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13 April 2010
Author: Giorgos Lazaridis
How Thermocouples (TCs) Work


A very simple thermocouple wire. One wire is made of Chromium and the other of Aluminum. Both are Nickel alloys.

In 1821 Thomas Johann Seebeck (1770-1831) noticed that, if two wires from different material are connected at their endings performing two different junctions, and these junctions do not have the same temperature, then placing these wires near by a compass, would cause the compass to move. At first he believed that the temperature is the reason to change the magnetic field. Later on, he understood that a current was flowing within the wires that generated a magnetic field and moved the compass. That called the Seebeck Effect

The Seebeck Effect is one of the three thermoelectric phenomena and the first to be discovered. Along with the Peltier effect and the Thomson effect, it explains how the temperature difference can create a voltage.

How Thermocouple works - A simple example

I have also a page with thermocouple experiments. Visit this link: Experimenting with Thermocouples.

The Seebeck effect is present whenever two dissimilar metals -of any material- performs a junction. Nevertheless, there are some metal pairs that have a predictable voltage according to temperature, and have also larger temperature gradients. These pairs are named with a letter, as for example the type-E thermocouple. The most popular type of thermocouple is the type-K. This type is made of Nickel-Chromium versus Nickel-Aluminum wires and has a very wide range of temperature gradients (from -200oC up to 1300oC) where the voltage changes almost linear.

The following drawing demonstrated a thermocouple. The left junction is heated. The right junction is kept at room temperature. Thus, a current flows within the wires:

As someone would expect, if the right junction does not exists, a voltage difference shall be created across the wire endings:

The three Thermo-Electric Laws

There are three empirical Thermo-Electric Laws to explain the operation of the thermocouple:

1. The Law of Homogeneous Circuits
An electric current cannot be sustained in a circuit of a single homogeneous metal

Look at the two following images. The two circuits are identical and the temperature in the junction is equal. Both will generate a voltage difference. On the right circuit, a second heater is placed under one of the two thermocouple wires and heat it up. The first law states that, because the second heater heats ONLY one wire and NOT the junction, the output voltage of both circuit will be the same. The output voltage is only affected by the junction temperature and not the temperature of the wires. That is because, any temperature change to a homogeneous wire will create no voltage.

2. The Law of Intermediate Metals
If two dissimilar metals performs a thermocouple, and a third dissimilar metal is introduced to the circuit, as long as the temperature along the entire length of the third metal is kept uniform, the output voltage will NOT be affected

Look at the following circuit. The blue wire of the thermocouple is joined with a third dissimilar metal. If the temperature of this wire across it's entire length is the same, then the output voltage will NOT be affected by this insert.

3. The Law of Intermediate Temperatures
If a thermocouple with 2 junctions with temperatures T1 and T2 produces a voltage difference V1, and voltage difference of V2 in temperatures T2 and T3, then voltage generated when the temperatures are T1 and T3 will be V1+V2.

This law is very important to understand how to acquire the temperature from a thermocouple. Usually, the temperature to voltage characteristic of a thermocouple (as you will see later on) is given for 0oC (32oF) reference temperature. Look at the following 2 circuits. The thermocouple is a K-type thermocouple, and the voltmeter uses copper wires. At the point that the voltmeter is connected to measure the voltage, two new thermocouple junctions are created! This is a major problem of thermocouples and will be discussed in details later on. At the left circuit, these new junctions are kept into 0oC (32oF). This way, the measured voltage can be directly converted into temperature, as the table to convert it is usually with reference temperature 0oC. But in real life, the reference temperature is not 0oC. Looking at the right circuit you can see a more realistic example. The reference temperature now it 20oC. The voltage cannot be directly converted into temperature as the reference junction (where the voltmeter is connected) is not at 0oc. According to the 3rd law, if we know the reference temperature, then we can calculate the measured temperature, by adding the measured voltage and another voltage named VREF. This VREF is the voltage that would be created by the thermocouple, if the non-zero reference voltage was measured by a thermocouple junction with reference temperature 0oC. I know it is complicated. Frankly, it is not!

Compensating the reference temperature

An example: For the left circuit, suppose that T1 = 0oc (first reference temperature). According to the measured voltage, we can directly convert to temperature for T2. In our example, for a K-type thermocouple, the measured voltage would be 4.095mV, that corresponds to 100oC. But a measuring instrument is rarely placed in 0oC. On the second circuit (right side), the measuring instrument is placed in a room with 20oC. Although the temperature T2 is still 100oC, the measured voltage will be now 3.297 mV! Why? Because the junctions where the voltmeter is connected is NOT in 0oC, thus they introduce another opposite current to the circuit!

