Higher current through a PT100 results in a bigger signal and better S/N ratio. OTOH, it also creates self heating which results in lack of accuracy. However, if the PT100 is in a fixed environment, for example is fixed to a metal heat sink, and is calibrated in place is that sufficient to remove the inaccuracy due to the self heating?
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\$\begingroup\$ Lets put it that way: you should be able to produce a steady state for all ambient temperatures, when energy flows are in equilibrium. The answer of your question now depends on whether or not the sensor reading and real temperature relationship is (mostly) bijective. \$\endgroup\$– PlasmaHHMar 2, 2015 at 11:14
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\$\begingroup\$ Why not use a thermistors? Better resolution feasable, but needs more calculation. \$\endgroup\$– user95218Dec 23, 2015 at 13:30
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\$\begingroup\$ With calibrated PT100 I can get an accuracy of 10mK across the 0-80 degC temperature range of interest. The circuit can resolve to 100 uK, limited by noise. \$\endgroup\$– Dirk BruereDec 23, 2015 at 13:58
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1\$\begingroup\$ because a 4wire PT100 with current monitoring and voltage monitoring far exceeds Thermistors for accuracy. \$\endgroup\$– user16222Dec 23, 2015 at 15:00
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If you read some specifications, the self-heating of a platinum sensor can be typically 0.2 K/mW at 0 °C.
So, at 10mA (which is quite a lot for a PT100) and a 0 °C ambient, a PT100 will be producing a warming power of 10mW and this will produce an error of 2 °C but heatsinking has to be considered. This document from Honeywell explains it fairly well: -
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\$\begingroup\$ True, but I am talking about doing a multipoint calibration and correction across (say) 5-85 degC with the PT100 attached to its heat sink. Is this enough to compensate for the heating or is there some other process I have to take into account? \$\endgroup\$ Mar 2, 2015 at 11:05
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1\$\begingroup\$ I think you should be OK but what you have to remember is that with a constant current excitation, the dissipated power due to that current increases as temperature increases due to the basic resistance rising. With a constant voltage and fixed resistor powering the sensor you can probably, with the right values obtain a more constant power delivery to the PT100. There is also some "lag" in the system if ambient temperature is fluctuating. All things to consider OR, just run at the recommended current (more like 1mA) and filter the output to reduce noise. \$\endgroup\$– Andy akaMar 2, 2015 at 11:14
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1\$\begingroup\$ The reason this S/N ratio matters to me is that I need mK resolution, so every method I can use to increase S/N is going to be needed. The only factor in my favor is that the probe will be measuring a fixed thermal mass. \$\endgroup\$ Mar 2, 2015 at 11:55
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\$\begingroup\$ Postscript - the eventual resolution was 200uK, accuracy +/-5mK \$\endgroup\$ Aug 1, 2018 at 10:20
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\$\begingroup\$ @DirkBruere I guess you must have used a bonded metal packaged PT100 to draw all the heat away? \$\endgroup\$– Andy akaAug 1, 2018 at 10:25