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schematic

simulate this circuit – Schematic created using CircuitLab

Here I am trying to find or derive expressions for both Vo1 and Vo2 in terms of Vi1 and Vi2. This is how I approach this. All resistors are the same. So I see that I can do KVL following I1 since there's no currents flowing into op amps. So -Vi1 + I1xR + I1xR + Vo2 = 0 and this yields: Vo2 = Vi1 - I1x(2R)

Vo1: Because of virtual ground I concluded that Vi2 is seen mid-point between the two Rs at the output of op-amps. Consequently Vo1 = Vi2 - (I/2)xR.

Are my expressions for both Vo1 and Vo2 correct?

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  • \$\begingroup\$ Your expressions are not purely in terms of the input voltages. Hint : what is the upper opamp In+ voltage? \$\endgroup\$ Sep 20 '16 at 17:01
  • \$\begingroup\$ Where's ground? \$\endgroup\$
    – Chu
    Sep 20 '16 at 17:04
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    \$\begingroup\$ @Brian. I think it's Vi2 @ opamp In+ \$\endgroup\$
    – Patrick
    Sep 20 '16 at 17:06
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    \$\begingroup\$ So, follow that through to its conclusion. Can you eliminate the currents from those expressions? \$\endgroup\$ Sep 20 '16 at 17:10
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    \$\begingroup\$ @Brian. I came up with Vo2 = 2Vi2 - Vi1. I replaced I1 by (Vi1 - Vi2)/R in my expression. \$\endgroup\$
    – Patrick
    Sep 20 '16 at 17:20
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YOu should be able to figure of Vo2, given only what you know about op-amps, feedback, and the currents. Think about what the voltage must be on the positive input of the top op amp. Once you have that, you have I1. Once you have I1, you have Vo2.

That should give you your current through both feedback resistors, which will lead you to Vo1

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So, as @neonzeon mentioned, there is a problem with Vi2 and Vo1 and Vo2. Namely, that Vi2 needs to be halfway between those two voltages.

\$ V_{i2} = \frac{V_{o1} + V_{o2}}{2} \$

However, if we assume that all inputs to all op amps are the same (the op amps will attempt to make sure that this is true by changing the outputs), we can see if this is a problem. This sets the voltage at the \$+\$ input of the upper op amp, which allows us to find the current:

\$ IR = V_{i1} - V_{i2} \$

This allows us to find the voltage at \$V_{o2}\$

\$ V_{i2} - V_{o2} = IR \$

And combined with the first equation, this gets us \$V_{o1}\$ and \$V_{o2}\$:

\$ 2V_{i2} = V_{i1} + V_{o2} \$ \$ 2V_{i2} = V_{o1} + V_{o2} \$

Which means \$V_{o1} = V_{i1}\$, and \$V_{o2} = 2V_{i2} - V_{i1}\$

As it turns out, the restriction at the beginning can be maintained along with the rest of the op amp rules.

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  • \$\begingroup\$ We try to encourage solution of homework problems, as opposed to solving them \$\endgroup\$ Sep 21 '16 at 0:21
  • \$\begingroup\$ @ScottSeidman: 1) Nothing that is in here is new. It is in the comments and other answers. I just put it together and made it clearer for other people who come here. 2) While I understand the sentiment, the problem was interesting to me, and I'm not his teacher. If he gets problems done for him the only person he is hurting is himself. I happened to enjoy working this out (which is the whole reason I'm on this site), and I'm not going to refrain from answering out of some misguided sense of protecting the OP from himself. \$\endgroup\$ Sep 21 '16 at 1:08
  • \$\begingroup\$ It's more out of a sense of discouraging the perception that students that don't want to work can find immediate help here than pedagogical concerns abut an individual student, as we could be overwhelmed by repetitive questions at too basic a level-- so you don't quite see the point. We're protecting the stack and not the OP, as has been taken up repeatedly in meta. I think it's particularly important at the beginning g of semesters, like now. \$\endgroup\$ Sep 21 '16 at 10:20
  • \$\begingroup\$ @ScottSeidman: Well, the point is fairly moot. As I mentioned, I didn't actually give him the answer, I just consolidated work that is elsewhere and made it pretty. \$\endgroup\$ Sep 22 '16 at 17:03
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EDIT: the node with the op-amp negative inputs is at Vi2. Also, since the op-amp inputs effectively draw no current, that node is not only at Vi2, but also exactly halfway between Vo1 and Vo2.

EDIT: as several have pointed out, the feedback action will force "halfway between Vo1 and Vo2" and also the positive input of OA1 to Vi2. Peer review is a wonderful thing. Circuit

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  • \$\begingroup\$ So you're saying that I can't derive an expression of Vo in terms of Vi1 or Vi2. I assumed that because of virtual ground. V+ = V- in an ideal op amp. Please enlighten me. \$\endgroup\$
    – Patrick
    Sep 20 '16 at 17:40
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    \$\begingroup\$ Of course all the negative input terminals are at Vi2. \$\endgroup\$ Sep 20 '16 at 17:46
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    \$\begingroup\$ ... as are the positive input terminals. \$\endgroup\$ Sep 20 '16 at 17:47
  • \$\begingroup\$ You are assuming that the two statements you have made are mutually exclusive... but they aren't. That node is at Vi2, which is also halfway between Vo1 and Vo2. \$\endgroup\$ Sep 20 '16 at 19:04
  • \$\begingroup\$ @AndrewSpott and Scott: Ah! Of course both of you are right - peer review is a wonderful thing. \$\endgroup\$
    – neonzeon
    Sep 22 '16 at 15:08

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