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Measuring current accurately?

Discussion in 'General Electronics Discussion' started by eem2am, Feb 16, 2012.

  1. eem2am

    eem2am

    414
    0
    Aug 3, 2009
    Hello,

    Please can you tell if our method of measuring a 500KHz triangular current waveform is accurate?

    We are just measuring the voltage across a series sense resistor.


    This is the shape of current waveform we are trying to measure the average value of:-
    http://i39.tinypic.com/2vwsht1.jpg

    So we are trying to measure the average inductor current in each inductor of a dual buck converter.
    (we need to do this to assess what actual maximum current the load (FPGA) is actually drawing)



    We are measuring the current by using a 2mR series resistor. (its a 1% surface mount sense resistor)
    -we measure the voltage across the sense resistor with a Hewlett Packard HP34401A digital meter.
    (obviously wires (twisted pair) are taken from the sense resistor terminals to the meter)

    HP34401A Digital meter User Guide:-
    http://www.ee.buffalo.edu/courses/elab/hp34401.pdf


    -the total load current, we know is around 15 to 18A , but we need to know it exactly.
    -anyway, each inductor current is obviously a triangle current waveform with a few amps peak to peak and 7 to 9A average.


    The buck is actually a dual-buck converter supplying an FPGA

    Vin = 12V
    Vout = 0.9V
    Each FET switches at 500KHz (there’s interleaving)
    Inductors are 470nH
    Iout(max) = What we are measuring here!!!!

    So please do you know if our sense resistor voltage being read by the HP34401A is going to give us an accurate way of measuring the current?


    NOTE:
    Before taking the measurements, we got the software engineer to code the FPGA with code that will make it draw the maximum current.
    The dual buck and the FPGA are actually in the environmental chamber at 80degC when we measure the current.
    (obviously we cannot put a current probe into the chamber at 80degrees C……even if we could, we’d still have to break the pcb track and bring a wire through the current probe’s “jaws” which would not be advisable at this high frequency)

    HP34401A Digital meter datasheet:-
    http://www.metrictest.com/catalog/pdfs/product_pdfs/hp_34401a.pdf

    HP34401A Digital meter User Guide:-
    http://www.ee.buffalo.edu/courses/elab/hp34401.pdf
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,191
    2,693
    Jan 21, 2010
    TL;DR

    Have you thought of looking at the voltage across the 2 milliOhm resistor on an oscilloscope?

    If your digital meter can't measure peak voltage then you have to make assumptions about the shape of the waveform. If it can read peak voltage you need to ensure that you're measuring signal and not noise. Being able to see it provides (at the very least) confirmation of what you're measuring.
     
  3. daddles

    daddles

    443
    3
    Jun 10, 2011
    Don't assume the resistance is independent of the current. It's only when you know this function R(i) that you can use the device as a current shunt. Unless the resistor is e.g. made of a material like a near-zero temperature coefficient material like Manganin, this variation of resistance with current will mess up your measurements (and most resistors change resistance when they heat up).

    I don't understand how you are measuring the current with the HP meter -- the 500 kHz triangle wave is far beyond the RMS bandwidth of the meter. A digital scope is an obvious tool here.

    Could you eliminate the shunt resistor and measure the voltage across the inductor to figure out the current?
     
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