# Measuring "high" voltage

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

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!

-Henk

2. ### Phil AllisonGuest

** That all depends on:

1. The frequency.

2. The waveform.

3. The source impedance.

Care to elaborate ??

....... Phil

3. ### NuclearFirestormGuest

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. ### Phat BytestardGuest

You have to use an HV probe. You have to make sure that your ground
points are in the right place.

5. ### Winfield HillGuest

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 EdwardsGuest

A voltmeter always loads the voltage to be measured and the voltage

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. ### Phat BytestardGuest

Not if a proper high voltage probe is utilized. In such a case, the

11. ### Phat BytestardGuest

Not if an HV probe is used. In such a case, the probe resistance is
the value which needs to be known.

12. ### Reg EdwardsGuest

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. ### Phat BytestardGuest

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 BaerGuest

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 BaerGuest

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. ### Paul E. SchoenGuest

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 AllisonGuest

** 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. ### Winfield HillGuest

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

.. 100:1 ac probe with response down to 800Hz.
.. 2pF ______ _
.. --||---)______|_|--scope
.. \ 1M 25pF
.. \
.. '--- cable, 175pF = 6 feet 50-ohm coax

20. ### Phat BytestardGuest

One that has a max input of 1500 V AC.

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