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%Dokumentinnstillinger:---------------------------------
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%Ved å google flitting kan du finne ut hva de forskjellige tingene her betyr, og hvordan du kan gjøre eventuelle endringer.
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\documentclass[11pt,norsk]{elsys-design}
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\usepackage{subfig}
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\tikzset{opampdownlbl/.style={
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below,
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draw=none,
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append after command={
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(\tikzlastnode.north) edge ([shift={(-5pt,0pt)}]\tikzlastnode.north)
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edge ([shift={(+5 pt,0 pt)}]\tikzlastnode.north)
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}},
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opampuplbl/.style ={
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above,
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draw=none,
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append after command={
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(\tikzlastnode.south) edge ([shift={(-5pt,0pt)}]\tikzlastnode.south)
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edge ([shift={(+5pt,0pt)}]\tikzlastnode.south)
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}}
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}
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\heading{Designnotat}
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\title{Trekant-oscillator}
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\author{Øyvind Skaaden}
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\version{2.0}
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\date{\today}
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\begin{document}
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\maketitle
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%Automatisk generert innholdsfortegnelse:------------------
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\toc
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%Selve rapporten:------------------------------------------
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\section{Problembeskrivelse}
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\label{sec:innledning}
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Vi vil ta for oss design av et system som vist i \figref{pic:blokkskjema}
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\begin{figure}[htbp]
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\centering
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\includegraphics[width=0.8\textwidth]{pics/blokkskjema.png}
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\caption{Blokkskjema for trekant-oscillator.}
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\label{pic:blokkskjema}
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\end{figure}
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Systemet skal kun ha en driftsspenning, og skal produsere et trekantsignal med en ønsket frekvens $f_0 $, eller periode $T_0 = \frac{1}{f_0} $ på utgangen $v_2$. Det skal kun være et avvik på $\Delta f_{max} = 10\ 000 $ ppm. Amplituden er ikke viktig i dette systemet.
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\section{Prinsipiell løsning}
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\label{sec:prinsipielllosning}
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Det er flere måter å generere en trekantpuls, men i dette designet skal operasjonsforsterkere brukes, opamp fra nå av. Det vil bli tatt utgangspunkt i et teknisk notat \cite{notat}.
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Vi tar utgangspunk i figur 1 i det tekniske notatet, se \figref{fig:trekantkrets}.
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\begin{figure}[htbp]
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\centering
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\begin{circuitikz}
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\draw
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(0,0) node[op amp,yscale=-1](opamp){}
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(opamp.+) to [short,-] ++(-1,0)
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to [R,l_=$R_1$,-] ++(0,-2)
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to [short,-] ++(0,-1) coordinate(v2v3)
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(opamp.+) to [short,*-] ++(0,1.5) coordinate(leftR)
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to [R,l=$R_2$] (leftR-|opamp.out)
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to [short,-*] (opamp.out) node[below]{$v_1$} coordinate(V1)
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(opamp.-) to node[ground]{} ++(0,-1)
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(opamp.up) ++(0,-.5) node[opampdownlbl]{$-V$} -- (opamp.up)
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(opamp.down) ++(0,.5) node[opampuplbl]{$+V$} -- (opamp.down)
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(opamp.+) to node[below]{$v_3$}(opamp.+)
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(5,-0.5) node[op amp](opamp){}
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(opamp.-) to[R,l_=$R_3$,-] (V1)
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(opamp.-) to[short,*-] ++(0,1.5) coordinate(leftC)
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to [C,l^=$C_1$] (leftC-|opamp.out)
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to [short,-*] (opamp.out)
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to [short,-o] ++(1,0) node[right]{$v_2$}
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(opamp.+) to node[ground]{} ++(0,-1)
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(opamp.out) to [short,-] ++(0,-2)
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to [short,-] (v2v3)
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;
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\end{circuitikz}
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\caption{Skjema for prinsipiell løsning.}
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\label{fig:trekantkrets}
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\end{figure}
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I denne kretsen vil punktet $v_2 $ ha trekantpulsform, sentrert rundt $0$V, fordi vi kretsen ``integrerer'' en konstant spenning.
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Vi kan se på det som å integrere en konstant, det vil bli en linje med stigningstall lik konstanen som blir integrert.
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Vi bruker dette signalet videre, slik at kretsen bli selv-drivende, slik at vi får et pulstog og ikke bare en enkelt puls.
