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CIRCUIT ANALYSIS: Non-Ideal OpAmp with 2 resistors - Find Thevenin Equivalent

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

    There is a non-inverting op-amp below.

    http://img291.imageshack.us/img291/8242/chapter5lastproblemea1.jpg [Broken]

    The op-amp is NOT ideal. We assume that [itex]R_i\,=\,\infty[/itex], [itex]R_0\,>\,0[/itex] and A is finite.

    Find the general Thevenin equivalent circuit seen at the terminals.

    2. Relevant equations

    KCL, v = i R

    3. The attempt at a solution

    I changed the diagram to use a model given for a non-ideal op-amp.

    http://img144.imageshack.us/img144/6144/chapter5lastproblempartyx5.jpg [Broken]

    Now I add some voltage and current markers.

    http://img166.imageshack.us/img166/5948/chapter5lastproblempartsk9.jpg [Broken]

    [tex]V_d\,=\,-V_{IN}[/tex] <-----Right?


    Now, the current equations)




    KCL at [itex]V_0[/itex])


    Solving for [itex]V_0[/itex])


    KCL at [itex]V_2[/itex])


    Solving that equation for [itex]V_0[/itex])


    But which do I use? Are they both right? ONe wrong? Or all wrong?
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Feb 5, 2007 #2
    I think you only did a partial KCL at node 2. Summing currents into the node you should get:

    [tex]\frac{V_d-V_2}{R_i} + \frac{V_0-V_2}{R_2} + \frac{0-V_2}{R_1} = 0 [/tex]

    and for the V0 node:

    [tex]\frac{AV_d - V_0}{R_0} + \frac{V_2 - V_0}{R_2} = 0[/tex]

    Now you can solve for V_0 and use the V_in=V_d constraint, though you seems to think is otherwise. You might have a reason that I don't see, but I think V_d = V_in, and not the negative.
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