Electronics > Metrology

Questions on pt100 probes

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Yuu:
Hello all,

I'm going to put together a simple pt100 circuit with an ADS122C04 soon and I had some questions about pt100 probes.
What is the best way to program resistance interpretations? There's really three things I could do:
1. utilize 0.3850 ohm/C
2. utilize polynomial fit (1 but more complex)
3. use a lookup table

Downloading a pt100 csv table from fluke and comparing that to values obtained using 0.3850 ohm/C, I calculate at times an end temp difference of 0.38 C which is significant.
So if I want this to be as accurate as possible over my temp range (0 to 100C) then it's clear options 2/3 are more accurate.

Here's another thing I don't understand. They give these probes classifications and mine is a class A meaning
$$\pm 0.15 C \pm 0.002 |T_C|.$$
Does that uncertainty in temp assume the user is using the resistance measurements in one way or another? In other words, are they assuming you're using a linear fit or polynomial fit for ideal pt100 curve?

mendip_discovery:
You might want to look up IEC 60751 which is the standard that defines them.

0.15 + (0.002 x Temp °C) = Tolerance
so
0.15 + (0.002 x 100 °C) = 0.35 °C

Classes are,
AA = ± (0.1 + 0.0017 x °C)
A = ± (0.15 + 0.002 x °C)
B = ± (0.3 + 0.005 x °C)
C = ± (0.6 + 0.01 x °C)

Formulas for the calculating,
For the range - 200 °C to  0 °C:
$$R_{t} = R_{0}[1+At+Bt^{2}+C(t-100 °C)t^{3}$$
For the range of 0 °C to 850 °C:
$$R_{t} = R_{0}(1+At+Bt^{2})$$
where
\$R_{t} \$ is the resistance at the temperature \$t \$;
\$R_{0} \$ is the resistance at \$t = 0 °C\$

The constants in these equations are:
A = 3.9083 x 10-3 °C-1
B = -5.775 x 10-7 °C-2
C = -4.183 x 10-12 °C-4

EDIT: To clean up text and to fix errors.

Yuu:
Ahh okay. So they make their table with the following polynomial fit for ideal pt100 curve:
$R_t = R_0 (1 + At + Bt^2)$
and that's also how they "interact" with their tolerance for the 0 to 850 C range.

Meaning, given a measured resistance, you calculate temp using either a lookup table (generated from above) or via math. Then, for that calculated temp, the actual temperature should be $$\pm (0.15 + 0.002 \cdot |T|)$$ for Class A. Where that uncertainty stems from the fact the probe isn't an ideal pt100 probe (e.g. may contain impurities) and its resistance curve may drift from the ideal curve.

mendip_discovery:
I would think of it as the probe tolerance, much like a resistor rather than an uncertainty. However, it would make up a part of an uncertainty if you were to try and account for most of the errors in the system. Probe + reader etc.

EC8010:
Exactly what I was going to say. It's a resistor. I had a couple of Farnell 219-1840 Class A PT100 probes measured for resistance at the triple point of water and whilst they were within 0.4% of 100 ohm, they weren't spot-on. That tolerance is probably the major player in the uncertainties.

Interestingly, the triple point of water is determined acoustically - at that temperature there are crackling noises. One day, that might be the deciding point in a pub quiz. Probably need to be a pub near a scientific establishment, though.  ;D