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IEC 60751 coefficients // −200 to 850 °C

Pt100 / Pt1000RTD Calculator

Convert temperature and resistance with the Callendar–Van Dusen relationship, then see how uncompensated two-wire copper leads can shift the indicated temperature.

Conversion

°C

Lead-wire estimate

m
mm²

The model treats 3- and 4-wire compensation as ideal. Real 3-wire accuracy depends on matched lead resistance and the input circuit. Temperature and connection resistance are not modeled.

Calculated result

TRUE TEMPERATURE

100.000 °C

SENSOR RESISTANCE

138.5055 Ω

ONE LEAD

0.7000 Ω

INDICATED R

139.9055 Ω

INDICATED TEMP

103.693 °C

LEAD ERROR

+3.693 °C

T ≥ 0: R(T) = R₀(1 + A·T + B·T²)
T < 0: R(T) = R₀(1 + A·T + B·T² + C·(T − 100)·T³)
A = 3.9083×10⁻³, B = −5.775×10⁻⁷, C = −4.183×10⁻¹²

How the Pt100 conversion works

A Pt100 has a nominal resistance of 100 Ω at 0 °C; a Pt1000 has 1000 Ω. The calculator uses the standard 0.00385 platinum characteristic coefficients and numerically solves the inverse resistance-to-temperature conversion across −200 to 850 °C.

Why a linear shortcut is not enough

The familiar approximation of about 0.385 Ω/°C for a Pt100 is useful near 0 °C, but the relationship is not perfectly linear across the complete range. Use the full relationship or the transmitter/input module’s certified conversion when calibration accuracy matters.

Two-wire lead error

In a two-wire circuit, both copper leads are in series with the sensor, so the input interprets their resistance as additional temperature. A Pt1000 is proportionally less sensitive to the same absolute lead resistance, while a properly implemented three- or four-wire measurement compensates much of the lead effect.

This is not a calibration certificate

The result is an ideal reference. Sensor tolerance class, self-heating, immersion depth, thermal gradients, transmitter error, input accuracy, lead mismatch and calibration uncertainty remain separate contributors.