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Op Amp Question

  1. Mar 1, 2016 #1
    1. The problem statement, all variables and given/known data
    Calculate the input voltage to an Op-Amp circuit. The circuit is exactly an ideal non-inverting amplifier except the inputs have been switched in that the supply is connected to the inverting input and the middle of the potential divider is connected to the non-inverting input.

    2. Relevant equations
    I would be completely comfortable solving this problem if it was not for the inputs having been switched.

    3. The attempt at a solution
    Again, had the inputs not been switched I'd use:-

    Vin = R1/(R1 + R2) x Vout

    The best thing I can come up with is that the answer is the same but negative as opposed to positive.

    Any help much appreciated.
     
  2. jcsd
  3. Mar 1, 2016 #2

    gneill

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    Staff: Mentor

    Take a look at the thread: Schmitt Trigger question , and see if your circuit looks anything like the one discussed there.
     
  4. Mar 1, 2016 #3
    Hi gneill,

    Yes that is the exact question funny old thing and I have the exact same answer using the formula I stated above. That thread seemed to go off on a tangent a little.

    I get 1.69v using R1/(R1+R2) x Vout.

    Shouldn't it be -1.69v?
     
  5. Mar 1, 2016 #4

    gneill

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    Staff: Mentor

    Nope.

    Follow the thread through to the end. The circuit does not behave linearly past certain threshold values (hence the name of the thread).
     
  6. Mar 1, 2016 #5
    I'm a bit confused by that thread. I've read it through and I understand that the difference between Vin and V+ will determine the polarity of the output voltage. The output is given as 2.4V. From that I can deduce that Vin is less than the 1.69V I calculated earlier. The question states that Vin is constant however so that leaves me no further forward. Surely the answer is not as simple as <1.69V? Although it does make sense as any value less than 1.69V will give an output of 2.4V.
     
    Last edited: Mar 1, 2016
  7. Mar 1, 2016 #6

    gneill

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    Staff: Mentor

    The question is a tricky one to answer because in reality the circuit would tend to latch at a supply rail value when the input passes certain threshold values (one positive, the other negative), and not move from there until the opposite threshold is passed.

    Here they are specifying a particular value for the output that is (presumably) not equal to either of the supply voltage values. So the circuit can't have a history of actual operation that would have placed it in either of the "latch" states. In this case I suppose you need to assume that the given output voltage is correct and find a value for the input voltage that would hold the output at this (unstable) equilibrium.

    You've done this by finding the +1.69 V value for Vin.

    If Vin were to increase to and exceed the threshold value for the circuit then the output would "latch" to the lower supply rail until the input decreased below the lower threshold value (at which point the output would rise and latch to the upper supply rail). Note that the threshold values will be some fraction of the supply voltages, not the 2.4 V or 1.69 V values. The fraction depends on gain of the circuit as set by the resistors.
     
  8. Mar 1, 2016 #7
    So you think the question is poor but the answer is correct?

    Thank you very much for your help with this. It's much appreciated.
     
  9. Mar 1, 2016 #8

    gneill

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    Staff: Mentor

    That sums it up nicely :smile:
     
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