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--- projects:electronics:weller:wxp80_reverse_engineer:analog_measurements [2018/05/12 23:28] – [PTC resistance and voltages WXP80] admin
+++ projects:electronics:weller:wxp80_reverse_engineer:analog_measurements [2018/05/15 00:39] (current) – [WXP80 analog measurements] admin
@@ Line 1: / Line 1: @@
 ====== WXP80 analog measurements ======
-This page shows the analog measurements of the WXP80 heating element connected to an operation amplifier. From the results a formula is calculated to model this.
+This page shows the analog measurements of the WXP80 heating element connected to an operational amplifier. From the results a formula is calculated to model this.
 ===== PTC resistance and voltages WXP80 =====
@@ Line 32: / Line 32: @@
 Based on the observations we assume that the relationship between temperature and resistance is linear. We can thus find a first order formula which approaches the data pretty well. A general function looks like:\\
 $R = aT + b \tag{1}$\\
-where a is the slope and b the intercept of the vertical-axis.
+where $\begin{align*}a\end{align*}$ is the slope and $\begin{align*}b\end{align*}$ the intercept of the vertical-axis.\\
-A formula to calculate the slope is:\\
+The so called normal equations for the estimated slope $a$ and intercept $b$ are:\\
+$$\begin{align*}n\ a + ST\ b = SR \end{align*}$$
-$a = \frac{n (\sum\limits_{i=1}^n T_iR_i) - (\sum\limits_{i=1}^n T_i)(\sum\limits_{i=1}^n R_i)}{n (\sum\limits_{i=1}^n T_i^2) - (\sum\limits_{i=1}^n T_i)^2} \tag{1}$\\
+$$\begin{align*}ST\ a + STT\ b = SRT \end{align*}$$
-where n is the number of samples (in our case 24 samples), T the temperature and R the resistance of TH1.
+with $\begin{align*}n\end{align*}$ the number of data\\
+$$\begin{align*}ST = \sum\limits_{i=1}^n T_i \end{align*}$$
-The intercept of the vertical axis is given with following equation:\\
+$$\begin{align*}STT = \sum\limits_{i=1}^n T_i^2 \end{align*}$$
+$$\begin{align*}SR = \sum\limits_{i=1}^n R_i \end{align*}$$
-$b = \frac{(\sum\limits_{i=1}^n T_i^2)(\sum\limits_{i=1}^n R_i) - (\sum\limits_{i=1}^n T_i)(\sum\limits_{i=1}^n T_iR_i)}{n (\sum\limits_{i=1}^n T_i^2) - (\sum\limits_{i=1}^n T_i)^2} \tag{2}$\\
+$$\begin{align*}SRT = \sum\limits_{i=1}^n T_iR_i \end{align*}$$
+with solution
+$$\begin{align*}a= \frac{n\ SRT- SR\ ST}{n\ STT - ST^2} \end{align*}$$
+$$\begin{align*}b= \frac{1}{ST} (SR-n\ a) \end{align*}$$
+or alternatively
+$$\begin{align*}b=\frac{STT\ SR - ST\ STR}{n\ STT - ST^2} \end{align*}$$
 With the data from the table above, the least squares trendline is:\\
-$R = 0.465T + 98.17 \tag{3}$\\
+$R = 0.465T + 98.17 \tag{2}$\\
-where T is the temperature in °C and Rv the resistance of TH1 in Ω.
+where $\begin{align*}T\end{align*}$ is the temperature in °C and $\begin{align*}R\end{align*}$ the resistance of TH1 in Ω.
-Please note that libre office offers an easy method to acquire this equation directly, using the 'show equation' option. It will show equation (3) right away:\\
+Please note that libre office calc offers an easy method to acquire this equation directly, using the 'show equation' option. It will show equation (2) right away:\\
 The opamp is configured as a DC coupled non-inverting amplifier with a bias. To calculate the gain (G) and bias (Vb), we use following equations:\\
-$G = 1 + \frac{R6}{R4 + \frac{R3 \cdot R5}{R3 + R5}} = 1 + \frac{1.0 \cdot 10^6}{33\cdot10^3 + \frac{33\cdot10^3 \cdot 47}{33\cdot10^3 + 47}} = 31.1 \tag{5}$\\
+$G = 1 + \frac{R6}{R4 + \frac{R3 \cdot R5}{R3 + R5}} = 1 + \frac{1.0 \cdot 10^6}{33\cdot10^3 + \frac{33\cdot10^3 \cdot 47}{33\cdot10^3 + 47}} = 31.1 \tag{3}$\\
-where R3 = 4.3kΩ, R4 = 33.0kΩ, R5 = 47Ω and R6 = 1.0MΩ\\
+where $\begin{align*}R3\end{align*}$ = 4.3kΩ, $\begin{align*}R4\end{align*}$ = 33.0kΩ, $\begin{align*}R5\end{align*}$ = 47Ω and $\begin{align*}R6\end{align*}$ = 1.0MΩ\\
-The bias voltage is a simply voltage divider with R3 and R5:
+The bias voltage is a simple voltage divider with $\begin{align*}R3\end{align*}$ and $\begin{align*}R5\end{align*}$:
-$V_b = V_{dd} \frac{R5}{R5 + R3} = 4.66 \frac{47}{47 + 4300} = 0.0501 \tag{6}$\\
+$V_b = V_{dd} \frac{R5}{R5 + R3} = 4.66 \frac{47}{47 + 4300} = 0.0501 \tag{4}$\\
-where Vdd is 4.66v\\
+where $\begin{align*}V_{dd}\end{align*}$ = 4.66v\\
 \\
 We now are able to calculate a transfer function using the PTC resistor value to find the opamp output voltage relative to the temperature. (Alternatively we could calculate the transfer function using the least squares method of the output voltage data from the table above.)\\
@@ Line 70: / Line 76: @@
   * Fluke 87 multimeter
+==== References ====
+  * [[https://ocw.mit.edu/courses/media-arts-and-sciences/mas-836-sensor-technologies-for-interactive-environments-spring-2011/readings/MITMAS_836S11_read02_bias.pdf]]
+  * least square trendline