I was a bit puzzled about the 'rule' of (+1 digit).
Let's see how the (1% + 1 digit) rule arises.

No doubts, the (1%) is a calibration error, as estimated by the manufacturer at the time the device was built.
It shows you by how much the reading may differ from the "true" value as a result of the "propagation of errors" due to inner circuit components.

The (1 digit) error appears as a 'safety feature' (and is typical to digital instruments). It appears because of the Analog-to-Digital conversion in the instrument. Without going into details: it is an error of counting in the A-D process.
Effectively, this rule prevents us from underestimating the accuracy error.

Example:  Say, on the "20V"scale we measure U= 0.65 V with 1% accuracy.
Then the errors are:

(1% of the reading) = 0.0065V

(1 digit) is 0.01V.

I believe the error to be quoted here is 0.01V. Or maybe even 0.02V.

The first impression is that 1% error is too small to be displayed within those 2 digits after the dot, available on this scale. Hence, it's clear that the 1-digit rule should be called in here. It might be that someone would expect to see the composed error quoted as 0.02 V because (1% ) + (1digit)= 0.0065 + 0.01= 0.0165V and we round it up.

My feeling is that 0.02V would be an overestimation, since instrument's electronics just "rounds up" to the next available digits for display, and doesn't really know how to add 1 digit on top of the rounding process :-).

On the other hand my explanation might not be the right one. I'm not that sure that I have a " strong case ", and therefore I advice you not to take for granted what I say here :-)
Let's see how the situation described above looks like if we follow the rule of thumb: "better overestimate the error than underestimate it...."

The "true" value of voltage is in between (0.65 V -1%) and up to (0.65 V + 1%). In other words, the real value might be anywhere in the interval (0.65-0.0065 , 0.65+0.0065) V= (0.6435 , 0.6565)V.
To show this with our measurement, we must think as the instrument would "think", i.e. display the above interval with only 2 digits after the dot: (0.64, 0.66)V. I just rounded up/down to the nearest displayable digit - but always in the direction of increasing the error.
As you see, the error would be indeed 0.01V as I stated at the very beginning.

It's true that someone might prefer to quote in this case the error as 0.02V. This "oversafe" quotation is also fine; perhaps it's even better than the "liberal" value I was trying to argue for above. At the end, we are more interested in getting the right mean value, not in estimating the error "too precise" :-)

 

More Examples:

Let us analyse the calibration example presented on the UPSCALE web-page

Hopefully, by "inverse engineering" we can find out the right rule for error quotation :-)

The Fluke-8000 voltmeter (digital instrument, used for the calibration of a less accurate analogue voltmeter) is said to have an accuracy error of (0.1% of the reading + 1 digit).
The reading error is 1/2 digit, as expected for a digital instrument.

This is the example discussed by Dr. Harrison in Sect.12 (link above). Let us analyze the values from table (sect. 12, see web-page).

Reading 1:
3.18 V   (I guess, it was on a "10V" or a "20V" scale, with 2 digits after the dot.)
0.1% of the reading is: 0.00318V.
1 digit is 0.01V
1/2digit is 0.005V.

The error quoted is 0.01V.

It looks like the error (0.1% + 1 digit) = 0.01318V has been rounded down to 0.01V. Observe that the 0.005V reading error has been ignored!

Reading 2:
5.13 V
0.1% of the reading is: 0.00513V.
1 digit is 0.01V
1/2digit is 0.005V.

The error quoted is 0.02V.

It looks like the error (0.1% + 1 digit) = 0.01513 V has been rounded up to 0.02V. Again, "no word" on the reading error of 0.005V.

Reading 3:
9.61 V
0.1% of the reading is: 0.00913V.
1 digit is 0.01V
1/2digit is 0.005V.

The error quoted is 0.02V.

It looks like the error (0.1% + 1 digit) = 0.01931 V has been rounded up to 0.02V. The reading error 0.005 has been (again!?) neglected!

Moral: we better overestimate the error a bit, than underestimate it.

Summary:

There is some room of maneuver with the error quotation and rounding!