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Signalet vil bevege seg til opampen til venstre bli til et fikantsignal (fordi vi forsterker til metning av opamp), som igjen vil drive opampen til høyre, som genererer trekantsignalet. Det er mulig å se hvordan spenningene $v_1$ og $v_2$ er i forhold til hverandre i \figref{graph:pyGraph}.
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Her er motstandene $R_1 $ og $R_2 $ en spenningsdeler mellom $v_1$ og $v_2$ som fører til inngangspenningen $v_3 $. $v_3 $ har formen til $v_2$, men vil ha en mindre amplitude grunnet spenningsdeleren.
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Opampen som ligger etter $v_3 $ er en komparator. $v_1$ vil gå mot driftsspenning $+V$ dersom $+$ inngangen eller $v_3$ er større enn $-$ (jord), og bevege seg mot $-V$ dersom $v_3$ er lavere enn jord (negativ spenning).
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Den høyre delen, med en opamp, en motstand $R_3 $ og kondensator $C_1 $ er en integrator, som sett i \figref{fig:integrator} eller i \cite{integrator}. Integratoren tidsintegrerer signalet som kommer ut fra $v_1$ og inn på $-$ på opampen. Det vil si at utgangssignalet $v_2$ vil bevege seg mot spenningen på $v_1$, over en periode som er lik tidskonstanten $\tau = R_3 \cdot C_1$.
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\begin{figure}[ht]
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\centering
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\begin{circuitikz}
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\draw
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(5,-0.5) node[op amp](opamp){}
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(opamp.-) to[R,l_=$R_3$,-o]++(-2,0) coordinate(Vi) node[left]{$v_1$}
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(opamp.-) to[short,*-] ++(0,1.5) coordinate(leftC)
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to [C,l^=$C_1$] (leftC-|opamp.out)
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to [short,-*](opamp.out)
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to [short,-o] ++(1,0) node[right]{$v_2$}
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(opamp.+) to node[ground]{} ++(0,-1)
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;
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\end{circuitikz}
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\caption{Skjema for en integrator.}
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\label{fig:integrator}
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\end{figure}
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Dersom vi integrerer en firkantpuls vil vi få en trekantpuls.
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Vi vet fra \cite{integrator} at spenningen $v_2$ i integrator som i \figref{fig:integrator} er gitt ved (\ref{eq:integrator}).
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\begin{align}
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v_2 = -\frac{1}{R_3 \cdot C_1}\int_0^t v_1 \ \text{d}t
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\label{eq:integrator}
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\end{align}
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Der $v_1$ er inngangsspenningen.
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Her vil $\tau$ si hvor lang til det tar for spenningen $v_2$ når metning/driftsspenning dersom inngangsignalet $v_1$ er en firkantpuls.
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Vi ønsker at trekantpulsen på $v_2$ både skal gå opp mot positiv driftsspenning og ned mot negativ driftsspenning på en periode.
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Ettersom det integratoren bruker en tid $\tau$ for å bevege seg enten opp eller ned, må den totale tiden for en periode, eller én enkelt trekantpuls være $ T = 2\tau $.
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Dermed må tiden $ \tau = \frac{1}{2} T $.
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Vi ønsker å gå både opp og ned fordi vi ønsker å få et symetrisk og kontinuerlig trekantsignal.
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Da må vi starte og stoppe på samme sted, sånn at pulstoget ikke blir forskjøvet opp eller ned i spenning.
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Vi kan velge $R_1$ og $R_2$ slik at spenningen $v_3$ ikke blir for stor. Visse opamper kan ha spenningsområder der de begynner å få uønsket opførsel, vi må bare passe på at vi ikke overstiger denne terskelspenningen, $V_T$. Vi må derfor opprettholde (\ref{eq:terskel}).
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\begin{align}
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|v_3| < V_T \label{eq:terskel}
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\end{align}
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Vi har dermed formelen (\ref{eq:tidskonstant}).
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\begin{align}
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\tau &= \frac{1}{2} T = \frac{1}{2f} \label{eq:tidskonstant}
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\end{align}
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Der $f$ er ønsket frekvens, $T$ er ønsket periode og $\tau$ er tidskonstanten til integratoren.
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$R_1 $ og $R_2 $ kan velges slik at $v_3$ holder seg innenfor en terskelspenning, og at $R_2 > R_1$.
