I am Trying to read in a K-Type Thermocouple to a 0-5V Arduino Uno.

I Can read my Thermocouple directly using a Fluke meter on temperature setting giving me the correct value of 21°C, but when I switch to mV setting I read 0 mV on my Fluke. Similarly on my Arduino I am reading 0mV or some garbage hits (0.3xx mV, 1.9xx mV) every 10 readings or so that I assume is noise. My room is 70°F or 21°C so I expect to see 0.838mV according to the K-Type Thermocouple table

I have made a test load that produces 0-54mV

Test Load

The INA121P Instrumentation Op-Amp works as expected giving a 50x gain on my test load when I use Rg = 1 kΩ resistor between pins 1 & 8.

So as stated before when I replace my test load with my K-Type Thermocouple, I read 0mV on Vo (expected should be 0.838 mV x 50 = 41.9 mV)

Schematic K-Type Thermocouple & INA121P

I have not included my Cold-Junction Tref in the Circuit yet, but I have my thermistor reading properly using the Steinhart-Hart equation in other tests.

In my code I will be dividing off the gain to get back to Vtc in mV. I am using the K-Type Coefficients to curve fit my thermocouple.

My overall questions are:

  1. Until I add my cold junction should I expect to be reading 0mV on my TC?
    Answer: Yes since TC and Tref are at the same Temp no differential occurs. Thanks @TimWescott
  2. How can I get rid of the "noise" is 0.01 μF in the ballpark or should I change that value?
  3. Would you recommend any changes to my approach?
  • \$\begingroup\$ Need datasheet for your thermocouple. \$\endgroup\$ – GSLI Aug 14 '19 at 18:26
  • \$\begingroup\$ What voltage are you expecting? What type of thermocouple is it? K? \$\endgroup\$ – mkeith Aug 14 '19 at 18:27
  • 1
    \$\begingroup\$ @pipe Sorry First ever post. I thought the Thermocouples and Op-Amps were specific. \$\endgroup\$ – posop Aug 14 '19 at 18:29
  • 1
    \$\begingroup\$ Your problem is likely that the thermocouple is allowed to "float" up to the rail. Try tying one end of the thermocouple to your "ground" to keep it in the input common-mode range of the amplifier. \$\endgroup\$ – John Birckhead Aug 14 '19 at 18:56
  • 2
    \$\begingroup\$ @posop Thanks for updating the title, it's a great first post. You should see the other ones we get... \$\endgroup\$ – pipe Aug 14 '19 at 19:46

You need a DC path for the in-amp bias currents, for example you could ground the junction or connect one lead or both to ground through a relatively high value resistor (the thermocouple and leads are usually well under 100 ohms, so any resulting error should be minimal.

In order to get predictable filtering you will need to add some series impedance to the thermocouple leads. Try 1K on each lead, 100nF between the leads and 10nF to ground on each input.

You should probably think about biasing the in-amp output above ground, since it's quite possible for the "hot" junction to be at a lower temperature than the cold junction, leading to a negative voltage. You'll also want to clamp the voltage appropriately for the Arduino (read the datasheet for the MCU to get the specs- you'll also need the MCU supply voltage).

It's usually desirable to run a bit of current through the thermocouple junction in order to detect breaks. There's a trade-off between the resulting error from that current times the loop resistance of the thermocouple probe and wires vs. the current if DC current is used. You could also periodically pulse it from the MCU to detect a broken sensor or connections, but your front end might take some time to recover.

| improve this answer | |
  • \$\begingroup\$ I will wire up paragraph 1 & 2 tonight. Correct me if I'm wrong, but I think I have in effect biased my circuit with the 2 820 Ω resistors. I can get negative values out of my instrumentation amp, but that is taken care of with the K-Type Thermocouple polynomial. negative Vo values indicate a negative temperature. \$\endgroup\$ – posop Aug 14 '19 at 19:29
  • \$\begingroup\$ How does your ADC deal with voltages below ground or that are too high? You need to get it into digital form before you can apply digital cold junction compensation and only then the linearization function. The mV from the sensor will be negative any time the "hot" junction is cooler than the reference junction (usually the terminals at the instrument, which you're going to be measuring with your thermistor). \$\endgroup\$ – Spehro Pefhany Aug 14 '19 at 22:14
  • \$\begingroup\$ You are absolutely correct, I wired things up with your suggestions and Vo will not drop below 0V. Can you recommend an easy way to bias my circuit? I am not 100% sure I need to go negative but it would be nice for things to work properly. \$\endgroup\$ – posop Aug 14 '19 at 23:48
  • \$\begingroup\$ Easiest way is to connect the reference on the in-amp to a buffered voltage a bit above ground. It needs an op-amp buffer (just a voltage follower) because it's not high impedance like the other inputs. Then subtract that voltage off digitally. \$\endgroup\$ – Spehro Pefhany Aug 14 '19 at 23:51

You misunderstand how a thermocouple works. A thermocouple generates a voltage that is more or less proportional to the difference in temperature between the junctions. If your "hot" junction is at 21C and your "cold" junction is also at 21C, then the difference voltage will be zero.

So reading 0V from a thermocouple at thermal equalibrium is exactly correct.

This is obvious from thermodynamics, because a thermocouple is a heat engine -- and heat engines cannot generate power when there is no difference in temperature.

Note that the table assumes a cold junction of 0C -- not 21C.

You need to measure the thermocouple voltage at your board, and you need to independently measure the temperature at the point where the thermocouple is attached to your board (or whatever point that your thermocouple leads transition to a pair of leads made of copper). Then you need to calculate the temperature difference that the thermocouple is measuring between your board, then you need to add that to the measured board temperature.

| improve this answer | |
  • \$\begingroup\$ I think this answers Question #1. Until I add my cold-junction compensation I will read 0mV --> 0°C. Makes perfect sense. Now on to the noise. \$\endgroup\$ – posop Aug 14 '19 at 19:40

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.