Merge branch 'main' of https://github.com/IAS-Uni-Siegen/PE_course

exercise/fig/ex04/Fig_currentI1PeriodTask1.tex

@@ -0,0 +1,30 @@
+\begin{solutionfigure}[htb]
+    \centering
+    \begin{tikzpicture}
+    \begin{axis}[
+        width=7cm, height=4.5cm,
+        grid=both,
+        major grid style={line width=.2pt,draw=gray!50},
+        minor grid style={line width=.1pt,draw=gray!20},
+        xlabel={$t$ / µs},
+        ylabel={$i_\mathrm{1}(t)$ / A},
+       % title={$i_\mathrm{L}$ for minimum output power},
+        xmin=0, xmax=40,
+        ymin=-0, ymax=2,
+        xtick={0, 20, 40},
+        ytick={0, 1,2},
+        ]
+        % Einschaltverhalten graph
+        \addplot[
+            thick,
+            mark=none,
+            color=black,
+        ] coordinates {
+            (0,0) (2.5, 1.257) (2.5, 0) (20, 0)(22.5,  1.257) (22.5,  0) (40, 0)
+        };
+    \end{axis}
+    \end{tikzpicture} 
+    \hspace{1cm} % Abstand zwischen den beiden Diagrammen
+    \caption{Display of the current $i_\mathrm{1}(t)$.}
+    \label{fig:currentSecondarySideTask1}
+    \end{solutionfigure}

exercise/fig/ex04/Fig_currentI2PeriodTask1.tex

@@ -0,0 +1,30 @@
+\begin{solutionfigure}[htb]
+    \centering
+    \begin{tikzpicture}
+    \begin{axis}[
+        width=7cm, height=4.5cm,
+        grid=both,
+        major grid style={line width=.2pt,draw=gray!50},
+        minor grid style={line width=.1pt,draw=gray!20},
+        xlabel={$t$ / µs},
+        ylabel={$i_\mathrm{2}(t)$ / A},
+       % title={$i_\mathrm{L}$ for minimum output power},
+        xmin=0, xmax=40,
+        ymin=-0, ymax=8,
+        xtick={0, 20, 40},
+        ytick={0, 2,4,6,8},
+        ]
+        % Einschaltverhalten graph
+        \addplot[
+            thick,
+            mark=none,
+            color=black,
+        ] coordinates {
+            (0,0) (2.5, 0) (2.5, 6.28) (12.7, 0) (22.5, 0)(22.5, 6.28) (32.7, 0) (40, 0)
+        };
+    \end{axis}
+    \end{tikzpicture} 
+    \hspace{1cm} % Abstand zwischen den beiden Diagrammen
+    \caption{Display of the current $i_\mathrm{2}(t)$.}
+    \label{fig:currentSecondarySideTask1}
+    \end{solutionfigure}

exercise/fig/ex04/Fig_voltageTransistorPeriodTask1.tex

@@ -8,7 +8,7 @@
         minor grid style={line width=.1pt,draw=gray!20},
         xlabel={$t$ / µs},
         ylabel={$u_\mathrm{T}(t)$ / V},
-        title={$i_\mathrm{L}$ for minimum output power},
+       % title={$i_\mathrm{L}$ for minimum output power},
         xmin=0, xmax=40,
         ymin=-100, ymax=600,
         xtick={0, 20, 40},

exercise/fig/ex04/Fig_voltageUsPeriodTask1.tex

@@ -0,0 +1,30 @@
+\begin{solutionfigure}[htb]
+    \centering
+    \begin{tikzpicture}
+    \begin{axis}[
+        width=7cm, height=4.5cm,
+        grid=both,
+        major grid style={line width=.2pt,draw=gray!50},
+        minor grid style={line width=.1pt,draw=gray!20},
+        xlabel={$t$ / µs},
+        ylabel={$u_\mathrm{S}(t)$ / V},
+       % title={$i_\mathrm{L}$ for minimum output power},
+        xmin=0, xmax=40,
+        ymin=-100, ymax=60,
+        xtick={0, 20, 40},
+        ytick={-100, -50, 0, 50},
+        ]
+        % Einschaltverhalten graph
+        \addplot[
+            thick,
+            mark=none,
+            color=black,
+        ] coordinates {
+            (0,0) (0,-76.4) (2.5,-76.4) (2.5, 15) (12.7, 15) (12.7, 0) (20, 0) (20, 0) (20, -76.4) (22.5, -76.4)(22.5, 15) (32.7, 15) (32.7, 0) (40, 0)
+        };
+    \end{axis}
+    \end{tikzpicture} 
+    \hspace{1cm} % Abstand zwischen den beiden Diagrammen
+    \caption{Display of the voltage $u_\mathrm{S}(t)$.}
+    \label{fig:voltageSecondarySideTask1}
+    \end{solutionfigure}

exercise/main.tex

@@ -1,7 +1,7 @@
 \documentclass[solution]{../course_template/exerciseClass}
 \title{Power Electronics}
 
