Measuring "high" voltage

** Groper Alert !!

** That all depends on:

1. The frequency.

1. The waveform.

2. The source impedance.

Care to elaborate ??

...... Phil

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.

On 29 Jul 2006 20:27:13 -0700, " snipped-for-privacy@hotmail.com" Gave us:

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

I could tell him how to accurately measure 10kV at 600kHz, that's high voltage. But he's probably not interested...

--
 Thanks,
    - Win

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:

1. The frequency = ~28 kHz

1. The waveform = sine wave?

2. 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

Phil Allis>

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.

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.

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.

--
Reg

On Sun, 30 Jul 2006 20:28:06 +0100, "Reg Edwards" Gave us:

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.

On Sun, 30 Jul 2006 20:28:06 +0100, "Reg Edwards" Gave us:

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

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.

---- Reg.

is

On Sun, 30 Jul 2006 22:32:41 +0100, "Reg Edwards" Gave us:

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.

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..

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.

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

** 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

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

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

Thanks!

-Henk

Paul E. Schoen 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

--
 Thanks,
    - Win

On 30 Jul 2006 19:49:46 -0700, " snipped-for-privacy@hotmail.com" Gave us:

One that has a max input of 1500 V AC.

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