millivots differentail input voltages op-amps

Discussion in 'General Electronics Chat' started by dar2525, May 23, 2006.

  1. dar2525

    Thread Starter Member

    Jun 17, 2005
    21
    0
    I am trying to observe the output of a comparator for small differential input voltages of -3mV to 3mV. I want to verify how accurate my comparators are.

    I tried using a power supply, by setting the non-inverting to 2.5V and adjusting the inverting until I measure a voltage difference of millivolts, but the measurments are not steady. I probally will need a more precision voltage supply for this.

    I was thinking of using potentiometers to adjust the non-inverting and inverting inputs until I get stable differential input millivolts. Will this work? Have anyone else tried this. I am using ALD2301A for my comparators.

    Thanks in advance
     
  2. Nik

    Well-Known Member

    May 20, 2006
    55
    3
    IMHO, you're into 'Kelvin Country'.

    I reckon you'll need a discrete resistor ladder and a very stable voltage supply to precisely pick off the tiny steps you'll need...
     
  3. dar2525

    Thread Starter Member

    Jun 17, 2005
    21
    0
    Thanks. I can adjust my voltage supply to 5V. I think I will try some very precise potentiometers
     
  4. n9352527

    AAC Fanatic!

    Oct 14, 2005
    1,198
    4
    Maybe a 10 turns pot? And something to get rid of the noise. Not really worth the trouble IMHO or the cost.
     
  5. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
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    You can set this up as an op amp and let it measure its own offset voltage. See below.
     
  6. dar2525

    Thread Starter Member

    Jun 17, 2005
    21
    0

    Thanks, what does IMHO mean?
     
  7. dar2525

    Thread Starter Member

    Jun 17, 2005
    21
    0

    Thanks fpr replying. Will this be measuring the input or output offset voltages? Will the output be the offset voltages? Why is negative feedback needed? Can you please explain to the best of your ability why negative feedback is involved?

    Thanks in advance
     
  8. kubeek

    AAC Fanatic!

    Sep 20, 2005
    4,590
    777
    It is connected as amplifier with gain -1000, so on the output is -1000*Voffset, becasue + and - inputs are on the same potential, so the opamp sees only the offset.
     
  9. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    655
    Thanks, kubeek. I've been out of town. To clarify - the output is equal to -1000*input_offset.
     
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