I have a side-project to read temperatures from 4 thermocouples (K type, around 500F for coffee roaster data logger), and no previous analog experience at these levels.

I've identified two possible solutions, but I'd like to hear feedback / suggestions / alternatives.

The first method would be to feed the four thermocouples through an analog mux (say HEF4052, though haven't done as much research on this- is finding a mux for low-freq uV signals), and then to a typical thermocouple IC (MAX6675).

The second method would be to feed the thermocouples into a 4 channel differential ADC, probably with an integrated PGA (AD7708). Then have an on-PCB temp sensor to handle cold junction compensation. Another option'd be analog mux to a single channel ADC, etc.

Are there advantages to one approach over the other? It seems like the first would be fewer components, and a bit more reliable, though a bit more expensive then the second option.

I've also seen circuits using an RC filter on both thermocouple lines- is that worthwhile? What kind of TC would be worthwhile on these kind of signals? Something for blocking 50/60Hz? Just installation specific?

My primary concern is that this is the first time working with signals measured in uV, so I'm not sure what is and isn't important (beyond basics like keeping digital and analog seperated).

EDIT: One more question for the ADC approach: is it important to measure the signals differentially, or could the low end of all 4 thermo couples be tied to ground, then be measured in a single ended fashion? ADS1118 seems like it has alot of what's wanted, but is only two-differential channel (and the datasheet shows differential measurements at that). More plainly: what's the advantage of doing differential measurement vs. single ended with one end tied to ground? Just isolation from the rest of the system?


2 Answers 2


If it were my design, I would use AD7708 and a separate temperature sensor on the PCB for ice point compensation.

Since 50/60Hz is a common mode signal on both TC lines, it's canceled by a differential amplifier (in the front end or in the integrated A/D). RC filters reduce higher frequency noise, which can be rectified by the amplifier, which would cause an offset in TC reading. Depending on the amount of noise in your environment, this may or may not be a problem.

The approach with an external mux is viable to. Check that the mux doesn't inject too much noise. A mechanical relay would be a really low noise mux.

edit: On floating thermocouples vs. common leads.
The topology with a common TC lead could work. But there are caveats. Different TCs can be a slightly different potential (common mode). If there is a shared lead, a current may be flowing through it, which can skew reading. If thermocouples are spread around spatially, they may see different noise. The noises will be added on the common lead, but will appear separately on the separate leads. 50/60Hz noise cancellation would not be as good.

update: Related post on floating vs. grounded thermocouples.

  • \$\begingroup\$ Most industrial data acquisition units (HP 34970A, etc.) use the mechanical relay approach, so +1 to you. \$\endgroup\$ Jan 13, 2012 at 17:15

If this is for a one-off system, you could do worse than to use a MAX6675 per thermocouple. While this would be too expensive for a production run, for one system and the price of a couple more of these, you'll save a whole lot of engineering. Amplification, compensation, & A/D will already have been done for you. A comment on this page suggests that you can buss the clock and data lines and only need a chip-select per device. It doesn't get a whole lot simpler than that.

  • \$\begingroup\$ Valid point, and it is a one off, but part of this is I'd like to get a better grasp for basic analog (what are the considerations, why would this not work, etc). \$\endgroup\$
    – CoderTao
    Jan 13, 2012 at 16:53
  • \$\begingroup\$ Fair enough! I hope you'll come back and add an addendum to your question describing what you learned. Good luck. \$\endgroup\$
    – JRobert
    Jan 13, 2012 at 16:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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