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17 April 2010
Author: Giorgos Lazaridis
Experimenting with Thermocouples

I am preparing a very interesting project with a soldering iron. For this project, i will need to measure the temperature of the solder tip. I could use a thermistor for this, yet, there will be a small problem... The thermistor will be tosted! Therefore, i will use something that can handle the soldering iron temperature. This "something" is called "a thermocouple". I have prepare a complete theory for "How Thermocouples Work". I strongly suggest you read this theory before, if you do not know how they work.

Before i start building my project, i had to run some experiments with thermocouples, as i was not very familiar with the subject. I got myself (thanks Kourt!) a pair of K-type thermocouple wire for this reason. The reason for these experiments is first of all to get the beginner's experience with thermocouples, and also to find the way that i will measure the temperature using a uController and the thermocouple junction.




Before i begin...

I have the table of the K-type thermocouple voltage to temperature, measured in mili-volts. You may need to use it yourself:

K-Type thermocouple voltage to temperature table


The first thermocouple experiment: Do they really work?

It may seem funny, but i always have this question when i first get in contact with new staff for me. I name this "the joy of discovering". This is how the first experiment goes: I connect a thermocouple pair to copper wires, and then to a multimeter with a 200 mV scale. Then i heat up the junction with a lighter to see if i get some readings:







The second thermocouple experiment: Does similar metals act like dissimilar metals? (Proving the first law of thermocouples)

According to the theory of thermocouples, this is out of the question. Yet, i had to see with my very own eyes:







The third thermocouple experiment: Proving the second law of thermocouples

In this experiment, i introduce a third dissimilar metal to the thermocouple pair like an extension. According to the theory of thermocouples, if the third metal has the same temperature across it's length, it will not affect the measurement of the thermocouple. So, i extended one cable of the thermocouple with a copper cable to see if the measurement is changed. Then, i changed the temperature of the new wire to see what happens:







The fourth thermocouple experiment: What happens if the temperature of the thermocouple to copper junction is changed?

This is extensively discussed in the theory of thermocouples. A temperature change in the thermocouple-to-copper junction will alter the readings, that is why all thermocouple pairs needs a hardware or software temperature compensation at this point. Look what happens:







The fifth thermocouple experiment: Measuring the boiling water and something... hotter!!!

The distiled water boils at 100oC. In this experiment, i measure the temperature of the boiling water with the thermocouple, and yet i get the result way below the 100oC. It proves that the measurement is absolutely depended from the thermocouple to copper junction temperature. Then, i have some more fun with something much hotter!







Another way to calculate the temperature from the voltage

In the theory of thermocouples, i explained how can someone solve a polynomial to calculate the temperature from the Seebeck voltage. Yet, this polynomial requires a lot of multiplications. For a PIC, this means a lot of time. Let's see something interesting. I made with open office spreadsheet a graph. The x axis had the voltage and the y axis had the temperatures from 0 to 1370oC, with a 10oC step. Surprisingly, this is what i got:






The open office spreadsheet to calculate the slope and offset for the line

In a first glance, this is a straight line. Well, actually it is not. The next step was to find out if this line could be useful to me. Using the open office spreadsheet again, i applied the least squares method to calculate the line. You can use my "least squares calculator". From there, i got the slope (a) and the offset (b) of the above line. Using these numbers and the line equation (line equation solver), i can calculate any Y (temperature) given an X (voltage).

What i actually did was this. For temperatures from 0 to 1300oC with a 10oC step, i calculated the temperature from the voltage using both the polynomial equation and the line equation. For each result, i calculated also the percentage error. The following screenshot shows this spreadsheet. I have mark with blue box the values that i am mostly interested at, for my coming project:





The column with the red values, is the column with the error from the polynomial. The column with the green values has the error from the line equation. For the first 20 degrees, the error is unacceptable. As the temperature rises though, the error is decreased to very satisfying values. To tell you the truth, i calculated the slope and offset of the line for temperatures from 100 to 350oC, and that is why i get so good approach. For a wider range, i would not have these results.

And because a picture is like 1000 words, look at the following graph:





The yellow line is the error from the line equation and the red is the error from the polynomial approach. Both error values are displayed in %. The temperature range (x axis) is form 0 to 1300oC. I think i will prefer this simply line equation (Y=aX + b)...













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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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