Question on Opamp homework problem: Find Iout between these two Opamps

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james weaver
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Homework Statement
find Io
Relevant Equations
ohms law
Hi, so my objective is to find Io, and I've attached a picture of my work. Here is my question: At the point Vo1, if I use the equation for finding the output voltage for a non-inverting op amp I end up getting this:

$$V_{0{1}}=\left( 1+\frac {r_{f}} {r_{1}} \right)V_{s1} = \left( 1+\frac {40k} {10k} \right)2 = 10v$$

However, I can calculate the current ##I_{3}## by using ohms law:

$$\frac{V_{s1}-V_{A}} {10k}=\frac {2v} {10k}=0.2mA$$

From there, I start at Vs1 and add the voltage drops until I reach Vo1, which gives me: ##2-50k*I_{3}=-8v##

My instinct is to go with the -8, since ##I_{3}## would have to be reversed if it were 10v. Can you please help explain to me which one is right and why? Thanks.

circ.png
 
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First, I think the polarity of the top amp is backwards.

james weaver said:
I start at Vs1 and add the voltage drops until I reach Vo1, which gives me: 2−50k∗I3=−8v
Nope, the voltage at A is zero (virtual ground assumption), So V01 = 0 - 50KΩ⋅I3.
Alternatively, if you want to start from VS1, you have to count both the 2V from VS1 minus the 2V drop across R1, which cancel each other. So, V01 = VS1 - R1⋅I3 - RF⋅I3.
 
DaveE said:
Nope, the voltage at A is zero (virtual ground assumption),
[edit] Or if you assume the circuit is correctly drawn: [/edit]
Since point "A" is fed with a positive battery voltage AND the positive output from its non-inverting configuration, where is the negative source to put it at zero volts?
 
Tom.G said:
the positive output from its non-inverting configuration
Which is an unlikely HW problem without a good solution since we don't have the PS voltages. So...

DaveE said:
First, I think the polarity of the top amp is backwards.
However, you can solve it the other way if you like.
 
However, the equation he wrote is for a positive gain amp. So it's not just the polarity that's messed up here.
 
To make it clear: The upper opamp has positive feedback and will not work as desired. It is in full saturation.
The other opamp (at the bottom) is wired as a positive gain amplifier (for Vs1) with - of course ! - negative feedback. For Vs2 it will work as an inverting amplifer.
Together, it forms an amplifier which amplifies the difference of both signals (with a different gain factor).