Op-Amp Offset Voltage

Discussion in 'Homework Help' started by yeezhihao, Mar 6, 2008.

  1. yeezhihao

    Thread Starter Member

    Sep 30, 2007
    11
    0
    Hi

    Can someone explain to me why in this circuit, I would get an Output offset voltage in magnitude of volts?

    I heard it was something along the lines that since input is 0v the equivalent circuit is that both capacitors can be removed together with the feedback line. And since 22m is huge, no current will flow to ground bla bla bla..........but I just cannot put everything together. Can someone please be kind enough to explain to me??

    Thanks in advance!
    Mark

    [​IMG]
     
  2. ecb123

    New Member

    Mar 6, 2008
    7
    0
    check the specs on your op amp-bias current, op-amps need certain requirments
    -22M too big-your circuit acts like a simple diff amp with these values.read up on op-amps
     
  3. yeezhihao

    Thread Starter Member

    Sep 30, 2007
    11
    0
    Hi ECB123

    Would you care to provide me with more details?
    Thanks a lot!

    Rgds
    Mark
     
  4. rwmoekoe

    Active Member

    Mar 1, 2007
    172
    0
    inputs have bias current (pretty small though). this current requires to be passed through equal resistor values for each inputs, to cancel offset at the differential inputs.
    in your case, one input gets the 22megs while the other gets only 22k//27k or 12k! imagine the different in voltage at the inputs, times the openloop gain of the opamp... no wonder the volts at the output are there.
     
  5. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    You show the resistors as 22m ohms which is 0.022 ohms which is almost a dead short.
    I think you mean 22M which is 22 million ohms.

    The lousy old 741 opamp has too much input bias current to use resistors with such a high value. You need an opamp with FET inputs. They have no input bias current, just a little leakage current.
     
  6. nomurphy

    AAC Fanatic!

    Aug 8, 2005
    567
    12
    You're making a high-pass filter with Fc = 0.33 Hz, is that really necessary?

    But, at least lower the resistance values and raise the cap values by a factor of 100. That would be 220K and 2.2uF.

    With a FET op amp you might get away with a factor of 10, that would be 2.2Meg and 0.22uF.


    (I prefer 22Meg or 0.022, over 22M or 22m, to help avoid confusion.)
     
  7. yeezhihao

    Thread Starter Member

    Sep 30, 2007
    11
    0
    Hi rwmoekoe

    Thanks for the explanation.
    I get what you mean totally.

    Thank u for your help.
     
  8. yeezhihao

    Thread Starter Member

    Sep 30, 2007
    11
    0
    Hi Audioguru

    Yes, I meant 22Mega. Sorry for the confusion.
    Thanks for your recommendation.

    Rgds
     
  9. yeezhihao

    Thread Starter Member

    Sep 30, 2007
    11
    0
    Hi nomurphy

    Yes, its a high pass filter with cutoff at 0.036.
    As I'm building a ECG aquisition device I would like to make the signal as clean as possible, thats why I have this HPF even though such low frequency noise many not be common.

    Thanks for your recommendation.
    I have increased my cap to micro values.

    Rgds
     
  10. rwmoekoe

    Active Member

    Mar 1, 2007
    172
    0
    yeezhihao,
    you don't have to change the values of the existing components.
    just add a 22meg res between the inverting input and the 27k // 22k node. the input
    will 'see' the impedance as 22meg (plus 12k :) ) also.
    or, calculate a new combination to replace the 27k // 22k that results in a 22meg or near as parallel. (56meg // 47meg = 26meg). well, it's better the first one though, 'cept for the component count, if that matters
     
Loading...