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small changes to clear out error

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Johnny Hsu 2025-03-23 01:16:23 +01:00
parent 925f4aea7e
commit 19c75e3883
2 changed files with 29 additions and 31 deletions
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temperature_measurement_error/AMS_T_error.pdf
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temperature_measurement_error/AMS_T_error.tex
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@ -16,8 +16,6 @@
\usepackage{float}
\usepackage{multicol}
\usepackage{amsmath, amssymb}
\usepackage{csvsimple}
\usepackage{pgfplotstable}
\pagestyle{fancy}
\fancyhf{}
@ -36,21 +34,21 @@ Since the Look Up Table provided by Vishay is used for the calculation in the AM
As the characteristic curve is not linear, it is relatively trivial to find the absolute maximum measurement error, therefore, the maximum error at \SI{60}{\celsius} is calculated here.
Our Voltage measurement system is based on a NTC (NTCLE413E2103F102L from Vishay) and a 10k 0.1\% resistor (name as $R_1$ here) froming a voltage divider, and the output voltage is then feed to an ADC after passing through a RC-filter.
Our Voltage measurement system is based on a NTC (NTCLE413E2103F102L from Vishay) and a 10k 0.1\% resistor (named $R_1$ here) froming a voltage divider, and the output voltage is then feed to an ADC after passing through a RC-filter.
To calculate the error, the highest possible measured voltage at \SI{60}{\celsius} is worked out here, since according to the design of our voltage divider, the lower the temperature, the higher the voltage. As shown in Fig. \ref{fig:vref2}, the supply voltage $V_{REF2}$ of the voltage divider can have a maximum value of \SI{3.006}{\volt}, while the total measurement error of the GPIO is $\pm$ \SI{2.8}{mV}. (Fig \ref{fig:aux}) In addition, the maximum resistance from the NTC can be \SI{3086.8}{\ohm} according to the LUT (Tab. \ref{tab:lut}). the maximum possible voltage recorded is therefore:
To calculate the error, the highest possible measured voltage at \SI{60}{\celsius} is worked out here, since according to the design of our voltage divider, the lower the temperature, the higher the voltage. As shown in Fig. \ref{fig:vref2}, the supply voltage $V_{REF2}$ of the voltage divider can have a maximum value of \SI{3.006}{\volt}, while the total measurement error of the GPIO is $\pm$ \SI{2.8}{mV}. (Fig. \ref{fig:aux}) In addition, the maximum resistance from the NTC can be \SI{3086.8}{\ohm} according to the LUT (Tab. \ref{tab:lut}). the maximum possible voltage recorded is therefore:
\begin{align*}
& V_{REF2} * \frac{R_{NTC}}{R_{NTC}+R_1} + V_{err} \\
= & 3.006V * \frac{3086.8}{3086.8+9990} + 0.0028V \\
& V_{REF2} \cdot \frac{R_{NTC}}{R_{NTC}+R_1} + V_{err} \\
= & 3.006V \cdot \frac{3086.8}{3086.8+9990} + 0.0028V \\
\approx & 712.4V
\end{align*}
to find the largest possible error, the lowest possible matching temperature should be calculated, that theoretically can produce the same voltage output. The calculation is as below:
\begin{align*}
& V_{REF2} * \frac{R_{NTC}}{R_{NTC}+R_1} + V_{err} \\
= & 2.994V * \frac{R_{NTC}}{R_{NTC}+10010} - 0.0028V = 712.4V \\
& V_{REF2} \cdot \frac{R_{NTC}}{R_{NTC}+R_1} + V_{err} \\
= & 2.994V \cdot \frac{R_{NTC}}{R_{NTC}+10010} - 0.0028V = 712.4V \\
& R_{NTC} \approx 3141.6
\end{align*}
@ -76,7 +74,7 @@ since the LUT is used to match the voltage to the temperature, and the nominal r
\label{tab:lut}
\begin{tabular}{||c c c c c c||}
\hline
Temp. [\SI{}{\celsius}] & $R_{nom}$ [\Omega] & $R_{min}$ [\Omega] & $R_{max}$ [\Omega] & \Delta R/R [\%] & \Delta T [\SI{}{\celsius}] \\
Temp. [\SI{}{\celsius}] & $R_{nom} [\Omega]$ & $R_{min} [\Omega]$ & $R_{max} [\Omega]$ & $\Delta R/R [\%]$ & $\Delta T [\SI{}{\celsius}]$ \\
\hline\hline
58 & 3214.99 & 3145.6 & 3284.4 & 2.16 & 0.69 \\
58.1 & 3204.88 & 3135.6 & 3274.2 & 2.16 & 0.69 \\