How to compensate for the cold junction with a diode?

Question

Electronics

Can you please explain how this circuit compensates for the cold junction?

Answer 1

It's elementary, Watson. The forward drop across the diode has a temperature coefficient of approximately 2 mV/degree. Using the divider R2, R3, this value is reduced to a level equal to the cold junction voltage, but with the opposite sign. This means that as far as the thermocouple output drops from the action of the cold junction, the same (ideally, of course) the voltage from R3 raises it. And the constant component from the diode and the divider is reduced to zero using R6.
You just need to remember that due to the nonlinearity of the characteristics of the diode and its limited pace. range, the compensation will be fundamentally inaccurate.