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Homework Help: CIRCUIT ANALYSIS: 8 Resistors, 1 IVS, 1 Differential Op-Amp - Find Vo

  1. Feb 5, 2007 #1
    1. The problem statement, all variables and given/known data

    The circuit below is a differential amplifier driven by a bridge, find [itex]V_0[/itex].

    http://img476.imageshack.us/img476/8827/chapter5problem48fi2.jpg [Broken]


    2. Relevant equations

    KCL, v = i R, Ideal Op-Amp relationships


    3. The attempt at a solution

    I added 4 voltage markers.

    http://img259.imageshack.us/img259/8314/chapter5problem48part2fs0.jpg [Broken]

    By the Ideal Op-Amp relationships, we know that there is 0 current at both input terminals of the Op-Amp.

    KCL at [itex]V_2[/itex])

    [tex]\frac{0.005\,-\,V_2}{40000}\,=\,\frac{V_2}{60000}\,+\,\frac{V_2\,-\,V_4}{20000}[/tex]

    [tex]18\,V_2\,-\,12\,V_4\,=\,0.003[/tex] < ------ Equation 1

    KCL at [itex]V_4[/itex])

    [tex]\frac{V_2\,-\,V_4}{20000}\,=\,\frac{V_4}{80000}[/tex]

    [tex]4\,V_2\,-\,5\,V_4\,=\,0[/tex] <----- Equation 2

    Now, using equations 1 and 2, I get [itex]V_2\,=\,0.003571\,V[/itex] and [itex]V_4\,=\,0.002857\,V[/itex]. Does that seem right?

    By the relationships of an ideal Op-Amp, [itex]V_3\,=\,V_4[/itex], so [itex]V_3\,=\,0.002857\,V[/itex].

    KCL at [itex]V_1[/itex])

    [tex]\frac{0.005\,-\,V_1}{10000}\,=\,\frac{V_1}{30000}\,+\,\frac{V_1\,-\,V_3}{20000}[/tex]

    [tex]11\,V_1\,-\,3\,V_3\,=\,0.03[/tex] <----- Equation 3

    KCL at [itex]V_3[/itex])

    [tex]\frac{V-1\,-\,V_3}{20000}\,=\,\frac{V_3\,-\,V_0}{80000}[/tex]

    [tex]4\,V_1\,-\,3\,V_3\,+\,V_0\,=\,0[/tex] <----- Equation 4

    Using equations 3 and 4, I get [itex]V_0\,=\,0.004363\,V[/itex].

    [tex]V_0\,=\,4.363\,mV[/tex] <------ Is that right?
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Feb 5, 2007 #2
    Bump - Bump:)
     
  4. Feb 5, 2007 #3
    Looks good. Quick question though, why don't you use linear algebra instead of messy substitutions?
     
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