exclude fig from ex02 in fig folder

IAS-Uni-Siegen · Nov 5, 2024 · 2faf416 · 2faf416
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diff --git a/exercise/fig/ex02/Fig_diode_switchOff.tex b/exercise/fig/ex02/Fig_diode_switchOff.tex
@@ -0,0 +1,62 @@
+\begin{figure}[htb]
+    \centering
+    \begin{tikzpicture}
+        \tikzmath{
+            real = \t0, \t1, \t2 \x1;
+            \x1 = 3;
+            \t1 = 6;
+            \t2 = 8;
+            \t0 = (\t1-\x1)/0.71;
+        }
+        \begin{axis}[
+            xlabel={$t$},
+            axis lines=middle,
+            ymin=-1.5, ymax=1.2,
+            xmin=0, xmax=10,
+            xtick={\t0,\t1,\t2},
+            xticklabels={$t_0$,$t_1$,$t_2$},
+            ticklabel style ={yshift=0.2cm,anchor=south},
+            yticklabels={},
+            width=12cm,
+            height=5cm,
+            thick,
+            smooth,
+            no markers,
+            grid
+        ]
+        \draw[white,pattern=north east lines, pattern color=signalorange] (\t0,0) -- (\t1,-1) -- (\t1,0);
+        %
+        \node[signalorange, fill=white, inner sep = 1pt, anchor = south] at (axis cs:\t0+1.3,-0.45) {$Q_1$};
+        %
+        \draw[white,pattern=north west lines, pattern color=signalgreen] (\t1,0) -- (\t1,-1) -- (\t2,0);
+        %
+        \node[signalgreen, fill=white, inner sep = 1pt, anchor = south] at (axis cs:\t1+0.6,-0.45) {$Q_2$};
+        %
+        \draw[thick,signalred] (0,0.7) -- (\x1,0.7) -- (\t1,-1) -- (\t2,0);
+        %
+        \node[signalred, fill=white, inner sep = 1pt, anchor = south] at (axis cs:2,0.8) {$i_{\mathrm{D}}$};
+        % 
+        \draw[thin, signalred] (3.5,0.6) -- (4,0.7);
+        \node[signalred, fill=white, inner sep = 1pt, anchor = south] at (axis cs:4.5,0.55) {$\frac{\mathrm{d}i_{\mathrm{D}}}{\mathrm{d}t}$};
+        % 
+        \draw[thin, signalred] (7,-0.6) -- (7.8,-0.85);
+        \node[signalred, fill=white, inner sep = 1pt, anchor = south] at (axis cs:8,-1) {$\frac{\mathrm{d}i_{\mathrm{rr}}}{\mathrm{d}t}$};
+        %
+        \draw[thin, signalred] (\t1-0.05,-1.02) -- (\t1-0.3,-1.2);
+        %
+        \node[signalred, inner sep = 1pt, anchor = south] at (axis cs:\t1-0.4,-1.45) {$\hat{i}_{\mathrm{rr}}$};
+        %
+        \node[black, fill=white, inner sep = 1pt, anchor = south] at (axis cs:2,-0.7) {$Q_{\mathrm{rr}} = Q_{1} + Q_{2}$};
+        %
+        \draw[thick, signalblue, dashed] (0,0.1) -- (\t0,0.1) -- (\t0,-0.1) -- (\t1,-0.1) -- (\t1,-1.3) -- (\t2,-1.3) -- (\t2,-1) -- (10,-1);
+        %
+        \draw[thin, signalblue] (8.1, -1.3) -- (8.5,-1.3);
+        % 
+        \node[signalblue, fill=white, inner sep = 1pt, anchor = south] at (axis cs:8.8,-1.5) {${U}_{\mathrm{RB}}$};
+        %
+        \node[signalblue, fill=white, inner sep = 1pt, anchor = south] at (axis cs:2,0.15) {${u}_{\mathrm{D}}$};
+        \end{axis}
+    \end{tikzpicture}
+    \caption{Turn-off behavior of a fast silicon diode.}
+    \label{fig:TurnOffSiliconDiode}
+\end{figure}
diff --git a/exercise/fig/ex02/sFig_boost_voltage_efficiency.tex b/exercise/fig/ex02/sFig_boost_voltage_efficiency.tex
@@ -0,0 +1,58 @@
+\begin{solutionfigure}[htb]
+    \centering
+    \begin{subfigure}{0.45\textwidth}
+        \centering
+        \begin{tikzpicture}
+            \begin{axis}[
+                xlabel={$D$},
+                ylabel={$U_{\mathrm{1}}/U_{\mathrm{2}}$},
+                axis lines=left,
