\documentclass[a4paper,10pt]{article} \usepackage[utf8x]{inputenc} \usepackage{graphicx} \usepackage{fancyhdr} \usepackage{lastpage} \usepackage{color} \definecolor{Blue}{rgb}{0.0,0.0,0.7} \definecolor{Red}{rgb}{0.7,0.0,0.0} \definecolor{Green}{rgb}{0.0,0.5,0.0} \usepackage{hyperref} \usepackage{ifthen} \usepackage{algorithm} \usepackage{algorithmic} \usepackage{multirow} \usepackage{textcomp} %opening \title{Capacitive Mains Inputs; choosing capacitor and resistor combinations for 120 V a.c. 240 V a.c and 50 and 60Hz} \author{R.P. clark} \begin{document} \maketitle \begin{abstract} Calculations to work out capacitance values to drive an opto-coupler to detect mains voltage for 50 to 60 Hz. \end{abstract} \begin{figure}[h] \centering \includegraphics[width=200pt]{./images_sw_doc/opto.jpg} % opto.jpg: 854x388 pixel, 72dpi, 30.13x13.69 cm, bb=0 0 854 388 \caption{Opto-coupled mains input circuit} \label{fig:opto} \end{figure} \section{Opto coupler circuit} This circuit is used to detect mains voltage via a capacitor and a resistor forming a potential divider so that a lower voltage can be used to drive an opto-isolator that protects the processor reading the signal. \section{Calculations} A potential divider using a capacitor and a resistor is used to lower mains voltage to levels that can drive a typical opto-coupler input ($\approx 2V$). A potential divider using a capacitor and resistor means using the complex identity for the capacitors reactance, $X$. $$ X = \frac{-j}{\omega C } $$ The ${\omega C }$ term is dependent on frequency and is equivalent to $2.\pi.f$ . Using a potential divider to determine the voltage over the resistor gives: $$ V_{out} = V_{in} \times \frac{R}{R-\frac{j}{2.\pi.f.C}} $$ The equation above leaves a complex divisor. To get a complex number as the numerator, the denominator and numerator must be multiplied by the conjugate of the denominator, thus: $$\frac{R}{R-\frac{j}{2.\pi.f.C}} \equiv \frac{R \times \Big({R+\frac{j}{2.\pi.f.C}}\Big) }{\Big({R-\frac{j}{2.\pi.f.C}}\Big) \times \Big({R+\frac{j}{2.\pi.f.C}}\Big) } $$ This leaves a real number as the denominator, i.e. $ R^2 + {\frac{1}{2.\pi.f.C}}^2$. The resulting complex number, $X$, $$X = \frac{R \times \Big({R+\frac{j}{2.\pi.f.C}}\Big) }{R^2 + {\frac{1}{2.\pi.f.C}}^2}$$ or, \begin{equation} X =\frac{R^2 + \Big({R\frac{j}{2.\pi.f.C}}\Big) }{R^2 + {\frac{1}{2.\pi.f.C}}^2} \label{eqn:genpotdivcapres} \end{equation} can now be evaluated for phase and magnitude. Equation~\ref{eqn:genpotdivcapres} can be generally applied to potential dividers in figure~\ref{fig:opto}. \subsection{Example calculation} At 50Hz with 240 V a.c. applied, with R at 1000 Ohms and C at 47 nF $$\frac{1000^2 + \Big({1000\frac{j}{2.\pi.50.47e-9}}\Big) }{R^2 + {\frac{1}{2.\pi.50.47e-9}}^2}$$ $$\frac{1000^2 + \Big({1000 \times 67726j}\Big) }{1000^2 + {67726}^2}$$ $$\frac{1000^2 + \Big({67726000j}\Big) }{4.5877 \times 10^9}$$ This gives a complex number $$ \frac{1000^2 + {67726000j} }{4.5877 \times 10^9}$$ i.e. $$(216 \times 10^{-6} + 14.76\times 10^{-3} j ) \;.$$ This complex number has a magnitude of 0.0147 and an argument of 89.15 degrees (which is expected as most of the reactance comes from the capacitor). So with 240 V a.c. applied (RMS) the opto would see a signal with $0.0147*240 = 3.54V (RMS)$ \clearpage \section{ploting the voltage at the opto-coupler} \begin{figure}[h] \centering \includegraphics[width=400pt]{./RMS_volts_to_opto.png} % RMS_volts_to_opto.png: 640x480 pixel, 72dpi, 22.58x16.93 cm, bb=0 0 640 480 \caption{RMS voltage seen at opto-coupler for 50 to 60 Hz range} \label{fig:rmstoopto} \end{figure} \clearpage \subsection{plotting the voltage at the opto-coupler: gnuplot scripts} { \tiny \begin{verbatim} ######################################################## # p=3.14159265358979323844 # # 47nF C=47e-9 # # 1k Ohms R=1000 # define complex operator j={0,1} set xlabel "Hertz" set ylabel "Resistance" # x is the frequency set xrange[50:60] # z(x) is the reactance z(x)=(j/(2*p*x*C)) # denominator d(x)=(R*R+z(x)*z(x)) # numerator n(x)=(R*R+R*z(x)) plot abs(z(x)) title "reactance over capacitor" !sleep 4 set ylabel "denominator value (abs)" plot abs(d(x)) !sleep 4 set ylabel "numerator value (abs)" plot abs(n(x)) !sleep 4 v(x)=abs((n(x))/(d(x))) # gives large numbers h(x)=arg((n(x))/(d(x))) set ylabel "voltage to opto-coupler (RMS)" plot 240*v(x) title "240 V a.c", 120*v(x) title "120 V a.c" !sleep 4 set terminal png set output "RMS_volts_to_opto.png" plot 240*v(x) title "240 V a.c", 120*v(x) title "120 V a.c" #set angles degrees #set label "phase change in mains over opto" #plot 240*h(x) title "240 V a.c", 120*h(x) title "120 V a.c" #!sleep 4 # \end{verbatim} } % % % Putting some numbers in this, 47nF for the capacitor, 1k for R and 50 Hz at 240V, means ${2.\pi.f.C} = 14.765 \times 10^{-6}$. % % $$ V_{out} = 240 \times \frac{1000}{ 1000 - \frac{j}{14.765 \times 10^{-6}} } $$ % or % % $$ V_{out} = 240 \times \frac{1000}{1000 - j \times 67.726 \times 10^3 }$$ % % % % % To get a complex number as the numerator, the denominator and numerator must be multiplied by % % its conjugate, thus: % % % % % $$\frac{1000}{1000 - {j} \times 67.726 \times 10^3 } $$ % % % $$ % % % % % \equiv \frac{ 1000 \times (1000 + {j} \times 67.726 \times 10^3) }{ (1000 - {j} \times 67.726 \times 10^3) \times (1000 + {j} \times 67.726 \times 10^3)} $$ % % $$ % typeset in {\Huge \LaTeX} \today. \end{document}