Voltage divider puzzle

[charliecoutas]

Some years ago I made a potential divider for the company I worked for: they wanted to measure up to 1000V with a 10:1 divider. The input resistance of the voltmeter is 10Meg.

Two resistors in series: 7.0Megs and 840K. The measurement is across the 840K with the 10Meg input voltmeter. This means that the lower value resistor looks like 840K in parallel with 10M, which is 775K.  Simply put: 1000V across 7Meg in series with 775K gives a current of 1.29mA. And 1.29mA in 775K is 100V give or take a bit. The meter shows 99.8V. So it works fine on DC.

Now measure a 1000V RMS voltage at 50Hz. The voltmeter (high quality) has the same input resistance on AC as DC. But it shows 62.5V RMS, not 100V.

Can anybody explain why please?  We thought that inductance/capacitance might be to blame so we jacked the frequency up to 200Hz, same result, 62.5V.

Charlie

[david]

Sounds like it was calibrated as average as opposed to RMS.

Cheers,
David

[david]

Sorry - I've read that wrong.   If you were on AC RMS measurement average would be about 90V.
Was this a pure sinewave or close to it?  If it had an odd form factor it is much harder to predict.

Cheers,
David

[david]

I give up.  Your attenuator looks fine and reads correctly for DC and should similarly work for AC.  You've shown it's not reactive and besides it would need something like 4.7nF across the meter to cause that error at 50Hz and it would be much greater at 200Hz. 
It's really suggesting that the 1000V AC signal was not 1000V or that there is a significant distortion of the AC source.  Was the same meter used to measure the input 1000V and the output 100V?

Cheers,
David

[charliecoutas]

Thanks for your thoughts David. The source s a hugely expensive calibration gadget, I think it was made by Fluke. So it's probably a pretty pure sine wave. Yes, it is RMS.

What if the input resistance of the (again, high quality) meter was 1Meg, instead of 10Meg?

I'll do the maths when I get back from a Saturday night meal out!

Charlie

[GDeSantis]

Assuming an input of 1,000 VAC RMS, per the attached Excel file:

- With a meter input resistance of 1 Meg, the output would be 61.2V RMS
- With a meter input resistance of 10 Meg, the output would be 99.7V RMS

[david]

Hmmmm.  If it was 1Meg rather than the 10Meg you indicated then the 1000V DC would also have read incorrectly or are you suggesting that the meter was 10Meg for DC and 1Meg for AC measurements?

Cheers,
David

[Fanie]

Hi Charlie !

You did not mention what type of a resistor you used ?  A wire wound resistor may have inductance (higher impedence) as well and not only resistance.
If I have it right, higher voltages tend to flow on the outside of a conductor, and that may add to the inductance effect.  At low voltages (pic level  ) the inductance effect should be a lot less, if any.

You could try to make it up using a string of carbon resistors and see if you get better results...

[david]

Fanie - when did you last see a 7 Megohm wirewound resistor?  It would need shiploads of inductance to add enough megohms to cause such an error at 50Hz.


Cheers,
David

[ken_k]

I believe David solved this dilemma.
The input impedance of the meter changes when measuring AC.
This could be verified by measuring the meter input impedance.

[david]

Would you expect a "high quality" meter to have a different input impedance for DC and AC?  I thought all that went away years ago but the numbers are stacking up that way.

Cheers,
David

[Fanie]

Fanie - when did you last see a 7 Megohm wirewound resistor?  It would need shiploads of inductance to add enough megohms to cause such an error at 50Hz.

You may be right, they seem to make the wirewounds to only 100k ohm...

[JonW]

I have encountered this with high-voltage probes when measuring high impedances.  It will be the meter input Z

DC 10M, AC 1M

[charliecoutas]

I don't know what the meter AC input resistance is but I just assumed it was 10M like the DC input. It's just that it must be more than coincidence that you get 61V if it's 1M!

I can't check until Tuesday, I don't work there any longer and I'm in Bletchley Park all day  tomorrow. If this is all down to some stupid mistake on my part, forgive me.

Charlie

Fanie: We have wirewound rats down here in the south of England. They tend to live up in the palm trees. Good to see you on the forum again.

[JonW]

Working in Bletchley Park must be fantastic; very jealous!

[rick.curl]

Hi Charlie-
What is the manufacturer and part number of the "high quality" voltmeter?  Is it Fluke?

-Rick

[charliecoutas]

Rick and all:

I got hold of my old boss just now. The two voltmeters that were "showing the fault" were Siglent SDM3055 and Fluke 8845A. Looking at the specs both have an AC input resistance of 1Meg and DC input resistance of 10Meg. Problem solved. Thanks everybody.

JonW: It's great. We are in The National Museum of Computing at Bletchley Park. We have working models of several old machines: Colossus, ICL 2966, EDSAC, Harwell WITCH plus others. The problem is that the people who keep these machine running are all old geezers; there aren't any youngsters that seem interested in this stuff.

[charliecoutas]

Sorry JonW, I skipped over your post #12 where you showed the two input R's. You can share the prize with me.

[JonW]

Fluke Manual

Page 21/22

[charliecoutas]

Yes indeed JonW, thanks. I'm surprised, I'd have expected a meter to have the same input resistance AC or DC.

So I am designing a version of a potential divider with an op-amp buffer on the output. I think one range my ex-boss wants is 10KV AC/DC down to 1 volt (via a buffer). This will involve quite a few high value high voltage resistors and I would like to use a pcb. What sort of leakage currents do I need to consider with a standard FR4 pcb please?

Obviously I can space things out as far as I like, it's the leakage that bothers me.

Charlie