Circuit is:
http://img213.imageshack.us/img213/5882/scan0024f.jpg
and I'm told that all Op Amps are ideal and operating in the linear region

and they want vo...

I'm going to use the v+=v- method..

I'm thinking:
vb/1k - va/1k = i (the current at the first dot heading into the first op-amp) = .0001 A

Then the output voltage of the first op-amp is:
0 - (.0001*10,000) = -1 V

i know the voltage and resistance heading into the second op-amp..
BUT there's no current going into either terminal of any op-amp.

this is as far as i've gotten.

 

I'm thinking:
vb/1k - va/1k = i (the current at the first dot heading into the first op-amp) = .0001 A

Then the output voltage of the first op-amp is:
0 - (.0001*10,000) = -1 V

The voltages vb and va will add and not subtract. The output voltage from the first stage will be -3V.

The second stage is a unity gain buffer and the last stage is an inverting amplifier.

 

Did you mean to say that the currents add and don't subtract? In any event I agree then that the output voltage of the first Op Amp is -3V...

I know that the in/out relation for an inverting amp is given by:
Vout = -VinR2/R1 if that helps..

So at the output of the first Op Amp I have
V = -3V
I = .0003 A

The voltage heading into the positive terminal of the second Op Amp is:
-3V - (.0003*1000) = -3 - .3 = -3.3 V?

This is also the voltage into the v- terminal according to the v+=v- method

So at the output of the second Op Amp I have:
V = -3.3 V
I = 0 A

At the input to the third Op Amp:
I = V/R = -3.3V / 2000 = -0.00165 A

The output of the final (third) Op Amp is then:
0 - (-0.00165 * 10,000) = +16.5V?

 

The voltage heading into the positive terminal of the second Op Amp is:
-3V - (.0003*1000) = -3 - .3 = -3.3 V?

Hmm.. I just noticed this wouldn't make sense because the input current to the Op Amp is supposed to be 0... yet I have a current at the output of the first Op Amp... a bit confusing..

 

Hmm.. I just noticed this wouldn't make sense because the input current to the Op Amp is supposed to be 0... yet I have a current at the output of the first Op Amp... a bit confusing..

Excellent! You are learning fast. It turns out that the current in the first stage will flow into the output terminal of the first OPAMP. Since no current flows into the positive terminal of the second OPAMP, there is no voltage drop on the 1K resistor, and the second stage is just a buffer stage with gain of one.

 

Ahhh I see now...

So:

Since I have a -3V at the + terminal of the 2nd op amp.. that's also at v-

I = V/R = -3/2000 = -0.0015 A flowing UP away from the v- terminal of the 3rd op amp....

Vout is then 0 (since the + terminal of the 3rd op amp is grounded) minus the voltage drop across the resistor and since we know the current going into the resistor

0 - (-0.0015*10,000) = +15 V

I didn't realize, the book had an answer in the back for this one.. it says 1.5... I was fairly careful with my calculations in getting 15V... is the book wrong?

 

ihaveaquestion,

... is the book wrong?

I think the book is wrong and you are right.

Ratch

 

Thanks Ratch, anyone else have an opinion on it?

 

Thanks Ratch, anyone else have an opinion on it?

I agree with Ratch.