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--- projects:electronics:weller:wxp80_reverse_engineer:analog_measurements [2018/05/12 23:21] – 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 30: / Line 30: @@
 \\
-Since the is linear, we can find a first order formula which approaches the data pretty well. A general function looks like:\\
+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\\
+$$\begin{align*}ST = \sum\limits_{i=1}^n T_i \end{align*}$$
+$$\begin{align*}STT = \sum\limits_{i=1}^n T_i^2 \end{align*}$$
+$$\begin{align*}SR = \sum\limits_{i=1}^n R_i \end{align*}$$
+$$\begin{align*}SRT = \sum\limits_{i=1}^n T_iR_i \end{align*}$$
+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
+$$\begin{align*}b=\frac{STT\ SR - ST\ STR}{n\ STT - ST^2} \end{align*}$$
-The intercept of the vertical axis is given with following formula:\\
+With the data from the table above, the least squares trendline is:\\
-$b = \frac{(\sum\limits_{i=1}^n T_i^2)(\sum\limits_{i=1}^n R_i) - (\sum\limits_{i=1}^n T_i)(\sum\limits_{i=1}^n T_iR_i)}{n (\sum\limits_{i=1}^n T_i^2) - (\sum\limits_{i=1}^n T_i)^2} \tag{2}$\\
+$R = 0.465T + 98.17 \tag{2}$\\
+where $\begin{align*}T\end{align*}$ is the temperature in °C and $\begin{align*}R\end{align*}$ the resistance of TH1 in Ω.
-With the data from above, the so called least squares trendline results in:\\
+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:\\
-$R = 0.465T + 98.17 \tag{3}$\\
+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:\\
-where T is the temperature in °C and Rv the resistance of TH1 in Ω.
-Another method to acquire this line is using the 'show equation' option in Libre office. It will show the formula (3) right away:\\
+$G = 1 + \frac{R6}{R4 + \frac{R3 \cdot R5}{R3 + R5}} = 1 + \frac{1.0 \cdot 10^6}{33\cdot10^3 + \frac{33\cdot10^3 \cdot 47}{33\cdot10^3 + 47}} = 31.1 \tag{3}$\\
+where $\begin{align*}R3\end{align*}$ = 4.3kΩ, $\begin{align*}R4\end{align*}$ = 33.0kΩ, $\begin{align*}R5\end{align*}$ = 47Ω and $\begin{align*}R6\end{align*}$ = 1.0MΩ\\
-The opamp is configured as a DC coupled non-inverting amplifier with a bias. To calculate the gain (G) and bias (Vb), we use following formulas:\\
+The bias voltage is a simple voltage divider with $\begin{align*}R3\end{align*}$ and $\begin{align*}R5\end{align*}$:
-$G = 1 + \frac{R6}{R4 + \frac{R3 \cdot R5}{R3 + R5}} = 1 + \frac{1.0 \cdot 10^6}{33\cdot10^3 + \frac{33\cdot10^3 \cdot 47}{33\cdot10^3 + 47}} = 31.1 \tag{5}$\\
+$V_b = V_{dd} \frac{R5}{R5 + R3} = 4.66 \frac{47}{47 + 4300} = 0.0501 \tag{4}$\\
-where R3 = 4.3kΩ, R4 = 33.0kΩ, R5 = 47Ω and R6 = 1.0MΩ\\
+where $\begin{align*}V_{dd}\end{align*}$ = 4.66v\\
-The bias voltage is a simply voltage divider with R3 and R5:
-$V_b = V_{dd} \frac{R5}{R5 + R3} = 4.66 \frac{47}{47 + 4300} = 0.0501 \tag{6}$\\
-where Vdd is 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 in the table above.)\\
+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.)\\
 \\
 The transfer function is:
@@ 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