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[SilasElter] authored and [SilasElter] committed Nov 26, 2024
2 parents 2296e08 + ae0c0ae commit e42bf79
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7 changes: 4 additions & 3 deletions exercise/fig/ex04/FigTab_SingledEndedForwardConverter.tex
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% Parameter of the singled ended forward converter
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{table}[ht]
\begin{table}[htb]
\centering % Zentriert die Tabelle
\begin{tabular}{llll}
\toprule
Input voltage: & $U_{\mathrm{1}} = \SI{240}{\volt}\ldots\SI{360}{\volt}$ & Switching frequency: & $f_{\mathrm{s}} = \SI{48}{\kilo\hertz}$\\
Forward voltage of $D_{\mathrm{1}}$: & $U_{\mathrm{D1,f}} = \SI{0.4}{\volt}$ & & \\
Input voltage: & $U_{\mathrm{1}} = \SI{325}{\volt}$ & Output voltage: & $U_{\mathrm{2}} = \SI{5}{\volt}$ \\
Output power: & $P_{\mathrm{2}} = \SI{125}{\watt}$ & Switching frequency: & $f_{\mathrm{s}} = \SI{48}{\kilo\hertz}$ \\
Forward voltage of $D_{\mathrm{1}}$: & $U_{\mathrm{D1,f}} = \SI{0.4}{\volt}$ & Forward voltage of $D_{\mathrm{2}}$: & $U_{\mathrm{D2,f}} = \SI{0}{\volt}$ \\
\bottomrule
\end{tabular}
\caption{Parameters of the circuit.} % Beschriftung der Tabelle
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2 changes: 1 addition & 1 deletion exercise/fig/ex04/Fig_FlybackConverter.tex
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\draw[double, double distance=3pt, thick] let \p1=(l1.core west), \p2=(l2.core east) in (\x1/2+\x2/2, \y1) -- (\x1/2+\x2/2, \y2);
\end{circuitikz}
\end{center}
\caption{Flyback converter topology}
\caption{Flyback converter topology.}
\label{fig:flyback_converter_topology}
\end{figure}

2 changes: 1 addition & 1 deletion exercise/tex/exercise04.tex
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The transformer's leakage inductance can be neglected.

\input{./fig/ex04/FigTab_SingledEndedForwardConverter}

\vspace{-0.5cm}
\subtask{Calculate the turns ratio $N_\mathrm{3}$/$N_\mathrm{1}$ limiting the maximum transistor blocking voltage
to $\SI{600}{\volt}$.}
\subtask{What is the maximum permissible duty cycle of the power transistor in this case?}
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4 changes: 2 additions & 2 deletions lecture/tex/Lecture03.tex
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Expand Up @@ -818,8 +818,8 @@ \subsection{Flyback converter}
applies due to the non-conducting diode during $T''_\mathrm{off}$.} \onslide<2->{To find the input-to-output voltage ratio in DCM, we again utilize the current ripple balance:}
\begin{equation}
\begin{alignedat}{2}
\onslide<2->{\Delta i_\mathrm{\mathrm{m}} &= \frac{U_1}{L_\mathrm{m}}T_\mathrm{on} = \frac{U_1}{L}DT_\mathrm{s} \quad &&\mbox{(rising edge)},}\\
\onslide<3->{\Delta i_\mathrm{\mathrm{m}} &= \frac{N_1}{N_2}\frac{U_2}{L_\mathrm{m}}T'_\mathrm{off} = \frac{N_1}{N_2}\frac{U_2}{L}D'T_\mathrm{s} \quad &&\mbox{(falling edge)}.}
\onslide<2->{\Delta i_\mathrm{m} &= \frac{U_1}{L_\mathrm{m}}T_\mathrm{on} = \frac{U_1}{L_\mathrm{m}}DT_\mathrm{s} \quad &&\mbox{(rising edge)},}\\
\onslide<3->{\Delta i_\mathrm{\mathrm{m}} &= \frac{N_1}{N_2}\frac{U_2}{L_\mathrm{m}}T'_\mathrm{off} = \frac{N_1}{N_2}\frac{U_2}{L_\mathrm{m}}D'T_\mathrm{s} \quad &&\mbox{(falling edge)}.}
\end{alignedat}
\end{equation}
\onslide<4->{Solving for $D'$ yields
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