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

  • Thread starter VinnyCee
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Homework Statement



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.


Homework Equations



KCL, v = i R


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?

[tex]V_1\,=\,-A\,V_{IN}[/tex]

Now, the current equations)

[tex]I_1\,=\,\frac{V_1\,-\,V_0}{R_0}\,=\,\frac{-A\,V_{IN}\,-\,V_0}{R_0}[/tex]

[tex]I_2\,=\,\frac{V_0\,-\,V_2}{R_2}[/tex]

[tex]I_3\,=\,\frac{V_2}{R_1}[/tex]

KCL at [itex]V_0[/itex])

[tex]I_1\,=\,I_2\,\,\longrightarrow\,\,\frac{-A\,V_{IN}\,-\,V_0}{R_0}\,=\,\frac{V_0\,-\,V_2}{R_2}[/tex]

Solving for [itex]V_0[/itex])

[tex]V_0\,=\,\frac{-R_2\,A\,V_{IN}\,+\,R_0\,V_2}{R_0\,+\,R_2}[/tex]

KCL at [itex]V_2[/itex])

[tex]I_2\,=\,I_3\,\,\longrightarrow\,\,\frac{V_0\,-\,V_2}{R_2}\,=\,\frac{V_2}{R_1}[/tex]

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

[tex]V_0\,=\,\frac{R_2\,V_2\,+\,R_1\,V_2}{R_1}[/tex]

But which do I use? Are they both right? ONe wrong? Or all wrong?
 
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Answers and Replies

  • #2
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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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