In the experiment for calibrating a platinum resistance thermometer, we are always approximating the resistance of the platinum thermometer by $$R(T) \approx R_0 (1+\alpha T),$$ taking the reference at $0^o C$ and using Celsius scale.

We measure temperature by a thermometer of least count $0.5^o C$, and resistance of the platinum thermometer by a Carey Foster bridge which is sensitive up to $0.005\Omega$.

We work in the range of $0^o C-100^o C$.

Qusetion: How can we justify (without saying that, experimentally it is observed...) the usage of the approximation for $R(T)$ for this temperature range given the precision of our apparatus?

What I’m basically asking is the justification for ignoring the higher order terms in the Taylor expansion of $R(T)$ at $T_0$ for ranges as large as $100^o C$?

  • $\begingroup$ Well, you could look at the tables of resistance vs temperature for your particular resistor and see how linear it is. $\endgroup$ – Jon Custer Sep 20 at 13:59
  • $\begingroup$ I asked on theoretical grounds... $\endgroup$ – Atom Sep 20 at 14:00
  • $\begingroup$ Well, the theory is in the Taylor expansion. The check if it is reasonable is the actual data for you particular material. Deriving resistivity vs temperature for an arbitrary material across a large temperature range is, well, quite difficult. $\endgroup$ – Jon Custer Sep 20 at 14:02
  • $\begingroup$ @JonCuster So the range of validity comes purely from experiment? $\endgroup$ – Atom Sep 20 at 14:04
  • $\begingroup$ Indeed. You might be interested in tsapps.nist.gov/publication/get_pdf.cfm?pub_id=904572 - it would suggest that the range should be based on the particular calibration requested. $\endgroup$ – Jon Custer Sep 20 at 14:07

The response of a platinum resistance thermometer (PRT) is better approximated with the so-called Callendar-Van Dusen equation (this is defined in the standard IEC EN 60751):

$R(t) = \begin{cases} R_0\left[1+At+Bt^2+C(t-100\,{}^\circ\mathrm{C})t^3\right]& t<0 \\[5mm] R_0(1+At+Bt^2)& t\geqslant 0 \end{cases}$

where $t$ is the Celsius temperature and the nominal values of the coefficients are (a calibration can give more accurate values)

$\begin{aligned} &R_0 = 100\,\Omega \\ &A = 3.9083\times 10^{-3}/{}^\circ\mathrm{C}, \\ &B = -5.775\times 10^{-7}/{}^\circ\mathrm{C}^2, \\ &C = -4.183\times 10^{-12}/{}^\circ\mathrm{C}^4, \end{aligned}$

For $t$ from $0\,{}^\circ \mathrm{C}$ to $100\,{}^\circ \mathrm{C}$, the response is

$R(t) = R_0(1+At+Bt^2)$

If you use instead the linear approximation $ R(t) = R_0(1+\alpha t)$, you will measure the temperature $t'$ such that (assuming that $R_0$ is the same for the two equations; otherwise, modify accordingly)

$R_0(1+At+Bt^2) = R_0(1+\alpha t')$

The difference $\Delta t = t'-t$ is the measurement error that you get from using the linear approximation. Substituting $t' = t+\Delta t$ in the above equation and simplifying, you get

$At+Bt^2 = \alpha(t+\Delta t)$

from which

$\Delta t = \dfrac{1}{\alpha}\left[(A-\alpha)t+B t^2\right].$

The above equation represents a parabola, and substituting the value of $\alpha$ that you use you can find the maximum error along the whole measurement range. For instance, if you use $\alpha = 3.85\times 10^{-3}/{}^\circ \mathrm{C}$, the nominal average temperature coefficient from $0\,{}^\circ \mathrm{C}$ to $100\,{}^\circ \mathrm{C}$ (according to the EU standards, for a US PRT you may get $\alpha = 3.92\times 10^{-3}/{}^\circ \mathrm{C}$), the error $\Delta t$ is maximum at $t = 50{}^\circ \mathrm{C}$ and $(\Delta t)_\mathrm{max}\approx 0.38{}^\circ \mathrm{C}$. This specific case is represented in the figure below (own picture):

Pt100 transfer characteristic

Depending on your application and the accuracy class of the PRT, you can then evaluate whether you can consider the above error negligible or not.


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