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TIPD209: How to calculate compensation resistor R2

James Archer
Prodigy 10 points
Part Number: TIPD209

In the user guide it says to calculate the compensation resistor (R2) using the following formula:

R2 = ((Vexcite – Vcm)/Vsb)* α

This would result in the following value for R2 based on K-type type thermocouple, Platinum RTD and values for Vexcite and Vcm defined in the user guide:

R2 = ((5V – 2.6V)/40 μV/ °C)*0.00385 Ω/°C = 231 Ω

I don’t think this is correct, surely for the circuit to work R2 needs to be significantly larger than RTD (100 Ω @ 0°C). As expected in the simulation shown in Figure 4, R2 is defined as much lager value of 18.7K Ω.

Please explain how R2 was calculated to be 18KΩ?

Any help would be much apricated, I’m trying to adapt this circuit to my needs.

  • +1
    Prodigy 10 points

    I worked out the problem myself.

    Typically, an RTD α is defined in units Ω/Ω/°C, however your equation for calculating the compensation resistor R2 requires α to be in units Ω/°C.

    For the Platinum RTD that I am using α = 0.385 Ω/°C, when I plug this value into the equation for calculating the compensation resistor R2 I now get sensible a sensible value of 23.1k Ω. Also, when I simulate the circuit at different cold junction temperatures the temperature compensation works wonderfully.

    I think the user guide could be improved by specifying the type of thermocouple and RDT used in TI’s test simulations.

  • 0 in reply to James Archer
    TI__Guru 69891 points

    Hi James, 

    James Archer said:
    R2 = ((Vexcite – Vcm)/Vsb)* α

    Since you have figured out, I am not going to explain in detail. Enclosed is the simulation FYI. 

    The application note is using J type TC based on -50C and 500C generated thermoelectric voltage, where Seeback Coefficient is specified at 51uV/C or Kevin. There is a factor of 100 missing in the above equation, since α is 0.00391 Ω/Ω/°C (US) or 0.00385 Ω/Ω/°C (Euro) ITS-90 standard is closer to US standard. 

    R1=R2=187Ω will also work, but the current (~8.x mA) may be a bit high and may cause unwanted self heating. 18.7kΩ or similar will be better. 

    INA188 TC with Cold Junction Comp 12092022.TSC

    Enclosed is the TC simulation. Thanks for using our product. 

    Best,

    Raymond

  • 0
    Guru 140200 points

    Hi James,

    the idea behind R2 is to let a current flow through the PT100 element which creates a voltage drop change per °C which is identical to the output voltage change per °C of the thermocouple.

    Assume you want to carry out the cold-junction compensation at 25°C. Then a J-type thermocouple generates an output  voltage change of 52µV/°C. The PT100 element creates a change of resistance of 0.388Ohm per °C. So a current of 52µV / 0.388R = 134µA must flow through the PT100.

    When a voltage of 5V - 2.6V = 2.4V shall drop across the series circuit of R2 and PT100, the total resistance must be 2.4V / 134µA = 17k9. Because of that, R2 must be 17k8.

    Kai