Question

The resistance of a thermistor is 1 k\(\Omega\) at \(25^\circ\mathrm{C}\) and 500 \(\Omega\) at \(50^\circ\mathrm{C}\). Find the temperature coefficient of resistance (in units of \(^{\circ}\mathrm{C}^{-1}\)) at \(35^\circ\mathrm{C}\). (Round off the answer to three decimal places.)

Hint:

- For thermistors, \(R(T)\) is often modeled as \(R=Ae^{B/T}\); then the fractional TCR is \(\alpha=-B/T^2\).
- Remember to convert temperatures to kelvin when using this model; the final units of TCR are per \(^\circ\)C (same as per K).

Answer 1

Correct Answer: -0.03

Step 1: Model the NTC thermistor as \(R(T)=A e^{B/T}\) (with \(T\) in kelvin). From the two data points: \[ \ln\!\left(\frac{R_{25}}{R_{50}}\right)=B\!\left(\frac{1}{T_{25}}-\frac{1}{T_{50}}\right), \] where \(R_{25}=1000\,\Omega,~R_{50}=500\,\Omega,\; T_{25}=298.15\,\text{K},~T_{50}=323.15\,\text{K}\). Hence, \[ B=\frac{\ln(1000/500)}{\frac{1}{298.15}-\frac{1}{323.15}} =\frac{\ln 2}{\frac{25}{298.15\times 323.15}} \approx 2.672\times 10^{3}\ \text{K}. \] Step 2: The temperature coefficient is \[ \alpha(T)=\frac{1}{R}\frac{dR}{dT}=\frac{d(\ln R)}{dT}=-\frac{B}{T^{2}}. \] At \(35^\circ\mathrm{C}\), \(T=308.15\,\text{K}\): \[ \alpha(35^\circ\mathrm{C})=-\frac{2671.9}{(308.15)^{2}} \approx -2.81\times 10^{-2}\ \,^{\circ}\mathrm{C}^{-1}. \] Rounded to three decimal places, \(\alpha \approx -0.028\,^{\circ}\mathrm{C}^{-1}\).