Thermistors — β-Model and Steinhart–Hart
Thermistors — Nonlinear Resistive Temperature Sensors
Thermistors are semiconductor resistors whose resistance changes dramatically — and nonlinearly — with temperature. Unlike the gentle slope of a platinum RTD, a typical NTC thermistor may change resistance by a factor of ten across a 100 °C span, giving it excellent sensitivity for narrow-range measurements like body temperature or battery monitoring.
Following the treatment in Pallàs-Areny and Webster, we characterize the thermistor with the exponential β-parameter model and, when higher accuracy is required, the three-term Steinhart–Hart equation. You will also learn to linearize a thermistor with a single parallel resistor — a classic signal-conditioning trick that flattens the response around a chosen operating point.
By the end of the lesson you will be able to compute resistance from temperature, invert the relationship to extract temperature from a measured resistance, and choose a linearization resistor for a specific midpoint.
$$R(T) = R_0\,\exp\!\left[\beta\left(\frac{1}{T} - \frac{1}{T_0}\right)\right]$$
where $T$ and $T_0$ are in kelvin, $R_0$ is the resistance at $T_0$ (often 25 °C = 298.15 K), and $\beta$ is the material constant, typically 3000–4500 K.
$$\frac{1}{T} = a + b\,\ln R + c\,(\ln R)^3$$
The three coefficients $a$, $b$, $c$ are determined from three calibration points and give sub-millikelvin fits over a wide range. This form is used wherever NTC thermistors are trusted for medical or laboratory work.
An NTC has $R_0 = 10\,\text{k}\Omega$ at $T_0 = 298.15\,\text{K}$ and $\beta = 3950\,\text{K}$. Find $R$ at $T = 50\,^\circ\text{C}$.
$T = 323.15\,\text{K}$, so $1/T - 1/T_0 = (1/323.15) - (1/298.15) = -2.594\times10^{-4}$.
$R = 10000\cdot e^{3950(-2.594\times10^{-4})} = 10000\cdot e^{-1.025} = 10000\cdot 0.359 \approx 3.59\,\text{k}\Omega$.
The same thermistor reads $R = 2.0\,\text{k}\Omega$. What is $T$?
Take the log of the β equation: $\ln(R/R_0) = \beta(1/T - 1/T_0)$.
$\ln(2000/10000) = \ln(0.2) = -1.609$. So $1/T = 1/T_0 + (-1.609)/3950 = 3.3540\times10^{-3} - 4.074\times10^{-4} = 2.947\times10^{-3}$.
$T = 1/0.002947 \approx 339.4\,\text{K} \approx 66.3\,^\circ\text{C}$.
To flatten the response of the thermistor around 25 °C, we place a resistor $R_p$ in parallel. A well-known result (Pallàs-Areny §2.2) gives the optimum value $R_p = R_0\,(\beta - 2T_0)/(\beta + 2T_0)$. Evaluate for our thermistor.
$R_p = 10000\cdot (3950 - 2\cdot 298.15)/(3950 + 2\cdot 298.15)$
$= 10000\cdot (3950 - 596.3)/(3950 + 596.3) = 10000\cdot 3353.7/4546.3 = 7377\,\Omega \approx 7.4\,\text{k}\Omega$.
A 7.5 kΩ resistor in parallel with the thermistor makes the combined resistance nearly linear in a ±25 °C band around 25 °C.
Practice Problems
Show Answer Key
1. $\ln(R/10\text{k}) = 3500(1/273.15 - 1/298.15) = 3500(3.066\times10^{-4}) = 1.073$. $R = 10000\,e^{1.073} = 29.2\,\text{k}\Omega$.
2. $\ln(0.5) = 3500(1/T - 1/298.15)$. $-0.693/3500 = -1.98\times10^{-4}$. $1/T = 3.354\times10^{-3} - 1.98\times10^{-4} = 3.156\times10^{-3}$. $T = 316.8\,\text{K} \approx 43.7\,^\circ\text{C}$.
3. Negative Temperature Coefficient — resistance decreases as temperature rises.
4. Near 0 °C. Because the exponential derivative $|dR/dT| = \beta R/T^2$ is largest when $R$ is large.
5. To convert measured resistance to temperature with very high accuracy (three-term fit).
6. $R_p = 10000(3500 - 596)/(3500 + 596) = 10000 \cdot 2904/4096 \approx 7.09\,\text{k}\Omega$.