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




Some solutions to this problem
The Ice Bath Reference

Any thermocouple pair, when is at 0oC, will generate no EMF! So, by simply connecting the wires inside an ice bath, the readings from the K-junction are not altered!





Double Oven Reference

Two ovens are used to simulate the ice point reference. The wires of the thermocouple are joined in opposite polarization inside one oven, and then are connected with the copper wires to the other oven. By having different temperatures inside the ovens, the ice point reference can be simulated.







Out of the lab door - The Isothermal Block

A K-Type thermocouple with the copper junction inside an isothermal block

The above methods are valid and will return correct results. Yet, they are very difficult to be used in real life applications. Imagine for example, an automation in a machine to measure the temperature with a thermocouple. You will need a thermocouple, a display, a refrigerator or two ovens.... Therefore, other methods have been invented to compensate the reference temperature.

But before i explain these methods, i need to explain first the isothermal block. In environments where the temperature may vary from point to point, like the engine room of a ship, there may be situations where the connectors of the measuring equipment will not have the same temperature, even if they are just 2-3 cm away. Therefore, we need to ensure that bot junctions will be at the same temperature. For this reason, the isothermal blocks are used. The simplest isothermal blocks are covers made of special material usually enclosed in plastic housing. Inside there are the thermocouple-to-copper junctions. The special material will ensure, that both junctions are kept under the same temperature, very important in some situations. Some other isothermal blocks are more advanced, with multiple thermocouple wire inputs and with extra analog signal with the junction reference temperature, used for software compensation (as we will discuss later on).



Hardware compensation

This is a very common method to compensate the temperature at the copper junction of the thermocouple, also called "Electronic Ice Point Reference". According to this method, a thermistor is placed inside the isothermal block. The thermistor resistance will change according to the temperature inside the isothermal block. Using a DC voltage and a couple of resistors to control the gain, the circuit will add the voltage needed for the specific temperature inside the isothermal block!

The reason why this method is very popular is because the isothermal block is rather cheap, while it saves a lot of computing time and effort to calculate the temperature. This way, it is possible for a PLC for example to retrieve much more readings from different points faster. Yet there is a drawback to this. The major drawback is that for each thermocouple, a different hardware compensation must be used. This will increase the space needed, as well as the price of the system. Here is a very common circuitry for hardware compensation.







Software compensation

This method will calculate the temperature of the junction, by monitoring the temperature inside the isothermal block. For this reason, a thermistor is placed there where the copper connection is made. The signal from the thermistor is transmitted then directly to the PLC (or computer or controller) which it will convert it into the voltage that the junction would create in this temperature (see the 3rd law of thermocouples). This voltage will be added to the measured voltage of the thermocouple, and through the voltage to temperature table, the temperature is calculated. This is a very simple way of connecting thermocouples to controllers and will require only 1 thermistor for all the thermocouple connections inside the same isothermal block. Yet, the compute time, especially if a PLC is used, is usually slow.




If a thermistor is used for compensation anyway, why do we use the thermocouple?

This is gonna be a short paragraph with a long title. Thermistors are used in a narrow temperature range, usually bellow 200oC. Thermocouples can be used to measure temperatures as high as 1800oC (3272 F), and as low as -270oC (-454 F).




Converting from voltage to temperature - The EMF vs Temperature table

The thermocouple pairs are usually chosen in a way that they have stability, repeatability and -if possible- linear results. As the K-type is the most popular, we will work on this. First of all, you need to provide yourself the EMF vs. Temperature table for thermocouple. Each type has a different table of course. In this page you will find a complete list with these tables. To follow this article, download the Thermocouple Type K(° C).

Now, how to read these tables. I will get a small part (from 0oC to 90oC) to explain. Here is the example:





So, here how it goes. In the very left column there are the temperatures. Can be at oC of oF. The second column (witch i have masked with red color) shows the actual voltage that will be generated by the thermocouple. It can be in mVolts or uVolts! This table shows the voltage in mili-volts. For example, 3.27 mVolts corresponds to 80oC. All the other columns, are the intermediate temperatures (1 degree step) of the temperatures. For example, at 10oC the voltage will be 0.4mV, but at 16oC the voltage will be 0.64mV! At 69oC the voltage will be 2.81mV and so on. It is actually straight-forward to get the temperature from the voltage.



And what can do with this table after all???

There are actually temperature controllers that can do all the hard job for you, and furthermore can be programmed to do several tasks, such as activating relays in specific temperatures of ranges. The thermocouple is directly connected on them, and they usually do the junction temperature compensation internally. Using such a controller, the above table is just... useless. But, what if you want to make a temperature controller by yourself? Most possible you will use a microcontroller for this, and an analog to digital converter. By converting the voltage into a digital value, you can use this table to get the temperature.




