Thermocouples: laws and cold junction compensation
Seebeck, Peltier and Thomson effects, the three thermocouple laws, common thermocouple types, and cold-junction compensation using emf addition, with worked numericals.
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Why it matters
Thermocouples measure most industrial temperatures above a few hundred degrees — furnaces, boilers, gas turbines, kilns, plastic extruders. They are cheap and rugged, but a thermocouple measures only a temperature difference, so every reading depends on knowing the reference-junction temperature and on wiring the circuit according to the thermocouple laws.
Key ideas
Thermoelectric effects.
- Seebeck effect: in a circuit of two dissimilar conductors A and B with junctions at different temperatures, a net emf appears. Physically, each wire develops an emf along its temperature gradient; because A and B respond differently, the loop emf does not cancel. The net emf depends only on the two junction temperatures (for homogeneous wires).
- Peltier effect: current through a junction of dissimilar metals absorbs or releases heat (the basis of thermoelectric coolers).
- Thomson effect: current along a single conductor with a temperature gradient absorbs or releases heat. The Seebeck coefficient S = dE/dT (µV/°C) is not constant; the E–T relation is non-linear, so accurate work uses standard reference tables or polynomials (emf referred to 0 °C).
Laws of thermocouples.
- Homogeneous circuit: in a wire of uniform composition, temperature gradients along its length do not change the net emf; only the junction temperatures matter. (A wire with inhomogeneities breaks this.)
- Intermediate metals: a third metal inserted in the circuit adds no emf if both its junctions are at the same temperature. This is why copper meter leads, terminals and solder do not upset the reading if the connection block is isothermal.
- Intermediate (successive) temperatures:
E(T1, T3) = E(T1, T2) + E(T2, T3). This lets you use tables referred to 0 °C when the reference junction is at some other temperature. A corollary of the intermediate-metals law givesE_AB = E_AC + E_CBat the same temperatures, so any pair can be characterised against platinum.
Common types (letter codes standardised; approximate figures — take exact emf values from reference tables):
- K (chromel–alumel): general purpose, about −200 °C to 1250 °C, about 41 µV/°C.
- J (iron–constantan): about 0–750 °C, about 52 µV/°C; iron oxidises.
- T (copper–constantan): about −200 °C to 350 °C; good at low temperatures.
- E (chromel–constantan): highest sensitivity, about 68 µV/°C.
- N (nicrosil–nisil): more stable than K at high temperature.
- R, S, B (platinum–rhodium): up to about 1600–1800 °C, low sensitivity (about 6–10 µV/°C), expensive, very stable.
Cold (reference) junction compensation. Tables assume the reference junction is at 0 °C.
- Ice-point bath: hold the reference at 0 °C (laboratory standard).
- Hardware compensation: a bridge or circuit containing a temperature-sensitive element (RTD, thermistor, IC sensor) at the terminal block adds a voltage equal to E(T_ref, 0).
- Software compensation: measure the terminal-block temperature with a sensor, convert it to emf from the table, add it to the measured emf, and convert the sum back to temperature using the table. Always add voltages, not temperatures — the E–T curve is non-linear.
- Extension and compensating cables carry the thermocouple materials (or cheaper alloys with matching emf over a limited range) from the head to the instrument, so the reference junction is effectively moved to the instrument terminals. Ordinary copper cable in between moves the reference junction to the thermocouple head, whose temperature is unknown.
Thermopile: N thermocouples in series multiply the output by N (used in radiation pyrometers and heat-flux sensors).
Formulas
E = S·(T_hot − T_ref) (approximation over a small range)
E = emf (V); S = mean Seebeck coefficient (V/°C).
E(T_hot, 0) = E(T_hot, T_ref) + E(T_ref, 0) (law of intermediate temperatures — the CJC equation)
E(T, 0) = table emf for temperature T referred to 0 °C (mV).
E_AB(T1, T2) = E_AC(T1, T2) + E_CB(T1, T2) (intermediate-metals corollary)
E_pile = N·S·ΔT (thermopile of N junction pairs)
Worked examples
Example 1 — cold-junction correction with tables. A type-K thermocouple gives 11.209 mV with its reference junction at 25 °C. Reference-table data (0 °C reference): E(25 °C) = 1.000 mV, E(300 °C) = 12.209 mV. Find the hot-junction temperature.
- Law of intermediate temperatures:
E(T, 0) = E(T, 25) + E(25, 0). E(T, 0) = 11.209 + 1.000 = 12.209 mV.- From the table, 12.209 mV corresponds to 300 °C.
- Note the correction is made in millivolts. Converting 11.209 mV to a temperature first and then adding 25 °C is wrong in principle, because the E–T curve is not a straight line; the error is small for type K near room temperature but can be several degrees for other types and ranges.
Example 2 — uncompensated error (GATE level). Take S = 41 µV/°C (constant). A thermocouple reads 8.0 mV; its reference junction is at 30 °C, but the instrument assumes 0 °C. Find the indicated and true temperatures.
- Indicated:
T = 8.0 mV / 0.041 mV/°C = 195.1 °C. - Compensation emf:
E(30, 0) = 0.041 × 30 = 1.23 mV. - True:
E(T, 0) = 8.0 + 1.23 = 9.23 mV→T = 9.23/0.041 = 225.1 °C.
