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Am I doing this ideal op-amp problem correctly?

  1. Apr 20, 2014 #1
    1. The problem statement, all variables and given/known data
    BTCR7kT.png

    2. Relevant equations
    [tex]Vp=Vn[/tex]
    [tex]Ip=In=0[/tex]

    3. The attempt at a solution
    By voltage division:
    [tex]Vp = 8/(8+8) V3[/tex]
    [tex]Vp = 1/2V3[/tex]
    [tex]Vn = 1/2V3[/tex]

    By node-voltage method at the inverting input terminal:
    [tex](Vn-V1)/4 + (Vn-V2)/4 + (Vn- V0)/8 = 0[/tex]
    Multiply equation by 8
    [tex]2(Vn-V1) + 2(Vn-V2) + Vn-V0 = 0[/tex]
    [tex]V0 = Vn+2Vn-2V1+2Vn-2V2[/tex]
    [tex]V0 = 5Vn-2V1-2V2[/tex]
    Substitute in 1/2V3 for Vn

    [tex]V0 = 5/2V3-2V1-2V2[/tex]

    Is this the correct answer? I'm pretty sure about the V1 and V2 part, because it matches the summing amplifier circuit equation of -(Rf/R1*V1+Rf/R2*V2). However, for V3, I don't see how its resistor composition would yield an answer of 5/2. Could someone clarify?

    Thanks
     
    Last edited: Apr 20, 2014
  2. jcsd
  3. Apr 20, 2014 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    One easy way to work problems with multiple inputs is to take one input at a time, get the corresponding output, then add all three outputs. So for your worrisome V3 input the output is

    Vo3 = V+(1 + Rf/Ri) = 0.5V3(1 + Rf/Ri) = 0.5V3(1 + 8/2) = 2.5V3

    Also,
    Vo1 = -8/4 v1
    Vo2 = -8/4 v2

    Then by superposition,
    Vo = Vo1 + Vo2 + Vo3.

    Note that when you do that the other voltage sources are set to zero volts (grounded), not left 'floating'.
     
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