User manual

DI
LED
x
=
'i
PP-LED
SW
fxr
D
x
C
O
= =
kHz007925.2 xx:
17.6 mA
238.0A1 x
F6.6 P
'i
PP-LED
f
'i
r
SWPP-LEDD
xx
DI
LED
x
C
O
=
= F84.6 P=
kHz007mA17925.2 xx:
238.0A1 x
C
O
H331L P
=
A31.1
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
762.0mA247
2
x
x
¸
¸
¹
·
¨
¨
©
§
A1
762.0
A1
PP-
==
L
DV
IN
x
f1L
SW
x
kHz007H33 xP
238.0V42 x
mA247
=
'i
==
DV
IN
x
f
SW
x
238.0V42 x
PH
32.6
=
1L
PP-
'i
L
kHz007250 mA x
:
k4.21R1
=
:
k1R7R8
==
0.1R9
=
:
I
LED
= =
k0.11.24V :x
A0.1=
k4.121.0 :x:
R1R9 x
R81.24V x
1A x 12.4 k: x 0.1:
=
1.24V
= 1.0 k:
R8 =
I
LED
x R1 x R9
1.24V
:
===
1.0
1A
R9
I
LED
mV100
V
SNS
www.ti.com Design Procedure
6.4 Average LED Current
Solve for R9:
(10)
Assume R1 = 12.4 kΩ and solve for R8:
(11)
The closest standard resistor for R9 is 0.1Ω and the closest for R8 (and R7) is actually 1 kΩ therefore I
LED
is:
(12)
The chosen components from step 3 are:
(13)
6.5 Inductor Ripple Current
Solve for L1:
(14)
The closest standard inductor is 33 µH therefore the actual Δi
L-PP
is:
(15)
Determine minimum allowable RMS current rating:
(16)
The chosen component from step 4 is:
(17)
6.6 Output Capacitance
Solve for C
O
:
(18)
A total value of 6.6 µF (using 3 2.2 µF X7R ceramic capacitors) is chosen therefore the actual Δi
LED-PP
is:
(19)
Determine minimum allowable RMS current rating:
7
SNVA404B–July 2009–Revised May 2013 AN-1986 LM3429 Boost Evaluation Board
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