+                ymin=0, ymax=10,
+                xmin=0, xmax=1,
+                xtick={0,0.2,0.4,0.6,0.8,1.0},
+                ytick={0,1,2,3,4,5,6,7,8,9,10},
+                xticklabels={0,0.2,0.4,0.6,0.8,1.0},
+                yticklabels={0,1,2,3,4,5,6,7,8,9,10},
+                thick,
+                smooth,
+                no markers,
+                height=7cm,
+                width = 0.99\textwidth,
+                grid
+                ]
+                \addplot[signalblue, domain=0:0.95] {(1-x)/((1-x)^2)};
+                \node[signalblue, fill=white, inner sep = 1pt, anchor = south] at (axis cs:0.5,8) {ideal function};
+                \addplot[signalred, domain=0:1.0] {(1-x)/((1-x)^2+0.2/30)};
+                \node[signalred, fill=white, inner sep = 1pt, anchor = south] at (axis cs:0.5,7) {function with losses};
+                \end{axis}
+        \end{tikzpicture}
+    \end{subfigure}%
+    \begin{subfigure}{0.45\textwidth}
+        \begin{tikzpicture}
+            \begin{axis}[
+                xlabel={$D$},
+                ylabel={$\eta$},
+                axis lines=left,
+                ymin=0, ymax=1.1,
+                xmin=0, xmax=1,
+                xtick={0,0.2,0.4,0.6,0.8,1.0},
+                ytick={0,0.2,0.4,0.6,0.8,1.0},
+                xticklabels={0,0.2,0.4,0.6,0.8,1.0},
+                yticklabels={0,0.2,0.4,0.6,0.8,1.0},
+                thick,
+                smooth,
+                no markers,
+                height=7cm,
+                width = 0.99\textwidth,
+                grid
+                ]
+                \addplot[signalblue, domain=0:1] {1};
+                \node[signalblue, fill=white, inner sep = 1pt, anchor = south] at (axis cs:0.5,0.4) {ideal function};
+                \addplot[signalred, domain=0:1] {1/(1+(0.2/30)/(1-x)^2)};
+                \node[signalred, fill=white, inner sep = 1pt, anchor = south] at (axis cs:0.5,0.3) {function with losses};
+            \end{axis}
+        \end{tikzpicture}
+    \end{subfigure}
+    \caption{Voltage ratio (on the left) and the efficiency function (on the right) are visualized. Both functions are dependant on the duty cycle $D$.}
+    \label{fig:voltageRatioAndEfficiency}
+\end{solutionfigure}
diff --git a/exercise/fig/ex02/sFig_diode_sw_off_esb.tex b/exercise/fig/ex02/sFig_diode_sw_off_esb.tex
@@ -0,0 +1,19 @@
+\begin{solutionfigure}[htb]
+    \centering
+    \begin{circuitikz}[european currents,european resistors,american inductors]
+        \draw (0,2) to [open, o-o, v = $\hspace{2cm}u_2(t)$, voltage = straight] ++(0,-2)
+        to ++(-5.3,0);
+        \draw (0,2) to (2,2);
+        \draw (2,0) to (0,0);
+        \draw (2,2) to [R, l=$R$](2,0);
+        \draw (-6.375,2) ++(0.625,0) node [cuteopenswitchshape, anchor = out, rotate=180] (S) {}
+        let \p1 = (S.mid) in (S.in) to  [short, i=$i_\mathrm{D}(t)$] ++(1,0)
+        to [inductor, l=$L_{\mathrm{c}}$, v = $u_\mathrm{L}(t)$, voltage = straight] ++(2,0)
+        to [short] ++(1,0)
+        to [short, -o, i=$i_2(t)$] (0,2) 
+        (S.mid) to [short, o-*](\x1,0);
+        \draw (-1.5,2) to [capacitor, *-*, l=$C$, i>^=$i_\mathrm{C}(t)$] ++(0,-2);
+    \end{circuitikz}
+    \caption{Equivalent circuit diagram of the diode switch-off event.}
+    \label{fig:DiodeSwitchOff}
+\end{solutionfigure}
diff --git a/exercise/tex/exercise02.tex b/exercise/tex/exercise02.tex
@@ -277,64 +277,8 @@
 
