RTD Self Heating Error Calculator
The measuring current heats an RTD slightly above the process temperature. This calculator finds the power dissipated and the resulting error from the excitation current and the sensor dissipation constant.
- R(t) from Callendar–Van Dusen = 138.51 Ω
- P = I² × R = 0.1385 mW
- ΔT = P ÷ E = 0.1385 ÷ 5 = 0.0277 °C
Self Heating Formula
ΔT = P ÷ E
I is the excitation current, R(t) the RTD resistance at the process temperature (IEC 60751 Callendar–Van Dusen), and E the dissipation constant in mW/°C from the sensor datasheet. E depends strongly on the medium: it is much higher in flowing water than in still air.
Worked example
A Pt100 at 100 °C (138.51 Ω) with 1 mA excitation dissipates 0.139 mW. With a dissipation constant of 5 mW/°C the error is about +0.028 °C; at 5 mA it would be 0.69 °C.
Frequently Asked Questions
What is RTD self heating?
The temperature rise of the RTD caused by the measuring current flowing through its resistance.
What is a dissipation constant?
The power needed to raise the sensor 1 °C above its surroundings, in mW/°C. It is given for a stated medium and flow.
Does a Pt1000 self heat more?
At the same current, yes, because P = I²R. Pt1000 is therefore used with lower currents.
Learn more on the blog
Thermocouple and RTD installation precautions
A practical, field focused guide on instrumentationblog.in that explains the theory behind this calculator and how engineers apply it on real plants.
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