Closed loop gain using small signal analysis for Op-amp

Discussion in 'Homework Help' started by u-will-neva-no, Nov 5, 2011.

  1. u-will-neva-no

    Thread Starter Member

    Mar 22, 2011
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    2
    Hi again everyone...I am really bad at this type of analysis. In particular, I have an issue forming equations (using KCL) from the small signal model.

    The question and the small signal model is attached.

    I want to form a KCL so I would have
     I_i = I_o from my drawing. From the solution provided to me, I know that I am already wrong.

    The solution I am trying to get to is:
    \frac{v_o -  A_0_l v_d}{R_o} + \frac{v_0 -v_i}{Ri} = 0

    and  v_d = v_i - v_o

    I may have 'i' wrong with 'I' subscript because it is not clear on the solution. This is the formula would like help with if possible. Thank you!
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    For the current direction as show in your diagram we can write one nodal equation for Vo node.

    \frac{Vin - Vo}{Ri} - \frac{Vo - Aol*Vd}{Ro} = 0

    And if we solve this for Vo/Vin we get this result

    \frac{Vo}{Vin} = \frac{Aol*Ri+Ro}{Ri + Aol*Ri + Ro} =\frac{Aol*Ri+Ro}{(Aol+1)*Ri + Ro}

    And if Aol = ∞ then Vo/Vin = 1
     
    Last edited: Nov 6, 2011
  3. u-will-neva-no

    Thread Starter Member

    Mar 22, 2011
    230
    2
    Hi Jony, am I correct to say that if the current flows in one direction then voltage across it points in the opposite direction so then I would subtract the voltage that the arrow point at. I will attach an image to show what I am on about. What confuses me is what voltage I subtract from it.

    On my diagram I have the arrow head pointing opposite to the current flow so I would have \frac{v_I - v_o}R_i for I_i and \frac{v_o - A_ol*v_d}{R_o} for I_0 I just want to see if my thinking is correct and I am never fully told how to analyse a circuit.

    I generally look at an equation and try to make sense of it but I want to be able to go from a circuit and form the equation... Please let me know how you came to your equation if my thinking is incorrect!
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Firs when we apply the nodal analysis we write the KCL for the node .

    In your diagram you assume that Iin is flow into the node and Iout flow out from the node. So we have
    Iin - Iout = 0 (KCL current entering the node is equal to the current thats leaving the node)

    Iin = (Vin - Vout)/Rin ---> becaues you assume that Iin is flow into the Vo node, so Vin voltage must be greater then Vout.

    And we have a similar situation for Iout current.

    Iout = (Vout - Aol*Vd) / Rout ----> we assume that Vout > Aol*Vd so the

    current can flow out from the node.
    http://www.youtube.com/watch?v=x6qmVkB_bLs&feature=channel_video_title
     
    u-will-neva-no likes this.
  5. u-will-neva-no

    Thread Starter Member

    Mar 22, 2011
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    2
    Thanks Jony130, what you said makes sense!
     
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