\documentclass{report}

\input{preamble}
\input{macros}
\input{letterfonts}
\usepackage{circuitikz}

\title{\Huge{Power Electronics}}
\author{\huge{Aidan Sharpe}}
\date{}

\begin{document}

\maketitle
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% \pdfbookmark[<level>]{<title>}{<dest>}
\pdfbookmark[section]{\contentsname}{toc}
\tableofcontents
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\chapter{}
\dfn{Power Electronics}
{
    Any electronics between the power source and the load.
}

\dfn{Diode}
{
    A non-linear, polarized circuit component. Its positive terminal is called the anode, and its negative terminal is called the cathode. If the voltage at the anode is greater the voltage at the cathode, the diode will conduct. Otherwise, the diode behaves as an open circuit.
}

An ideal diode, when conducting, will have no voltage drop, so it will behave like a direct short. An ideal diode, when not conducting, will allow no current to flow, so it will behave like an open circuit.

For any periodic signal, the average value is given by:
$$V_\text{avg} = {1\over T} \int\limits_0^T x(\omega t) dt$$
Where:
\begin{itemize}
    \item[$\omega$] is the angular frequency of the signal, $x(\omega t)$
    \item[$T$] is the period of the signal, $x(\omega t)$
\end{itemize}

Since a half-wave rectifier only allows current to flow for half of a period, the average voltage out is only a fraction of the input voltage.

\section{Half-Wave Recifier with Resistive Load and Inductive Filter}
Since inductors act like a flywheel for current, when an inductor is placed in series with the half-wave rectifier, the voltage across the load and inductor is not always positive. Adding this inductor, however, allows current to flow through the load for more time during each period. With a properly selected inductance, positive current may be able to flow throught the load for the entire period.

\section{Freewheeling with Resistive Load and Inductive Filter}
Simply by adding a second diode, the inductor will slowly dissipate into the resistor between the positive cycles. This slow dissipations is called \emph{freewheeling}. Increasing the size of the inductor will result in less ripple in the output voltage.

\begin{center}
\begin{circuitikz}[american voltages]
\draw
(0,0) to [short] (1,0) to [empty diode, l=$D_1$] (3,0) (3,-2) to [empty diode, l=$D_2$] (3,0)
(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}
\noindent
For the first half-cycle ($0 < \omega t < \pi$):
\\
\noindent
$D_1$ is on, $D_2$ is off, and $i_L = i_R$.
$$V_s \sin(\omega t) = L {d i_L \over dt} + i_l R$$
$$i_L(t) = {V_s \over Z}\sin(\omega t - \varphi) + Ae^{-Rt/L}$$
For the second half-cyle ($\pi < \omega t < 2\pi$):
$$V_L + V_R = 0$$
$$L {d i_L \over dt} + i_L R = 0$$
$$i_L(t) = Be^{-R(\omega t - \pi)/\omega L}$$
Since $i_L(\pi)$ must be the same for both the first and the second half-cycles, the two equations must be equal at $t=\pi$.
$$i_L(\pi)=B$$
$$i_L(0^+)=A-{V_s \over Z^2} \omega L$$

\section{Full Bridge with Resistive Load and Inductive Filter}

\begin{center}
    \begin{circuitikz}[american voltages]
        \draw
        (0,0) to [short] (3,0) to [american inductor, l=$L_1$] (6,0)
        (0,-4) to [empty diode] (0,-2) to [empty diode] (0,0)
        (3,-4) to [empty diode] (3,-2) to [empty diode] (3,0)
        (0,-2) to [vsourcesin] (3,-2)
        (6,0) to [american resistor, l=$R_1$] (6,-4)
        (0,-4) to [short] (6,-4);
    \end{circuitikz}
\end{center}

\section{Periodic Steady State}
The periodic steady state is when a system returns to the same steady state at the end of each cycle. 

