Measuring 800V AC signal with 16-bit DAQ card within 1% accuracy

[Jonathan Mc]

I'm trying to measure high voltages with a DAQ card. Obviously I'm going to have to attenuate the voltage, but I need to make sure that I can measure accurately (within 1%)

The code width of the DAQ is
Vcw = 20Vpp/2^16 = 305.176uV

For this code width to be within 1%, the minimum voltage I can read is 30.517mV (30.517mV +\- 1% or +\- Vcw)

Now here is where I need guidance. Is my thought process correct?

The maximum voltage that I will read is 800Vrms (1131.37Vp), and the minimum voltage I can read accurately is 30.517mVp

If I take the ratio of those two I calculate that I need about 37000 steps to measure 1131.37V.

The DAQ 2^16 (65536) steps.

So I have more than enough steps in order to keep from needing to amplify anything, yes?

And if the above is correct, realistically I will only mess with 0-120Vrms, which means that I will only ever see a little over 1v in the DAQ. But I feel like that is a waste of resolution. I MUST design the circuit to assume max input voltage of 800V though.

Any thoughts or suggestions?

 

[anorlunda]

This is not an answer to your question, but if you are measuring samples, then only peak voltage matters, not RMS.

 

[Windadct]

A number of options here - but the simple way is http://www.ohmcraft.com/precision-resistors/leaded-resistors/high-voltage-leaded-dividers-hvd/?gclid=Cj0KCQiAnuDTBRDUARIsAL41eDqThjEzZj17tNaQ7AS-IwZ7pXhCwsUJvbvl7w-c2XetHWDGgDt8-uIaAuE4EALw_wcBr - you may / probably want Isolation as well - so an Isolation amp..

1% only tell part of the story - if you ONLY need voltage, then a transformer MAY get you there and provide the isolation. A trans will typically create a phase shift, and variation over frequency range.

 

[Averagesupernova]

Attenuation will cause you to lose resolution. My 2 cents.

 

[Borek]

Perhaps there is more to it, but unless there is some convoluted physics behind AC measurements it looks to me quit easy conceptually.

For 1% accuracy you need approximately n bits such that \frac 1 {2^n} < 0.01, that means around 7 bits is about enough. 16 bits is way more, giving you an error in the \frac 1 {2^{16}} range (assuming you can trust all lower bits).

Sure, some kind of voltage divider will be required.

 

[sophiecentaur]

Attenuation will cause you to lose resolution. My 2 cents.

Why? As long as the signal is well above the noise level (and 800V qualifies for this) the ratio of the divider cannot affect precision. I don't know of an ADC that can handle a raw 800V (with a supply rail of +/-12V).

 

[rbelli1]

If you really do need to use much of your dynamic range on a wide variety of voltages than you only need to add voltage protection to your ADC input and have multiple values of resistor on the bottom half of your divider.

You should have some form of voltage protection anyway.

When the "wrong" setting is selected you will see some severe clipping of the signal. You can automate the resistor switching with suitable transistors or FETS.

BoB

 

[berkeman]

I'm trying to measure high voltages with a DAQ card. Obviously I'm going to have to attenuate the voltage, but I need to make sure that I can measure accurately (within 1%)

The code width of the DAQ is
Vcw = 20Vpp/2^16 = 305.176uV

For this code width to be within 1%, the minimum voltage I can read is 30.517mV (30.517mV +\- 1% or +\- Vcw)

Now here is where I need guidance. Is my thought process correct?

The maximum voltage that I will read is 800Vrms (1131.37Vp), and the minimum voltage I can read accurately is 30.517mVp

What is the bandwidth of your signal? AC Mains 50/60Hz?

And further to the safety comments by @rbelli1 do you know what this is, and why it will help your project a lot...?

https://www.alliedelec.com/b-k-prec...MI_fLb45-Q2QIVSbnACh2W0wA5EAQYASABEgLcDfD_BwE