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Instrumentation

RTD Resistance and Temperature

Convert between platinum RTD resistance and temperature using the IEC 60751 Callendar-Van Dusen relationship, with lead resistance error shown.

Inputs

Only affects a two-wire connection. Three- and four-wire connections compensate for it.

Calculated reference results

Temperature
100.01C
Temperature
212.02F
Resistance
138.5100ohm
Lead compensation
Three wire, compensated

The arithmetic

  1. Solving R = R0 (1 + A t + B t squared) for t with R0 = 100 ohm
  2. t = 100.012 C

What this does not account for

  • A two-wire RTD connection adds the lead resistance directly to the measurement. On a Pt100 that is roughly 2.6 degrees C per ohm, so a long run produces a large and completely invisible error.
  • This uses the IEC 60751 alpha 385 coefficients. Sensors built to other curves, and thermistors, do not follow this relationship.
  • A three-wire connection compensates only if the two current-carrying legs have equal resistance. Mismatched conductors or a bad termination reintroduce the error.

How this is calculated

A platinum RTD follows the Callendar-Van Dusen equation: above 0 degrees C, resistance equals R0 multiplied by 1 plus A times temperature plus B times temperature squared, with A of 3.9083e-3 and B of -5.775e-7 for a standard 385 alpha sensor. A Pt100 reads 100 ohms at 0 degrees C and about 138.5 ohms at 100 degrees C.

Formulas

R(t) = R0 (1 + A t + B t squared) for t at or above 0 C

  • A = 3.9083e-3
  • B = -5.775e-7
  • R0 — resistance at 0 degrees C

Assumptions and limitations built into this calculator

  • IEC 60751 platinum sensor with alpha of 0.00385.
  • Ideal sensor with no self-heating and no tolerance class error.
  • Lead error is applied only to the two-wire case.

Frequently asked questions

Why does my two-wire RTD read high?
Because the lead resistance is added to the sensor resistance and the transmitter cannot tell them apart. On a Pt100 each ohm of lead resistance is roughly 2.6 degrees C. Use three or four wires, or mount the transmitter at the sensor.
Is Pt1000 better than Pt100?
For long two-wire runs, yes, because lead resistance is a much smaller fraction of the sensor resistance. The trade-off is compatibility, since many inputs expect Pt100.

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