When I look some multimeter manuals it says stuff like "0,2% +5", and in videos I hear this as "0,2% plus 5 digits"...but what does it mean in practice? What is this +5 saying?
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The 5 digits is 5 least-significant digits in the display on that specific range. That corresponds to a percentage of full scale, which is the more modern way to indicate accuracy.
The 0.2% should be (but is not always) a percentage of reading.
The total inaccuracy is the sum of the two.
In other words, if the basic DC accuracy of a 4-1/2 digit meter on the 2 volt range is stated as 0.2% + 5 digits, and you are reading a 0.5V signal, the error specification is 0.002 * 500mV + 0.5mV or 1.5mV (+/-0.3% of reading).
A reading of 500.0mV would represent an input voltage somewhere between 498.5mV and 501.5mV.
Edit: Maybe if I show this in a table it will be a bit more clear
Let's assume the range is +/-1.9999 Volts (4.5 digit meter) and the specification is +/-0.2% + 5 digits.
Naturally, you'd normally switch to a lower range (if one is available) once you won't go over the lower range. So if there's a +/-199.99mV range available you'd switch for the 100mV example, but for a 200.1mV input you'd be stuck with +/-0.45% error.
answered Apr 6, 2014 at 23:24
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\$\begingroup\$ Small elaboration of Spehro's answer: a 4 1/2 digit meter, on 2V scale, reads up to 1.9999 (the leading '1' is the half-digit). The lowest order digit thus counts increments of 0.1mV. So 5 'lowest order digits' would be up to +0.5mV. The 0.2% is of the signal under test, so in this case 0.002 * 0.5V = 1.0mV. Assuming +/- for both error components, that's how you arrive at Spehro's answer. \$\endgroup\$– gwidemanCommented Apr 7, 2014 at 2:00
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\$\begingroup\$ Ok, I think I am getting it, but it's still a little confusing...why do they sow it like this, and not just a percentage? Why I have to do the math if they already know it in the first hand? And if I got it right, just get the lowest digit (0.1mV), multiply by the number in the manual (5) and take the percentage out of this (0,2%)...is this correct? \$\endgroup\$ Commented Apr 7, 2014 at 3:20
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\$\begingroup\$ @SpehroPefhany, if you could elaborate more the math and explain things more clearly, this will make thing a lot easier for begginers reading this :) \$\endgroup\$ Commented Apr 7, 2014 at 3:23
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\$\begingroup\$ @mFeinstein The reason it's shown like this is that it's different for every reading. If you try to read 500uV on the 2 volt scale in my example, the error could be 100% (5 'counts'). You get the most accurate reading with a high number of counts (which should be intuitive). \$\endgroup\$ Commented Apr 7, 2014 at 3:24