Temperature sensors: thermocouple, RTD, thermistor

Why it matters

Almost every mechatronic product watches a temperature somewhere: motor windings, battery packs, extruder barrels, coolant loops, reflow ovens. Choosing between a thermocouple, an RTD and a thermistor decides the range, accuracy, cost and the signal conditioning you must design. Getting the conversion from voltage or resistance back to temperature wrong gives errors of tens of degrees that look perfectly plausible on a display.

Key ideas

Thermocouple (Seebeck effect). Two dissimilar metals joined at the measuring ("hot") junction, with the free ends at a reference ("cold") junction. A temperature gradient along each conductor produces an EMF, and because the two metals have different Seebeck coefficients a net voltage appears that depends on the temperature difference between the two junctions, not on the hot-junction temperature alone.

• The output is small: about 41 µV/°C for type K (chromel–alumel) near room temperature, about 52 µV/°C for type J, about 40 µV/°C for type T. Exact values come from the standard reference tables (IEC 60584 / NIST ITS-90 tables), because the EMF is not perfectly linear in temperature.
• Cold-junction compensation (CJC): tables are referred to 0 °C. If the reference junction sits at ambient, its temperature is measured (usually by a thermistor or IC sensor on the terminal block) and the equivalent EMF is added back.
• Law of intermediate temperatures: E(T₁, T₃) = E(T₁, T₂) + E(T₂, T₃) — the basis of CJC.
• Law of intermediate metals: a third metal (copper leads, solder) adds no error as long as both its junctions are at the same temperature. Extension or compensating cables keep the right alloy back to the reference point.
• Strengths: very wide range (roughly −200 °C to above 1200 °C for K; higher for R, S, B), rugged, small junction gives fast response, self-powered, cheap. Weaknesses: µV-level signal, needs CJC and linearisation, typical accuracy only ±1–2 °C.

RTD (resistance temperature detector). A pure metal whose resistance rises almost linearly with temperature — almost always platinum (Pt100 = 100 Ω at 0 °C, Pt1000 = 1000 Ω).

• Industrial platinum has α ≈ 0.00385 /°C (the IEC 60751 value), so a Pt100 changes by about 0.385 Ω/°C.
• Over wide ranges the Callendar–Van Dusen equation is used; for 0–100 °C the linear form is adequate for most problems.
• Excitation is a small constant current (≈1 mA) to limit self-heating. Lead resistance adds directly to a 2-wire reading; 3-wire connections cancel matched lead resistance in a bridge, and 4-wire (Kelvin) connections remove it completely.
• Strengths: accurate (±0.1–0.3 °C), stable, repeatable, near-linear. Weaknesses: needs excitation, slower and costlier than a thermocouple, range typically −200 to about 600 °C.

Thermistor. A sintered metal-oxide semiconductor (NTC: resistance falls steeply as temperature rises) or a doped ceramic/polymer PTC device (resistance rises sharply above a switching temperature, used for protection).

• NTC sensitivity is large, about −3 to −5 %/°C near 25 °C versus +0.385 %/°C for platinum, so thermistors resolve small changes cheaply.
• Response is strongly non-linear (exponential); it is described by the β model over a limited range or the Steinhart–Hart equation for precision.
• Range is narrow (typically −50 to 150 °C), interchangeability is modest, and self-heating matters because the element is small.

Dynamic response. A bare sensor in a fluid behaves as a first-order system with time constant τ = m·c / (h·A). A sheath or thermowell adds mass and slows it, so a fast junction can still give a slow reading.

Choosing. High temperature or fast, rugged and cheap → thermocouple. Best accuracy and stability over a moderate range → RTD. Highest sensitivity at low cost over a narrow range (battery packs, motor windings, HVAC) → NTC thermistor.

Formulas

E ≈ S·(T_h − T_c)

• E: thermocouple EMF (V); S: Seebeck coefficient of the pair (V/°C); T_h, T_c: hot and cold junction temperatures (°C). Linear approximation only; use the reference table for accurate work.

E(T_h, 0) = E(T_h, T_c) + E(T_c, 0)

• Cold-junction compensation by the law of intermediate temperatures; all EMFs in V, from the table of the thermocouple type.

R_T = R₀·(1 + α·T)

• R_T: RTD resistance at T (Ω); R₀: resistance at 0 °C (Ω); α: temperature coefficient (/°C, 0.00385 for industrial Pt); T in °C. Valid for platinum over about 0–100 °C; beyond that use Callendar–Van Dusen.

R_T = R₀·exp[β·(1/T − 1/T₀)]

• NTC thermistor β model. R₀: resistance at reference T₀ (usually 25 °C = 298.15 K); β: material constant (K, typically 3000–4500 K from the data sheet); T and T₀ in kelvin.

