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Op Amp Problem

  1. Feb 6, 2016 #1
    1. The problem statement, all variables and given/known data

    Hello, I have attached the problem below. I am trying to find the value for v1 - v2 and iA. The given values are V, RA, and R for the other resistors. There are no numbers.

    2. Relevant equations

    It is given that the op amp is ideal, therefore v- = v+, and i- = i+ = 0 for both op amps.

    3. The attempt at a solution

    I attempted the solution by using node analysis. iA = V/(3R + RA) (assuming that none of the current flows into the output of the op amp). The v1 value at the output of the first op amp should be v- of the second op amp. This becomes v+, and we can write in v1 at the point below v2. Therefore, the current in that case is v2-v1/R. We equate these two values for the current and get that v1-v2= -V*R/(3R + RA). However, my professor stated that this solution is incorrect. How do I proceed?
     

    Attached Files:

  2. jcsd
  3. Feb 6, 2016 #2

    phyzguy

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    It's correct to assume that no current flows into the inputs of the op amp, but not correct to assume no current flows into (or out of) the op amp output. Try again without making that assumption.
     
  4. Feb 6, 2016 #3

    gneill

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    Staff: Mentor

    You'll need to show your work in more detail. Saying you got an expression that is incorrect doesn't tell us why it is incorrect, nor can we point out errors in your method or steps.

    That said, I'm suspicious about the voltage expressions that you've penciled in on the inputs to the top op-amp in your figure. How did they come about?
     
  5. Feb 6, 2016 #4
    Sorry, I've attached the problem with work. The blue pen is what is given and the pencil is my own work. That value for v- that I had written was wrong. The op amp is ideal so we assume that v-=v+.

    I will be trying the problem without making the assumption that no current flows into the op amp. However, I am not sure how I could find how much current is going into the output of the second op amp.
     

    Attached Files:

  6. Feb 7, 2016 #5

    LvW

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    Shivy, are you sure that your input signal has no ground reference?
     
  7. Feb 7, 2016 #6
    Input signal has no ground reference? I thought that I connected to ground right after RA.
     
  8. Feb 7, 2016 #7

    LvW

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    Yes - I know.
    However, my question was if the source V1 is really a floating source (and the only ground reference is through the whole resistor chain.)
    OK - for my opinion, the best way for solving the task is to use the superposition theorem for the three voltage sources within the circuit (V1 and the two opamp outputs). This is a straight way for finding the solution.
    (By the way, the final solution is a very simple expression)
     
  9. Feb 7, 2016 #8
    I believe it is V with reference to the ground. We haven't talked about the difference between floating and ground sources, so I'm not too sure. I'm fairly certain that the only ground reference is at the bottom of the chain. I also assumed that the output voltages from the Op Amp is 0 since we are assuming an ideal op amp where v- = v+.
     
  10. Feb 7, 2016 #9

    LvW

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    Shivy - OK, I got it.
    V1 is a source without any internal reference to ground (like a battery) -and the only reference is realized by the connected circuit.
    However, of course you must NOT assume that the opamps output voltage is zero.
    There is always a very small differential input voltage to the opamps - however, it is so small that, normally, this tiny voltage (µV range) is neglected.
    That means, we are treating the opamp as ideal and calculate with (V- = v+). However, the output voltage of the opamps always is finite.
    And the task is to find the difference (Vout1-Vout2).
    Again my recommendation: Use the superposition principle.
     
  11. Feb 7, 2016 #10
    I am attempting to use superposition, and have hit some road blocks. I don't believe my circuit diagrams are correct. I have assumed that the op amps can be replaced by a short. Perhaps this is wrong and I need a VCVS.

    I am beginning to rethink a major in electrical engineering...
     

    Attached Files:

  12. Feb 7, 2016 #11

    LvW

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    No - opamps cannot be replaced by a short.
    The procedure is as follows (you have three voltage sources in your circuit):
    1.) Use superposition for computing all three parts which form - for example - the voltage Vin- at the inv. terminal for opamp1.
    2.) Then, do the same for the non-inv. terminal Vin+.
    3.) As a final step, set Vin-=Vin+.
    4.) You have two unknown voltge sources (opamp outputs) in the equations. And you have two equations for solving the system.
     
    Last edited: Feb 7, 2016
  13. Feb 8, 2016 #12

    LvW

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    Shivy G - for my opinion, your circuit has very interesting properties and I wonder where you have found it?
    So - would you please give me some information about the source of this circuit? Any book ?
     
  14. Feb 8, 2016 #13
    My friend gave the problem to me and told me it's for an online course he's taking. It's not out of a book unfortunately.

    I have tried to solve the problem but am still unable to get it, and I doubt I can do it at this point. Would you mind sharing a solution?
     
  15. Feb 8, 2016 #14

    LvW

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    Shivy - let`s start:
    Following the superposition theorem, we assume - as the first of three steps - the output voltages of both opams to be zero (ground).
    Now we have V1 and four equal resistors conncted to V1. It is a simple task to use the voltage divider rule for computing the voltages (against ground) for the two nodes which are connected to inverting and the non-inv. input terminal for opamp1. It is important to have the correct sign for the voltages.
    What is your result?
    (As a 2nd step we can compute the same voltages when only one of the opamp output voltages is active).
     
  16. Feb 8, 2016 #15
    Where are the four equal resistors connected to V1? We don't replace the OpAmp with a short, but do we replace it with nothing at the point?

    If possible, would you like to talk over Google Hangouts? I think we can get the problem done more quickly in that way, and it'd be easier to talk to each other.
     
  17. Feb 8, 2016 #16

    LvW

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    You cannot replace "the opamp" - look what I wrote: The OUTPUT of both opamps is set to zero.
    I am not familiar with "hangouts".
     
  18. Feb 8, 2016 #17
    Google Hangouts is an extension of google + to talk over voice and video. I think it'd be easier to communicate, since I am utterly clueless with this problem. I will mark this as solved for now, since I think I need to review quite a bit of material before tackling this problem again. Thank you for your help. If you want to talk over hangouts, private message me and we can set up a time.
     
  19. Feb 8, 2016 #18

    phyzguy

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    ShivyG, this problem just isn't that hard. Write the current out of the top op amp as i1, and the voltage at the output of the top op amp as v1, the voltage at the output of the bottom op amp is v2, and the current at that op amp is i2, just as you did in post #4. Now, starting with the top resistor and working downwards, tell me how much current is flowing through each resistor. Remember no current flows into the op amp inputs. Answer those for me and we'll take it from there.
     
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