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Measuring "high" voltage

Discussion in 'Electronic Design' started by [email protected], Jul 30, 2006.

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  1. Guest

    Any idea on how I can measure 1200 Vrms accurately? The multiimeters I
    have here at the office seem to be giving wrong results!

    Thank you for your thoughts!

    -Henk
     
  2. Phil Allison

    Phil Allison Guest

    ** Groper Alert !!


    ** That all depends on:


    1. The frequency.

    2. The waveform.

    3. The source impedance.


    Care to elaborate ??




    ....... Phil
     
  3. Just putt a massive resistance across the voltage, and use as a voltage
    divider... Maybe, it all depends on frequency, source impedence and
    what you mean by accurate.
     
  4. You have to use an HV probe. You have to make sure that your ground
    points are in the right place.
     
  5. Phil Allison wrote...
    I could tell him how to accurately measure 10kV at 600kHz,
    that's high voltage. But he's probably not interested...
     
  6. Guest

    Phil,

    Yes - the groper would love to elaborate the best that he can!! :)

    I'm an using the CXA-M10A-L inverter from TDK to power the backlight of
    my CSTN panel. The open voltage should be 1200 Vrms. Below is the short
    datasheet:

    http://power.tdk.com/dcac/brochure/pdf/CXA-M10A-L (CTR-0742-A) PRODUCT DRAWING.pdf

    1. The frequency = ~28 kHz

    2. The waveform = sine wave?

    3. The source impedance = The only resistance/impedance I see in the
    datasheet is for the load for their testing.

    Thank you for yor time and any information!

    -Henk
     
  7. Guest

    Thanks for the response. I'm taking a stab at this here guys so please
    correct me if I am wrong:

    My multimeter does have an ~AC mode but I am not sure on how it is
    calcuating the AC voltage. ( I need to dig up the manual hope I can
    find it! ) If it is just taking the peak value instead of the rms
    value it will always be wrong because the majority of the time the sine
    wave is not at its peak value. Does that sound about right? I want to
    measure 1200 Vrms.

    NuclearFirestorm - do you mind saying a few words on how the frequency,
    source impedence or whatever else will effect my measurement?

    Thanks.
     
  8. Guest

    Thanks for the response. I'm taking a stab at this here guys so please
    correct me if I am wrong:

    My multimeter does have an ~AC mode but I am not sure on how it is
    calcuating the AC voltage. ( I need to dig up the manual hope I can
    find it! ) If it is just taking the peak value instead of the rms
    value it will always be wrong because the majority of the time the sine
    wave is not at its peak value. Does that sound about right? I want to
    measure 1200 Vrms.

    NuclearFirestorm - do you mind saying a few words on how the frequency,
    source impedence or whatever else will effect my measurement?

    Thanks.
     
  9. Reg Edwards

    Reg Edwards Guest

    A voltmeter always loads the voltage to be measured and the voltage
    falls on load.

    The multimeters in the office are probably all giving the correct
    answers - that is they all correctly indicate the voltage which
    appears across their input terminals.

    If the high voltage to be measured has a high internal resistance then
    the resistance of the meter causes the high voltage to fall somwhat.

    What you could use is a voltage divider consisting of two resistors in
    the ratio of the order of 100 or 1000 to 1.

    The high value resistor should have a value say 10 times the source
    resistance of the high voltage to be measured. The low value resistor
    should have a value 1/10th of the meter used.

    You will need Ohm's Law, a knowledge of the meter resistance, and a
    bit of arithmetic to work it out.

    As I say, your multimeters are probably giving you the correct
    answers. But you need to know what the internal resistances are of
    the meter and of the voltage to be measured in order to calculate what
    the open=circuit voltage of the source is.
     
  10. Not if a proper high voltage probe is utilized. In such a case, the
    loading is ten or 100 Gig Ohm depending on the probe.
     
  11. Not if an HV probe is used. In such a case, the probe resistance is
    the value which needs to be known.
     
  12. Reg Edwards

    Reg Edwards Guest

    What I have said is perfectly correct.

    But you seem to know enough about it to sort out the errors without
    bothering newsgroups with vague questions. Swat up on Ohm's Law and do
    a little arithmetic.
     
  13. Nope. Also, not quoting what you are claiming is rather stupid as
    well.
    Bugger off, bother boy.
    Dumbass. That's what an HV probe does. It is a very high resistance
    presented to the load so that the meter's internal resistance does not
    present a load to the supply being probed.

