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Op-Amp trick questions

  1. Jun 28, 2012 #1

    Femme_physics

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    1. The problem statement, all variables and given/known data

    http://img502.imageshack.us/img502/4082/vddvss.jpg [Broken]

    I'm asked to

    1) Develop a formula for

    Av = Vout/Vin

    And

    2) If the LED was flipped the other way, what must we do for for the rest of the circuit so current would still flow through it


    3. The attempt at a solution

    1) I'm really confused about this one. I thought to just do:

    http://img23.imageshack.us/img23/6084/avavavv.jpg [Broken]


    2) You just turn the op-amp from non-inverter to inverter
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jun 28, 2012 #2

    gneill

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    Staff: Mentor

    Somethings fishy about your derivation for Av.

    Some questions to ask yourself:

    - What potential will appear at the op-amp's inverting input?
    - What current will flow through R1 (and in what direction)? Where must that current come from?
    - So what current will flow through R2 (and in what direction)?

    Given the current flowing through R2 and the potential at the op-amp inputs, what then must be the potential at the op-amp's output? (KVL)

    Note that the op-amp is powered by a dual voltage supply. What are the limits on the possible range of Vout? How might you alter the polarity of Vout without changing the basic layout of the circuit? (Hint: how does vout depend upon Vin?)
     
  4. Jun 28, 2012 #3

    Femme_physics

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    Should ignore the denominator - should be just Vin


    2 volts

    At non-inverting?

    Vout = R2 x Vin / R1 + R2 = 1.66 V
    Vout = 1.66 volts

    Using KVL:

    1.66 -R2I1 - R1I1 = 0

    I1 =0.1333 mA

    The current will come from the op-amp and flow towards the ground.


    Same current, I1.

    The current will come from the op-amp and flow towards the ground as well.

    I did it via the non-inverter formula already

    The limits are +18 volts -18 volts... I can alter the polarity by making the limits to be -8 volts and -18 volts
     
  5. Jun 28, 2012 #4

    gneill

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    I think you're complicating the analysis unnecessarily :smile: If the potential at the inverting input is Vin, what is the potential across R1? Then use Ohm's law to find the current through R1 (in terms of Vin). Using that current, use Ohm's law again to find the potential across R2 (in terms of Vin). If you know the potentials across both resistors you can find the potential at Vout.

    Changing the op-amps power supply values isn't going to be helpful, since Vout is trying to satisfy the requirements of the feedback network and Vin. There's a niftier way to affect the change. Consider what you might do treating Vin as something you can play with...
     
  6. Jun 28, 2012 #5

    Femme_physics

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    Ahh I set Vin = -2 insteaf of +2

    VR1 = 2 volts
    R1 = 2k

    IR1 = VR1/R1 = 2/2000 = 1 mA
     
  7. Jun 28, 2012 #6

    gneill

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    Yup. Easy-peasy.

    Keep the calculation in terms of a variable Vin. You can also keep the resistors as variables until the end if you want Av as a general formula. You can then plug in the resistor values to find Av for this particular circuit.

    So, what's the expression for the current? Then, what's an expression for the potential at Vout?
     
  8. Jun 28, 2012 #7

    Femme_physics

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    Last edited by a moderator: May 6, 2017
  9. Jun 28, 2012 #8

    gneill

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    Last edited by a moderator: May 6, 2017
  10. Jun 28, 2012 #9

    Femme_physics

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    Vout/Vin = (1+R2/R1)
     
  11. Jun 28, 2012 #10

    gneill

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    Yes, that looks good :smile:
     
  12. Jun 28, 2012 #11

    Femme_physics

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    Thanks gneill, I appreciate the mentoring. They're gonna have some funky questions at the final electronics test this year and trying to get ready for them... appreciate your time and knowledge
     
  13. Jun 28, 2012 #12

    gneill

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    It's always a pleasure to help. Good luck in your test!
     
  14. Jun 28, 2012 #13

    Femme_physics

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