1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
|
\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}
|