aol.
    Any larger/smaller output voltage causes a small reduction/increase for Vdiff thereby correcting this deviation from the equilibrium.

    7.) In this example, the input difference voltage, of course, is NOT zero. It never will be zero - however, the diff. voltage is so small (in our case app. 0.2mV) that in can be neglected (assumed to be zero for calculations) in many cases.
     
  6. Feb 4, 2016 #5

    meBigGuy

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    There is phase delay and frequency response around the loop that will determine the transient response to a step or impulse input. The iterations you describe are a way to mentally approach an ideal steady state solution but are not how the circuit operates.
     
  7. Feb 4, 2016 #6
    I may be having a lapse, where is the phase delay introduced in this op-amp circuit?
     
  8. Feb 4, 2016 #7

    meBigGuy

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    Sorry, maybe I went off topic a bit. Or at least am guilty of introducing "real components" to an ideal discussion.

    If we assume an all ideal opamp and components with no parasitics, then there is no phase delay. The transient response would be ideal and there would be no phase shift.
    I was just reacting to the word "transient". The "transient" or iterative behavior the OP was referring to is not a real circuit phenomena with ideal or real components. It's just a step-wise mental process.
     
  9. Feb 5, 2016 #8

    LvW

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    Sorry, but I disagree.
    The questioner speakes about "iterations" - and he describes his attempt to explain how the opamp with feedback arrives at a stable equilibrium (and a fixed gain).
    But he has made an error while assuming that the amplifier from the beginning (t=0) would amplify with (1+R1/R2)
    The basic content of my contribution was to show - using a simple example - what really happens:
    I have described how such a circuit behaves after switch-on (t=0) of the suplly voltages.
    How can you say this is "not how the circuit operates"?
    Something wrong? Do you have another explanation how the circuit arrives at the stable working point?
    Perhaps the questioner should decide if my desription was helpful or not.
     
  10. Feb 5, 2016 #9

    jim hardy

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    Usually the feedback is faster than the opamp
    but minor overshoot is not atypical
    from LM358 datasheet
    upload_2016-2-5_3-36-30.png

    i wonder if original poster is writing a simulation program?
    What is his time increment per iteration?
    For most practical purposes he'd be better off to just solve the gain equation.

    system ignoring preview buttons and logging me out again
     
  11. Feb 5, 2016 #10

    LvW

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    You have applied a voltage step at the input - assuming the opamp has already a working point in the linear region (supply voltages on for a long time), correct?
    But - this was NOT my intention.
    I have explained what happens (and I think THIS was the original question) after power switch-on for the circuit.
    This is a different problem.
     
  12. Feb 5, 2016 #11

    jim hardy

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    Ahhh... okay !

    old jim
     
  13. Feb 5, 2016 #12
    You may be confusing supply with Signal, as the Diagram only includes signal inputs and has neglected +/- supply connectors. With this we can presume the supply is already on and only need to account for the signal.

    If we look at the chart shown by Jim Hardy, we can see 1/2 of Vout is going through R1, and 1/4 through R2, leaving 1/4 at -Vin. This -1/4 is added to 4/4 at Vin, which will have Vout at 3/4. The main purpose of a Negative feedback loop is to keep Vin below the saturation point of the Op-Amp and to allow a little more control over Vout or at least reduce the impact of flutter related to the slew rate.

    Edit to add: Even at power on, I wouldn't expect -Vin to be significantly higher than Vin. If it ever exceeded Vin, it shouldn't exceed by more than the slew rate?
     
  14. Feb 5, 2016 #13

    LvW

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    OK - it is possible that I have misunderstood the core of the question because the questioner spoke about "iterations". And such "iterations" cab be observed (assumed) when the opamp starts with non-linear operations and, finally, arrives at an equilibrium in the linear range.
    Nevertheless, perhaps it helps to know how the circuit reaches its equilibrium.
     
    Last edited: Feb 5, 2016
  15. Feb 5, 2016 #14
    Yes, I should have included that in my first response to avoid the iteration debate. :D

    Still, much has been shared in this thread. :-)
     
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