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I am trying to design a 3 wire RTD circuit. I've been looking into either a Wheatstone bridge or a constant current source with ratiometric measurements. I would like to know which is the better approach in terms of, linearity, precision, stability, drift etc.. What are the advantages and disadvantages for both approaches and the complexity involved in the design.

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A Wheatstone bridge does not offer perfect cancellation of leadwire resistance changes, and the linearity is somewhat affected. Keep in mind that the 3-wire connection strategy does not perfectly correct in any case since the wires may differ in resistance between them.

Constant current sources are more complex and expensive and may add significant drift if they are badly designed. With a good design the voltage output will be ratiometric to a reference voltage, which may also be used for an ADC reference.

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  • \$\begingroup\$ If I am connecting with a 3 wire pt100 RTD and a Wheatstone bridge. The resistor on the left branch should be 100 ohms(to balance the bridge at 0degrees), to the RTD. But how much should the other two resistors be on the right side of the branch? I would be doing a voltage excitation on the bridge input. \$\endgroup\$
    – UKNP
    Dec 6, 2022 at 16:30
  • \$\begingroup\$ @UKNP Depends how much current you want to put through the sensor (more current = more output = good, but more self-heating = bad). Somewhere in the range of 0.2 to 1mA is typical, but if you're measuring air temperature vs. a solid or flowing liquid the answers may vary somewhat since the heat loss from the sensor (and thus self-heating error) will vary with the thermal resistance etc. \$\endgroup\$ Dec 6, 2022 at 18:56
  • \$\begingroup\$ My concern is the values of the two identical resistors that would be on the opposite side of the sensor on the bridge. Does the current really matter on that branch? How much resistor values would you suggest I go with? I'm not too worried about self heating, since I will be turning the sensor on/off for very short periods of time. I am measuring air temperature. \$\endgroup\$
    – UKNP
    Dec 6, 2022 at 23:52
  • \$\begingroup\$ If you have a 5V power supply then 4K99 will give you a bit under 1mA. If you have more complex questions it might be better to open a new question. \$\endgroup\$ Dec 7, 2022 at 0:28

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