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RTD (Pt100) & Thermistor Calculator

Callendar–Van Dusen for platinum, Beta or Steinhart–Hart for thermistors, both ways.

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  • Free preview: the names of the table’s first rows (up to 3), with the temperature or resistance you convert, the working and every other figure hidden.
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Sensor

Sensor type
Convert
Ω
IEC 60751 platinum, α = 0.00385 /°C.
Sets the temperature range each class covers.
Table range
°C
°C
°C

Result

Temperature —

How it was worked out

    Next steps Done. What next?

    All engineering & electronics calculators

    Results are estimates from the formulas shown, not a professional design or certification. Have a qualified engineer verify anything safety-critical (structures, electrical installations, gas or pressure systems).

    About the RTD (Pt100) & Thermistor Calculator

    Convert the resistance of a temperature sensor into its temperature, or a temperature into the resistance to expect, for two kinds of sensor. Platinum RTDs (Pt100, Pt500, Pt1000 or any R0) use the Callendar–Van Dusen equation with the coefficients of IEC 60751, below 0 °C as well as above. NTC thermistors use the Beta model from R25 and β, or the Steinhart–Hart equation, whose three coefficients the calculator can also fit to three points from a datasheet table.

    For an RTD you also get its sensitivity, the IEC 60751 tolerance of class AA, A, B or C at that temperature (and whether the class covers it for a wire-wound or film element), and the error that two-wire leads add, worked out from their resistance or from the cable’s length and size. A resistance table over any range, in °C, °F or kelvin, completes the result.

    Without a pass, the free preview runs on your own reading: it shows the notes that apply and the temperatures of the table’s first rows, with the temperature or resistance you convert, every other figure and the working hidden; the full result, Copy result and the CSV table come with a Pro pass.

    How to use it

    1. Choose Platinum RTD or NTC thermistor, and whether you have a resistance or a temperature to convert.
    2. For an RTD, pick Pt100, Pt500 or Pt1000 (or enter R0), its tolerance class and construction, and how it is connected. With two wires, enter the resistance of each lead, or work it out from the cable length and conductor size.
    3. For a thermistor, enter R25 and β from the datasheet, or the Steinhart–Hart coefficients A, B and C, or three temperature and resistance points to fit them.
    4. Read the temperature or resistance, the sensitivity and the working; open Table range for the table you need, then copy the result or download the table as CSV — with a pass, or after unlocking this result.

    Examples

    A Pt100 reading
    Input
    Pt100, 110 Ω (four wires)
    Result
    25.68 °C
    Pt100 at 100 °C
    Input
    Temperature → resistance, 100 °C
    Result
    138.506 Ω; class B ±0.80 °C (±0.303 Ω), class A ±0.35 °C
    Below 0 °C
    Input
    Pt100 at −50 °C
    Result
    80.306 Ω (the C term of Callendar–Van Dusen applies below 0 °C)
    Two-wire lead error
    Input
    Pt100 at 0 °C on 10 m of 0.5 mm² copper per lead
    Result
    0.345 Ω per lead, 0.690 Ω in the loop → the reading is 1.77 °C high unless corrected
    A 10 kΩ NTC thermistor
    Input
    R25 = 10 kΩ, β = 3950 K, reading 5 kΩ
    Result
    41.46 °C, falling 4.0 % per °C at that temperature

    Platinum RTDs: Callendar–Van Dusen (IEC 60751)

    • From 0 °C to 850 °C: R = R0·(1 + A·t + B·t²)
    • From −200 °C to 0 °C: R = R0·[1 + A·t + B·t² + C·(t − 100 °C)·t³]
    • A = 3.9083 × 10⁻³ °C⁻¹, B = −5.775 × 10⁻⁷ °C⁻², C = −4.183 × 10⁻¹² °C⁻⁴, giving a mean temperature coefficient α = 0.00385 /°C from 0 °C to 100 °C

    At or above 0 °C the calculator solves the quadratic exactly; below 0 °C it solves the quartic with Newton’s method. A Pt100 is 100 Ω at 0 °C and changes by about 0.39 Ω per °C; a Pt1000 changes ten times as much, which makes lead resistance ten times less important.