α_T = (1/R)·(dR/dT) = −β / T²

• Thermistor fractional sensitivity (/K) at absolute temperature T.

T(t) = T_f + (T_i − T_f)·e^(−t/τ) with τ = m·c / (h·A)

• First-order step response; T_i initial, T_f fluid temperature (°C); m mass (kg), c specific heat (J/kg·K), h convection coefficient (W/m²·K), A surface area (m²); τ in s.

Worked examples

Example 1 (standard) — Pt100 reading and lead error. A Pt100 (α = 0.00385 /°C) reads 119.4 Ω on a 4-wire meter. Find the temperature. If the same sensor were connected 2-wire with 0.5 Ω per lead, what error would appear?

1. R_T = R₀·(1 + α·T) → T = (R_T/R₀ − 1)/α.
2. T = (119.4/100 − 1)/0.00385 = 0.194/0.00385 = 50.39 °C.
3. 2-wire: leads add 2 × 0.5 = 1.0 Ω. Sensitivity = R₀·α = 0.385 Ω/°C, so error = 1.0/0.385 = 2.60 °C (reads high).
4. Answer: T ≈ 50.4 °C; a 2-wire connection would read about 2.6 °C high.

Example 2 (GATE level) — cold-junction compensation. A type K thermocouple (take S = 41 µV/°C as linear) gives 3.69 mV at the meter. The terminal block (reference junction) is at 30 °C. Find the hot-junction temperature.

1. EMF of the reference junction relative to 0 °C: E(30, 0) = 41 × 10⁻⁶ × 30 = 1.23 mV.
2. E(T_h, 0) = E(T_h, T_c) + E(T_c, 0) = 3.69 + 1.23 = 4.92 mV.
3. T_h = 4.92 × 10⁻³ / 41 × 10⁻⁶ = 120 °C.
4. Answer: T_h ≈ 120 °C. Ignoring CJC would give 3.69/0.041 = 90 °C — a 30 °C error. (The standard type K table gives 4.920 mV at 120 °C, so the linear model is good here.)

Example 3 (GATE level) — NTC thermistor. An NTC thermistor has R = 10 kΩ at 25 °C and β = 3950 K. It reads 3.5 kΩ. Find the temperature and the sensitivity at 25 °C.

1. From the β model: 1/T = 1/T₀ + ln(R/R₀)/β.
2. 1/T = 1/298.15 + ln(0.35)/3950 = 3.3540 × 10⁻³ − 2.6578 × 10⁻⁴ = 3.0882 × 10⁻³ K⁻¹.
3. T = 323.81 K = 50.66 °C.
4. Sensitivity at 25 °C: α = −β/T₀² = −3950/298.15² = −0.0444 /K.
5. Answer: T ≈ 50.7 °C; sensitivity ≈ −4.4 %/°C, more than ten times that of platinum.

Example 4 — response time. A sensor with τ = 5 s, initially at 25 °C, is plunged into 100 °C water. After 10 s it reads T = 100 + (25 − 100)·e^(−2) = 89.8 °C, still 10 °C low.

Common mistakes

• Using the thermocouple voltage as if the cold junction were at 0 °C. Always add E(T_c, 0) before converting.
• Converting temperatures (not EMFs) when compensating: you must add EMFs, then convert, because the curve is non-linear.
• Putting °C into the thermistor β equation. T and T₀ must be in kelvin.
• Writing R₀(1 + α·T) with R₀ taken at 25 °C. For an RTD, R₀ is at 0 °C.
• Ignoring lead resistance in 2-wire RTDs and self-heating from too much excitation current.
• Assuming a thermistor is "more accurate" because it is more sensitive; sensitivity and accuracy are different things.
• Using ordinary copper wire to extend a thermocouple, which creates a new junction at an uncontrolled temperature.

For GATE ME

Expect NAT problems on RTD resistance–temperature conversion, thermocouple EMF with a non-zero reference junction (law of intermediate temperatures), thermistor β-model calculations in kelvin, and first-order sensor response (time constant, time to reach a fraction of a step). MCQs test which sensor suits a given range or accuracy, the laws of thermocouples, NTC versus PTC behaviour, and 2/3/4-wire RTD connections. Practise the first-order response equation alongside this topic.

Quick check

1. A Pt100 (α = 0.00385 /°C) is at 100 °C. What is its resistance?
2. Why does a thermocouple need cold-junction compensation?
3. In the β equation, which unit must temperature be in?
4. Which RTD wiring scheme eliminates lead resistance completely?
5. A first-order sensor reaches what percentage of a step change after one time constant?

Answers: 1. 138.5 Ω. 2. Its EMF depends on the difference between the hot and reference junction temperatures, so the reference junction's own EMF must be added. 3. Kelvin. 4. 4-wire (Kelvin) connection. 5. About 63.2 %.