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How does the feedback work in an inverting op amp amplifier?

Discussion in 'Homework Help' started by Heidi, Dec 20, 2014.

  1. Tony Stewart

    Tony Stewart Well-Known Member Most Helpful Member

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    Heidi the assumption is that std Op Amps only have GBW products in the 1~10 Meg range.
    If you need more , we switch to design called a Video Amplifier which has different characteristics with much lower Aol & but high GBW product.
     
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  2. Heidi

    Heidi Member

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    Increasing Aol seems to increase both the differential input |vp-vn| and Vout.(?)

    |vn| = G*sqrt[1+(Aol/GBW)^2 * f^2] / sqrt[(1+G+Aol)^2 + (1+G)^2 * (Aol/GBW)^2 * f^2]

    |Vout| = G*AOL / sqrt[(1+G+Aol)^2 + (1+G)^2 * (Aol/GBW)^2 * f^2], for Vin=1.

    Assume G=5, f=10^6, GBW=10^6, Vin=1, I got the following results:
    upload_2015-3-19_10-47-17.png
     
  3. Heidi

    Heidi Member

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    Hi Tony,

    Thanks for mentioning that.

    Seems that I was just doing some mental or imaginary math activities, not practical, wasn't it?:D Do you have any suggestions?:) Thank you!
     
  4. dave

    Dave New Member

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  5. Tony Stewart

    Tony Stewart Well-Known Member Most Helpful Member

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    Yes

    1) stop calculating for more significant figures than is possible for R ratio controlled parts.

    2) When Aol is 120dB input differential voltage will be effectively zero while output is in linear range with feedback.

    3) Accept Design guidelines until you can correct your math.
    Acl = -Zf/Zin accuracy are dependant on ratio tolerances.

    Factories can laser trim if using COB or hybrids.

    Stability depends on phase or gain margin at Acl=1

    Input offset must consider bias current * impedance differences (+) and (-) + Vin offset for each IC specs.
    CMOS and FET inputs are low Ib
    Rail to Rail OA's are high impedance low current out compared to BJT
    Old BJT OA's may have no performance within 2V of Rails or have good performance just below Vee.

    Did I forget anything basic?

    160 thread responses tell me you still dont understand basics until this thread is closed.
     
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  6. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello again,

    Tony:
    We are looking at a particular theoretical aspect of this particular op amp, so we want accurate numbers when we have to actually use numbers.

    Heidi:
    I see the phase changing at the output and perhaps when the feedback resistor is low the phase difference between Vin and Vout causes the addition of the two to generate a smaller error voltage (vn). That would in turn make a smaller output because the internal gain is fixed. With larger feedback resistor the output will not add to the input as much, and with low outputs for both conditions there will be a difference in vn.
    In other words, if Vout is low and out of phase it creates a lower (or higher) vn but when Vout is low and more in phase (with the input) it has less (or more) of an effect on vn which changes the output.
    This is keeping in mind that the output and input combine to form vn through the same voltage divider action caused by the input and feedback resistors, and with a constant output but different phase we would see more (or less) apparent gain because out of phase voltages add to less than in phase voltages, and with larger feedback resistance the output voltage affects vn less than with smaller feedback resistance.
     
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  7. Heidi

    Heidi Member

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    Hi MrAl,

    It's like what you said, I think that is the most important point to keep in mind. But now we are dealing with AC voltage source which is more complicated than the situation you presented in your post #47. So one way I think to check out if I really understand what you have been trying to show me is to derive a formula of the voltage vn using the input output combination idea, if what it predicts matches the result by using LTspice, then at least in theory, I understand how feedback works in the ideal amplifier we've been talking about.

    Considering the contributions from both input and output, the voltage vn will be
    vn = (R2+R+jR2*Aol*f/GBW)/[R1+R2+R+Aol*R1+j(R1+R2)*Aol*f/GBW]

    In order to show it applies to any circuit parameters, I arbitrarily set
    R1=1, R2=2, R=1, Aol=10, GBW=100, f=1, Vin=1, so
    vn= (3+j0.2)/(14+j0.3)
    or
    vn= 0.2147120863*sin(360*t + 2.5864960022)

    Compared with the simulation, the predicted values are pretty much close to those in LTspice:
    upload_2015-3-20_23-26-8.png
    upload_2015-3-21_0-23-46.png

    I want to especially thank you, MrAl, for your patience and your guide all the way through. I think I've had a better understanding on how the feedback works. Certainly I still have much much more to learn, both in theory and real electronic world. We all do, don't we?:D See you!
     

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  8. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again Heidi,

    Well you are welcome, and on my end i got to review my theory too. It's funny how much you can forget over the years.
    I also realize we may have gotten a little deeper into this then you wanted to right now, but for me, when something doesnt work out to what i expect it to be, i like to dig in a little to find out why. Without looking into it sometimes i cant tell if it was just a mistake or it really works that way. Very often when something comes up that doesnt look right, it's a simple mistake and fixing it solves everything. In this case the difference is real so that's another story altogether. As long as we ask ourselves questions we'll always find out more, and when we can find proofs we can be more sure of the results.
     
  9. JoeJester

    JoeJester Active Member

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    I know the two of you had a very good detailed and analytical discussion. I hink it should be a sticky note ....
     

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