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A question on ideal op-amp (finding a current).

  1. Jul 13, 2012 #1
    1. The problem statement, all variables and given/known data

    I had an exam question on ideal op-amp asking to find a number of unknowns. I found them all except that I'm skeptic of my attempt in finding the output current which is symbolized on the circuit (io), so I have come to ask to throw my skepticism away. See the circuit below in the link including my attempt of solution:

    http://tinypic.com/view.php?pic=2evrzap&s=6


    2. Relevant equations


    3. The attempt at a solution

    As shown on the image, my attempt was: ((va-vo)/R6)) - (vo/R7)= i0
     
    Last edited: Jul 13, 2012
  2. jcsd
  3. Jul 13, 2012 #2

    rude man

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    The equation you wrote in your handwriting is correct, but you need two more equations. You have 3 unknowns: io, va and vo.

    Hint: two of the resistors can be ignored completely.
     
  4. Jul 13, 2012 #3
    I found these three unknowns, then I substituted in the 'io' equation and it resulted to zero amps. I don't remember the values that were given in the exam for the resistors and the DC voltage source, but is it correct that 'io' may equal zero in this question specifically?
     
  5. Jul 13, 2012 #4

    rude man

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    No. io has to flow into the op amp (i0 > 0) for a positive voltage input, and vice-versa for a negative input. Only for Vin = 0 is io = 0.

    Assume a positive input: then Va and therefore Vb must be < 0 (why?). But Vb can only be < 0 if the op amp draws current into itself (io > 0).

    The value of the dc source does not alter the above statement (unless of course it was zero).
     
    Last edited: Jul 13, 2012
  6. Jul 13, 2012 #5
    That's what actually resulted for 'io' when I substituted in the equation, which is zero amps. I've become worried that I may have some error made in finding one of the voltages, probably the output voltage 'vo'. However, yes, both Va and Vo are negative voltages because it's an inverting op-amp, and that's what actually resulted for me in the exam. I probably forgot a minus sign somewhere when I substituted the resulted values into the equation.
     
    Last edited: Jul 13, 2012
  7. Jul 25, 2012 #6
    I got the exam paper back and I found the error that I made. When I found the output voltage (Vo), I wrote the nodal equation at the output node while I don't have the value of the output current (Io), so it was missing in the equation, and for this reason, I got the value of Vo wrong. I should have done the nodal equation at node Va to get the value of Vo.
     
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