\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}