-\includeonly{tex/exercise03}
+\includeonly{tex/exercise04}
 
 \begin{document}
     \include{tex/exercise01}

exercise/tex/exercise04.tex

@@ -29,28 +29,29 @@
 \subtask{The input voltage is $U_\mathrm{1}=\SI{760}{\volt}$ at rated power at the output. What is the peak value $\hat I_\mathrm{1}$ of the primary current $i_\mathrm{1}$? What is the peak value $\hat I_\mathrm{2}$ of the secundary current $i_\mathrm{2}$? Calculate the duty cycle of the transistor for this operating case.}
 
 \begin{solutionblock}
+To determine the current  $\hat I_\mathrm{1}$, the equation for determining the output power \eqref{eq:output power ex04} is primarily used. The unknown energy of the inductance of the primary winding side \eqref{eq:energy primary inductance ex04} is inserted into this equation.
 \begin{equation}
-    P_\mathrm{2} = W_\mathrm{L} f_\mathrm{p}
+    P_\mathrm{2} = W_\mathrm{L} f_\mathrm{p}, \label{eq:output power ex04}
 \end{equation}
 
 \begin{equation}
-    W_\mathrm{L} = \frac{1}{2}L_\mathrm{1}\hat I_\mathrm{1}^2 
+    W_\mathrm{L} = \frac{1}{2}L_\mathrm{1}\hat I_\mathrm{1}^2. \label{eq:energy primary inductance ex04}
 \end{equation}
-
+After substituting, you get this equation:
 \begin{equation}
-    \hat I_\mathrm{1} = \sqrt{\frac{2P_\mathrm{2}}{L_\mathrm{1}f_\mathrm{p}}}= \sqrt{\frac{2\cdot\SI{30}{\watt}}{\SI{760}{\micro\henry}\cdot\SI{50}{\kilo\hertz}}}=\SI{1.257}{\ampere}
+    P_\mathrm{2} = \frac{1}{2}L_\mathrm{1}\hat I_\mathrm{1}^2 f_\mathrm{p}.\label{eq:output power with IDach ex04}
 \end{equation}
-
+\eqref{eq:output power with IDach ex04} must be converted to $\hat I_\mathrm{1}$:
 \begin{equation}
-    \hat I_\mathrm{1} = \hat I_\mathrm{2}
+    \hat I_\mathrm{1} = \sqrt{\frac{2P_\mathrm{2}}{L_\mathrm{1}f_\mathrm{p}}}= \sqrt{\frac{2\cdot\SI{30}{\watt}}{\SI{760}{\micro\henry}\cdot\SI{50}{\kilo\hertz}}}=\SI{1.257}{\ampere}.
 \end{equation}
-
+In an ideal transformer, the energy is maintained between the primary and secondary sides (neglecting losses). To determine the peak value $\hat I_\mathrm{2}$, the known current $\hat I_\mathrm{1}$ and the number of turns of the primary and secondary side can be used. Because of this background, the following equation can be used:
 \begin{equation}
-    \hat I_\mathrm{2} = \hat I_\mathrm{1} \frac{N_\mathrm{1}}{N_\mathrm{2}} = \SI{1.257}{\ampere} \cdot \frac{60}{12} = \SI{6.28}{\ampere}
+    \hat I_\mathrm{2} = \hat I_\mathrm{1} \frac{N_\mathrm{1}}{N_\mathrm{2}} = \SI{1.257}{\ampere} \cdot \frac{60}{12} = \SI{6.28}{\ampere}.
 \end{equation}
-Because of CCM the duty cycle $D$ is expressed .....
+Because of CCM the duty cycle $D$ is expressed as
 \begin{equation}
-    \frac{U_2}{U_1} = \frac{D^2}{2} \frac{\Delta i_\mathrm{m,max}}{\overline{i}_2} \label{eq:Duty cycle ex04}
+    \frac{U_2}{U_1} = \frac{D^2}{2} \frac{\Delta i_\mathrm{m,max}}{\overline{i}_2}. \label{eq:Duty cycle ex04}
 \end{equation}
 Because of the unknown $\Delta i_\mathrm{m,max}$ this value has to be calculate first as
 \begin{equation}
@@ -67,12 +68,46 @@
 \subtask{The input voltage is  $U_\mathrm{1}=\SI{382}{\volt}$ at nominal load. Calculate and sketch the following voltage and current curves for this operating case over one cycle period: $u_\mathrm{T}(t), u_\mathrm{s}(t), i_\mathrm{2}(t), i_\mathrm{1}(t)$ (Note: corresponds to the switch-on time of the transistor).}
 
 \input{./fig/ex04/Fig_voltageTransistorPeriodTask1.tex}
-
+\input{./fig/ex04/Fig_voltageUsPeriodTask1.tex}
+\input{./fig/ex04/Fig_currentI2PeriodTask1.tex}
+\input{./fig/ex04/Fig_currentI1PeriodTask1.tex}
 
