Finding V_o and i_o for an Ideal Op Amp Circuit with Multiple Unknowns

[Rozenwyn]

Find V_o and i_o. Assume an ideal op amp.

http://img236.imageshack.us/img236/2426/633bb0.jpg

I tried applying KCL at v1 and v2 nodes but then got stuck because I had 2 equations with 3 unknowns.

\frac{12-v_1}{4} = \frac{v_1-v_2}{8} and

\frac{v_1-v_2}{8}+\frac{-10-v_2}{20}= i_o

Where i_o is the output current of the op amp and not the current flowing thru the 8 Ohm resistor ? I missed a few lectures on Op Amps and now I am trying to catch up from the book but I am having a hard time with it.

 

[chroot]

Remember that an op-amp actively drives its output so as to make its inputs equal. This results in what is known as the virtual short principle, which essentially just says that the two inputs of an op-amp are always at the same potential. The non-inverting terminal is obviously at -2V, so v1 is also at -2V.

- Warren

 

[Rozenwyn]

Ok I took v_2 = -2 and did the calculations but I still don't get the right answers.
The answers given in the textbook are v_o = -30V and i_o=3.5mA

Here are my calculations:

\frac{12-^-2}{4}= \frac{-2-v_2}{8} \ \longrightarrow 28 = -2-v_2 \ \longrightarrow v_2 = -30V

\frac{-2-^-30}{8} + \frac{-10-^-30}{8} = i_o \ \longrightarrow i_o = \frac{28+8}{8} = 4 mA

From these v_o = v_2 - v_1 = -30 -^-10= -20V.

These are wrong. However if we take the i_o to be the current between nodes v2 and v1 then; \frac{-2-^-30}{8} = \frac{28}{8} = 3.5mA. This is rigth but v_o is still wrong?

 

[doodle]

Why is v2 = -2V? Why is the voltage of the node at the bottom-right -10V? And how is v0 = v2-v1?

 

[teknodude]

Look at where V0 is. In this case V0 = V2

I'm a bit confused as to why there is a -10V at the bottom since there's a ground there.

Also you really don't want to use the node at V2, cause you also need to account for the current coming out of the op amp. I believe there is one coming out, but someone correct me on that. Been awhile since i did some circuits.

 

[chroot]

Sorry, I guess the node labelled "ground" is actually at -10V, therefore the voltage at the noninverting input of the op-amp is -12V, and thus v2 = -12V too.

The ground symbol should really not be used to mean anything other than 0V though, so I consider this to be an abuse of notation.

- Warren

 

[doodle]

You can't have ground at -10V and then have the node at the top-left be +12V. I think the poster have to reexamine the labels she/he attached to the circuit.

Sorry, I guess the node labelled "ground" is actually at -10V, therefore the voltage at the noninverting input of the op-amp is -12V, and thus v2 = -12V too.

I think you meant v1 = -12V and not v2 = -12V.

 

[chroot]

You can't have ground at -10V and then have the node at the top-left be +12V. I think the poster have to reexamine the labels she/he attached to the circuit.

Indeed, I didn't notice that. The diagram doesn't make sense.

I think you meant v1 = -12V and not v2 = -12V.

Sorry, yes, that's what I meant.

- Warren

 

[Rozenwyn]

Oh it's my bad. I disregarded ground for a minute and thought the Voltage Sources would create a potential of -10V over there. ( -12 + 2 V ). Meh.
Then as technodude says, v0 = v2 and the problem is solved. Thanks all. Sorry to confuse you with a bad diagram.