Op Amp Negative Feedback Algebra

Discussion in 'Math' started by KCHARROIS, Dec 26, 2013.

  1. KCHARROIS

    Thread Starter Member

    Jun 29, 2012
    292
    1
    Hello,

    I've been reading Op-Amps and Linear Integrated Circuits by Ramakant and I'm stuck yet again with an algebraic equation to solve the negative feed back gain with non inverting input.

    So far I understand I got as far as to this formula below but can't solve for Vo and I don't know where to start to solve for Vo.

    Vo = A (Vin - R1*Vo/R1 + Rf)

    Thanks
     
  2. Shiv123

    New Member

    Dec 26, 2013
    1
    1
    I'm guessing the formula should be Vo = A (Vin - R1*Vo/R2 + Rf)?

    In this case take A to the left by division:
    Vo/A = Vin - R1*Vo/R2 + Rf

    then take R1*Vo/R2 to left by addition:
    Vo/A + R1*Vo/R2 = Vin + Rf

    Now, factor out Vo:
    Vo(1/A + R1/R2) = Vin + Rf

    the take (1/A + R1/R2) to the right by division:
    Vo = (Vin + Rf)/(1/A + R1/R2)

    there you have it....i hope this is what you wanted. :)
     
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  3. KCHARROIS

    Thread Starter Member

    Jun 29, 2012
    292
    1
    Genius!

    Thanks
     
  4. KCHARROIS

    Thread Starter Member

    Jun 29, 2012
    292
    1
    Never mind I'm not sure where you get the R2 from but its not suppose to be in there.

    Sorry... Thanks though
     
  5. WBahn

    Moderator

    Mar 31, 2012
    17,743
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    There are hundreds (probably thousands) of op amp circuits. The equation that applies depends on the circuit. Don't make us guess -- post the circuit.

    There's no need, however, to see the circuit to know that the equation you have above is wrong. Look at the units!

    (Vin - R1*Vo/R1 + Rf)


    This is (volts) - (volts) + (ohms)

    Can't be done.

    Without the circuit, we can't begin to even guess where you've gone wrong because we are not mind readers.

    Well, actually that's not competely true. I think I can guess what you meant the equation to say, but engineering is not about guessing -- that gets people killed. Take the time to express your equations properly and we can go from there.
     
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  6. KCHARROIS

    Thread Starter Member

    Jun 29, 2012
    292
    1
    Ok the circuit is like the one attached...

    So the book goes like this:

    Closed loop voltage gain: Af = Vo/Vin

    however: Vo = A(V+ - V-)

    But: V+ = Vin and V- = Rg*Vo / Rg+Rf

    Therefor putting these formulas together you get: Vo = A(Vin - Rg*Vo/Rg + Rf)

    And then solving for Vo you get: Vo = A(Rg + Rf)vin/ Rg + Rf + ARg

    I can't personaly solve how they got to the last equation algebraically with

    Vo = A(Vin - Rg*Vo/Rg + Rf)???

    Thanks
     
  7. #12

    Expert

    Nov 30, 2010
    16,284
    6,797
    Vo = Vin [1+(Rf/Rg)]

    I can't solve their equation either. I think in terms of voltage matching at the inputs, and this works.
     
  8. WBahn

    Moderator

    Mar 31, 2012
    17,743
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    A huge part of your problem is that you are being sloppy, sloppy, sloppy with your math.

    V- = Rg*Vo / Rg+Rf

    By order of operations -- remember multiplication and division before addition and subtraction? -- this is

    V- = ((Rg*Vo) / Rg) + Rf

    Which is not what you want. You want

    V- = Rg*Vo / (Rg + Rf)

    You need to start writing your expressions correctly, otherwise you will forever have serious problems with the algebra.

    Because you are being so sloppy with the math and because you are skipping too many steps at once. Slow down and take it stop by step.

    Vo = A(V+ - V-)

    Vo = A(Vin - (Rg*Vo / (Rg+Rf)))

    Vo = A*Vin - A*(Rg*Vo / (Rg+Rf))

    Vo + A*(Rg*Vo / (Rg+Rf)) = A*Vin

    Vo + Vo *A*(Rg/(Rg+Rf)) = A*Vin

    Vo (1 + A*(Rg/(Rg+Rf)) = A*Vin

    Af = Vo/Vin = A/(1 + A*(Rg/(Rg+Rf))

    Af = A*(Rg+Rf)/((Rg+Rf) + A*Rg)

    Af = (Rg+Rf)/((Rg+Rf)/A + Rg)

    if A >> (Rg+Rf)

    Af ≈ (Rg+Rf)/Rg

    Af ≈ 1 + (Rf/Rg)
     
    LvW likes this.
  9. KCHARROIS

    Thread Starter Member

    Jun 29, 2012
    292
    1
    All's I can say is thanks and that yes your right my math is sloppy it does need a lot of work.
     
  10. WBahn

    Moderator

    Mar 31, 2012
    17,743
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    As long as you recognize it and take steps to improve it, you will be fine.

    A key part is matching the pace at which you crank through the math to your ability to do so accurately. That's something everyone has to always be on the guard for and I frequently have to stop myself and force myself to slow down and do it carefully and correctly. It's just that, through LOTS of mistakes, I have learned to usually -- usually -- catch when I am running ahead of myself so that I CAN slow down.
     
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