 
 The voltage ratio dependant on the duty cycle $D$ is shown in \autoref{fig:voltageRatioAndEfficiency} on the left side. Both functions for the loss-free and lossy operation are visualized. Moreover, on the right side of \autoref{fig:voltageRatioAndEfficiency} the resulting efficiency cures are shown.
-\begin{solutionfigure}[htb]
-    \centering
-    \begin{subfigure}{0.45\textwidth}
-        \centering
-        \begin{tikzpicture}
-            \begin{axis}[
-                xlabel={$D$},
-                ylabel={$U_{\mathrm{1}}/U_{\mathrm{2}}$},
-                axis lines=left,
-                ymin=0, ymax=10,
-                xmin=0, xmax=1,
-                xtick={0,0.2,0.4,0.6,0.8,1.0},
-                ytick={0,1,2,3,4,5,6,7,8,9,10},
-                xticklabels={0,0.2,0.4,0.6,0.8,1.0},
-                yticklabels={0,1,2,3,4,5,6,7,8,9,10},
-                thick,
-                smooth,
-                no markers,
-                height=7cm,
-                width = 0.99\textwidth,
-                grid
-                ]
-                \addplot[signalblue, domain=0:0.95] {(1-x)/((1-x)^2)};
-                \node[signalblue, fill=white, inner sep = 1pt, anchor = south] at (axis cs:0.5,8) {ideal function};
-                \addplot[signalred, domain=0:1.0] {(1-x)/((1-x)^2+0.2/30)};
-                \node[signalred, fill=white, inner sep = 1pt, anchor = south] at (axis cs:0.5,7) {function with losses};
-                \end{axis}
-        \end{tikzpicture}
-    \end{subfigure}%
-    \begin{subfigure}{0.45\textwidth}
-        \begin{tikzpicture}
-            \begin{axis}[
-                xlabel={$D$},
-                ylabel={$\eta$},
-                axis lines=left,
-                ymin=0, ymax=1.1,
-                xmin=0, xmax=1,
-                xtick={0,0.2,0.4,0.6,0.8,1.0},
-                ytick={0,0.2,0.4,0.6,0.8,1.0},
-                xticklabels={0,0.2,0.4,0.6,0.8,1.0},
-                yticklabels={0,0.2,0.4,0.6,0.8,1.0},
-                thick,
-                smooth,
-                no markers,
-                height=7cm,
-                width = 0.99\textwidth,
-                grid
-                ]
-                \addplot[signalblue, domain=0:1] {1};
-                \node[signalblue, fill=white, inner sep = 1pt, anchor = south] at (axis cs:0.5,0.4) {ideal function};
-                \addplot[signalred, domain=0:1] {1/(1+(0.2/30)/(1-x)^2)};
-                \node[signalred, fill=white, inner sep = 1pt, anchor = south] at (axis cs:0.5,0.3) {function with losses};
-            \end{axis}
-        \end{tikzpicture}
-    \end{subfigure}
-    \caption{Voltage ratio (on the left) and the efficiency function (on the right) are visualized. Both functions are dependant on the duty cycle $D$.}
-    \label{fig:voltageRatioAndEfficiency}
-\end{solutionfigure}
+
+    \input{fig/ex02/sFig_boost_voltage_efficiency.tex}
 