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\section{Realisering og test}
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\label{sec:realisering}
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Trekantgeneratoren i dette designet har en ønsket frekvens på $f_0 = 3000$Hz. Vi kan da regne ut en tau ut i fra dette tallet.
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\begin{align}
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\tau = \frac{1}{2 \cdot 3000 \text{Hz}} = \frac{1}{6} \cdot 10^{-3} \ \text{s} = \frac{1}{6}\ \text{ms} \approx 1.67\ \text{ms}
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\end{align}
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Siden $\tau = R_3\cdot C_1 $ kan vi velge en av verdiene $R_3 $ eller $C_1 $ for å finne den andre.
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Setter $C = 68$nF. Da blir
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\begin{align*}
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R &= \frac{\tau}{C} \\
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R &= \frac{\frac{1}{6} \cdot 10^{-3} \ \text{s}}{68 \cdot 10^{-9}\text{F}} \\
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R &\approx 2451\Omega
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\end{align*}
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Opampen som brukes i kretsen er en LF353P opamp. Den kan oppføre seg ulineært dersom spenningene inn på $+$ og $-$ har en amplitude på over $4.2\text{V}$, $V_T = 4.2$V.
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Vi har driftsspenninger $+V=5$V og $-V=-5$V. Bruker Kirchoffs' spennignslov og spenningsdeling og velger $R_1 = 4\text{k}7\Omega $ og $R_2 = 10\text{k}\Omega $ slik at spenningen $|v_3| < 4.2\text{V}$, og da opprettholdes (\ref{eq:terskel}).
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Vi har da komponentverdier som sett i Tabell \ref{tab:kompnenter}.
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\begin{table}[htbp]
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\centering
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\caption{Utregnede komponentverdier.}
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\begin{tabular}{|c|c|}
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\hline\hline
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Komponent & Verdi \\\hline
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\hline
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$R_1$ & $4.7\text{k}\Omega$ \\
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$R_2$ & $10\text{k}\Omega$ \\
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$R_3$ & $2.5\text{k}\Omega$ \\
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$C_1$ & $68$nF \\
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OpAmp & LF353P \\ \hline
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\hline
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\end{tabular}
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\label{tab:kompnenter}
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\end{table}
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Kobler opp kretsen etter figur \ref{fig:trekantkrets} med verdier fra tabell \ref{tab:kompnenter}.
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Siden motstanden $R_3 = 2\text{k}5\Omega$ ikke er en standard motstand bruker vi flere andre typer. Resultatet er følgende
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$$ R_3 = 2\text{k}2\Omega + 270\Omega = 2470\Omega$$.
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Etter målinger med denne motstandsverdien, finner vi at frekvensen var litt for rask. Etter videre testing finner vi at ved motstandsverdien $R_3 = 2509\Omega$ er frekvensen $f_0 = 2.999$kHz.
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Nye motstandsverdier er beskrevet i (\ref{eq:motstandsverdi}).
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\begin{align}
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R_3 = 2.2\text{k}\Omega + 270\Omega + 39\Omega = 2509\Omega \label{eq:motstandsverdi}
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\end{align}
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Alle komponentverdier er i Tabell \ref{tab:kompnenterFerdig}.
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\begin{table}[ht]
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\centering
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\caption{Utregnede komponentverdier.}
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\begin{tabular}{|c|c|}
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\hline\hline
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Komponent & Verdi \\\hline
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\hline
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$R_1$ & $4.7\text{k}\Omega$ \\
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$R_2$ & $10\text{k}\Omega$ \\
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$R_3$ & $2.2\text{k}\Omega + 270\Omega + 39\Omega = 2509\Omega$ \\
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$C_1$ & $68$nF \\
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OpAmp & LF353P \\ \hline
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\hline
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\end{tabular}
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\label{tab:kompnenterFerdig}
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\end{table}
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Skjema for ferdig krets i \figref{fig:trekantkretsFerdig}.
