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An OPAMP with input offset voltage

  1. May 26, 2013 #1
    1. The problem statement, all variables and given/known data
    Capture.jpg


    2. Relevant equations
    Superposition
    Non-inverting OPAMP: V_out = (1 + R/R)V_in = 2*V_in2
    Inverting OPAMP: V_out = -V_in
    V_os = 2.5mV

    3. The attempt at a solution

    1. V_in enabled, V_os disabled

    V_out1 = -V_in

    2. V_os enabled, V_in disabled

    To get V_out2, use the non inv. equation where V_in2 is equal to V_1 + V_os
    where V_1 is the node between R and R at the middle bottom of the circuit.
    V_out2 = 2*(V_1 + V_os)

    3. V_out = V_out1 + V_out2




    Is this the correct approach?

    Thanks in advance :)
     
    Last edited: May 26, 2013
  2. jcsd
  3. May 26, 2013 #2

    gneill

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    Staff: Mentor

    You may want to take a closer look at the gain of the circuit; The resistor values in the network joining Vout and +Vin may not be identical to those of the prototypical circuit from which your gain formula was derived.

    The Vos is usually modeled as a fixed voltage source in series with one of the op-amp inputs, typically the + input. The circuit is not particularly difficult to analyze with a bit of nodal analysis, voltage division, etc.

    attachment.php?attachmentid=59030&stc=1&d=1369575714.gif
     

    Attached Files:

  4. May 26, 2013 #3

    Do you mean that neither the inverting nor non-inverting equations will work in this case and that I need to derive new gain equations using nodal analysis?
     
  5. May 26, 2013 #4

    gneill

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    Staff: Mentor

    The equations would need adjusting to reflect the actual circuit's values. To adjust the equations, you'll have to know the influence of each resistance on the equation. It would likely be simpler/safer to just analyze the circuit from scratch.
     
  6. May 26, 2013 #5
    Or do you mean that I need to modify my gain equation in step 2 by finding V_3 (by doing voltage division) and plugging that in to the equation?

    edit: nvm, I need to read that first
     
  7. May 26, 2013 #6
    I'm pretty sure that step 2 is correct. The professor did an example in class that was similar enough to that case. But in step 1, I have input voltage at both OPAMP terminals so I do not know what to do. I'd rather take the OPAMP as a black box with a characteristic equation if possible.

    In case 1, can't I just use V_3 as a reference (ground) and then modify my equation to become
    V_out = -(V_in - V_3)?
    The prototypical inverting gain equation assumes that V_+ is ground, so if I take my reference at V_3 the above equation should work?
     
  8. May 26, 2013 #7

    gneill

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    Staff: Mentor

    The problem is, if you take V3 as the ground reference then all the points that are shown as connected to ground will have to take on some other potential (-V3), and you'll have to adjust Vin accordingly as well (since Vin is a ground-referenced value). Lot's of things are interconnected...

    Why don't you try it both ways if you can and compare results?
     
  9. May 26, 2013 #8
    I'll do that. I was just hoping that there was an intuitive way to do it since I'd rather not derive formulas when writing exams unless I have to :S
     
    Last edited: May 26, 2013
  10. May 26, 2013 #9

    gneill

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    Staff: Mentor

    I think you're likely to run into more "non-standard" circuits than textbook standard ones. Cookie-cutter formulas are limited in applicability. You're much better off being able to quickly analyze a given circuit on-the-fly.
     
  11. May 26, 2013 #10
    I did the derivation and my "intuitive" formula for part 1 was incorrect (by a factor of 3).
    Part 2 was good but I think that I will derive formulas directly from now on.
    Thanks for the help
     
  12. May 26, 2013 #11

    gneill

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    Staff: Mentor

    Glad to help! Good luck in your studies :smile:
     
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