PT100 / PT1000 RTD Calculator

Q: What resistance is a PT100 at 100 °C?

138.5055 Ω. The round number people remember is 138.5 Ω. A PT1000 at the same temperature is 1,385.055 Ω, exactly ten times as much.

Q: What is the Callendar–Van Dusen equation?

The polynomial IEC 60751 uses to define platinum resistance against temperature: R = R₀(1 + At + Bt²) above 0 °C, with a third term C(t − 100)t³ added below it. A = 3.9083×10⁻³, B = −5.775×10⁻⁷ and C = −4.183×10⁻¹² for the standard α = 0.003851 sensor.

Q: Why can't temperature be worked out directly below 0 °C?

Because the equation becomes a quartic in t, and while quartics do have a closed-form solution it loses precision badly at these coefficient sizes. Every practical implementation iterates. This page starts from the quadratic root and takes four Newton steps, which is enough to reach the limit of double precision over the whole −200 to 0 °C range.

Q: How much error does two-wire wiring add to a PT100?

About 2.56 °C for every ohm of total loop resistance near 0 °C, and always in the direction of reading too high. Two 0.5 Ω leads are therefore worth roughly 2.6 °C. Use three-wire if the instrument supports it, four-wire if accuracy matters, or a PT1000, which suffers only 0.256 °C per ohm.

Q: What is the difference between class A and class B?

The tolerance and the range it applies over. Class A is ±(0.15 + 0.002|t|) °C from −100 to +450 °C for a wire-wound element; class B is ±(0.3 + 0.005|t|) °C from −196 to +600 °C. At 100 °C that is ±0.35 °C against ±0.8 °C. Classes AA and C sit either side.

Q: How much excitation current should I use?

As little as the instrument can resolve with. 1 mA is conventional for a PT100 and 0.1 mA for a PT1000, because both dissipate a comparable power. The figure that matters is the self-heating error above: if it is a significant part of your tolerance class, the current is too high.

Temperature to resistance and resistance back to temperature for a platinum RTD, using the Callendar–Van Dusen equation of IEC 60751 — with the class AA/A/B/C tolerance at your reading, the self-heating error from your excitation current, and what two-wire leads cost you.

Last updated 22 Sep 2026

PT100 and PT1000 RTD

°C ↔ Ω, IEC 60751

SensorA PT1000 is the same curve scaled by ten, so lead resistance matters ten times less.

Which way?

Temperature (°C)IEC 60751 defines the sensor from −200 to +850 °C.

Measured resistance (Ω)The resistance of the element itself, with any lead resistance already taken off.

Tolerance classClass B is what an unspecified industrial probe usually is.

Excitation current (mA)1 mA is the usual compromise; 0.1 mA for a PT1000. Enter 0 to leave self-heating out.

Self-heating coefficient (°C per mW)From the probe’s data sheet — roughly 0.05 °C/mW for a sheathed probe in stirred water and ten times worse in still air.

Wiring

Resistance of one lead wire (Ω)One way, not the round trip. About 0.5 Ω for 25 m of 0.5 mm² copper.

The usual way an RTD is read: a constant current through the element, and the voltage across it measured separately. The dots are the excitation current you entered. Lead resistance is not drawn — with four-wire sensing it is genuinely not in the measurement, and with two-wire it adds straight onto the element, which is the figure reported below.

138.5ΩExample

A PT100 at 100 °C, class B, 1 mA excitation, three-wire

Callendar–Van Dusen, IEC 60751

t ≥ 0 °C: R_t = R₀(1 + At + Bt²) t < 0 °C: R_t = R₀(1 + At + Bt² + C(t − 100)t³)
inverse above 0 °C: t = [−A + √(A² − 4B(1 − R_t/R₀))] ÷ 2B
inverse below 0 °C: no closed form — Newton from that root, t ← t − f(t)/f′(t), clamped to −200…0 °C
class tolerance: ±(a + b|t|) °C two-wire error: 2R_lead ÷ (dR/dt)

R0: 100 Ω for a PT100, 1,000 Ω for a PT1000, both at 0 °C
A: 3.9083 × 10⁻³ per °C
B: −5.775 × 10⁻⁷ per °C² — the curvature, and why a straight line is wrong by tens of degrees at 600 °C
C: −4.183 × 10⁻¹² per °C⁴, active only below 0 °C
alpha: 0.003851 — the mean slope from 0 to 100 °C, the number that names this curve. It is not a coefficient of the equation