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\begin{figure}[ht]
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\centering
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\begin{circuitikz}
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\draw
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(0,0) node[op amp,yscale=-1](opamp){}
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(opamp.+) to [short,-] ++(-1,0)
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to [R,l_=$4.7\text{k}\Omega$,-] ++(0,-2)
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to [short,-] ++(0,-1) coordinate(v2v3)
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(opamp.+) to [short,*-] ++(0,1.5) coordinate(leftR)
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to [R,l=$10\text{k}\Omega$] (leftR-|opamp.out)
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to [short,-*] (opamp.out) node[below]{$v_1$} coordinate(V1)
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(opamp.-) to node[ground]{} ++(0,-1)
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(opamp.up) ++(0,-.5) node[opampdownlbl]{$-V$} -- (opamp.up)
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(opamp.down) ++(0,.5) node[opampuplbl]{$+V$} -- (opamp.down)
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(opamp.+) to node[below]{$v_3$}(opamp.+)
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(8.5,-0.5) node[op amp](opamp){}
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(opamp.-) to[R,l_=$39\Omega$,-] ++(-2,0)
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to[R,l_=$270\Omega$,-] ++(-2,0)
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to[R,l_=$2.2\text{k}\Omega$,-] (V1)
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(opamp.-) to[short,*-] ++(0,1.5) coordinate(leftC)
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to [C,l^=$68$nF] (leftC-|opamp.out)
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to [short,-*] (opamp.out)
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to [short,-o] ++(1,0) node[right]{$v_2$}
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(opamp.+) to node[ground]{} ++(0,-1)
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(opamp.out) to [short,-] ++(0,-2)
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to [short,-] (v2v3)
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;
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\end{circuitikz}
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\caption{Skjema for den ferdige kretsen med komponentverdier.}
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\label{fig:trekantkretsFerdig}
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\end{figure}
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Med dette koblet opp som vist i figur \ref{pic:ferdigKrets}.
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\begin{figure}[ht]
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\centering
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\subfloat[Ferdig fungerende krets.]{\includegraphics[width=0.45\textwidth]{pics/RenKrets.png}\label{pic:renKrets}}
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\hfill
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\subfloat[Ferdig fungerende krets med navn.]{\includegraphics[width=0.45\textwidth]{pics/KretsMedNavn.png}\label{pic:kretsMedNavn}}
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\caption{Ferdig krets, med og uten navn.}
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\label{pic:ferdigKrets}
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\end{figure}
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Kretsen ble også målt med oscilloscop for å sjekke frekvensen. Se \figref{graph:pyGraph}.
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\begin{figure}
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\centering
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\includegraphics[width=\textwidth]{grafer/Trekantpuls3k.png}
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\caption{Måligner av frekvens og utseende på spenninger $v_2$ (trekantpuls) og $v_1$ (firkantpuls).}
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\label{graph:pyGraph}
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\end{figure}
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Målt frekvens er på $f = 2.9999$kHz eller $f = 2999.9$Hz. Dette er innenfor avviket på $\Delta f = 10\ 000$ ppm unna $3000$Hz.
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\clearpage
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\section{Konklusjon}
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\label{sec:konklusjon}
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Målet var $3000$Hz, og kretsen klarte å produsere en trekantpuls som hadde en frekvens på $2999.9$Hz. Noe som også er godt innenfor avviket på $10\ 000$ ppm.
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\section{Takk}
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Takk til Ulrik Bredland for bra samarbeid og gode diskusjoner.
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%Bibliografi: Legg til flere elementer ved å legge til flere \bibitem:--------
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\phantomsection
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\addcontentsline{toc}{section}{Referanser}
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\begin{thebibliography}{99}
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\bibitem{notat}
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Torstein Bolstad,
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\emph{Teknisk notat: Trekantgenerator},
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NTNU,
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TTT4260 Elektronisk sysdemdesign og -analyse,
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2019.
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\bibitem{integrator}
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Wikipedia contributors. (2019, April 4). Op amp integrator. In Wikipedia, The Free Encyclopedia. Retrieved 13:02, April 28, 2019, from \url{https://en.wikipedia.org/w/index.php?title=Op_amp_integrator&oldid=891001432}
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\end{thebibliography}{}
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\clearpage
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\appendix
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%Tillegg. Flere tillegg legges til ved å lage flere sections:-----------------
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\end{document}
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@ -0,0 +1,248 @@
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%Dokumentinnstillinger:---------------------------------
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%Ved å google flitting kan du finne ut hva de forskjellige tingene her betyr, og hvordan du kan gjøre eventuelle endringer.
