Lots of updates

6th-Semester-Spring-2024/Power-Electronics/Notes/power-electronics-notes.tex

@@ -115,5 +115,42 @@ The term, $a_0 \over 2$ is the \emph{DC offset} of the signal. While the values
 $$f(t) = A_0 + \sum_{n=1}^\infty A_n \cos(n \omega t + \varphi_n)$$
 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}