The resistance of the sensor plus 2 of the leads is subtracted from the resistance of 2 leads. I do not understand how the circuit accomplishes this. These calculations and adjustments must be done by the controller. If anyone can offer some insight into this it would be appreciated.
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You have to look at the fact that the controller cannot sense the resistance directly to understand how it does the subtraction. (A digital controller could actually calculate both resistances and then subtract, but an analog circuit can also get the same result.)
The following may not be the way a particular controller does it, but it is one way it could be done:
1. Start by driving a known current, I, in the same direction through both R2 and R3, with the sum of the two currents coming over the return path via R1.
Now measure the voltages at terminals 1, 2, and 3.
2. Now note that V
12 is equal to 2I(R1 + Rb) + IR2 and V
23 is equal to I(R2 + R3).
For this three terminal network to deliver the result stated, you have to be able to assume that R1 is equal (or very close to) R2 and R3, call it R. This will be the case if they are the same diameter conductor run in a single 3 wire cable with any intervening connectors also having matched resistances on all contacts.
3. The equations in 2 then become V
12 = 3IR + 2IRb and V
23 = 2IR.
If we run these two voltages into an op amp with the input and feedback resistors set up so that the output voltage is (1/2)V
12 - (3/4)V
23/I, then the output voltage will be directly proportional to IRb. At this point either take advantage of the fact that I is known and regulated or just use another op amp-based divider circuit to get Rb. From there, do the calculation (or analog manipulation) to find the corresponding value of T.
To use one controller with two RTD sensors that have different length leads, you will want to use a three pole relay unless you are willing to parallel the connection of one of the leads.