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\documentclass[11pt,norsk]{elsys-design}
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\usepackage{subfig}
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\tikzset{opampdownlbl/.style={
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below,
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draw=none,
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append after command={
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(\tikzlastnode.north) edge ([shift={(-5pt,0pt)}]\tikzlastnode.north)
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edge ([shift={(+5 pt,0 pt)}]\tikzlastnode.north)
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}},
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opampuplbl/.style ={
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above,
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draw=none,
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append after command={
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(\tikzlastnode.south) edge ([shift={(-5pt,0pt)}]\tikzlastnode.south)
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edge ([shift={(+5pt,0pt)}]\tikzlastnode.south)
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}}
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}
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\heading{Designnotat}
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\title{Trekant-oscillator}
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\author{Øyvind Skaaden}
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\version{1.0}
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\date{\today}
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\begin{document}
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\maketitle
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%Automatisk generert innholdsfortegnelse:------------------
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\toc
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||||
%Selve rapporten:------------------------------------------
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\section{Problembeskrivelse}
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\label{sec:innledning}
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Vi vil ta for oss design av et system som vist i \figref{pic:blokkskjema}
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|
||||
\begin{figure}[htbp]
|
||||
\centering
|
||||
\includegraphics[width=0.7\textwidth]{pics/blokkskjema.png}
|
||||
\caption{Blokkskjema for trekant-oscillator.}
|
||||
\label{pic:blokkskjema}
|
||||
\end{figure}
|
||||
|
||||
Systemet skal kun ha en driftsspenning, og skal produsere en trekantpuls med frekvens $f_0 $, eller periode $T_0 = \frac{1}{f_0} $. Det skal kun være et avvik på $\Delta f_{max} = 10\ 000 $ ppm. Amplituden er ikke viktig i dette systemet.
|
||||
|
||||
\section{Prinsipiell løsning}
|
||||
\label{sec:prinsipielllosning}
|
||||
|
||||
Det er flere måter å generere en trekantpuls, men i dette designet skal operasjonsforsterkere brukes, opamp fra nå av. Det vil bli tatt utgangspunkt i et teknisk notat \cite[notat]{notat} og oppgave 3 på øving 3 \cite[øving 3]{oving3} for å beskrive en prinsipiell løsning.
|
||||
|
||||
I både det tekniske notatet og øvingen er kretsen i \figref{fig:trekantkrets}.
|
||||
|
||||
\begin{figure}[htbp]
|
||||
\centering
|
||||
\begin{circuitikz}
|
||||
\draw
|
||||
(0,0) node[op amp,yscale=-1](opamp){}
|
||||
(opamp.+) to [short,-] ++(-1,0)
|
||||
to [R,l_=$R_1$,-] ++(0,-2)
|
||||
to [short,-] ++(0,-1) coordinate(v2v3)
|
||||
(opamp.+) to [short,*-] ++(0,1.5) coordinate(leftR)
|
||||
to [R,l=$R_2$] (leftR-|opamp.out)
|
||||
to [short,-*] (opamp.out) node[below]{$v_1$} coordinate(V1)
|
||||
(opamp.-) to node[ground]{} ++(0,-1)
|
||||
(opamp.up) ++(0,-.5) node[opampdownlbl]{$-V$} -- (opamp.up)
|
||||
(opamp.down) ++(0,.5) node[opampuplbl]{$+V$} -- (opamp.down)
|
||||
(opamp.+) to node[below]{$v_3$}(opamp.+)
|
||||
|
||||
(5,-0.5) node[op amp](opamp){}
|
||||
(opamp.-) to[R,l_=$R$,-](V1)
|
||||
(opamp.-) to[short,*-] ++(0,1.5) coordinate(leftC)
|
||||
to [C,l^=$C$] (leftC-|opamp.out)
|
||||
to [short,-*](opamp.out)
|
||||
to [short,-o] ++(1,0) node[right]{$v_2$}
|
||||
(opamp.+) to node[ground]{} ++(0,-1)
|
||||
(opamp.out) to [short,-] ++(0,-2)
|
||||
to [short,-](v2v3)
|
||||
;
|
||||
\end{circuitikz}
|
||||
\caption{Skjema for prinsipiell løsning.}
|
||||
\label{fig:trekantkrets}
|
||||
\end{figure}
|
||||
|
||||
I denne kretsen vil punktet $v_2 $ ha trekantpulsform, sentrert rundt $0$V. Dette signalet blir brukt til å drive kretsen.
|
||||
Her er motstandene $R_1 $ og $R_2 $ en spenningsdeler for inngangspenningen $v_3 $. $v_3 $ har formen til $v_2$, men lavere grunnet spenningsdeleren.
