A three-wire PT100 connection does not cancel lead resistance automatically. The measurement circuit must sense one current-carrying lead's voltage drop and subtract it from the RTD-plus-lead measurement. The result is exact only when the two current-carrying leads have equal resistance.
Three-Wire Measurement Principle
Drive the excitation current through the single lead and one of the two leads connected to the opposite PT100 terminal. Use the third lead as a high-impedance sense connection. Because the sense input ideally draws no current, its own lead resistance does not create a significant voltage drop.
| Measurement | Result | Purpose |
|---|---|---|
| Excitation-return lead to sense lead | −I × RLB | Measures the return-lead drop with the polarity required for subtraction |
| Sense lead to excitation-source lead | I × (RRTD + RLA) | Measures the PT100 plus the source-lead drop |
Lead-Resistance Compensation Math
Let I be the excitation current, RRTD the PT100 resistance, and RLA and RLB the resistances of the two current-carrying leads. Adding the two measured voltages gives:
Vcorrected = I(RRTD + RLA) - I(RLB)
= I(RRTD + RLA - RLB)
Rmeasured = Vcorrected / I
= RRTD + RLA - RLBIf RLA equals RLB, the lead terms cancel and Rmeasured equals RRTD. If they differ, the remaining resistance error is RLA − RLB. A four-wire measurement removes the need to assume equal current-lead resistance.
Diagnosing the 3 V and 200 Ω Circuit
The stated 3 V excitation and 200 Ω bottom resistor do not define the complete measurement circuit. A schematic and the measured voltage nodes are required to determine whether the circuit performs the two measurements above. A 200 Ω resistor is not proven to be the fault merely because it differs from 100 Ω; its effect depends on the circuit topology and the resistance-to-temperature calculation.
- Verify that the PCB separately measures the return-lead voltage drop and the PT100-plus-source-lead voltage, with the polarities shown above. A single uncompensated divider-voltage measurement cannot perform the stated subtraction by wiring alone.
- Measure the resistance of the two current-carrying leads individually. Their difference becomes the residual resistance error after three-wire compensation.
- Account for changing excitation current if the PT100 and 200 Ω resistor form a series path. Under that assumption, I = 3 V ÷ (200 Ω + RRTD + RLA + RLB), so the calculation must use the actual current rather than treating it as constant.
- Compare corrected resistance with the directly connected PT100. If the difference follows RLA − RLB, lead mismatch explains the residual error; otherwise, inspect the measurement polarity, current calculation, and divider transfer function.
FAQ
Does a three-wire PT100 cancel cable resistance automatically?
No. The electronics must measure one current-carrying lead's voltage drop and subtract it from the PT100-plus-other-lead measurement.
What error remains when the two PT100 leads have different resistance?
The inferred resistance is RRTD + RLA − RLB. The residual error is therefore the resistance difference between the two current-carrying leads.
Is the 200 Ω bottom resistor causing the PT100 error?
The available information does not establish that. Verify the schematic, excitation current, measured nodes, and compensation calculation before replacing the resistor.
