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RTD Calculator (Pt100, Pt1000)

Convert platinum RTD resistance to temperature and back using the IEC 60751 Callendar–Van Dusen equation. See the tolerance for each accuracy class and the error caused by lead resistance in 2 wire circuits.

Pt100, Pt500, Pt1000IEC 60751Class AA, A, B, C2, 3, 4 wire
Resistance at the sensor157.3251 Ω
Reading at the instrument157.3251 Ω
Sensitivity0.3735 Ω/°C
Lead wire error≈ 0 °C (if leads are equal)Lead resistance is compensated

Tolerance at 150 °C (IEC 60751)

Accuracy classTolerance (°C)Tolerance (Ω)
Class AA± 0.355 °C± 0.1326 Ω
Class A± 0.450 °C± 0.1681 Ω
Class B± 1.050 °C± 0.3922 Ω
Class C± 2.100 °C± 0.7844 Ω
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The Callendar–Van Dusen Equation

IEC 60751 defines the resistance of an industrial platinum RTD with α = 0.00385 Ω/Ω/°C using the Callendar–Van Dusen equation:

0 to 850 °C: R(t) = R0 × (1 + A·t + B·t²)
−200 to 0 °C: 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

R0 is the resistance at 0 °C (100 Ω for Pt100, 1000 Ω for Pt1000). For temperatures above 0 °C the equation can be solved directly for temperature:

t = (−A + √(A² − 4B(1 − R/R0))) ÷ 2B

Below 0 °C the extra C term makes a direct solution impractical, so the calculator solves it numerically.

Worked example: 2 wire lead error

A Pt100 at 150 °C has a resistance of 157.325 Ω and a sensitivity of 0.3735 Ω/°C. It is connected with 2 wires of 1.0 mm² copper over 50 m.

  1. Resistance of one wire: 0.0172 × 50 ÷ 1.0 = 0.860 Ω.
  2. Both wires are in series with the sensor: 2 × 0.860 = 1.720 Ω.
  3. Error = 1.720 ÷ 0.3735 = +4.61 °C, far larger than any tolerance class.
Key insight: use 3 wire connections for normal plant work and 4 wire for laboratory or custody transfer accuracy. A Pt1000 reduces 2 wire error tenfold because its sensitivity is ten times higher.

IEC 60751 Tolerance Classes

ClassTolerance formulaAt 0 °CAt 100 °C
Class AA± (0.1 + 0.0017 × |t|) °C± 0.10 °C± 0.27 °C
Class A± (0.15 + 0.002 × |t|) °C± 0.15 °C± 0.35 °C
Class B± (0.3 + 0.005 × |t|) °C± 0.30 °C± 0.80 °C
Class C± (0.6 + 0.01 × |t|) °C± 0.60 °C± 1.60 °C

The temperature range over which each class applies depends on the construction (wire wound or thin film). Check the sensor datasheet before relying on a class outside 0 to 100 °C.

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Frequently Asked Questions

What is the resistance of a Pt100 at 0 °C and 100 °C?

A Pt100 is exactly 100 Ω at 0 °C and 138.5055 Ω at 100 °C, which gives the average coefficient α = 0.00385 Ω/Ω/°C.

How do I convert Pt100 ohms to temperature quickly?

A rough rule is 0.385 Ω per °C near room temperature, so 110 Ω is about 26 °C. For accurate work use this calculator, which applies the full Callendar–Van Dusen equation.

What is the difference between 2, 3 and 4 wire RTDs?

In 2 wire circuits the lead resistance adds directly to the sensor. A 3 wire circuit cancels it if all leads are equal. A 4 wire circuit measures the voltage across the sensor alone, eliminating lead resistance completely.

Does this work for Pt1000 sensors?

Yes. Select Pt1000; the same IEC 60751 equation is used with R0 = 1000 Ω, so every resistance is ten times the Pt100 value.

What about RTDs with α = 0.00392?

Some older American and Japanese sensors use a different platinum curve. This calculator follows IEC 60751 (α = 0.00385), which is the standard for almost all industrial RTDs today.

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