铂电阻传感器计算

6.一个铂电阻传感器被用来测量0到200摄氏度的温度，已知当温度为 T ∘ C T^{\circ}C T∘C时传感器对应电阻为 R T Ω R_T\Omega RT​Ω， R T R_T RT​满足的方程为 R T = R 0 ( 1 + α T + β T 2 ) R_T=R_0(1+\alpha T+ \beta T^2) RT​=R0​(1+αT+βT2)，并且 R 0 = 100.0 R_0=100.0 R0​=100.0， R 100 = 138.50 R_{100}=138.50 R100​=138.50， R 200 = 175.83 Ω R_{200}=175.83\Omega R200​=175.83Ω，试计算：

（a） α \alpha α和 β \beta β的值

（b）在100℃时的全量程的非线性度百分比

解：（a）将 R 0 = 100 Ω R_0=100\Omega R0​=100Ω， R 100 R_{100} R100​， R 200 = 175.83 Ω R_{200}=175.83\Omega R200​=175.83Ω代入方程 R T = R 0 ( 1 + α T + β T 2 ) R_T=R_0(1+\alpha T+ \beta T^2) RT​=R0​(1+αT+βT2)得

{ R 0 = 100 100 ( 1 + α × 100 + β × 10 0 2 ) = 138.50 100 ( 1 + α × 200 + β × 20 0 2 ) = 175.83 \left\{\begin{array}{l}R_{0}=100 \\ 100\left(1+\alpha \times 100+\beta \times 100^{2}\right)=138.50 \\ 100\left(1+\alpha \times 200+\beta \times 200^{2}\right)=175.83\end{array}\right. ⎩⎨⎧​R0​=100100(1+α×100+β×1002)=138.50100(1+α×200+β×2002)=175.83​

解得

{ R 0 = 100 Ω α = 138.50 ∘ C − 1 β = 175.83 ∘ C − 2 \left\{\begin{array}{l}R_{0}=100\Omega \\ \alpha=138.50{}^{\circ}C^{-1} \\ \beta =175.83{}^{\circ}C^{-2}\end{array}\right. ⎩⎨⎧​R0​=100Ωα=138.50∘C−1β=175.83∘C−2​

（b）传感器理想电阻值的一次线性表达式为

R IDEAL = R 200 − R 0 200 − 0 × T + R 0 = 0.379 T + 100 R_{\text{IDEAL}}=\frac{R_{200}-R_0}{200-0} \times T+R_0=0.379T+100 RIDEAL​=200−0R200​−R0​​×T+R0​=0.379T+100

注意只要终点和起点的斜率即可，

当 T = 10 0 ∘ C T=100^{\circ}C T=100∘C时， R I D E A L . 100 = 137.9 Ω R_{IDEAL.100}=137.9\Omega RIDEAL.100​=137.9Ω

于是最大非线性度为

P = R 100 − R I D E A L . 100 R 200 − R 0 × 100 % = 0.76 % P=\frac{R_{100}-R_{IDEAL.100}}{R_{200}-R_{0}}\times100\%=0.76\% P=R200​−R0​R100​−RIDEAL.100​​×100%=0.76%

标签：铂电阻,138.50,R0,传感器,200,175.83,计算,100,Omega
来源： https://blog.csdn.net/xingzhan2142/article/details/118246978