1 files changed, 0 insertions, 140 deletions
diff --git a/doc/power/power_supply.tex b/doc/power/power_supply.tex
deleted file mode 100644
index 5b23d7b..0000000
--- a/doc/power/power_supply.tex
+++ /dev/null
@@ -1,140 +0,0 @@
-\documentclass{article}
-\usepackage{amsmath}
-\usepackage{hyperref}
-\usepackage{IEEEtrantools}
-
-\title{Power Supply}
-\author{Sam Anthony}
-
-\begin{document}
-
-\maketitle
-
-Switching Regulator Component Selection \& Sizing --- Phil's Lab \#71: \\
-\url{https://youtu.be/FqT_Ofd54fo}
-
-
-\section{Requirements}
-
-Input voltage range:
-\begin{equation}
-	9V \le V_{In} \le 16V
-\end{equation}
-\begin{equation}
-	V_{In,nom} = 12.7V
-\end{equation}
-
-Nominal output voltage:
-\begin{equation}
-	V_{Out,nom} = 5V
-\end{equation}
-
-Maximum load current:
-\begin{equation}
-	I_{Load,max} = 250 mA
-\end{equation}
-
-\section{IC selection}
-
-Diodes Inc. AP64100Q: \\
-\url{https://www.diodes.com/assets/Datasheets/AP64100Q.pdf}
-
-Switching frequency:
-\begin{equation}
-	100 kHz \le f_{sw} \le 2.2 MHz
-\end{equation}
-Select a frequency of 2MHz:
-\begin{equation}
-	f_{sw} = 2 MHz
-\end{equation}
-\begin{equation}
-	RT[k \Omega] = \frac{100000}{f_{sw}[kHz]} = 50 k\Omega
-\end{equation}
-
-Max. high-side current:
-\begin{equation}
-	I_{Peak,lim} = 2.2 A
-\end{equation}
-
-
-\section{Maximum switching current}
-
-Efficiency:
-\begin{equation}
-	\eta \approx 0.85
-\end{equation}
-
-Duty cycle:
-\begin{equation}
-	\delta = \frac{V_{Out}}{V_{In} \cdot \eta}
-\end{equation}
-
-\begin{equation}
-	\delta(V_{In,min}) \approx 0.654
-\end{equation}
-
-\begin{equation}
-	\delta(V_{In,nom}) \approx 0.464
-\end{equation}
-
-\begin{equation}
-	\delta(V_{In,max}) \approx 0.368
-\end{equation}
-
-Inductor ripple current estimated using average recommended inductor value from datasheet:
-\begin{equation}
-	L_{avg} = \frac{6.8 \mu H + 33 \mu H}{2} = 19.9 \mu H
-\end{equation}
-
-\begin{equation}
-	\Delta I_L = \frac{(V_{In} - V_{Out}) \cdot \delta}{f_{sw} \cdot L_{avg}} \\
-\end{equation}
-
-\begin{equation}
-	\Delta I_L(V_{In,min}) = \frac{(9V - 5V) \cdot 0.654}{2MHz \cdot 19.9 \mu H} = 65.7 mA
-\end{equation}
-
-\begin{equation}
-	\Delta I_L(V_{In,nom}) = \frac{(12.7V - 5V) \cdot 0.464}{2MHz \cdot 19.9 \mu H} = 86.4 mA
-\end{equation}
-
-\begin{equation}
-	\Delta I_L(V_{In,max}) = \frac{(16V - 5V) \cdot 0.368}{2MHz \cdot 19.9 \mu H} = 102 mA
-\end{equation}
-
-Available output current:
-\begin{IEEEeqnarray}{rCl}
-	I_{Avail,min} & = & I_{Peak,lim} - \frac{\Delta I_{L,max}}{2} \\
-	& = & 2.2A - 51 mA \\
-	& = & 2.15 A > I_{Load,max}
-\end{IEEEeqnarray}
-
-Max power supply current:
-\begin{equation}
-	I_{Supply,max} = I_{Load,max} + \frac{\Delta I_{L,max}}{2} = 301mA
-\end{equation}
-
-\section{Inductor selection}
-
-\begin{equation}
-	L = \frac{V_{Out} \cdot (V_{In} - V_{Out})}{\Delta I_L \cdot f_{sw} \cdot V_{In}}
-\end{equation}
-
-Estimate ripple current as $20\%$--$40\%$ of maximum supply current:
-\begin{equation}
-	\Delta I_{L,approx.} = 0.30 \cdot 301 mA = 90.3 mA
-\end{equation}
-
-\begin{equation}
-	L(V_{In,min}) = \frac{5V \cdot (9V - 5V)}{90.3mA \cdot 2MHz \cdot 9V} = 12.3 \mu H
-\end{equation}
-
-\begin{equation}
-	L_(V_{In,nom}) = \frac{5V \cdot (12.7V - 5V)}{90.3mA \cdot 2MHz \cdot 12.7V} = 16.8 \mu H
-\end{equation}
-
-\begin{equation}
-	L(V_{In,max}) = \frac{5V \cdot (16V - 5V)}{90.3mA \cdot 2MHz \cdot 16V} = 19.0 \mu H
-\end{equation}
-
-\end{document}