+\subtask{The input voltage is  $U_\mathrm{1}=\SI{382}{\volt}$ at nominal load. Determine the mean value $\overline i_\mathrm{T}$ and the effective value of the current $i_\mathrm{T, rms}$ through the transistor. Determine the mean value $\overline i_\mathrm{D}$ and the effective value of the current $i_\mathrm{D, rms}$ through the diode. What is the maximum reverse voltage load $u_\mathrm{T, max}$ of the transistor? What is the maximum reverse voltage load $u_\mathrm{D, max}$ of the diode?}
+\begin{solutionblock}
+    \begin{equation}
+        \overline{i}_\mathrm{T} = \frac{1}{T_\mathrm{S}}\frac{1}{2}\hat I_\mathrm{1}T_\mathrm{on}=\frac{1}{\SI{20}{\micro\s}}\cdot\frac{1}{2}\SI{1.257}{\ampere}\cdot\SI{2.5}{\micro\s}=\SI{78.53}{\milli\ampere}
+    \end{equation}
+    \begin{equation}
+        i_\mathrm{T,rms}^2=\frac{1}{T_\mathrm{S}} \int_{0}^{T_\mathrm{S}} i_\mathrm{1}^2(t) \,dt \ = \frac{1}{T_\mathrm{S}}\int_{0}^{T_\mathrm{on}} \frac{\hat I_\mathrm{1}^2t^2}{T_\mathrm{on}^2} \,dt \ =  \frac{\hat I_\mathrm{1}^2T_\mathrm{on}}{3T_\mathrm{on}}
+    \end{equation}
+    \begin{equation}
+        i_\mathrm{T,rms} = \hat I_\mathrm{1} \sqrt{\frac{T_\mathrm{on}}{3T_\mathrm{S}}}= \SI{1.257}{\ampere}\cdot\sqrt{\frac{\SI{2.5}{\micro\s}}{3\cdot\SI{20}{\micro\s}}}= \SI{256.58}{\milli\ampere}
+    \end{equation}
+
+    \begin{equation}
+        \overline{i}_\mathrm{D} = \frac{1}{T_\mathrm{S}}\frac{1}{2}\hat I_\mathrm{2}T_\mathrm{off}=\frac{1}{\SI{20}{\micro\s}}\cdot\frac{1}{2}\SI{6.28}{\ampere}\cdot\SI{12.7}{\micro\s}=\SI{2}{\ampere}
+    \end{equation}
+
+    \begin{equation}
+        i_\mathrm{D,rms} = \hat I_\mathrm{2} \sqrt{\frac{T_\mathrm{off}}{3T_\mathrm{S}}}= \SI{6.28}{\ampere}\cdot\sqrt{\frac{\SI{12.7}{\micro\s}}{3\cdot\SI{20}{\micro\s}}}= \SI{2.89}{\milli\ampere}
+    \end{equation}
+
+    \begin{equation}
+        u_\mathrm{T,max} = U_\mathrm{1} + \frac{N_\mathrm{1}}{N_\mathrm{2}}U_\mathrm{2}= \SI{382}{\volt}+\frac{60}{12}\cdot\SI{15}{\volt}= \SI{457}{\volt}
+    \end{equation}
+
+    \begin{equation}
+        u_\mathrm{D,max} = U_\mathrm{2} + \frac{N_\mathrm{2}}{N_\mathrm{1}}U_\mathrm{1}= \SI{15}{\volt}+\frac{12}{60}\cdot\SI{382}{\volt}= \SI{91.4}{\volt}
+    \end{equation}
 
-\subtask{The input voltage is  $U_\mathrm{1}=\SI{382}{\volt}$ at nominal load. Determine the mean value $\overline i_\mathrm{T}$ and the effective value of the current $i_\mathrm{T, rms}$ through the transistor. Determine the mean value $\overline i_\mathrm{D}$ and the effective value of the current $i_\mathrm{D, rms}$ through the diode. What is the maximum reverse voltage load $u_\mathrm{T, max}$ of the transistor? What is the maximum reverse voltage load $u_\mathrm{D, max}$ of the diode? Calculate the fluctuation range $\Delta i_\mathrm{C, pp}$ of the current $i_\mathrm{C}$ in the output capacitor.}
+\end{solutionblock}
 
-\subtask{The input voltage is  $U_\mathrm{1}=\SI{382}{\volt}$ at nominal load.How much energy is transferred from the input to the output per switching period $\Delta E$ and what is the resulting average power? What happens if there is no ideal voltage source on the output side but an unloaded capacitor and the circuit is operated with $D>0$?}
+\subtask{The input voltage is  $U_\mathrm{1}=\SI{382}{\volt}$ at nominal load. How much energy is transferred from the input to the output per switching period $\Delta E$ and what is the resulting average power? What happens if there is no ideal voltage source on the output side but an unloaded capacitor and the circuit is operated with $D>0$?}
+
+\begin{solutionblock}
+
+At each switching interval, energy is pushed into the capacitor, the voltage of which continues to rise until a component fails due to a lack of dielectric strength.    
+\end{solutionblock}
 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 %% Task 2: Forward converter with asymmetric half-bridge