|
||||
|
||||
Opampen som ligger etter $v_3 $ er en komparator og $v_1$ vil da gå mot driftsspenning $+V$ dersom $+ $ inngangen eller $v_3 $ er større enn $-$ (jord) og mot $-V$ dersom $v_3$ er lavere enn jord (negativ spenning).
|
||||
|
||||
Den høyre delen, med en opamp, en motstand $R$ og kondensator $C$ er en integrator. Den vil integrere inngangssignalet med hensyn på tidskonstanten $\tau$ som oppstår mellom $R$ og $C$.
|
||||
|
||||
Dersom vi integrerer en firkantpuls vil vi få en trekantpuls.
|
||||
|
||||
Vi vet fra øving 3 at spenningen $v_o $ i integrator som i \figref{fig:integrator} er gitt ved
|
||||
|
||||
|
||||
\begin{align}
|
||||
v_o = -\frac{1}{\tau}\int v_i \ \text{d}t
|
||||
\label{eq:integrator}
|
||||
\end{align}
|
||||
|
||||
Der $v_i$ er inngangsspenningen.
|
||||
|
||||
\begin{figure}[htbp]
|
||||
\centering
|
||||
\begin{circuitikz}
|
||||
\draw
|
||||
(5,-0.5) node[op amp](opamp){}
|
||||
(opamp.-) to[R,l_=$R$,-o]++(-2,0) coordinate(Vi) node[left]{$v_i$}
|
||||
(opamp.-) to[short,*-] ++(0,1.5) coordinate(leftC)
|
||||
to [C,l^=$C$] (leftC-|opamp.out)
|
||||
to [short,-*](opamp.out)
|
||||
to [short,-o] ++(1,0) node[right]{$v_o$}
|
||||
(opamp.+) to node[ground]{} ++(0,-1)
|
||||
;
|
||||
\end{circuitikz}
|
||||
\caption{Skjema for en integrator.}
|
||||
\label{fig:integrator}
|
||||
\end{figure}
|
||||
|
||||
Her vil $\tau$ si hvor lang til det tar for spenningen $v_o $ når metning/driftsspenning. Vi ønsker at denne både skal gå opp og ned på en periode. Dermed må $ T = 2\tau \Leftrightarrow \tau = \frac{1}{2} T $. Vi ønsker å gå både opp og ned fordi vi ønsker å få en symetrisk og kontinuerlig trekantpuls. Da må vi starte og stoppe på samme sted.
|
||||
|
||||
Vi kan velge $R{_1}$ og $R{_2}$ slik at spenningen $v{_3}$ ikke overstiger en terskelspenning der opampen begynner å oppføre seg merkelig.
|
||||
|
||||
Vi har dermed formelen
|
||||
|
||||
\begin{align}
|
||||
\tau &= \frac{1}{2} T = \frac{1}{2f} \nonumber \\
|
||||
&\Updownarrow \nonumber \\
|
||||
f &= \frac{1}{2\tau} = \frac{1}{2RC}
|
||||
\end{align}
|
||||
|
||||
Der $f$ er ønsket frekvens, $T$ er ønsket periode og $\tau$ er tidskonstanten til integratoren.
|
||||
|
||||
$R_1 $ og $R_2 $ kan velges slik at $v_3$ holder seg innenfor en terskelspenning, og at $R_2 > R_1$
|
||||
|
||||
|
||||
\section{Realisering og test}
|
||||
\label{sec:realisering}
|
||||
|
||||
Trekantgeneratoren i dette designet har en ønsket frekvens på $3000$Hz. Vi kan da regne ut en tau ut i fra dette tallet.
|
||||
|
||||
$$ \tau = \frac{1}{2 \cdot 3000 \text{Hz}} = \frac{1}{6} \cdot 10^{-3} \ \text{s} = \frac{1}{6}\ \text{ms} $$
|
||||
|
||||
Siden $\tau = RC $ kan vi velge en av verdiene $R$ eller $C$ for å finne den andre.
|
||||
|
||||
Setter $C = 68$nF. Da blir
|
||||
|
||||
\begin{align*}
|
||||
R &= \frac{\tau}{C} \\
|
||||
R &= \frac{\frac{1}{6} \cdot 10^{-3} \ \text{s}}{68 \cdot 10^{-9}\text{F}} \\
|
||||
R &\approx 2451\Omega
|
||||
\end{align*}
|
||||
|
||||
Vi kan runde dette opptil $2500\Omega = 2\text{k}5\Omega$
|
||||
|
||||
Opampen som brukes i kretsen er en LF353P opamp. Den oppfører seg litt merkelig dersom spenningene inn på $+$ og $-$ er over $|4.2\text{V}|$.