You have to round the (combination of accuracy and reading) errors
to one or at most two significant digits!

 

Choosing the right range

Let us return to Lab Test specifications for the accuracy error:
(1% of the reading + 1 digit).

Example:

Using the "20kOhm" scale:
  R= 7.63 kOhm.
  Reading error: 0.005 kOhm
  Accuracy/calibration error: 0.0763 kOhm + 0.01 kOhm.
Error to be quoted: 0.08 kOhm, or even 0.09 kOhm!
  Final answer: R = ( 7.63 ± 0.08 ) kOhm.
The relative error in the final answer is E = 0.08/7.63 = 1.05 % !

Please, observe that the above reading for R has 3 significant digits. The resistance was measured on the "20kOhm" scale.

What happens if we are not careful with the choice of the range?

Assume we measured the above resistance on the "200 kOhm" scale.

The reading would be: R = 7.6KOhm.

In this example you see what the 'reading error' of the digital instrument means: it shows up because of the circuits of the instrument, which do some rounding (up or down)!
There is no room for that "end 3 digit", so it is simply thrown away: 7.63 --> 7.6

This is allowed/supposed to happen since on the "200kOhms" scale the reading error is 0.05 kOhm > 0.03 kOhm. The instrument's electronics just does its job of rounding!

(1% + 1 digit) is: 0.076 kOhm + 0.1 kOhm.
Reading error is: 0.05 kOhm.

The error to be quoted here is 0.2kOhm (common sense reasons).

Final answer: R = (7.6 ± 0.2)kOhm.

The relative error in the final answer is E = 0.2/7.6 = 2.6 % !

Summary:
As you see, because of the unproper choice of scale the precision dropped by a factor of 2.

Instead of having the error dictated by the 1% accuracy of the instrument, we ended up with an error twice as big.

Examples:

a)     U = 4.23 V (measured on the 20V scale)

(1%) = 0.0423 V;
(1 digit) = 0.01 V
Error to be quoted: 0.05 V. (major contribution from the 1% error!)

Final result: U = (4.23 ± 0.05) V

 

U= 4.2 V (measured on the 200V scale)

(1%) = 0.042 V;
(1 digit) = 0.1 V
Error to be quoted: 0.15 V (major contribution from the 1-digit error!).

Final result: U = (4.20 ± 0.15) V

As you see the error increased by a factor of 3 because we didn't work on the appropriate scale!

b)

U = 15.34 V (on the 20V scale)

(1%) = 0.1534 V;
(1 digit) = 0.01 V
Error to be quoted: 0.16 V.

Final result: U = (15.34 ± 0.16) V

 

U= 15.3 V (on the 200V scale)

(1%) = 0.153 V;
(1 digit) = 0.1 V

Error to be quoted: 0.25 V
(or maybe even 0.30 V, because the reading error of 0.05V enters here).

Final result: U = (15.3 ± 0.3) V

Again, because of the improperly chosen scale the relative error blows up by a factor of 2

Moral:  As argued above, you need to do the measurements on a proper scale: the more significant digits you have the better.
Indeed, if you have 3 or 4 significant digits then the 1-digit and 1/2-digit errors are negligible compared to the 1% accuracy error!

Important note:   There is a "dark side" of taking multi-digit data.
When it comes the time to plot them by hand you'll have a hard time ...

You cannot accurately plot 4 digit values by hand, on a letter-size graph paper (if you have a wide range of (V,I) values).
Some plotting errors show up, because you have to round your values to 2 or 3 digits to be able to plot.

For example, on graph paper, assuming you have voltage from 0 to 20V, the 1V unit gets represented as 1cm.Then, 0.1V is 1mm and that's about the smallest voltage difference you can display. In this case 1.114V and 1.075V are indistinguishable on the plot. Both get represented as 1.1V !
You may, of course, display 0.5mm or even 0.3mm on the graph paper. To me, it looks like taking 3-digit measurements is a good compromise.

 

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Last revised: July 02, 2003
© Sorin Codoban, 2003