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I need to know the measurement accuracy of Keysight N6705C power analyzer. I came across this document (N6700 Modular Power System Family.pdf) that states the error in terms of a percentage plus a voltage/current value. For example: 0.1% + 20 mV. Does this mean that if I'm measuring a 100V, I might see something between 100.120 and 99.88? Or maybe 100.02 * (99.9% -100.1%). Or maybe it means that for voltages over 20 mV the accuracy is 0.1%? I searched for it a little, but I couldn't find an answer. ChatGPT says it's the first case, but I can't really trust that. I would appreciate your input.

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2 Answers 2

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It means that if you measure zero volts (short the probes) the display might show a voltage of up to 20 mV instead of 0 volts. If you are measuring 100 volts, the display might show 100.12 volts.

Does this mean that if I'm measuring a 100V, I might see something between 100.120 and 99.88?

If the specification is +/- 0.1% +/- 20 mV then, yes. In your question you said 0.1% + 20 mV by the way.

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  • \$\begingroup\$ Ok, thank you. So, for 0.1% + 20 mV it should mean 100V would be measured as 100V to 100.12. Correct? \$\endgroup\$
    – MNaz
    Commented Jun 6 at 17:41
  • \$\begingroup\$ Just read my answer, the first paragraph. \$\endgroup\$
    – Andy aka
    Commented Jun 6 at 21:07
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The N6776A can measure 100V and the voltage measurement accuracy is stated as

0.1% + 100 mV

That is assumed to be

a) the inaccuracy

b) +/- are assumed in front of the sum of errors

So a 100.00V voltage could result in a reading (under the specified conditions - such as supply voltage, temperature or temperature range, and minimum warm-up time ) of as high as 100.2V or as low as 99.8V.

With reputable companies when there is a % plus a fixed value (or number of 'counts' or 'digits') the % is assumed to be 'of reading'. If there is only a '%' then it usually means the percentage of the full-scale range. So a 0.1% error on a 100V range could represent a 1% error if your signal is only 10V. This is not that.

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