To calculate the T2, we need to compensate T3! According to the 3rd law, we need to add to the measured voltage, the voltage that would be generated if T2 was T3 and T3 was 0oC. In this case, the voltage can be directly found from the table (as we are talking about 0oC reference temperature) and would be 0.798 mV. So, we add this value to the measured voltage and the total voltage is 0.798+3.297 = 4.095 mV, and this corresponds to 100oC!

The Problem of measuring temperature with thermocouples

Any junction between two dissimilar metals, will perform a thermocouple! That is where the whole problem begins. Suppose that we have a K-type thermocouple. We will need to connect both wires of the pair to copper wires, either because we want to extend the wires, or because we want to connect the multimeter. These connections will create another thermocouple pair. The following drawing demonstrate this circuit:

The red and blue wires comes from the K-type thermocouple. The orange wires are copper wires. Thus, due to the first law, T4 and T5 will not create any difference as the wires are always copper. So, we can re-write the circuit as follows:

The problem is that, at the points where the thermocouple wires are connected with the copper wires, two new thermocouples are created! Thus, these thermocouples will create EMF (Electro Motive Forces) to the circuit and alter the readings.



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  • At 25 October 2015, 6:59:54 user Giorgos Lazaridis wrote:   [reply @ Giorgos Lazaridis]
    • @Joseph the output is DC so you cannot step-it up with a transformer. You can amplify it with an OP-AMP though.
      You can indeed use pipes to increase the temperature, but... why?

  • At 23 October 2015, 5:49:14 user Joseph wrote:   [reply @ Joseph]
    • Is it possible to use some form of transformer to step-up the voltage?
      One other idea: Is it possible to use black painted pipes to create a lot of heat using solar as the power?

  • At 8 October 2014, 2:56:20 user Kelvin wrote:   [reply @ Kelvin]
    • If a thermocouple with 2 junctions with temperatures T1 and T2 produces a voltage difference V1, and voltage difference of V2 in temperatures T2 and T3, then voltage generated when the temperatures are T1 and T3 will be V1 V2.

      The formulae should be: V1-V2

      T2-T1 = V1 (100'C - 0'C)
      T2-T3 = V2 (100'C - 20'C)

      Substitute: T3-T1 = V1-V2 (20'C - 0'C)

  • At 13 September 2013, 13:15:28 user Nhuan wrote:   [reply @ Nhuan]
    • @Giorgos Lazaridis
      Many Thanks Giorgos, I'll keep that in mind.

  • At 13 September 2013, 6:35:55 user Giorgos Lazaridis wrote:   [reply @ Giorgos Lazaridis]
    • @Nhuan I suppose (but have not test) it will be as if you have 2 TCs connected in parallel. The lower temperature will appear. The higher temperature with higher voltage will be consumed as heat dissipation in the closed branch of the two shorted points (as if you have 2 voltage sources of different voltage connected in parallel).

  • At 12 September 2013, 20:49:35 user Nhuan wrote:   [reply @ Nhuan]
    • Hope you can understand my "illustration" below...
      What is the effect, if any, if the thermocouple is shorted at any point (e.g."B") before the tip ("A")?

      A B [measurement?]

  • At 5 August 2013, 1:50:32 user none wrote:   [reply @ none]
    • Excellent explanation.

  • At 13 April 2013, 9:43:39 user Giorgos Lazaridis wrote:   [reply @ Giorgos Lazaridis]
    • @Awsome just twist them together...

  • At 11 April 2013, 9:51:48 user Awsome wrote:   [reply @ Awsome]
    • Ok I am trying to build this exact circuit I know you used Nickel Chromium and Nickel Aluminum wires but what is the thermocouple copper junction made of or what do I do to create one?


  • At 25 March 2013, 10:31:50 user Debjit wrote:   [reply @ Debjit]
    • Your website and you both are fabulous,keep up the good work...
      And thanks for these articles..

  • At 17 October 2012, 2:03:05 user Jackie wrote:   [reply @ Jackie]
    • Fabulous article!

