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Analysis for calculating Ideal Opamp current without resistor

  1. Dec 19, 2012 #1
    1. The problem statement, all variables and given/known data
    opamp.png


    2. Relevant equations
    Can anyone help me how to find the current 3.6mA analytically?
    I have tried but I didn't get any right answer.
    Thanks
     
  2. jcsd
  3. Dec 19, 2012 #2
    If I used nodal anaylsis, I could get two equations, but what equation should I use for a node between 4k and 12k resistor?
     
  4. Dec 19, 2012 #3

    gneill

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    Hi Hadi, Welcome to Physics Forums.

    Hint:

    Notice that the junction of R1 and R3 connect to the (-) input of the op-amp. What does that tell you about the currents through R3 and R1?

    Now suppose that the op-amp is pulling some current I from the top node. Where can that current come from? Doesn't it all end up coming from the 12V supply? So if you did happen to know what I is, could you write an expression for the voltage at the (-) terminal?
     
    Last edited: Dec 19, 2012
  5. Dec 19, 2012 #4
    With nodal analysis you are solving for voltages and you already have that voltage (Vo) so a useful node equation cannot be written there. Similarly, you can't write a useful node equation at the top of V1.

    In the case of V1, you could change the combination of V1/R4 into a norton equivalent so that V1 becomes a current source and then a useful node equation could be written. Notice that this process removes one of the nodes -- in particular the one that can't have a useful node equation written.

    In other words, solve a circuit without writing KCL equations at nodes where voltages are specified. op-amps place special conditions on the circuit so use that to reduce the number of unknowns. For example, in this circuit you know the voltage at the - terminal of the op-amp so that unknown is gone.
     
    Last edited: Dec 19, 2012
  6. Dec 19, 2012 #5

    rude man

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    You really have only 2 unknown nodes. The output is one of them.

    The voltage at the - input of the op amp should be obvious.

    olve for the two unknown nodes using KVL.
     
  7. Dec 20, 2012 #6
    Ok, thanks. I got it.
    I only need two nodes, between (R1, R2, and R4) and between (R1, R3 and (-) terminal).
     
  8. Dec 20, 2012 #7

    rude man

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    No, your second unknown node is the output of the amplifier.

    It should be obvious by inspection what the voltage of the second node you cite is!
     
  9. Dec 21, 2012 #8
    He's doing a nodal analysis. You can't write useful nodal equations at a node connected to a voltage source (the top of V1 or the output of the opamp Vo) and that's what he's saying.
     
  10. Dec 21, 2012 #9

    rude man

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    Vo is his other unknown node. It's a node and it's unknown.
    Let v2 = voltage at R1/R2/R4 junction.

    Taking advantage of knowledge of the voltage at the - input,
    G2(v2 - Vo) + Io = G3*Vo where in general Gi = 1/Ri, i = 1,2,3 or 4 and Io = current out of op amp.

    Maybe this is not what you call a "nodal equation" but it sure solves the problem ...
     
  11. Dec 21, 2012 #10

    rude man

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    Oh? And what would it be?
     
  12. Dec 21, 2012 #11
    It doesn't solve the equation, it introduces another unnecessary unknown. By introducing Io you now have to add another equation by writing KCL at the ground node.

    You can look up modified nodal analysis to see what is done about voltage sources. Either circuit transformation is done (norton equivalent to change a voltage source to a current source) or auxiliary equations are added to make up for additional unknowns. But MNA is mainly theoretical or for computer solvers.

    Nodal analysis gets you a matrix equation: I=GV. In the circuit here, writing KCL at the top of V1 (or the output of Vo but the voltage source is hidden there), you'd introduce the current I1 flowing through V1 and that would be equal to the current through R4. What does that do to the matrix equation? It means I has a redundant row and the two rows can be added together and one can be eliminated. After doing that, G is no longer square and you have to introduce another equation. That equation is V1=12V. Substituting that into I=GV, one row of V can be eliminated. So now you have a solvable square matrix after eliminating the current variable through the voltage source V1 and after removing the node equation at the top of V1 (this happened with V1=12V). So in the end, the node equations are written without including the node attached to the voltage source.

    This is why you can't get useful nodal equations in nodal analysis at voltage source nodes. You can choose to write the maximum number of simultaneous equations by introducing Io, I1 and KCL at the ground node, but that's more for computers.


    It would be Vo, the symbol that will appear in the final solution along with Vi -- we're not trying to eliminate either of them.
     
    Last edited: Dec 21, 2012
  13. Dec 21, 2012 #12

    gneill

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    Since it's the current that we're looking for, I don't see any harm in using it as a variable :smile:

    If you assume that current I is being sunk by the op-amp, it can only come from the 12V voltage source. That gives you one series resistor and a parallel section. You know what the voltage has to be where the op-amp connects in the chain, indicated "V" in the diagram:

    attachment.php?attachmentid=54154&stc=1&d=1356122344.gif

    You can use the current divider rule to find the portion of I which passes though the 8k resistor, and taking a KVL walk from the battery to the point marked V has a known potential drop. Solve for I.
     

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  14. Dec 21, 2012 #13

    The Electrician

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    It's not that hard to deal with voltage sources in nodal analysis.

    Label the nodes: V1 is the top of the 12 volt source.
    V2 is the junction of R1, R2 and R4.
    V3 is the - input of the opamp.
    Vo is the output of the opamp.

    The first method would be to perform a source transformation on the 12 volt source and R4. They are replaced with a current source of value 12/2000 in parallel with a 2000 ohm resistor, and we don't deal with the V1 node directly. Instead of a KCL equation at the Vo node, we use a constraint equation. Let the opamp have a finite gain of A. Then the constraint equation is Vo=-A*V3. In the solution take the limit as A -> ∞ to account for an ideal opamp . This solution in matrix form looks like this:

    attachment.php?attachmentid=54156&stc=1&d=1356125387.png

    The second method is to divide the constraint equation through by A, giving a new constraint equation -V3=0. Now we get a solution that doesn't involve A, and we don't have to take a limit to get a final solution for an ideal opamp:

    attachment.php?attachmentid=54157&stc=1&d=1356125387.png

    The third method includes V1 explicitly and adds another constraint equation V1=12. The solution for all 4 nodes now includes the result V1=12, which we already knww:

    attachment.php?attachmentid=54158&stc=1&d=1356125387.png


    And, of course, the current the OP asked for is simply the sum of the currents through R2 and R3, which are easily calculated since we know the voltages across each of them, or it's the current through R4.
     

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    Last edited: Dec 21, 2012
  15. Dec 21, 2012 #14

    rude man

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    Well, we are trying to solve for Io ... "unnecessary unknown"??

    3 equations, 3 unknowns: G4(V1 - v2) = G2(v2-Vo) + G1v2 plus my other two. Solve for v2 (optional), Vo (optional) and Io. Simple. Using determinants you can restrict your effort to just finding Io.
     
    Last edited: Dec 21, 2012
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