Consider the following, simplified schematic for a single, two-wire RTD sensor:
simulate this circuit – Schematic created using CircuitLab
The two wires from the RTD sensor plug into pins 1 and 2 on J1. A bridge is used, consisting of resistors R1, R2, and R3. U1 is an LTC2418 24-bit low-latency ∆Σ ADC (but the specific ADC should not actually matter for the purposes of this question).
There is nothing strange about pin 1 on J1: that connects to the positive (+) wire from the two-wire RTD (RTD1_P). However, pin 2 on J2—the negative (−) wire from the RTD—connects straight to ground, instead of to RTD1_N.
I am told that this circuit works properly. Is that claim correct?
If so, it implies that multiple RTD sensors can be added (using additional channels on the ADC), with all RTDs connected to a shared ground (pin 2 on J1). Thus, for n RTDs, only n+1 wires are required on the external connector, rather than n×2. This is a significant advantage for my application. Again, is this correct and feasible? And if it is, what are the limitations (if any) involved in doing it this way, as opposed to a more standard configuration that is fully differential, requiring n×2 wires?
(Nota bene: I understand that RTDs can also be connected in three-wire and four-wire configurations, as described here and/or on Wikipedia. I understand the limitations/compromises involved in selecting a two-wire configuration. For the purposes of this question, please assume a two-wire configuration.)
Best Answer
Given the current thru each divider is 2.5/200 = 12.5 milliamps, if the Ground wire on the connector is 10 milliOhms resistance, the error change for each added RTD will be 0.01 ohms * 0.0125 amps = 125 microVolts ---- per added divider.
Examine your error budget and decide what is permitted.