RIPPLE
L peak
OUT
I
I
I
2
-
=
+
(
)
OUT
RIPPLE
2
sw
IN
V
V
D
I
L
F
-
´
=
´
IN
OUT
OUT
mi
I
n
sw
OUT
sw
OUT
N
(V
V
)
D
(V
V
)
D
L
0.4 F
I
0.2 F
I
-
´
-
´
£
£
´
´
´
´
5
sw
40200
R
0.6
F
=
-
System Overview
23
JAJU324B – March 2015 – Revised July 2017
翻訳版
—
最新の英語版資料
http://www-s.ti.com/sc/techlit/TIDU832
Copyright © 2015–2017, Texas Instruments Incorporated
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2.3.1.3.1.1 Passive Components Calculations
The first step is to decide and set the switching frequency of the regulator. In the TIDA-00204, the
switching frequency (F
SW
) is set to 500 kHz thanks to R
5
.
indicates that R
5
should be 79.8 k
Ω
. 80.6 k
Ω
is then used as value for R
5
.
(1)
with
•
F
SW
in kHz and R
5
in k
Ω
The default frequency is also 500 kHz, due to the internal oscillator precision of 20% it was chosen to use
a resistor for the 500-kHz frequency. This gives a 10% precision. Secondly, during EMI test it can be
useful for fine tuning to move the frequency up or down to change the harmonic spectrum of the
converter.
Once the switching frequency is set, the size of inductance needed it chosen. For most buck converters,
this value is chosen to achieve the wanted peak-to-peak ripple current that flows in the inductor along with
the DC load current. A higher inductance gives lower ripple current, which then gives a lower output
voltage ripple with the same output capacitors. An inductance that gives a ripple current of 20% to 40% at
the maximum current is normally a good starting point. The minimum inductor value is calculated based
on input voltage range (17 to 60 V), output voltage (5 V), maximum output current (2 A) and the switching
frequency (500 kHz) as set in the previous step. The duty cycle (D) can be approximated as D = V
OUT
/V
IN
,
assuming no loss power conversion.
(2)
The calculation gives that the inductor needs to be between 9 and 23
μ
H; for this design, 15
μ
H was
chosen.
Besides the inductor size, the rated saturation current needs to be specified as per
.
(3)
(4)
The inductor peak current is 2.305 A according to
and
. A good practice is to use an inductor with
the saturation current around 1.5 to 2 times higher than calculated.
The LM46002 has both valley current limit and peak current limit. During an instantaneous short, the peak
inductor current can be high due to a momentary increase in duty cycle. The inductor current rating should
be higher than the peak current limit. Select an inductor with a larger core saturation margin and
preferably a softer roll off of the inductance value over load current.
