Thermocouple Simulation Test Circuit and Voltage Comparator Circuit

Question

I have a challenge that I would like to share with you. (at least a challenge for me)

The resistor R8 (10 turn potentiometer) can be adjusted from ~0 to 1K Ohm to output voltages from 3mv to 30mv. The divider labeled (400F) outputs constant 8.3mv that equals 400ºF for a K Type Thermocouple. All that worked great.

So later I decided to add a fancy LED (D2). The LED should turn ON when the temperature is above 400F and OFF when below. I suspect using a LM741 as a comparator is a poor design because of precision and low voltages. One option would be an AD790 comparator http://www.analog.com/static/imported-files/data_sheets/AD790.pdf A comparator in the .25mv offest range would be suficient for the precision I need.

Is there a simpler (cheaper) way to achieve this with precision and low voltage in mind?

*UPDATE - HERE IS THE FINAL DESIGN http://iddevlabs.blogspot.com/2014/09/diy-thermocouple-testing-circuit.html

Answer 1

The LM741 has mV of offset. If you want to get an accurate comparison you could use a precision op-amp as a comparator. For example, an ADA4051.

If you're going to build this, I would like to see the source impedance a bit lower, maybe use a 10 ohm 1% resistor rather than 165 ohms and make the resistor that is now 100K something odd in the tens of K-ohms. Generally thermocouple instruments assume a source impedance in the 100 ohm range, at most, and most have noticeable error at 100 ohms.

Note that you are not doing cold-junction compensation (or, more accurately, cold junction emulation), so you'll not be able to get a good simulation of a thermocouple-- you'll get one (approximately) degree of error for every degree your terminal strip temperature is different from reference temperature. The 8.3(16)mV voltage you're using assumes a 32°F temperature- I hope you're warmer than that. At a 70°F CJC temperature, 8.3mV would give you about 438°F equivalent.

Answer 2

The approach you are proposing might look okay in a simulation. But it's very unlikely to work in real life (i.e. it will not work). 8.3mV is a weak signal. You can easily have more than 8.3mV of noise in your system. The noise will be tripping your comparator, and you will be having false detections all the time.

In real life practice, a thermocouple signal (tens of μV/°C) is amplified before something useful can be done with it. The amplification is usually done with an instrumentation amplifier. There are also specialized thermocouple front end ICs.

You should also look into Cold Junction Compensation of thermocouples. It's also sometimes called Ice Point Compensation.

There many application notes on thermocouple signal conditioning. Just search for these terms.