Worked example

A PT100 at 100 °C, class B, 1 mA excitation, three-wire
Above 0 °C only the quadratic applies: R = 100 × (1 + 0.0039083 × 100 − 5.775×10⁻⁷ × 100²)
= 100 × (1 + 0.39083 − 0.005775) = 138.5055 Ω — the figure at the top of every PT100 table
Sensitivity there is R₀(A + 2Bt) = 0.3793 Ω/°C, so a milliohm of measurement error is about 2.6 millikelvin
Class B tolerance is ±(0.3 + 0.005 × 100) = ±0.80 °C, which is ±0.3034 Ω at this point
1 mA through 138.5055 Ω dissipates 138.5 µW; at 0.05 °C/mW that is 0.0069 °C of self-heating — negligible here, but not in still air

PT100 and PT1000 at the temperatures worth knowing

Every value computed from the Callendar–Van Dusen equation with the IEC 60751 coefficients. Note how the sensitivity falls as the temperature rises — 0.39 Ω/°C at 0 °C but only 0.33 Ω/°C at 600 °C — which is the curvature the B coefficient describes.
Temperature	PT100 (Ω)	PT1000 (Ω)	Sensitivity (Ω/°C, PT100)	Class B tolerance
-200 °C	18.5201	185.201	0.4323	±1.300 °C
-100 °C	60.2558	602.558	0.4053	±0.800 °C
-50 °C	80.3063	803.063	0.3971	±0.550 °C
0 °C	100.0000	1,000.000	0.3908	±0.300 °C
25 °C	109.7347	1,097.347	0.3879	±0.425 °C
50 °C	119.3971	1,193.971	0.3851	±0.550 °C
100 °C	138.5055	1,385.055	0.3793	±0.800 °C
200 °C	175.8560	1,758.560	0.3677	±1.300 °C
400 °C	247.0920	2,470.920	0.3446	±2.300 °C
600 °C	313.7080	3,137.080	0.3215	±3.300 °C
850 °C	390.4811	3,904.811	0.2927	±4.550 °C

IEC 60751 tolerance classes and where each one is valid

The range is part of the class. A thin-film class AA element is only class AA between 0 and +150 °C; outside that band the specification simply does not apply, and a manufacturer quoting class AA over a wider range is quoting something else.
Class	Tolerance	Wire-wound range	Thin-film range	In practice
Class AA	±(0.10 + 0.0017\|t\|) °C	-50 to +250 °C	0 to +150 °C	±0.10 °C at 0 °C, ±0.27 °C at 100 °C
Class A	±(0.15 + 0.002\|t\|) °C	-100 to +450 °C	-30 to +300 °C	±0.15 °C at 0 °C, ±0.35 °C at 100 °C
Class B	±(0.30 + 0.005\|t\|) °C	-196 to +600 °C	-50 to +500 °C	±0.30 °C at 0 °C, ±0.80 °C at 100 °C
Class C	±(0.60 + 0.01\|t\|) °C	-196 to +600 °C	-50 to +600 °C	±0.60 °C at 0 °C, ±1.60 °C at 100 °C

Why the two halves of the curve are solved differently

A platinum resistance thermometer works because platinum’s resistivity rises with temperature in a way that is repeatable, nearly linear, and — crucially — the same for every manufacturer, because IEC 60751 fixes the curve. A PT100 is 100 Ω at 0 °C and 138.5055 Ω at 100 °C; a PT1000 is the identical curve multiplied by ten. That average slope over the first hundred degrees, 0.003851 per °C, is the α that names the sensor. It is not a coefficient of the equation, and using it as one is the commonest mistake made with RTDs.

How wrong is the straight line? At 600 °C the linear form R₀(1 + αt) gives 331.033 Ω where the standard gives 313.708 Ω — 17.325 Ω out, which at that point is about 53.9 °C. Even at 200 °C the straight line is 3.14 °C out. The quadratic term exists for a reason.

Above zero the inversion is algebra; below zero it is not. For t ≥ 0 the equation is a quadratic in t, so going from resistance back to temperature is the quadratic formula: t = [−A + √(A² − 4B(1 − R/R₀))] / 2B, taking the root that gives t = 0 at R = R₀. Below 0 °C the C(t − 100)t³ term makes it a quartic, and although quartics have a closed form, it is unusable in practice — catastrophic cancellation eats the precision long before the answer appears. So this page iterates instead, and says so. It starts from the quadratic root (which is already within a couple of degrees at −100 °C), applies four Newton steps, and clamps each one into −200…0 °C. The slope dR/dt is strictly positive over that whole interval — between 0.39083 and 0.43234 Ω/°C for a PT100 — so the function is monotonic, there is exactly one root, and Newton converges from any starting point on the correct side. Two steps already reach a nanokelvin; four is margin, not necessity. That was checked here against a 60-step bisection and against a sweep of the forward equation every 0.02 °C.

