Calibration Error calculation

Hello friends;
Good day.I have a presure transmitter of 0 to 100 inches of water column and give me 4 to 20 mA.
Today i calibrate it. I check at zero pressure it gives me 4.1 mA and at 100 it gives me 20.5 mA.
So i apply a pressure o , 25 , 50, 75 and 100 inches of water column. and set the 4 to 20 mA at zero and full range.
before calibrate i note the following value
at 25"w.c It gives me 25.3
at 50" w.c it gives me 50.2
at 75" w.c it gives me 75.2
and at 100 it gives me 100.3
How can i calculate the %age error of that transmitter?
Regards

Indicated readings of 100.3 and 25.3 are both 0.3"w.c. have the highest deviation above the respective applied pressures, which are from a source that I assume you trust as 'accurate'.

0.3/100 = 0.003 = 0.3% uncertainty

However, you don't have the indicated value at 0 applied pressure. That figures into the calculation, as well. (you say it's 4.1mA but is the displayed value?)

1) Dan's math is correct. Note that the error is approximately constant, so you can calculate the error as percent of reading or percent full scale. If you calculate as percent of reading the value will be higher at lower pressures. This may be important if you are using it for control and the nominal pressure is at the lower end of the range.

2) It looks to me like the zero is calibrated incorrectly, and causes an offset in the readings. Checking zero should always be the first step in calibration.

if calibration is done correct 0zero and 100% should be exact.
with the numbers you gave
average error is 0.3/25=1.2/100 +0.2/50=0.4/100 0.2/75=0.3/100 0.3/100
total 2.2/400=0.55 %
however on a real meter we always say a deviation off full scale + deviation of reading.
your sensor is very good.

shooter said:

if calibration is done correct 0zero and 100% should be exact.

Click to expand...

There's no such thing as exact where measurement is concerned. You will always be fighting repeatability, hysteresis, and temperature stability to name a few. It is true that if you do a two point calibration on a primarily linear measurement and use the high and low end of the scale for the two points of calibration that those points should have the virtually no error as a result of the "non linearity" of the device assuming you do not make adjustments to minimize the peak error deviation by splitting the difference of your measurement curve with your linear approximation.

Most curves will have a simple "belly" to them. Therefore it in usually more advantageous that the 0% point and the 100% point not to be made to lie on the curve.

the error is not in the x direction but in the Y direction.
yes i do realise all books are wrong on this point.
normally you put the x and read the y and its error to what it should be.
and yes i do know when both are same unit it does not matter, it is principle.

I neglected to state that my error calculation was 0.3% full scale (FS).
A full scale error is a larger error as a "percentage of reading". the lower the reading is on the scale. That's a given.

It's also pertinent that there are far fewer instruments whose accuracy is stated as % of reading, most are stated as % FS.

A ±0.3 units FS error at 100 units is 99.7 to 100.3, 0.3% of reading error.
A ±0.3 units FS error at 50 units is 49.7 to 50.3, 0.6% of reading error
A ±0.3 units FS error at 30 units is 29.7 to 30.3, 1% of reading error
A ±0.3 units FS error at 20 units is 19.7 to 20.3, 1.5% of reading error
A ±0.3 units FS error at 10 units is 9.7 to 10.3, 3% of reading error
A ±0.3 units FS error at 5 units is 4.7 to 5.3, 6% of reading error
A ±0.3 units FS error at 1 unit is 0.7 to 1.3. a 30% of reading error.