Measuring Temperature — Question 3

Question 3

A quadratic resistance calibration needs its full inverse

A resistance sensor obeys $R/R_0=1+a\Delta T+b(\Delta T)^2$, where $R_0=100\,\Omega$, $a=.0040\,\mathrm{K^{-1}}$, $b=1.0\times10^{-5}\,\mathrm{K^{-2}}$, and $\Delta T=T-300\,\mathrm K$. Its validated domain is $0\le\Delta T\le100\,\mathrm K$. A bath gives $R=121.0\,\Omega$.

1. Find the physical temperature and the error from ignoring the quadratic term.

2. At this operating point find $dR/dT$ and estimate the temperature half-width from a small resistance half-width $.10\,\Omega$. Label the result as a local linear estimate.

Question 3 - Solution

A quadratic resistance calibration needs its full inverse

(a) Method and calculation

$$10^{-5}(\Delta T)^2+.004\Delta T-.21=0.$$$$\Delta T=\frac{-.004+\sqrt{.004^2+4(10^{-5})(.21)}}{2\times10^{-5}}\,\mathrm K\approx46.982\,\mathrm K.$$Thus $T=346.982\,\mathrm K$. The other root is outside the domain. A linear inverse gives $52.5$ K, about $5.518$ K too high.

(b) Interpretation and further calculation

$$\frac{dR}{dT}=R_0(a+2b\Delta T)\approx.49396\,\Omega/\mathrm K.$$$$\delta T\simeq\frac{.10}{.49396}\,\mathrm K=.20244\,\mathrm K.$$This sensitivity estimate is not a substitute for exact endpoint inversion when a strict hard interval is required.

Physical check

The derivative is positive throughout the validated domain, ensuring a unique inverse there.