Find Thevenin Equiv. at a-b in Circuit Below

[VinnyCee]

Homework Statement

Find the Thevenin equivalent at terminals a-b in the circuit below.

http://img361.imageshack.us/img361/2250/chapter5problem36gq0.jpg

Homework Equations

KVL, KCL, v = i R, current and voltage equations for Operation Amplifier

The Attempt at a Solution

So I altered the diagram a bit.

http://img258.imageshack.us/img258/4382/chapter5problem36part2pr4.jpg

I_1\,=\,\frac{V_S}{R_1}

I_2\,=\,\frac{V_S\,-\,V_1}{R_2}

KCL at V_S)

I_1\,=\,-I_2\,\,\longrightarrow\,\,\frac{V_S}{R_1}\,=\,-\frac{V_S\,-\,V_1}{R_2}

V_1\,=\,\frac{R_2\,V_S}{R_1}\,+\,V_S

V_{Th}\,=\,V_1\,=\,\frac{R_2\,V_S}{R_1}\,+\,V_S

Is that right? If so, I will now put a test current of 1 Amp at the terminals to get the R_{Th}.

I_2\,=\,1

\frac{V_S\,-\,V_1}{R_2}\,=\,1\,\,\longrightarrow\,\,V_1\,=\,V_S\,-\,R_2

Now, using v = i R to get the Thevenin equivalent resistance.

R_{Th}\,=\,\frac{V_1}{1}\,=\,V_S\,-\,R_2

Does that seem correct?

 

[Mindscrape]

Seems fine to me.

 

[bdby99999]

do you have to put in a test current and does it have 2 be 1??

 

[blahblahblahh]

Hi, I know this is old but I have the exact same problem. I've done it the same as the guy in the first post up until finding V_th, but after that I'm confused.

When he applied a test current, shouldn't he have removed all of the independent sources? (Namely V_s).

This would reduce the node voltage equation down to

V_p = 0 = V_n

(V_n - V_th)/R_2 = I_(test)

(0 - V_th)/R_2 = I_(test)

-V_th/R_2 = I_(test)

R_th = V_th/I_(test) = -R_2

Which is invalid since it's a negative resistance, therefore you have to assume that R_th = 0?

Can somebody explain if this thinking is valid? Since in the thread starter's solution for some reason he doesn't take out the independent voltage source when applying the test current, can someone explain why he doesn't?

Or can someone in general just explain the last step for finding R_th since I understand the finding V_th its just that last part that I don't understand what the original poster did exactly or why he didn't take out the independent source.

PS: For some reason my sub notation isn't working completely right, I_(test), V_th, R_th are the test current, thevenin voltage, and thevenin resistance respectively.

 

[mikelurch]

I just ran across this problem I think Rth = 0 and Vth = Voc

I have a similar problem with a inverting op-amp.

The definition of an ideal op-amp is that the input has an infinite resistance and the output has an impedance of 0. The output of an ideal op-amp will always have the same voltage, it will not vary with the load.

Does anyone agree or disagree with my logic?