From voltage to temperature - Solving a polynomial

There is a possibility that a processor may use the above table to compare the measured voltage and calculate the temperature. There is also a mathematical way to solve this problem. The temperature can be calculated by solving a polynomial, and as usually, they are long. Here us the sum:





an for a K-type thermocouple can be found from the following table:


n an
a0 = 0.226584602
a1 = 24152.109
a2 = 67233.4248
a3 = 2210340.682
a4 = -860963914.9
a5 = 4.83506 x 1010
a6 = -1.18452 x 1012
a7 = 1.38690 x 1013
a8 = -6.33708 x 1013


If this looks Greek to you, i will show you an example. Suppose that the measured voltage (forget about the cold junction compensation) is 3.7 mVolts. First, we convert to volts, that is 0.0037 volts. Now to solve the polynomial:



T = a0 x V0 + a1 x V1 + a2 x V2 + a3 x V3 + a4 x V4 + a5 x V5 + a6 x V6 + a7 x V7 + a8 x V8

I do not find any particular reason to further solve the above anaconda, as requires only plain mathematics. The result is: T = 89.68oC. If you look at the table above, you will see that it is a good approximation to the real temperature provided by the table.




Common thermocouple problems
Connection problems:

We referred to this extensively in this article. Connecting a thermocouple to copper (or other metal) will create another thermocouple. We need to compensate the temperature on this junction.

Wire resistance:

To keep a good accuracy and response time, thermocouple wires are usually kept thin. Yet, thin wires means high resistance. Most temperature controllers are programmed to compensate this resistance. There are cases that thermocouples needs to be placed many meters away from the controller. In such case, the wire resistance needs to be taken into account. To solve this problem, we use a technique called "compensated ohms measurement". The controller first measures the offset voltage without the ohms source resistance. Then the ohms source resistance is switched on and the voltage across the resistance is measured again. The voltmeter compensates with software the offset voltage of the thermocouple and calculates the real thermocouple source resistance.

Decalibration:

This is the most annoying error, as the temperature reading appears to be valid, although it is wrong. This error occurs when the makeup of thermocouple wire is altered. This is usually caused by the diffusion of atmospheric particles into the metal when operating in extreme temperatures, or in chemicals.

Noise:

This is common to all electronic applications when operating into electrically noisy environments. Yet, thermocouples can suffer much greater influence from external noise, as the output voltage from a junction is measured in micro-volts. There are of course methods to minimize noise problems. First of all, an analogue filter at the input of the voltmeter will reduce dramatically external noise. Due to the fact that thermocouples will generate only DC voltage and external noise comes from AC or pulsed voltage, the use of capacitors will reduce such noises. Yet, the acquisition time will be reduced. Another simple method is the use of twisted pair wires, same as used in ethernet cables. To further reduce external noise we can use shielded grounded wire.Finally, a very attractive method to virtually eliminate noise is the integrated A/D convention. The thermocouple is connected into a transmitter (that can compensate the connection temperature as well). The transmitter will convert to digital and also average the signal so to reduce the noise. Then, it will transmit a clear signal onto far distances. This is by far the best way to eliminate noise when measuring in far distance.

Thermal Shunting

Thermocouples have some mass. To read the temperature, the thermocouple mass must be heated. To heat this mass, an amount of energy will be needed. If the heating source does not constantly provide energy (like a boiler does), then the thermocouple will "draw" an amount of energy from the source and thus the reading will not be accurate. On the other hand, The wires of the thermocouple itself and the housing will dissipate another amount of energy in the atmosphere. To average this problem, there are thermocouples with different wire thickness and different housings. Using the proper type will allow a better reading. For example, a thin wire with small housing is popper for measuring lower temperatures, with faster response time.

Common Mode Voltage

This problem refers to parasitic voltages. They can occur for example by inductive pickup near the voltmeter. Most often, such problems occur from bad insulation in connections or the thermocouple itself. If for example you are measuring the temperature of water in a water tank and the thermocouple is badly or no insulated, then possible some volts will exist between the tank (or metallic pipes) and the earth of the voltmeter. To avoid this problem, properly insulated thermocouples should be used, and precautions such as for noise should be taken.




Things to remember about thermocouples
Two dissimilar metals of any kind, can perform a thermocouple. There are certain metal pairs that have quite a linear behavior in temperature change and good temperature gradients.

The thermocouple will NOT measure the temperature of the hot junction. It will measure the temperature difference between the hot and the cold junction. If for example the hot junction is measuring boiling water and the thermocouple to copper junction is at 25 degrees, the measured temperature will be 75 degrees (100-25)

Thermocouple needs the “cold junction compensation” because of the reason previously described.

A temperature measurement (with thermistor or diode or embedded chip) is required there where the thermocouple wires are connected with the copper wires or the measuring instrument, for the cold junction compensation.

Thermocouples suffers from external noise due to the small magnitude of the Seebeck voltage. In harsh environments with electrical noise, appropriate filters must be applied, along with shielded wires.

The thicker the thermocouple junction, the higher temperatures it can measure, but the slower response it has.

If measuring in conductive environments like water, the thermocouple probe must be isolated.

If the cables of the thermocouple wires must be extended to reach the controller, then choose thermocouple extension wires of the same type as best solution.

Thermocouple wires can be also extended with copper wires. If the temperature across the whole length of the copper wires is the same, then the cold junction compensation can be performed on the controller. Otherwise, the compensation must be performed according to the temperature where the copper wires are connected to the thermocouple wires.












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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?

      Thanks


  • 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]
    • ANY ONE CAN YOU PLEASE SHERE ME AS DESCRIPTION ABOUT ASEMBLING PROCESS OF A PID TEMPRATURE INDICATER AND CONTROLER................

      EMAIL YVKELURU@GMAIL.COM
      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?

      Thanks,
      Rushikesh


  • 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?

      Thanks,
      Rushikesh


  • 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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