|
||||
|
||||
Vi har driftsspenninger $+V=5$V og $-V=-5$V. Velger derfor $R_1 = 4\text{k}7\Omega $ og $R_2 = 10\text{k}\Omega $ slik at spenningen $v_3 < |4.2\text{V}|$.
|
||||
|
||||
Vi har da komponentverdier
|
||||
|
||||
\begin{table}[htbp]
|
||||
\centering
|
||||
\begin{tabular}{|c|c|}
|
||||
\hline\hline
|
||||
Komponent & Verdi \\\hline
|
||||
\hline
|
||||
$R_1$ & $4\text{k}7\Omega$ \\
|
||||
$R_2$ & $10\text{k}\Omega$ \\
|
||||
$R$ & $2\text{k}5\Omega$ \\
|
||||
$C$ & $68$nF \\
|
||||
OpAmp & LF353P \\ \hline
|
||||
\hline
|
||||
\end{tabular}
|
||||
\caption{Utregnede komponentverdier.}
|
||||
\label{tab:kompnenter}
|
||||
\end{table}
|
||||
|
||||
Kobler opp kretsen etter figur \ref{fig:trekantkrets} med verdier fra tabell \ref{tab:kompnenter}.
|
||||
Siden motstanden $R = 2\text{k}5\Omega$ ikke er en standard motstand bruker vi flere andre typer. Resultatet er følgende
|
||||
$$ R = 2\text{k}2\Omega + 270\Omega + 39\Omega = 2509\Omega$$
|
||||
|
||||
Med dette koblet opp som vist i figur \ref{pic:ferdigKrets}
|
||||
|
||||
\begin{figure}[htbp]
|
||||
\centering
|
||||
\subfloat[Ferdig fungerende krets.]{\includegraphics[width=0.45\textwidth]{pics/RenKrets.png}\label{pic:renKrets}}
|
||||
\hfill
|
||||
\subfloat[Ferdig fungerende krets med navn.]{\includegraphics[width=0.45\textwidth]{pics/KretsMedNavn.png}\label{pic:kretsMedNavn}}
|
||||
\caption{Ferdig krets, med og uten navn.}
|
||||
\label{pic:ferdigKrets}
|
||||
\end{figure}
|
||||
|
||||
Kretsen ble også målt med oscilloscop for å sjekke frekvensen. Se figur
|
||||
|
||||
\begin{figure}
|
||||
\centering
|
||||
\includegraphics[width=\textwidth]{grafer/Trekantpuls3k.png}
|
||||
\caption{Måligner av frekvens og utseende på spenninger $v_2$ (trekantpuls) og $v_1$ (firkantpuls).}
|
||||
\label{graph:pyGraph}
|
||||
\end{figure}
|
||||
|
||||
Målt frekvens er på $2.9999$kHz eller $2999.9$Hz. Dette er innenfor avviket på $10\ 000$ ppm unna $3000$Hz.
|
||||
|
||||
\clearpage
|
||||
|
||||
\section{Konklusjon}
|
||||
\label{sec:konklusjon}
|
||||
|
||||
Ettersom kretsen var uhyre nærme målet på $3000$Hz, kan vi si at kretsen fungerte veldig bra. Målet var $3000$Hz, og kretsen klarte å produsere en trekantpuls som hadde en frekvens på $2999.9$Hz. Noe som også er godt innenfor avviket på $10\ 000$ ppm.
|
||||
|
||||
|
||||
\section{Takk}
|
||||
Takk til Ulrik Bredland for bra samarbeid og gode diskusjoner.
|
||||
|
||||
%Bibliografi: Legg til flere elementer ved å legge til flere \bibitem:--------
|
||||
\phantomsection
|
||||
\addcontentsline{toc}{section}{Referanser}
|
||||
\begin{thebibliography}{99}
|
||||
|
||||
\bibitem{notat}
|
||||
Torstein Bolstad,
|
||||
\emph{Teknisk notat: Trekantgenerator},
|
||||
NTNU,
|
||||
TTT4260 Elektronisk sysdemdesign og -analyse,
|
||||
2019.
