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Opamp GBW product and accuracy?

Discussion in 'General Electronics Discussion' started by eem2am, Dec 31, 2012.

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  1. eem2am

    eem2am

    422
    0
    Aug 3, 2009
    Hello,

    I have a 50 KiloHertz rectified sinewave current waveform which is flowing in a 1 milliohm current sense resistor (so its really a 100KHz current waveform).
    I wish to amplify only the average value of the current sense resitor voltage (x100)

    Here is the schematic

    Schematic
    http://i46.tinypic.com/29n9qh1.jpg

    Here is the current waveform

    100 KHz current waveform
    http://i46.tinypic.com/2vj74gj.jpg



    ...........I do only want to amplify the average value of the current waveform.

    But ....OPA335 GBW = 2MHz so with a gain of 100, i'll have about 20kHz bandwidth, the amplifier isn't fast enough to even "see" the waveform.


    ...is this going to be a problem?.....will i not get an accurate amplification of the average value of the current sense resistor voltage waveform?

    I put the capacitors C1 and C2 in there so that i could hopefully filter out the 100KHz.......so that i just amplify the average value, but is this the case?

    .....or do i need to actually somehow filter the voltage across the current seense resistor?
     
    Last edited: Dec 31, 2012
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,496
    2,837
    Jan 21, 2010
    A low pass filter is one solution. However, "average" is not a well defined term. The value you get is unlikely to be RMS and is even less likely to approximate it as the waveform becomes more complex. Would that be important?
     
  3. eem2am

    eem2am

    422
    0
    Aug 3, 2009
    thanks...it wouldnt be important ...the waveform is always going to be a rectified sinewave.

    I am figuring that there is an average value to that input, and the opamp will amplify that by x100.?
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,496
    2,837
    Jan 21, 2010
    Yes, C1 and C2 are already present to produce a low pass filter. Increasing their values so that the 3db point is at a frequency lower than your input frequency should do it.

    See here for more information.
     
  5. tedstruk

    tedstruk

    475
    7
    Jan 7, 2012
    first isolate your circuit, and ground each led outside of the circuit.
    then change those 1n clips to 16n clips.
    what this will do, is make each LED seek the most direct path, you need enough clip to cause each LED to act as a meter on its own.
    Only way I know to make a meter without a meter/guage.
    just ignore me... theorized this once!
     
  6. davenn

    davenn Moderator

    13,833
    1,950
    Sep 5, 2009
    tedstruk

    none of that makes any sense

    to stop people being led astray, keep your posts as accurate as possible
    if you know nothing on the particular subject DONT POST ANYTHING :)

    Even I live by that rule ... there's many subjects on here that I cannot give a good answer so I stay quiet and learn from those that do know what they are talking about :)

    cheers
    DAve
     
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