Resistance temperature detectors

Why it matters

The platinum RTD is the reference-quality industrial temperature sensor: Pt100 probes control pharmaceutical reactors, food pasteurisers, HVAC plants and laboratory baths where a few tenths of a degree matter. Getting that accuracy depends on three things you control — the R–T equation you use, how you wire the leads and how much current you push through the element.

Key ideas

Principle. The resistance of a pure metal rises with temperature because lattice vibrations scatter conduction electrons more. An RTD is a precisely made metal element whose resistance–temperature relation is standardised.

• Platinum is preferred: chemically inert, very stable and repeatable, nearly linear, usable from about −200 °C to 600–850 °C depending on construction. Industrial Pt100 (100 Ω at 0 °C) and Pt1000 elements follow IEC 60751 with mean coefficient α = 0.00385 Ω/Ω/°C between 0 and 100 °C.
• Nickel has a higher α (more sensitive) but is less linear and limited to about 300 °C; copper is very linear but limited to about 150 °C and has low resistivity, so elements are bulky.
• Construction: wire-wound (strain-free coil in ceramic, most stable) or thin-film (platinum film on ceramic, cheap, small, fast). Both are sheathed in a thermowell for process use, which slows response.

Resistance–temperature relation. Over 0–100 °C the linear form R_T = R0(1 + αT) is adequate. Over wider ranges use the Callendar–Van Dusen equation: R_T = R0(1 + A·T + B·T²) for T ≥ 0 °C, with an extra C term below 0 °C. The standard IEC coefficients are A = 3.9083 × 10⁻³ °C⁻¹ and B = −5.775 × 10⁻⁷ °C⁻² (take from the standard or the sensor datasheet). α is always referred to 0 °C — do not apply it from another base temperature.

Measurement and error sources.

• Lead resistance: in a 2-wire connection both leads add directly to the RTD (0.385 Ω is 1 °C for a Pt100). A 3-wire connection places one lead in the RTD arm and one in the adjacent bridge arm so equal leads cancel. A 4-wire connection drives a constant current through two leads and senses the voltage with the other two (which carry no current), eliminating lead resistance entirely.
• Self-heating: the measuring current dissipates P = I²R in the element, raising its temperature by ΔT = P/δ, where δ is the dissipation constant (mW/°C) — it depends on the probe and the medium (much larger in moving water than in still air). Keep current around 1 mA for a Pt100.
• Other errors: thermal response time (τ of seconds to tens of seconds in a thermowell), stem conduction (immersion depth too short), strain on the element, insulation leakage at high temperature.
• Tolerance classes in IEC 60751 (Class A, Class B, etc.) set the permitted interchangeability error.

Comparison with thermocouples. RTDs are more accurate, stable and linear, need no reference junction, and measure absolute temperature. Thermocouples are cheaper, faster, rugged, self-powered and reach much higher temperatures. Compared with thermistors, RTDs are less sensitive but far more linear and cover a much wider range.

Formulas

R_T = R0·(1 + α·T) (linear approximation, 0–100 °C) R_T = resistance at T (Ω); R0 = resistance at 0 °C (Ω); α = mean temperature coefficient (Ω/Ω/°C); T in °C.

R_T = R0·(1 + A·T + B·T²) (Callendar–Van Dusen, 0 °C ≤ T ≤ 850 °C) A = 3.9083 × 10⁻³ °C⁻¹; B = −5.775 × 10⁻⁷ °C⁻² (IEC 60751 platinum).

α = (R100 − R0) / (100·R0)

ΔT_SH = I²·R_T / δ ΔT_SH = self-heating error (°C); I = measuring current (A); δ = dissipation constant (W/°C).

T_err = ΔR_lead / (α·R0) (2-wire lead error) ΔR_lead = total lead resistance in the loop (Ω).

Worked examples

Example 1 — linear vs Callendar–Van Dusen. A Pt100 is at 200 °C. Find R from the CVD equation, R from the linear formula, and the temperature you would infer from the true resistance using the linear formula.

1. CVD: R = 100(1 + 3.9083 × 10⁻³ × 200 − 5.775 × 10⁻⁷ × 200²) = 100(1 + 0.78166 − 0.0231) = 175.86 Ω.
2. Linear: R = 100(1 + 0.00385 × 200) = 177.00 Ω.
3. Inferring T linearly from 175.86 Ω: T = (175.86 − 100)/(0.385) = 197.0 °C.

Answer: 175.86 Ω (true), 177.00 Ω (linear); linear inversion reads about 3 °C low at 200 °C.

Example 2 — leads and self-heating (GATE level). A Pt100 at 100 °C (R = 138.5 Ω) is connected by two leads of 0.5 Ω each. (a) Find the 2-wire temperature error. (b) The dissipation constant in the process fluid is 2 mW/°C. Find the self-heating error at 1 mA and at 5 mA.

1. (a) Lead resistance in series = 1.0 Ω. T_err = 1.0/(0.00385 × 100) = 2.6 °C (reads high).
2. (b) At 1 mA: P = (1 × 10⁻³)² × 138.5 = 0.1385 mW; ΔT = 0.1385/2 = 0.069 °C.
3. At 5 mA: P = (5 × 10⁻³)² × 138.5 = 3.46 mW; ΔT = 3.46/2 = 1.73 °C.

Answer: (a) +2.6 °C; (b) 0.07 °C at 1 mA, 1.73 °C at 5 mA — current matters as the square.

Example 3 — finding α. A thin-film RTD reads 200 Ω at 0 °C and 315.5 Ω at 150 °C. Mean α = (315.5 − 200)/(200 × 150) = 0.00385 °C⁻¹.

Common mistakes

• Applying α from a base other than 0 °C: given R at 25 °C, first find R0 = R25/(1 + 25α), then R at the new temperature.
• Using the linear formula far beyond 100 °C without saying it is an approximation.
• Ignoring lead resistance in a 2-wire circuit; 1 Ω is about 2.6 °C for a Pt100.
• Increasing the excitation to get more signal and creating self-heating error.
• Forgetting that a 3-wire connection only cancels lead resistance if the leads are equal.
• Confusing R0 (at 0 °C) with the resistance at room temperature.

For GATE IN

NAT questions: R at a temperature (linear or with a given quadratic coefficient), temperature from a measured R, α from two readings, lead-wire and self-heating errors, bridge outputs with an RTD arm. MCQs: why platinum, 2/3/4-wire comparisons, RTD vs thermocouple vs thermistor. Practise the bridge output with an RTD arm, which combines this topic with bridges.

Quick check

1. A Pt100 reads 119.25 Ω. Using α = 0.00385, what is the temperature?
2. Which wiring scheme eliminates lead resistance completely?
3. At 2 mA, R = 100 Ω and δ = 4 mW/°C, what is the self-heating error?
4. A Pt100 reads 110 Ω at 25 °C. Is that consistent with α = 0.00385? Answers: 1. 50 °C; 2. 4-wire; 3. 0.4 mW/4 = 0.1 °C; 4. no — a standard Pt100 should read 109.6 Ω at 25 °C, so this sensor has an offset of about 1 °C.