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.)

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    \$\begingroup\$ Can we assume that the 100\$\Omega\$ resistances are just placeholders? Because the ~16 mW of self-heating from having 1.25V across each RTD would be pretty horrific unless you're actually using the RTD element as a heater for some reason. Usually 0.1 to 1mA tops, not 12.5mA, for a Pt100 RTD. \$\endgroup\$ Commented Jan 9, 2020 at 3:01
  • \$\begingroup\$ @Spehro Hmm, good point. No, the 100 Ω resistance is from the actual circuit, not a placeholder. I believe the design is to turn off the 2.5 V source when measurements are not being obtained from the RTDs, since measurements are only obtained on demand, not continuously. I assume that would fix the heat-generation problem? But...even then, it may not be a sensible design, since that would be a significant amount of heat generation even for short bursts when the resistance values are being measured. You would recommend substituting, say, 5 kΩ resistors for R1, R2, and R3? \$\endgroup\$ Commented Jan 9, 2020 at 3:31
  • \$\begingroup\$ ☞ +1 for full latin use of N.B. \$\endgroup\$ Commented Jan 9, 2020 at 3:36
  • \$\begingroup\$ That's one way. It's more usual to use a current source. I think you can switch only one on at a time and use differential amplification so you'll get the same error as a single two-wire RTD. \$\endgroup\$ Commented Jan 9, 2020 at 4:47

2 Answers 2


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.


You have a 24-bit ADC which has a resolution of ~150nV. This means that worst case error (calculated by analogsystemsrf) 125uV will be about 833bits or the lower 9 or 10 bits of your ADC. This means that you may not be taking full advantage of your ADC.

I usually split grounds from thermistors or RTD's, it leave the headache of common mode error.


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