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

1. Feb 5, 2007

### VinnyCee

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 $R_i\,=\,\infty$, $R_0\,>\,0$ 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]

$$V_d\,=\,-V_{IN}$$ <-----Right?

$$V_1\,=\,-A\,V_{IN}$$

Now, the current equations)

$$I_1\,=\,\frac{V_1\,-\,V_0}{R_0}\,=\,\frac{-A\,V_{IN}\,-\,V_0}{R_0}$$

$$I_2\,=\,\frac{V_0\,-\,V_2}{R_2}$$

$$I_3\,=\,\frac{V_2}{R_1}$$

KCL at $V_0$)

$$I_1\,=\,I_2\,\,\longrightarrow\,\,\frac{-A\,V_{IN}\,-\,V_0}{R_0}\,=\,\frac{V_0\,-\,V_2}{R_2}$$

Solving for $V_0$)

$$V_0\,=\,\frac{-R_2\,A\,V_{IN}\,+\,R_0\,V_2}{R_0\,+\,R_2}$$

KCL at $V_2$)

$$I_2\,=\,I_3\,\,\longrightarrow\,\,\frac{V_0\,-\,V_2}{R_2}\,=\,\frac{V_2}{R_1}$$

Solving that equation for $V_0$)

$$V_0\,=\,\frac{R_2\,V_2\,+\,R_1\,V_2}{R_1}$$

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

Last edited by a moderator: May 2, 2017
2. Feb 5, 2007

### Mindscrape

I think you only did a partial KCL at node 2. Summing currents into the node you should get:

$$\frac{V_d-V_2}{R_i} + \frac{V_0-V_2}{R_2} + \frac{0-V_2}{R_1} = 0$$

and for the V0 node:

$$\frac{AV_d - V_0}{R_0} + \frac{V_2 - V_0}{R_2} = 0$$

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.