Datasheet

0.04:R
LIM
=
:04.0
===
6.13A
mV245mV245
I
LIM
R
LIM
:=== 041.0
6A
R
LIM
mV245mV245
I
LIM
C
O
= 4 x 10 PF
x
A1
=
I
LED
I
RMSCO-
=
1- 0.677
677.0
1.45A
x
1- D
MAX
D
MAX
=
DI
LED
x
=
'i
PP-LED
SW
fxr
D
x
C
O
2
= =
kHz04595.1 xx:
1 mA
467.0A1 x
F40 P
'i
PP-LED
f
'i
r
SWPP-LEDD
xx
DI
LED
x
C
O
=
2
= F39.6P=
kHz045mA195.1 xx:
467.0A1 x
C
O
H331L P
=
A89.1
12
1
1
I
RMSL
=
+
x
=
-
I
I
LED
RMSL
x
=
-
12
1
1
2
x
+
¸
¸
¹
·
¨
¨
©
§
Di
PPL
c
x
'
-
I
LED
D
c
533.0mA674
2
x
¸
¸
¹
·
¨
¨
©
§
A1
x
533.0
A1
PP-
==
L
DV
IN
x
f1L
SW
x
kHz045H33 xP
467.0V42 x
mA674
=
'i
==
DV
IN
x
f
SW
x
467.0V42 x
PH
32
=
1L
PP-
'i
L
kHz045700 mA x
LM3424
www.ti.com SNVS603B –AUGUST 2009–REVISED OCTOBER 2009
5. INDUCTOR RIPPLE CURRENT
Solve for L1:
(116)
The closest standard inductor is 33 µH therefore Δi
L-PP
is:
(117)
Determine minimum allowable RMS current rating:
(118)
The chosen component from step 5 is:
(119)
6. OUTPUT CAPACITANCE
Solve for C
O
:
(120)
The closest capacitance totals 40 µF therefore Δi
LED-PP
is:
(121)
Determine minimum allowable RMS current rating:
(122)
The chosen components from step 6 are:
(123)
7. PEAK CURRENT LIMIT
Solve for R
LIM
:
(124)
The closest standard resistor is 0.04 Ω therefore I
LIM
is:
(125)
The chosen component from step 7 is:
(126)
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