RTDs and the Callendar–Van Dusen Equation
Resistive Temperature Detectors (RTD)
Resistive Temperature Detectors, or RTDs, are among the most accurate and stable temperature sensors available. Built from a carefully wound platinum wire or deposited thin-film element, they exploit the nearly linear relationship between a metal's electrical resistance and its temperature to deliver reproducible measurements over a wide range.
In the tradition of Webster and Pallàs-Areny's Sensors and Signal Conditioning, we treat the RTD as a passive resistive transducer whose behavior is captured by the Callendar–Van Dusen equation. Understanding the transfer function, self-heating, and lead-wire effects is the key to turning a tiny resistance change into a clean, calibrated temperature reading.
This lesson develops the mathematics of the Pt100 and Pt1000, derives the linear approximation used in most industrial work, and walks through the practical three-wire and four-wire connection schemes that eliminate lead-resistance error.
For temperatures above 0 °C, the platinum RTD follows:
$$R(T) = R_0 \left[1 + A\,T + B\,T^2\right]$$
where $R_0$ is the resistance at 0 °C (100 Ω for Pt100, 1000 Ω for Pt1000), $A = 3.9083 \times 10^{-3}\,^\circ\text{C}^{-1}$, and $B = -5.775 \times 10^{-7}\,^\circ\text{C}^{-2}$.
For most industrial ranges the quadratic term is small, giving the widely used linear approximation:
$$R(T) \approx R_0\,(1 + \alpha\,T), \qquad \alpha \approx 3.85 \times 10^{-3}\,^\circ\text{C}^{-1}$$
- Excitation current must be kept small (≤ 1 mA) to avoid self-heating error $\Delta T_{\text{sh}} = I^2 R / \delta$, where $\delta$ is the dissipation constant in mW/°C.
- Lead resistance of a 2-wire connection adds directly to $R(T)$ and is the dominant error for long cables. Use 3-wire or 4-wire Kelvin connections for sub-0.1 °C accuracy.
- Sensitivity: $S = dR/dT = R_0 \alpha \approx 0.385 \, \Omega/^\circ\text{C}$ for a Pt100.
A Pt100 is immersed in a bath at $T = 25\,^\circ\text{C}$. Compute $R(T)$ using the full quadratic form, then compare with the linear approximation.
Step 1: Apply the Callendar–Van Dusen equation.
$R = 100\left[1 + 3.9083\times10^{-3}(25) + (-5.775\times10^{-7})(25)^2\right]$
$= 100\left[1 + 0.09771 - 3.61\times10^{-4}\right] = 100(1.09735) = 109.73\,\Omega$.
Step 2: Linear approximation with $\alpha = 3.85\times10^{-3}$:
$R_{\text{lin}} = 100(1 + 3.85\times10^{-3}\cdot 25) = 100(1.09625) = 109.63\,\Omega$.
Step 3: The difference is $0.10\,\Omega$, equivalent to $\approx 0.26\,^\circ\text{C}$. The quadratic term matters at high precision.
A Pt100 carries an excitation current $I = 2\,\text{mA}$ and has a dissipation constant $\delta = 5\,\text{mW}/^\circ\text{C}$ in still air. At $T = 100\,^\circ\text{C}$, estimate the self-heating error.
Step 1: $R(100) \approx 100(1 + 3.85\times10^{-3}\cdot 100) = 138.5\,\Omega$.
Step 2: Power dissipated: $P = I^2 R = (2\times10^{-3})^2 (138.5) = 5.54\times10^{-4}\,\text{W} = 0.554\,\text{mW}$.
Step 3: Temperature rise: $\Delta T = P/\delta = 0.554/5 = 0.11\,^\circ\text{C}$.
Acceptable for most industrial work; halving the current would drop the error to $0.028\,^\circ\text{C}$.
A Pt100 is measured at $R = 125.0\,\Omega$. What is the temperature, using the linear model?
Invert $R = R_0(1 + \alpha T)$:
$T = \dfrac{R/R_0 - 1}{\alpha} = \dfrac{1.250 - 1}{3.85\times10^{-3}} = \dfrac{0.250}{3.85\times10^{-3}} \approx 64.9\,^\circ\text{C}$.
Practice Problems
Show Answer Key
1. $R = 100(1 + 3.85\times10^{-3}\cdot 200) = 177.0\,\Omega$
2. $S = R_0\alpha = 1000\cdot 3.85\times10^{-3} = 3.85\,\Omega/^\circ\text{C}$
3. $T = (0.8 - 1)/3.85\times10^{-3} \approx -51.9\,^\circ\text{C}$
4. $I^2 R \le \delta \cdot \Delta T_{\max} = 4\,\text{mW/°C} \times 0.05\,^\circ\text{C} = 0.2\,\text{mW}$
5. The third wire carries the excitation return, cancelling the voltage drop in the two measurement leads that run side by side.
6. Platinum is chemically stable, highly linear, and has a well-defined temperature coefficient; it is the international standard (ITS-90).