  • At 5 September 2012, 17:44:39 user Ronnie wrote:   [reply @ Ronnie]
    • the very last line of explaining the third law, "... and this corresponds to 20 degrees Celsius", should actually be "... and this corresponds to 100 degrees Celsius".

  • At 8 August 2012, 22:37:45 user Giorgos Lazaridis wrote:   [reply @ Giorgos Lazaridis]
    • @Dinesh Dhumal 0V

  • At 8 August 2012, 5:31:20 user Dinesh Dhumal wrote:   [reply @ Dinesh Dhumal]
    • without heating the hot junction,at normal ambient temperature what will be the mv?that time what temperature should we read

  • At 10 July 2012, 5:31:35 user Giorgos Lazaridis wrote:   [reply @ Giorgos Lazaridis]
    • @Alan the and - goes to an amplifier, you may choose how to connect them (inverting/non-invertig). The PTC increases its resistance with temperature. To if you connect it in series with a supply, the current will drop.

  • At 9 July 2012, 12:52:46 user Alan wrote:   [reply @ Alan]
    • Giorgos

      Wow, that was relatively easy to follow for a layman. The thermocouple is polarized! Yes, so when coupling a thermocouple in a soldering iron to its control amp in its controller. The of the thermocouple goes to of the op-amp? and the - of thermocouple goes to the op-amp -. Is this correct? A PTC will increase the output (millivolts) as the temperature rises?

  • At 12 May 2012, 7:32:40 user VIJAY wrote:   [reply @ VIJAY]

      CONTACT 0-8008325089

  • At 27 February 2012, 10:26:32 user Nikos wrote:   [reply @ Nikos]
    • Very interesting web site!
      Actually I was trying to understand how is possible to depict the thermocouple mV indication on the measuring point using a commercial thermocouple meter/calibrator.Normally these devices depict mV difference between measuring point V1 and junction point V2. The only way to measure (not calculate!) the V1 voltage is by using electronic ice point devices which output the V1 voltage canceling out the error introduced by the cooper wires.

  • At 30 January 2012, 17:01:33 user Giorgos Lazaridis wrote:   [reply @ Giorgos Lazaridis]
    • @Johnny no it is not possible. one thermocouple can provide feedback to one controller. it is up to the controller to have more fuctions. a controller may have for example 2 outputs or more.

  • At 30 January 2012, 3:44:53 user Johnny wrote:   [reply @ Johnny]
    • Excellent! Could someone please let me know if it's possible to connect a single thermocouple but with two outputs? e.g. e thermocouple is connected to the heater and the other end connected to 2 temp displays/meter. Thanks very much.

  • At 3 November 2011, 20:28:26 user Blue Bamboo wrote:   [reply @ Blue Bamboo]
    • I have been using thermocouples for more than 10 years but did not understand the mechanism. After reading this article everything became so clear.

      Thanks Giorgos.

  • At 7 October 2011, 2:08:41 user Nick wrote:   [reply @ Nick]
    • This sentence in the "3." section should read 100* that corresponds to 4.095mV as it said above this statement.

      So, we add this value to the measured voltage and the total voltage is 0.798+3297 = 4.095 mV, and this corresponds to 20oC!

      Sorry to nitpick but it adds confusion.

  • At 21 September 2011, 18:23:37 user Kammenos wrote:   [reply @ Kammenos]
    • @dharamvir your answer is in the theory page (go to my theory pages and read about thermocouples)

  • At 21 September 2011, 17:38:11 user dharamvir wrote:   [reply @ dharamvir]
    • dear sir,

      how a maximum voltage i can get from a 500 dig. heat through which thermocouple.

      i want to generate a current fro fine thermocouple, my working hear is min 100 dig. to max. 500 dig.

      please guide me in detail

  • At 19 September 2011, 7:59:25 user SATISH TUMBARE wrote:   [reply @ SATISH TUMBARE]
    • Thanks first time i clearly undrstand T/C

  • At 22 April 2011, 16:47:48 user Kammenos wrote:   [reply @ Kammenos]
    • @vino that is exactly what you need to see, no reading. If you compensate the room temperature and add it to your reading, then you get the real temperature, which in your case is the room temperature. When you heat the junction, you get the added temperature to the room temperature.
      If for example room is 20 degrees and you heat the junction to 100 degrees, you must read 80 (100-20)

  • At 22 April 2011, 16:41:48 user vino wrote:   [reply @ vino]
    • without heating the hot junction of thermocouple,at room temperature can see any reading ,can\'t means why