\ex{Inductor PSS}
{
$$v_L = L {di_L \over dt}$$
$$\langle v_L \rangle = {1 \over T} \int\limits_0^T v_L dt = {1 \over T} \int\limits_0^T L {di_L \over dt} dt$$ 
$$\langle v_L \rangle = {L \over T} \lt[i_L(T) - i_L(0)\rt] = 0$$
}

\ex{Capacitor PSS}
{
    $$i_C = C {dv_C \over dt}$$
    $$\langle i_C \rangle = {1 \over T} \int\limits_0^T i_C dt = {1\over T} \int\limits_0^T C {dv_C \over dt} dt$$
    $$\langle i_C \rangle = {C \over T} \lt[v_C(T) - v_C(0)\rt] = 0$$
}

\section{Trigonometric Fourier Series}
For any periodic signal with period, $f(t) = f(t + nT), n \in \mathbb{Z}$, where $T$ is the period, a trigonometric series can be constructed to create an identical signal:
$$f(t) = {a_0 \over 2} + \sum_{n=1}^\infty (a_n \cos(n\omega t) + b_b \sin(n \omega t))$$
where $\omega = {2\pi \over T}$. The value of $a_0$ is twice the average value of the signal:
$$a_0 = {2 \over T} \int\limits_{t_0}^{t_0 + T} f(t) dt.$$
Additionally, the value of $a_n$ and $b_n$ are given by:
$$a_n = {2 \over T} \int\limits_{t_0}{t_0 + T} f(t)\cos(n \omega t) dt$$
$$b_n = {2\over T} \int\limits_{t_0}{t_0 + T} f(t)\sin(n \omega t) dt.$$
The term, $a_0 \over 2$ is the \emph{DC offset} of the signal. While the values of $a_n$ and $b_n$ are defined, since the only difference between $\sin$ and $\cos$ is a phase shift, the expression for the trigonometric Fourier series can be rewritten as:
$$f(t) = A_0 + \sum_{n=1}^\infty A_n \cos(n \omega t + \varphi_n)$$
where $A_n = \sqrt{a_n^2 + b_n^2}$ and $\varphi_n = -\arctan\lt({b_n \over a_n}\rt)$

\chapter{Three-Phase Power}
\section{Y and $\Delta$ Configurations of AC Voltage Sources}
\subsection{The Y Configuration}
The Y configuration is shaped like the letter Y with a neutral connection in the center.
$$V_a = V_an = V_s \sin(\omega t)$$
$V_b$ lags by 120$^\circ$.
$$V_b = V_bn = V_s \sin(\omega t - {2\pi \over 3})$$
$V_c$ lags by 240$^\circ$.
$$V_c = V_cn = V_s \sin(\omega t - {4\pi \over 3})$$

\subsection{The $\Delta$ Configuration}
The $\Delta$ configuration is shaped like the letter $\Delta$ with no neutral connection.
$$V_ab = V_a - V_b = V_s \lt[\sin(\omega t) - \sin(\omega t - {2\pi \over 3})\rt] = \sqrt{3} V_s \sin(\omega t + {\pi \over 6})$$
$$V_bc = V_b - V_c = \sqrt{3}V_s \sin(\omega t - {\pi \over 2})$$
$$V_ca = V_c - V_a = \sqrt{3}V_S \sin(\omega t + {5\pi \over 6})$$

\subsection{Y Configuration Power}
If a Y configuration of AC sources is connected to a Y configuration of resistors, the power through to each load resistor is:
$$P_a = {V_s^2 \over 2R_a} \lt[1 - \cos(2\omega t)\rt]$$
$$P_b = {V_s^2 \over 2R_a} \lt[1 - \cos(2\omega t - {4\pi \over 3})\rt]$$
$$P_c = {V_s^2 \over 2R_c} \lt[1 - \cos(2\omega t + {4\pi \over 3})\rt]$$
The total power transfer is:
$$P = 3{V_s^2 \over 2R}$$
where $R = R_a = R_b = R_c$.
\\
\\
\noindent
One benefit of using three-phase is constant power from the source. Another advantage of using three-phase power is that any harmonics divisible by either 2 or 3 cancel out. The first harmonic after the fundamental is the 5$^\text{th}$ harmonic, followed by the 7$^\text{th}$, 11$^\text{th}$, and 13$^\text{th}$. Harmonics are non-zero for $6n \pm 1, n \in \mathbb{Z}$.

\chapter{Silicon Controlled Rectifier (SCR)}
The SCR will be off even when $V_{AC} > 0$ until a pulse is applied to the gate. Once the pulse is applied, the device will stay on as long as $i_F > 0$.

\nt
{
	The voltage across the SCR can go negative as long as the current remains positive.
}


\end{document}
