Temperature-Voltage Curves Of All Thermocouple Types

Notice that the temperature in the graph is in °C, not even in °K.

0°C is the border between temperatures of water and ice under 1 atm pressure. Why do thermocouples give 0V voltage at a temperature of 0°C? Is there any relationship between water and thermocouples, are the thermocouples designed to behave like this, or is this just a coincidence?

  • \$\begingroup\$ @clabacchio: thermocouples with a supply voltage? I think you talk about a different type of temperature sensor. \$\endgroup\$
    – 0x6d64
    Jun 18, 2012 at 8:34
  • 1
    \$\begingroup\$ @0x6d64: whoops :) \$\endgroup\$
    – clabacchio
    Jun 18, 2012 at 8:36

4 Answers 4


The thermocouple voltage is from the difference in temperature between the sense junction and a reference junction; the reference junction always exists and may be the connection of the thermocouple wires to your voltmeter; note that brass, copper, plating materials all form additional intermediate junctions that cancel if they are at a uniform temperature. Thermocouple tables are based on the reference junction being at 0C (ice-point reference, a convention) so that is why the sense junction temperatures in these tables are all 0V at 0C. Omega Engineering website has great reference documentation on thermocouples that explains all about reference and intermediate junctions. http://www.omega.com/techref/

There are many temperature compensation methods. One is to put the reference junction in ice slush, a stable and known reference point. Another is to produce a voltage to compensate for the actual junction temperature so that the measured voltage will be the "table temperature" with no offsets. Another method is to use an above-ambient and stabilized junction temperature in which case a fixed offset can be used to compensate. While this may all sound complicated, thermocouples are frequently used for extreme temperatures, and the accurate measuring or stabilization of a reference junction can be done with lower temperature electronic methods.


The temperature on the X axis is the temperature difference between the two thermocouple junctions. That's not the absolute temperature. If both junctions are at 0oC, the thermocouple voltage would be 0V. If both junctions are at 100oC, the thermocouple voltage would be 0V still.

  • \$\begingroup\$ +1 and what is obvious, it's hard to get a difference less than zero. ;) \$\endgroup\$
    – kenny
    Jun 18, 2012 at 11:52

The voltage in a thermocouple is a function of the difference between the temperature of the "hot" and "cold" parts of the thermocouple. If there is no temperature difference then there is no voltage. See Wikipedia about thermocouple for details.

That is why that in order to have absolute readings one needs to employ "cold-junction compensation".


All the previous answers say that the voltage from a thermocouple is a function of the temperature difference between the hot and cold junctions. This is a somewhat crude approximation, and is not entirely true.

For example, if the hot junction is at 100°C and the cold junction is at 0°C, the voltage will not be exactly the same as if the hot junction is at 200°C and the cold junction is at 100°C. For a type K thermocouple, the former will be 4.096mV and the latter 4.042mV. The difference may appear to be small (about a 1% error), but it's not negligible in many cases.

By convention, thermocouple voltages are given with a cold junction temperature of 0°C (because it is easy to create a constant accurate temperature of 0°C/32°F with an ice bath). Now, the voltage from a thermocouple with both ends at 0°C must be 0uV (should be obvious from thermodynamics/conservation of energy), so the graphs will always pass through 0 (when 0 is selected as the cold-junction temperature!). Graphs made with °F usually pass through 32, sometimes 75°F.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.