  • At 14 April 2011, 21:21:01 user Kammenos wrote:   [reply @ Kammenos]
    • @ashwin well, no you cannot light an LED efficiently. You can use a peltier thermoelement though. Read here: http://pcbheaven.com/wikipages/The_Peltier_Thermo-Element

  • At 14 April 2011, 14:01:14 user ashwin wrote:   [reply @ ashwin]
    • its very good,......can we light a led light using this thermocouple

  • At 9 February 2011, 17:56:45 user Kammenos wrote:   [reply @ Kammenos]
    • "The Seebeck effect is present whenever two dissimilar metals -of any material- performs a junction."

      taken from the theory page (http://pcbheaven.com/wikipages/How_Thermocouples_Work/)
      It is a natural process, called Thermoelectric effect

  • At 8 February 2011, 13:54:34 user firdaus wrote:   [reply @ firdaus]
    • i have one question.. how the electric current can conduct from the heat? ..pls eply a.s.a.p i need the answer for my presentation..tq

  • At 6 February 2011, 8:39:42 user Kammenos wrote:   [reply @ Kammenos]
    • Rushikesh i suggest you get a T type thermocouple. It has a range of −185 to +300 Celsius. It is made of Copper and Constantan.

  • At 6 February 2011, 4:22:37 user Rushikesh wrote:   [reply @ Rushikesh]
    • Thanks for the article. I never had a idea of what a thermocouple is?
      But now I clear about its laws and how it works.
      I had one question:
      I want to but one thermocouple for my experiment which type I should go for? I will be using a temperature bet -15 to 50 Celsius.
      Please lemme know which company provide a accurate Thermocouple?


  • At 6 February 2011, 4:21:22 user Rushikesh wrote:   [reply @ Rushikesh]
    • Thanks for the article. I never had a idea of what a thermocouple is?
      But now I clear about its laws and how it works.
      I had one question:
      I want to but one thermocouple for my experiment which type I should go for? I will be using a temperature bet -15 to 50 Celsius.
      Please lemme know which company provide a accurate Thermocouple?


  • At 5 February 2011, 10:35:47 user Kammenos wrote:   [reply @ Kammenos]
    • Good question Harry! According to the 2nd law, if any intimidate material are in the same temperature across all its length, then the reading is NOT affected. Which means that you can add as many materials as you like without affecting the reading, but these materials must be in the same temperatures through all its length.

  • At 4 February 2011, 18:32:55 user Harry wrote:   [reply @ Harry]
    • A very nice and clearly explained article.

      One question though: The wires leaving from the reference junction are both copper all the way to the terminals of the voltmeter so no additional EMF is generated as they are homogeneous materials. That's all good...

      But especially in digital multimeters or data acquisition boards, the voltmeter terminals themselves might be made out of something else than copper, so can you really eliminate T4 and T5 from your diagram? Wouldn't you actually have two more junctions there? Or do you say that the terminal internal wire length is so small that it is isothermal, and thus you can treat it according to the law of intermediate metals and say it generates no emf.

      Or am I confused with something?

      Thanks for the article!r

  • At 12 October 2010, 18:41:22 user Kammenos wrote:   [reply @ Kammenos]
    • The info is the same regardless the metals, as long as they are different. I tried it once with copper wire and iron and worked. As for the formula, it applies as well, yet the "a" values are different. For this reason, when you get e thermocouple, you ask it by its type, for example "K-Type thermocouple". And then google for the "a" values and the corresponding voltage to temperature tables.

  • At 12 October 2010, 17:30:20 user kjh wrote:   [reply @ kjh]
    • thank you for your nice help does these information or equation at least work for Aluminum &Chromium thermocouple.?

  • At 12 October 2010, 17:23:39 user kjh wrote:   [reply @ kjh]
    • Oh iam sorry i got the voltage but why you take it to the 8 why you stopped over there !is it for accuracy purposes!

  • At 12 October 2010, 17:20:54 user kjh wrote:   [reply @ kjh]
    • very nice !
      but how did you plug in the voltage .0037 inside that equation (i mean which v is 0.0037v???!!

  • At 2 July 2010, 8:54:24 user Christopher wrote:   [reply @ Christopher]
    • Fantastic videos!! They really helps in understanding the principle of operation of thermocouples. Now, I will perform the experiments by myself:)Thank you!!!

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