Answer: indicated 195.1 °C, true 225.1 °C — the error equals the reference-junction temperature, 30 °C.
Example 3 — thermopile. Ten type-K junction pairs in series see a 5 °C difference. E = 10 × 41 µV/°C × 5 °C = 2.05 mV.
Common mistakes
- Thinking the junction itself "generates" the voltage — the emf arises along the wires in the temperature gradient; the junction temperatures set the net value.
- Adding the reference temperature to the indicated temperature instead of adding the reference emf to the measured emf.
- Joining the thermocouple to the instrument with copper wire and assuming the reference is at the instrument.
- Reversing polarity of extension wire — this creates two errors that add.
- Treating S as constant over a large range.
- Ignoring loading: a long thermocouple loop has resistance; use a high-impedance readout.
For GATE IN
Questions use the law of intermediate temperatures with table values or a constant S, ask for indicated vs true temperatures with an uncompensated reference, test the three laws as MCQs, and sometimes combine thermocouples with amplifiers (gain needed for a given mV/°C output). Practise always converting temperatures to emf before adding.
Quick check
- State the law of intermediate temperatures.
- A thermocouple with S = 50 µV/°C has hot junction 200 °C and reference 30 °C. Emf?
- Why do copper leads at an isothermal terminal block not affect the reading?
- Which thermocouple type is used above 1500 °C? Answers: 1. E(T1, T3) = E(T1, T2) + E(T2, T3); 2. 8.5 mV; 3. by the law of intermediate metals, a third metal with both junctions at the same temperature adds no emf; 4. platinum–rhodium types such as B, R or S.
Interview questions
All Sensors and Transducers interview questionsTry answering each one aloud before you open it.
1.What is a thermocouple and how does it work?Concept
A thermocouple is two dissimilar metal wires joined at a measuring (hot) junction, with the other ends at a reference (cold) junction. Because each wire develops a different thermoelectric emf along its temperature gradient (the Seebeck effect), a net emf appears that depends on the temperature difference between the two junctions, about 41 µV/°C for type K. The reference-junction temperature must therefore be known or compensated. Thermocouples are simple, rugged, self-powered and cover roughly −200 °C to 1800 °C depending on type.
2.Explain the Seebeck effect in the context of thermocouples.Concept
When a circuit of two dissimilar conductors has its junctions at different temperatures, a net emf appears: this is the Seebeck effect. The emf depends only on the junction temperatures (for homogeneous wires) and on the pair of materials. Its slope, the Seebeck coefficient, varies with temperature, so the emf–temperature relation is non-linear and accurate work uses standard reference tables referred to 0 °C.
3.What are the laws of thermocouples and why are they important?Concept
The laws of thermocouples include the Law of Homogeneous Materials, the Law of Intermediate Metals, and the Law of Successive or Intermediate Temperatures. These laws are important because they help in understanding and predicting the behavior of thermocouples in different configurations. They ensure that the thermocouple readings are accurate and consistent, even when additional materials or junctions are introduced in the circuit.
4.Why is cold junction compensation necessary in thermocouples?Concept
Cold junction compensation is necessary because the voltage generated by a thermocouple is based on the temperature difference between the hot junction and the reference junction (cold junction). Since the reference junction is usually at a different temperature than the standard reference temperature (0°C), compensation is needed to account for this difference and ensure accurate temperature readings.
5.How is cold junction compensation typically achieved in thermocouple circuits?Application
Cold junction compensation is typically achieved by using a temperature sensor, such as a thermistor or RTD, at the reference junction to measure its temperature. The measured temperature is then used to adjust the thermocouple voltage reading, effectively simulating a reference junction at 0°C. This adjustment ensures that the temperature reading is accurate.
6.What happens if the cold junction temperature is not compensated in a thermocouple measurement?Application
If the cold junction temperature is not compensated, the temperature reading from the thermocouple will be inaccurate. This is because the voltage generated by the thermocouple is based on the temperature difference between the hot and cold junctions. Without compensation, the reading will not account for the actual temperature at the cold junction, leading to errors.
7.Why are thermocouples preferred over other temperature sensors in high-temperature applications?Application
Thermocouples are preferred in high-temperature applications because they can withstand extreme temperatures, often up to 1800°C or more, depending on the materials used. They are also robust, have a fast response time, and are relatively inexpensive compared to other high-temperature sensors. These characteristics make them suitable for industrial and scientific applications where high temperatures are common.
8.A thermocouple produces a voltage of 2 mV when the temperature difference between its junctions is 50°C. What is the Seebeck coefficient of this thermocouple?Numerical
The Seebeck coefficient (S) can be calculated using the formula S = V / ΔT, where V is the voltage and ΔT is the temperature difference. Here, V = 2 mV and ΔT = 50°C. Therefore, S = 2 mV / 50°C = 0.04 mV/°C or 40 µV/°C.
9.Explain how the Law of Intermediate Metals is applied in thermocouple circuits.Application
The Law of Intermediate Metals states that the introduction of a third metal into a thermocouple circuit will not affect the voltage generated, as long as the junctions between the metals are at the same temperature. This law is applied in thermocouple circuits to allow for connections and extensions without affecting the measurement accuracy. It ensures that additional materials do not introduce errors, provided they are at a uniform temperature.
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