 \end{solutionblock}
 
@@ -420,68 +364,7 @@
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \subtask{Beside the conduction losses, also switching losses need to be considered in practice. In \autoref{fig:TurnOffSiliconDiode} the voltage and current waveforms are visualized for the turn-off event of a diode (reverse recovery effect). Therefore, calculate the turn-off losses for a fast diode with a commutation inductivity loop of $L_{\mathrm{c}} = \SI{500}{\nano\henry}$. The peak reverse recovery current is $\hat{i}_{\mathrm{rr}} = \SI{4}{\ampere}$ and the reverse recovery time is $t_{\mathrm{rr}} = t_2 - t_0 = \SI{46.6}{\nano\second}$.}
 
-\begin{figure}[htb]
-    \centering
-    \begin{tikzpicture}
-        \tikzmath{
-            real = \t0, \t1, \t2 \x1;
-            \x1 = 3;
-            \t1 = 6;
-            \t2 = 8;
-            \t0 = (\t1-\x1)/0.71;
-        }
-        \begin{axis}[
-            xlabel={$t$},
-            axis lines=middle,
-            ymin=-1.5, ymax=1.2,
-            xmin=0, xmax=10,
-            xtick={\t0,\t1,\t2},
-            xticklabels={$t_0$,$t_1$,$t_2$},
-            ticklabel style ={yshift=0.2cm,anchor=south},
-            yticklabels={},
-            width=12cm,
-            height=5cm,
-            thick,
-            smooth,
-            no markers,
-            grid
-        ]
-        \draw[white,pattern=north east lines, pattern color=signalorange] (\t0,0) -- (\t1,-1) -- (\t1,0);
-        %
-        \node[signalorange, fill=white, inner sep = 1pt, anchor = south] at (axis cs:\t0+1.3,-0.45) {$Q_1$};
-        %
-        \draw[white,pattern=north west lines, pattern color=signalgreen] (\t1,0) -- (\t1,-1) -- (\t2,0);
-        %
-        \node[signalgreen, fill=white, inner sep = 1pt, anchor = south] at (axis cs:\t1+0.6,-0.45) {$Q_2$};
-        %
-        \draw[thick,signalred] (0,0.7) -- (\x1,0.7) -- (\t1,-1) -- (\t2,0);
-        %
-        \node[signalred, fill=white, inner sep = 1pt, anchor = south] at (axis cs:2,0.8) {$i_{\mathrm{D}}$};
-        % 
-        \draw[thin, signalred] (3.5,0.6) -- (4,0.7);
-        \node[signalred, fill=white, inner sep = 1pt, anchor = south] at (axis cs:4.5,0.55) {$\frac{\mathrm{d}i_{\mathrm{D}}}{\mathrm{d}t}$};
-        % 
-        \draw[thin, signalred] (7,-0.6) -- (7.8,-0.85);
-        \node[signalred, fill=white, inner sep = 1pt, anchor = south] at (axis cs:8,-1) {$\frac{\mathrm{d}i_{\mathrm{rr}}}{\mathrm{d}t}$};
-        %
-        \draw[thin, signalred] (\t1-0.05,-1.02) -- (\t1-0.3,-1.2);
-        %
-        \node[signalred, inner sep = 1pt, anchor = south] at (axis cs:\t1-0.4,-1.45) {$\hat{i}_{\mathrm{rr}}$};
-        %
-        \node[black, fill=white, inner sep = 1pt, anchor = south] at (axis cs:2,-0.7) {$Q_{\mathrm{rr}} = Q_{1} + Q_{2}$};
-        %
-        \draw[thick, signalblue, dashed] (0,0.1) -- (\t0,0.1) -- (\t0,-0.1) -- (\t1,-0.1) -- (\t1,-1.3) -- (\t2,-1.3) -- (\t2,-1) -- (10,-1);
-        %
-        \draw[thin, signalblue] (8.1, -1.3) -- (8.5,-1.3);
-        % 
-        \node[signalblue, fill=white, inner sep = 1pt, anchor = south] at (axis cs:8.8,-1.5) {${U}_{\mathrm{RB}}$};
-        %
-        \node[signalblue, fill=white, inner sep = 1pt, anchor = south] at (axis cs:2,0.15) {${u}_{\mathrm{D}}$};
-        \end{axis}
-    \end{tikzpicture}
-    \caption{Turn-off behavior of a fast silicon diode.}
-    \label{fig:TurnOffSiliconDiode}
-\end{figure}
+    \input{fig/ex02/Fig_diode_switchOff.tex}
 
 \begin{solutionblock}
     Switching losses occur when the diode voltage or current is not zero. In \autoref{fig:TurnOffSiliconDiode} the voltage and current waveforms are shown. The voltage $u_{\mathrm{D}}$ is for the interval $t_0$ until $t_1$ small and, therefore, the switching losses. At the time step $t_1$ the diode is blocking and the reverse breakdown voltage $U_{\mathrm{RB}}$ applies. Due to this much bigger value, the switching losses increases significantly, resulting that only the interval between $t_1$ and $t_2$ is responsible for the switching losses. Hence, the losses are calculated as follows
@@ -504,25 +387,7 @@
     \end{equation}
 
     The equivalent circuit diagram of the diode switch-off event is shown in \autoref{fig:DiodeSwitchOff}.
-    \begin{solutionfigure}[htb]
-        \centering
-        \begin{circuitikz}[european currents,european resistors,american inductors]
-            \draw (0,2) to [open, o-o, v = $\hspace{2cm}u_2(t)$, voltage = straight] ++(0,-2)
-            to ++(-5.3,0);
-            \draw (0,2) to (2,2);
-            \draw (2,0) to (0,0);
-            \draw (2,2) to [R, l=$R$](2,0);
-            \draw (-6.375,2) ++(0.625,0) node [cuteopenswitchshape, anchor = out, rotate=180] (S) {}
-            let \p1 = (S.mid) in (S.in) to  [short, i=$i_\mathrm{D}(t)$] ++(1,0)
-            to [inductor, l=$L_{\mathrm{c}}$, v = $u_\mathrm{L}(t)$, voltage = straight] ++(2,0)
-            to [short] ++(1,0)
-            to [short, -o, i=$i_2(t)$] (0,2) 
-            (S.mid) to [short, o-*](\x1,0);
-            \draw (-1.5,2) to [capacitor, *-*, l=$C$, i>^=$i_\mathrm{C}(t)$] ++(0,-2);
-        \end{circuitikz}
-        \caption{Equivalent circuit diagram of the diode switch-off event.}
-        \label{fig:DiodeSwitchOff}
-    \end{solutionfigure}
+    \input{fig/ex02/sFig_diode_sw_off_esb.tex}
     The current $i_{\mathrm{D}}(t)$ is dependant on the voltage $U_2$ by:
     \begin{equation}
         \Delta i_{\mathrm{D}} = \frac{\mathrm{d}i_{\mathrm{D}}(t)}{\mathrm{d}t} = \frac{U_2}{L_{\mathrm{c}}} = \frac{\SI{60}{\volt}}{\SI{500}{\nano\henry}} = \SI{120}{\ampere\per\micro\second}.