|
||||
|
||||
\bibitem{oving3}
|
||||
ELSYS,
|
||||
\emph{Øving 3},
|
||||
NTNU,
|
||||
TTT4260 Elektronisk sysdemdesign og -analyse,
|
||||
2019.
|
||||
|
||||
\end{thebibliography}{}
|
||||
|
||||
\clearpage
|
||||
\appendix
|
||||
%Tillegg. Flere tillegg legges til ved å lage flere sections:-----------------
|
||||
|
||||
|
||||
|
||||
\end{document}
|
||||
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|
After Width: | Height: | Size: 154 KiB |
|
|
@ -0,0 +1,39 @@
|
|||
#!/usr/bin/env python3
|
||||
# -*- coding: utf-8 -*-
|
||||
"""
|
||||
Created on Sun Jan 13 16:30:58 2019
|
||||
|
||||
@author: oyvind
|
||||
"""
|
||||
|
||||
import csv
|
||||
import matplotlib.pyplot as plt
|
||||
|
||||
header = []
|
||||
data = []
|
||||
|
||||
filename = "Trekantpuls3k"
|
||||
|
||||
|
||||
with open(filename + ".csv") as csvfile:
|
||||
csvreader = csv.reader(csvfile)
|
||||
|
||||
header = next(csvreader)
|
||||
|
||||
for dataplot in csvreader:
|
||||
print(dataplot)
|
||||
values = [float(value) for value in dataplot]
|
||||
|
||||
data.append(values)
|
||||
|
||||
time = [p[0] * 1000 for p in data]
|
||||
ch1 = [p[1] for p in data]
|
||||
ch2 = [p[2] for p in data]
|
||||
|
||||
plt.figure(figsize=(12,5))
|
||||
plt.plot(time,ch1, time,ch2)
|
||||
plt.xlabel("Tid (ms)")
|
||||
plt.ylabel("Spenning (V)")
|
||||
plt.legend(["Trekantpuls","Firkantpuls"], loc="upper right")
|
||||
plt.savefig(filename + ".png", dpi=300)
|
||||
plt.show()
|
||||
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<</Size 161
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startxref
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45364
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%%EOF
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Binary file not shown.
|
After Width: | Height: | Size: 24 KiB |
|
|
@ -0,0 +1,57 @@
|
|||
#!/usr/bin/env python3
|
||||
# -*- coding: utf-8 -*-
|
||||
"""
|
||||
Created on Sun Apr 28 16:08:42 2019
|
||||
|
||||
@author: oyvind
|
||||
"""
|
||||
import matplotlib.pyplot as plt
|
||||
|
||||
freq = 1000
|
||||
|
||||
period = 10**3 / freq
|
||||
|
||||
cycles = 2
|
||||
|
||||
resolution = 1000
|
||||
|
||||
upperVolt = 1
|
||||
|
||||
lowerVolt = -1
|
||||
|
||||
|
||||
def GenTime():
|
||||
time = []
|
||||
for t in range(2 * resolution * cycles):
|
||||
time.append(t * period / (2 * resolution))
|
||||
|
||||
return time
|
||||
|
||||
def GenSquare():
|
||||
square = []
|
||||
for s in range(cycles):
|
||||
for h in range(resolution):
|
||||
square.append(upperVolt)
|
||||
for h in range(resolution):
|
||||
square.append(lowerVolt)
|
||||
return square
|
||||
|
||||
def GenCenter(times):
|
||||
center = []
|
||||
for i in times:
|
||||
center.append(0)
|
||||
return center
|
||||
|
||||
time = GenTime()
|
||||
square = GenSquare()
|
||||
center = GenCenter(time)
|
||||
|
||||
plt.figure(figsize=(12,5))
|
||||
plt.plot(time, square)
|
||||
plt.hlines(0, time[0], time[-1], color='k',linestyles='dashed')
|
||||
plt.title("Test")
|
||||
plt.xlabel("Tid [ms]")
|
||||
plt.ylabel("Spenning [V]")
|
||||
plt.legend(["Pulstog, " + r'$v_1$', "Utgående spenning, " + r'$v_2$', "Terskelspenning, " + r'$V_T$'], loc="upper right")
|
||||
#plt.savefig(filename + ".png", dpi = 300)
|
||||
plt.show()
|
||||
Loading…
Reference in New Issue