Temperature Transmitter mA Calculator
Check a temperature transmitter loop: convert process temperature to the expected 4 to 20 mA output, or a measured current back to temperature, with the RTD resistance to inject when simulating.
- mA = 4 + 16 × (T − LRV) ÷ (URV − LRV) = 10.800 mA
Five point check sheet
| Percent | Temperature | Output | Pt100 Ω |
|---|---|---|---|
| 0 % | 0.00 °C | 4.00 mA | 100.000 |
| 25 % | 50.00 °C | 8.00 mA | 119.397 |
| 50 % | 100.00 °C | 12.00 mA | 138.505 |
| 75 % | 150.00 °C | 16.00 mA | 157.325 |
| 100 % | 200.00 °C | 20.00 mA | 175.856 |
Temperature Loop Formulas
T = LRV + (I − 4) ÷ 16 × (URV − LRV)
RTD resistance uses the IEC 60751 Callendar–Van Dusen equation, so you can inject the exact ohms with a decade box or calibrator.
Worked example
A 0 to 200 °C transmitter at 85 °C should output 4 + 16 × 85 ÷ 200 = 10.8 mA. To simulate this with a Pt100, inject 132.80 Ω.
Frequently Asked Questions
How do I convert temperature to mA?
Find the percentage of the calibrated range and apply it to the 16 mA span above 4 mA.
What resistance should I inject to simulate 100 °C?
138.506 Ω for a Pt100 and 1385.06 Ω for a Pt1000, per IEC 60751.
Why does my reading drift with ambient temperature?
Check the transmitter ambient temperature effect, cold junction compensation for thermocouples and lead resistance for 2 wire RTDs.
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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