The three errors that actually get people. First, lead resistance. A two-wire PT100 puts both leads in series with the element, and at 0.3908 Ω/°C every ohm of loop resistance is 2.56 °C of reading, always high. Three-wire wiring cancels it if all three leads match; four-wire removes it by construction, because the sense pair carries no current. A PT1000 divides the same error by ten — 0.256 °C per ohm — which is the real reason to prefer one on a long run. Second, self-heating: the excitation current warms the element it is measuring, by the power times the probe’s self-heating coefficient. A milliamp through a PT100 at 100 °C is only 138.5 µW, but the same probe in still air instead of stirred water can have ten times the coefficient. Third, the tolerance class is a function of temperature and has a range of validity; a class B probe is ±0.3 °C at zero but ±0.8 °C at 100 °C and ±3.3 °C at 600 °C.

The divider or current source around the sensor is ordinary circuit work: see the voltage divider calculator for a resistive divider and the Ohm’s law calculator for the excitation current. For a thermistor instead of an RTD — far more sensitive, far less linear, and cheap — use the NTC thermistor calculator. To send the result somewhere over a pair of wires, the 4–20 mA loop calculator.

Frequently asked questions

What resistance is a PT100 at 100 °C?

138.5055 Ω. The round number people remember is 138.5 Ω. A PT1000 at the same temperature is 1,385.055 Ω, exactly ten times as much.

What is the Callendar–Van Dusen equation?

The polynomial IEC 60751 uses to define platinum resistance against temperature: R = R₀(1 + At + Bt²) above 0 °C, with a third term C(t − 100)t³ added below it. A = 3.9083×10⁻³, B = −5.775×10⁻⁷ and C = −4.183×10⁻¹² for the standard α = 0.003851 sensor.

Why can’t temperature be worked out directly below 0 °C?

Because the equation becomes a quartic in t, and while quartics do have a closed-form solution it loses precision badly at these coefficient sizes. Every practical implementation iterates. This page starts from the quadratic root and takes four Newton steps, which is enough to reach the limit of double precision over the whole −200 to 0 °C range.

How much error does two-wire wiring add to a PT100?

About 2.56 °C for every ohm of total loop resistance near 0 °C, and always in the direction of reading too high. Two 0.5 Ω leads are therefore worth roughly 2.6 °C. Use three-wire if the instrument supports it, four-wire if accuracy matters, or a PT1000, which suffers only 0.256 °C per ohm.

What is the difference between class A and class B?

The tolerance and the range it applies over. Class A is ±(0.15 + 0.002|t|) °C from −100 to +450 °C for a wire-wound element; class B is ±(0.3 + 0.005|t|) °C from −196 to +600 °C. At 100 °C that is ±0.35 °C against ±0.8 °C. Classes AA and C sit either side.

How much excitation current should I use?

As little as the instrument can resolve with. 1 mA is conventional for a PT100 and 0.1 mA for a PT1000, because both dissipate a comparable power. The figure that matters is the self-heating error above: if it is a significant part of your tolerance class, the current is too high.

Related calculators

References

IEC 60751:2022 (and EN 60751). Industrial platinum resistance thermometers and platinum temperature sensors. Coefficients quoted here from WIKA data sheet IN 00.29, Callendar–Van Dusen equations for the calibration of platinum resistance thermometers: R_t = R₀(1 + At + Bt²) for t > 0 °C, R_t = R₀(1 + At + Bt² + C(t − 100)t³) for t < 0 °C, with A = 3.9083 × 10⁻³ °C⁻¹, B = −5.775 × 10⁻⁷ °C⁻² and C = −4.183 × 10⁻¹² °C⁻⁴.
Temperature Developments International. IEC 60751 explained: tolerance classes, W and F designations. Class AA ±(0.1 + 0.0017|t|) °C, class A ±(0.15 + 0.002|t|), class B ±(0.3 + 0.005|t|), class C ±(0.6 + 0.01|t|), each valid only over its stated temperature range — −50 to +250 °C for a wire-wound class AA, 0 to +150 °C for a thin-film one.
Analog Devices. Complete RTD measurement systems — application material on two-, three- and four-wire RTD connections, excitation current and self-heating in the signal chain.
Horowitz P, Hill W. The Art of Electronics, 3rd ed. Cambridge University Press, 2015. §1.3.2 on the decibel and the two factors, and Appendix A on the oscilloscope — what a measured amplitude actually is.