Input offset current calculation

Discussion in 'General Electronics Chat' started by mentaaal, Dec 1, 2008.

  1. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
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    Hey guys trying to work out the input offset current for an opamp schematic used in a practical. Please see attached pdf.
     
  2. mik3

    Senior Member

    Feb 4, 2008
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    Because you have almost equal input resistances on both inputs (1M, the parallel combination of the 100K-100R is negligible) the output voltage equals

    Vo=Iio*R2

    This is derived after analysis of the circuit.
     
  3. Ron H

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    Apr 14, 2005
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    Your circuit has positive feedback. I don't think that is what you intended.
     
  4. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
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    Yup your right RonH just a mistake of drawing it in a hurry to show you guys. Thanks Mik3, though i dont see how you get the final result, if you have a few minutes could you show me how you arrived at it?
     
  5. mik3

    Senior Member

    Feb 4, 2008
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    I will try but ti find it analyze the circuit without the 1M resistors to find the voltage due to bias current of each input. Then substitute the bias current of each input with the average input bias current plus/minus the offset current. Then you will see that if the non inverting input had a resistor with a value of the parallel combination of the feedback resistor and the resistor on the inverting input, the output will depend only on the offset voltage times the feedback resistor.
     
  6. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
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    cool i'll give it a shot
     
  7. Ron H

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    Check this out...

    EDIT: This does not include amplifier offset voltage.
     
    Last edited: Dec 3, 2008
  8. mentaaal

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    Oct 17, 2005
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    RonH are you aware that you are a superstar?! As a second year engineering student I hang my head in shame at how easy you made that look!

    Thanks very much for going to the trouble, I really appreciate it, especially after my teacher confusing me today!

    And thank you Mike for explaining it to me as well.
     
  9. mik3

    Senior Member

    Feb 4, 2008
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    I disagree.

    Ronh, I think you have calculated Ibp and not Ioff.

    Ioff should equal [Vo/(10^9)]-Ibn
     
  10. Ron H

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    Did you actually look at how the equation was derived?
    I ran a simulation and verified it before my previous post.
     
    Last edited: Dec 2, 2008
  11. mik3

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    Feb 4, 2008
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    Yes I did. I don't agree with

    Vx/100=Vo/(100K+100)
     
  12. Ron H

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    You're right. I forgot to account for Ibn flowing through R2. The error is generally miniscule, but it's an error nonetheless. I will edit the annotated schematic.
     
  13. Ron H

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    OK, I changed the original drawing to reflect the correction. Thanks, mik3, for catching it.
     
  14. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
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    Thanks to both of you again!
     
  15. The Electrician

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    Oct 9, 2007
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    Isn't there another tiny error?

    Your 5th equation down is:

    Vout = (Vx/100 - Ibn)*100k

    Shouldn't this be:

    Vout-Vx = (Vx/100 - Ibn)*100k

    because the current in R2 is (Vx-Vout)/100k, not -Vout/100k

    This leads to a slight change in your equation 8, which becomes:

    Vout = 1.001e9*Ioffset - 1e5*Ibn
     
    Last edited: Dec 3, 2008
  16. Ron H

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    You are correct, of course. I went back and looked at my attachment and was stunned to see it was not what I thought I had posted. It is corrected now, I hope.:eek:
     
  17. Ron H

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    Look at the attachment again. I still had errors. I thought I had replaced the erroneous drawing yesterday. I should be correct now.:(
     
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