    It is Ohms law, and you have failed the test. If you knew anything
    at all about HV probes you would never have come back to post this
    utter crap.

    You are the one that needs to BONE UP, and do some math.

    Questions:

    What loading does a 10Meg Ohm meter present to a 1200 Volt supply?

    What loading does a 10Gig Ohm HV Probe monitored by a 10MegOhm meter
    present to the same source?

    Are those too "vague" for you to grasp?

    Ooops... You lose.
     
  14. Robert Baer

    Robert Baer Guest

    An "unloaded" CFL inverter puts out a sine wave where the rest of the
    input square wave power gets lost in the core.
    But a CFL is a non-linear load; crudely speaking a voltage regulating
    arc - so i doubt the voltage waveform is a sinewave.
    Most handhelds measure peak and read RMS of an assumed sine wave; one
    potential cause of error.
    Then there is a frequency response, which should be translated to
    accuracy (or error) VS frequency.
    So FFT the waveform and "overlay" an error correction curve for that
    part.
    Then inverse FFT back to a "corrected" waveform and mathematically
    apply a peak detector, and then "convert" to an RMS "reading".
    Ain't going to be much better than 5% methinks..
     
  15. Robert Baer

    Robert Baer Guest

    Err..you have an incorrect idea as to what is happening.
    It does not matter about "the majority of the time".
    Peak value is peak value whether it is a pulse or triangle or square
    or sine wave.
    If the waveform is not ("close" to) sine, then the RMS value reported
    by the meter will be incorrect, assuming the majrity of the waveform
    frequencies are inside the bandwith of the measuring circuitry.
    If you really want true RMS, then get a meter that actually measures
    true RMS.
    If you have a "high frequency" square wave, or pulses, or "RF" then
    you should consider using thermal bolometers or the like.
     
  16. Looking at the specs, this device is rated at 6 W, 400 VRMS, and the 1200
    volts open circuit is measured directly at the transformer secondary (see
    note 1-6). They specify a 1000:1 probe, and a true-RMS meter. If you are
    measuring this directly with an ordinary DMM, you are very likely exceeding
    its 1000 VDC and 600 VAC input limitations, and unless it is a high quality
    true RMS meter rated up to 30 kHz, you will get errors because the waveform
    is almost certainly not sinusoidal. The high voltage is required for
    initial "strike" voltage of the CCFL. If you use a high voltage probe or a
    resistive divider, you must make sure it is properly compensated, because
    on AC, there is a large capacitive component that can cause errors as well.
    You could add a high voltage diode and capacitor, and measure the peak DC
    output with a simple uncompensated divider.

    Paul
     
  17. Phil Allison

    Phil Allison Guest



    ** A ordinary DMM is not suitable at all - 28 kHz is way outside the AC
    range's capability.

    What you need is a bench scope fitted with a 100:1 divider probe rated at
    1500 volts or more.

    If the wave you see is close to square shaped ( which is as I expect) the *
    rms * voltage is 1/2 the peak to peak value as seen on the scope
    multiplied by 100.





    ........ Phil
     
  18. Guest

    Paul! Best post I have ever seen! Thank you! :)

    What type of true RMS meter would you recommend for purchase?

    Thanks!

    -Henk
     
  19. wrote...
    This is proper territory for a scope-probe waveform examination.

    1500V Tektronix 100:1 probes are available from the factory, or
    are often seen on eBay, but you can easily make a home-made probe
    by ignoring the dc resistive-divider aspect, and making an ac-only
    probe with a capacitive divider. For example, if a scope's 1M
    input impedance is paralleled with 200pF of capacitance, the ac
    "probe" will have a high-pass response down to ~800Hz. If the
    input capacitance is 2pF, then the division ratio will be 100:1

    Instead of struggling to achieve exactly 2.02pF, or some other
    capacitance value, a twisted-wire "gimmick" can be used to make
    an adjustable capacitor that can be adjusted at low voltages to
    calibrate the home-made probe.

    .. 100:1 ac probe with response down to 800Hz.
    .. 2pF ______ _
    .. --||---)______|_|--scope
    .. \ 1M 25pF
    .. \
    .. '--- cable, 175pF = 6 feet 50-ohm coax
     
  20. One that has a max input of 1500 V AC.

    Stop top posting, or be known as a Usenet retard.
     
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