Variations in output of opamp LM324

Discussion in 'Homework Help' started by prkpushp, Dec 17, 2012.

  1. prkpushp

    Thread Starter New Member

    Nov 20, 2012
    1
    0
    I am getting variations in output order to amplify the output of shunt resister.
    I am giving voltage across shunt resistor as input to the op amp LM324. But at different instant of time getting different outputs. I also replaced opamp many times but could not yield the required output. I am dividing the further output by voltage divider.

    When no input is provided to opamp then output voltage across voltage divider= 0.8V
    total opamp voltage= 0.24 when shunt voltage is zero.

    shunt voltage in mV and opamp Vltage is in Volts.

    It is not following the formula of output voltage V=-(Rf/R1)*Vin

    1st observation

    Shunt
    Voltage
    (i/p
    to
    opamp Output of apamp
    across voltage divider
    1.5 0.32 means output voltage of opamp =3*0.32=0.96V
    1.6 0.29
    2.0 0.36
    2.5 0.43
    3.0 0.51
    3.5 0.57
    4.0 0.64
    4.5 0.71
    5.0 0.78
    6.0 0.93
    7.0 1.09
    8.0 1.22
    10 1.53
    14.0 2.03
    15.0 2.57
    18.0 2.78
    19.0 3.09
    22.0 3.57


    After some time (around 15 min)
    Observation 2

    Shunt
    Voltage
    (i/p
    to
    opamp Output of apamp
    1.7 0.28
    2.0 0.30
    3.0 0.41
    3.5 0.46
    4.0 0.53
    4.5 0.58
    5.0 0.64
    5.5 0.69
    6.0 0.76
    7.0 0.88
    8.0 0.96
    10 1.09
    12.0 1.30
    18.0 2.52
    22.0 3.03

    input is in mV and output is in Volts.

    i have taken 5 observations in the interval of 15-30 min each. and i got 4 different values
    PLz reply asap.
     
    Last edited: Dec 17, 2012
  2. Sensacell

    Well-Known Member

    Jun 19, 2012
    1,130
    266
    Step 1: Google "Common Mode Rejection Ratio"

    Step 2: Throw away every LM324 you have.

    The LM324 is a really sad, old part, modern single supply opamps are so much better it's shocking.
     
  3. JohnInTX

    Moderator

    Jun 26, 2012
    2,345
    1,028
    This simple configuration will work to measure current only if the resistors are well matched AND the shunt resistor is VERY low i.e. a nearly 0 source resistance. Otherwise, you'll get the results you are observing.

    You need an instrumentation amplifier. Wikipedia
     
  4. #12

    Expert

    Nov 30, 2010
    16,284
    6,797
    I think because the gain is so high you might be picking up noise from nearby electrical devices or the chip might be oscillating.

    Presenting a more complete schematic might help us understand if the supply voltages are sufficient.
     
Loading...