94 lines
2.9 KiB
TeX
94 lines
2.9 KiB
TeX
\documentclass{article}
|
|
|
|
\title{Power Electronics Homework 1}
|
|
\author{Aidan Sharpe}
|
|
|
|
\usepackage{circuitikz}
|
|
\usepackage{amsmath}
|
|
\usepackage{listings}
|
|
|
|
\begin{document}
|
|
\maketitle
|
|
\date{}
|
|
|
|
\section{Initial Circuit}
|
|
\begin{center}
|
|
\begin{circuitikz}[american voltages]
|
|
\draw
|
|
(0,0) to [short] (1,0) to [empty diode, l=$D_1$] (3,0) to [open, v=$V_d$] (3,-2)
|
|
(3,0) to [short, american inductor, l=$L_1$, i=$i_L$] (6,0)
|
|
to [short, american resistor, l_=$R_1$, v^=$V_R$] (6,-2)
|
|
to [short] (0,-2)
|
|
(0,0) to [vsourcesin, v_=$V_s$] (0,-2);
|
|
\end{circuitikz}
|
|
\end{center}
|
|
|
|
$$V_s(t) = 110 \sqrt{2} \cos(2\pi f t)$$
|
|
$$f = 60 \text{[Hz]}$$
|
|
$$L_1 = 20 \text{[mH]}$$
|
|
$$R_1 = 10 [\Omega]$$
|
|
|
|
\section{Diode is "On"}
|
|
\begin{center}
|
|
\begin{circuitikz}[american voltages]
|
|
\draw
|
|
(0,0) to [closing switch] (3,0) to [open, v=$V_d$] (3,-2)
|
|
(3,0) to [short, american inductor, l=$L_1$, i=$i_L$] (6,0)
|
|
to [short, american resistor, l_=$R_1$, v^=$V_R$] (6,-2)
|
|
to [short] (0,-2)
|
|
(0,0) to [vsourcesin, v_=$V_s$] (0,-2);
|
|
\end{circuitikz}
|
|
\end{center}
|
|
|
|
$$V_d(t) = V_s(t) = V_L(t) + V_R(t)$$
|
|
$$V_L(t) = L_1 {d \over dt}i_L(t)$$
|
|
$$V_R(t) = R_1 i_L(t)$$
|
|
$$110 \sqrt{2} \cos(120\pi t) = 10 i_L(t) + 0.02 {d\over dt}i_L(t)$$
|
|
Apply Laplace Transform:
|
|
$$10 I_L(s) + 0.02 s I_L(s) + i_L(0) = {110 \sqrt{2} s \over 14400 \pi^2 + s^2}$$
|
|
Solve for $I_L(s)$:
|
|
$$I_L(s) (10 + 0.02 s) = {110 \sqrt{2} s \over 14400 \pi^2 + s^2}$$
|
|
$$I_L(s) = \frac{110\,\sqrt{2}\,s}{\left(\frac{s}{50}+10\right)\,\left(s^2+14400\,\pi ^2\right)}$$
|
|
Apply Inverse Laplace Transform:
|
|
$$i_L(t) = \frac{6875\,\sqrt{2}\,\cos\left(120\,\pi \,t\right)+1650\,\pi \,\sqrt{2}\,\sin\left(120\,\pi \,t\right)}{36\,\pi ^2+625}-\frac{6875\,\sqrt{2}\,{\mathrm{e}}^{-500\,t}}{36\,\pi ^2+625}$$
|
|
$$V_R = \frac{10\,\left(6875\,\sqrt{2}\,\cos\left(120\,\pi \,t\right)+1650\,\pi \,\sqrt{2}\,\sin\left(120\,\pi \,t\right)\right)}{36\,\pi ^2+625}-\frac{68750\,\sqrt{2}\,{\mathrm{e}}^{-500\,t}}{36\,\pi ^2+625}$$
|
|
|
|
|
|
\section{Diode Is "Off"}
|
|
\begin{center}
|
|
\begin{circuitikz}[american voltages]
|
|
\draw
|
|
(0,0) to [opening switch] (3,0) to [open, v=$V_d$] (3,-2)
|
|
(3,0) to [short, american inductor, l=$L_1$, i=$i_L$] (6,0)
|
|
to [short, american resistor, l_=$R_1$, v^=$V_R$] (6,-2)
|
|
to [short] (0,-2)
|
|
(0,0) to [vsourcesin, v_=$V_s$] (0,-2);
|
|
\end{circuitikz}
|
|
\end{center}
|
|
|
|
\section{LTSpice}
|
|
\begin{figure}[h]
|
|
\includegraphics[width=\textwidth]{simulation.png}
|
|
\end{figure}
|
|
|
|
\section{MATLAB}
|
|
\begin{lstlisting}[language=MATLAB]
|
|
syms s t
|
|
R_1 = 10;
|
|
L_1 = 0.02;
|
|
v_s = 110*sqrt(2)*cos(120*pi*t);
|
|
|
|
V_s = laplace(v_s);
|
|
I_L = V_s / (R_1 + s*L_1);
|
|
i_L = ilaplace(I_L);
|
|
|
|
v_o = i_L * R_1
|
|
\end{lstlisting}
|
|
|
|
\section{Large Inductor}
|
|
As the inductor gets proportionally larger than the resistor, the amplitude of the output voltage decreases and the phase shift increases.
|
|
\section{Large Resistor}
|
|
As the resistor gets proportionally larger than the inductor, the amplitude of the output approaches the amplitude of the input and the phase shift goes to zero.
|
|
|
|
\end{document}
|