    Tolerance classes

    IEC 60751 limits the deviation of a sensor from the table to ±(a + b·|t|):

    • Class AA (W 0.1 / F 0.1): ±(0.1 °C + 0.0017·|t|), wire-wound −50 °C to 250 °C, film 0 °C to 150 °C
    • Class A (W 0.15 / F 0.15): ±(0.15 °C + 0.002·|t|), wire-wound −100 °C to 450 °C, film −30 °C to 300 °C
    • Class B (W 0.3 / F 0.3): ±(0.3 °C + 0.005·|t|), wire-wound −196 °C to 600 °C, film −50 °C to 500 °C
    • Class C (W 0.6 / F 0.6): ±(0.6 °C + 0.01·|t|), wire-wound −196 °C to 600 °C, film −50 °C to 600 °C

    Outside its range a class does not apply: the calculator says so and shows the figure in amber.

    Two, three or four wires

    With two wires the instrument measures the sensor plus both leads, and every ohm of lead looks like about 2.6 °C on a Pt100. Copper’s resistance also changes with its own temperature, so the error moves with the cable. Three wires let the instrument subtract the lead resistance, which cancels when the leads are equal. Four wires measure the voltage at the sensor itself, so leads add no error. The cable helper uses annealed copper at 20 °C (1/58 Ω·mm²/m) and the standard AWG diameters.

    NTC thermistors: Beta and Steinhart–Hart

    • Beta model: 1/T = 1/T0 + ln(R/R0)/β, with T in kelvin. It is exact at T0 and drifts away from it, because β itself changes with temperature: that is why datasheets name the pair of temperatures β was measured between (B25/50, B25/85).
    • Steinhart–Hart: 1/T = A + B·ln R + C·(ln R)³ (Steinhart and Hart, Deep-Sea Research). Three coefficients follow a thermistor’s curve far better over a wide range. The calculator fits them through three points with the closed-form solution of the three equations, and inverts the equation for R as a cubic in ln R (Cardano’s formula, or the trigonometric solution when a fit gives a slightly negative C).
    • Temperature coefficient: α = (1/R)·dR/dT, about −4.4 % per °C at 25 °C for β = 3950 K.

    Limitations

    • Results are for the sensor itself. Self-heating from the measuring current, the instrument’s own accuracy and thermal contact with what you measure all add error.
    • RTD conversions follow IEC 60751 platinum (α = 0.00385 /°C). Platinum sensors to older or other curves (α = 0.003916 or 0.003925 /°C) need their own coefficients.
    • The lead helper assumes copper at 20 °C. The three-wire case assumes equal leads; the difference between unequal leads adds directly to the reading.
    • The Beta model and a three-point fit are only as good as the data behind them; for the best accuracy use points spread over the range you measure.

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    Frequently asked questions

    What do I get without a pass?

    Without a pass, RTD (Pt100) & Thermistor Calculator shows the names of the table’s first rows (up to 3), with the temperature or resistance you convert, the working and every other figure hidden. Until you unlock it, the result can’t be downloaded or copied. A Pro, Premium or Ultimate pass, a one-time payment that never renews, unlocks the full result. The pricing page lists the passes and their prices.

    How do I convert Pt100 resistance to temperature?

    Above 0 °C, solve R = 100·(1 + A·t + B·t²) for t: t = [−A + √(A² − 4B·(1 − R/100))] ÷ (2B), with A = 3.9083 × 10⁻³ and B = −5.775 × 10⁻⁷. For 110 Ω that gives 25.68 °C. Below 0 °C the extra C term needs an iterative solution, which the calculator does for you.

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

    100 Ω at 0 °C and 138.506 Ω at 100 °C; a Pt1000 is 1000 Ω and 1385.06 Ω. The average change is 0.385 Ω per °C for a Pt100.

    How much error do two-wire leads add?

    Their resistance in the loop divided by the sensor’s sensitivity: on a Pt100 every ohm of lead resistance reads as about 2.6 °C. 10 m of 0.5 mm² copper per lead (0.69 Ω in the loop) makes a Pt100 at 0 °C read 1.77 °C high. Use three or four wires, a Pt1000, or subtract the lead resistance.

    What is β for a thermistor?

    β (the B value) describes how steeply an NTC thermistor’s resistance falls with temperature: β = ln(R1/R2) ÷ (1/T1 − 1/T2) with T in kelvin. Datasheets give it between two temperatures, often 25 °C and 50 °C or 85 °C. A larger β means a steeper curve.

    How do I find Steinhart–Hart coefficients?

    Take three temperature and resistance pairs from the thermistor’s table, ideally near the ends and the middle of the range you measure, and choose Fit Steinhart–Hart to three points. The calculator solves for A, B and C and uses them for the conversion and the table.

    Quick answers and tool search

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