Datasheet

H331L P
=
A88.1
I
12
1
1
I
RMSL
RMSL
=
+
x
=
-
-
I
I
LED
RMSL
x
=
-
12
1
1
2
x
+
¸
¸
¹
·
¨
¨
©
§
Di
PPL
c
x
'
-
I
LED
D
c
533.0mA485
2
x
x
¸
¸
¹
·
¨
¨
©
§
A1
533.0
A1
PP-
==
L
DV
IN
x
f1L
SW
x
kHz007H33 xP
467.0V42 x
mA485
=
'i
==
DV
IN
x
f
SW
x
467.0V42 x
PH
32
=
1L
PP-
'i
L
kHz007500 mA x
:
k4.21R
CSH
=
:
k1RR
HSN
==
HSP
0.1R
NSS
=
:
I
LED
= =
k0.11.24V :x
A0.1=
k4.121.0 :x:
RR
CSHSNS
x
R1.24V
HSP
x
=
1.24V1.24V
=
R
HSP
:
=
k0.1
:
x
:
x
0.1k12.4A1
xx
RRI
SNSCSHLED
:
===
1.0
1A
R
SNS
I
LED
mV100
V
SNS
C
T
= 1 nF
R
T
= 35.7 k:
LM3429, LM3429-Q1
www.ti.com SNVS616G –APRIL 2009–REVISED MAY 2013
The chosen components from step 2 are:
(96)
3. AVERAGE LED CURRENT
Solve for R
SNS
:
(97)
Assume R
CSH
= 12.4 kΩ and solve for R
HSP
:
(98)
The closest standard resistor for R
SNS
is actually 0.1Ω and for R
HSP
is actually 1 kΩ therefore I
LED
is:
(99)
The chosen components from step 3 are:
(100)
4. INDUCTOR RIPPLE CURRENT
Solve for L1:
(101)
The closest standard inductor is 33 µH therefore the actual Δi
L-PP
is:
(102)
Determine minimum allowable RMS current rating:
(103)
The chosen component from step 4 is:
(104)
5. OUTPUT CAPACITANCE
Solve for C
O
:
Copyright © 2009–2013, Texas Instruments Incorporated Submit Documentation Feedback 33
Product Folder Links: LM3429 LM3429-Q1