From 1db5c3f58d907d2a071e1d414dc4e8ee6c1fbdb0 Mon Sep 17 00:00:00 2001 From: Sam Anthony Date: Tue, 14 Oct 2025 10:07:40 -0400 Subject: remove deprecated power supply doc --- doc/power/power_supply.tex | 140 --------------------------------------------- 1 file changed, 140 deletions(-) delete mode 100644 doc/power/power_supply.tex (limited to 'doc') 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} -- cgit v1.2.3