My DMM has a accuracy of 0.0030% of reading + 0.0030% of range, when the voltage range is set to 100mV. If a measurement with this DMM gives 0.1mV, then is the true input voltage between 0.1mV - 3.003uV and 0.1mV + 3.003uV? Or is there a different factor than the uncertainty that I have to account for?
Your calculation is correct but the result is not really the uncertainty of your measurement, it's only the maximum instrument error.
If you want to calculate the uncertainty following the ISO standard you need to go a bit further. In this document you have a very good example:
If you follow the example presented with your data you'll get to the uncertainty values you're looking for. Give it a go and post your results, then we can let you know if your answer is correct.
If accuracy is ±0.0030% of a 100mV range, ±0.0030% * 100 mV = ±0.00003 * 100 mV= ± 3 μV
So if your example was 0.1 mV = 100 μV ± 3 μV
- Accuracy can be specified differently as a % of reading with a number of counts.
- It can be defined in terms of each part of error; gain, offset, non-linearity
normally examples like yours are expressed in parts per million, ppm
Features 34460A 34461A 34465Anew 34470Anew Digits of resolution 6½ 6½ 6½ 7½ 1 year DCV accuracy 75 ppm 35 ppm 30 ppm 16 ppm
low cost DMM's might be specified in terms of ±% of reading and ±counts