Texas Instruments LM3409HV, User Manual

Page: 6 / 14

F 2 . 2 C2 C1
P = =
F 1 . 0
C3 P =
I
RMS IN-
kHz 404 A 5 . 1 x x
=
mA 446
=
ns 242 s 32 . 2 x
P
t t
OFF ON
x f I I
SW LED RMS IN
x x
=
-
1.75 C
MIN IN
=
x
=
-
F 07 . 4
P
C
IN
C
MIN IN
=
-
= =
F 32 . 2
P
V 44 . 1
A 5 . 1 x s 23 . 2
P
v
PP IN
'
-
t I
ON LED
x
1
t
ON
=
1
t
OFF
=
- - ns 242
=
s 23 . 2
P
kHz 404 f
SW
= R9 : 15 . 0
I
LED
=
2
308
- A 5 . 1
=
15 . 0 5 : x
V 24 . 1
I
LED
=
R9 5 x
-
2
i
PP L
'
-
V
ADJ
mA
=
R9
V
ADJ
x
-
I 5
MAX L
= =
:
15 . 0
V 24 . 1
x 5 A 65 . 1
I I
LED MAX L
+
=
-
2
A 1.5 + = A 65 . 1 =
2
i
PP L
'
-
308 mA
H 33 L1
P =
= =
H 33
P
308 =
242 ns V 42 x
L1
t V
OFF O
x
i
PP L
'
-
mA
=
H 9 . 33
P =
L1
=
i
PP L
'
-
t V
OFF O
x
ns 242 V 42 x
mA 300
Design Procedure
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6.3 Inductor Ripple Current
Solve for L1:
(5)
The closest standard inductor value is 33 µH therefore the actual Δ i
L-PP
is:
(6)
The chosen component from step 2 is:
(7)
6.4 Average LED Current
Determine I
L-MAX
:
(8)
Assume V
ADJ
= 1.24V and solve for R9:
(9)
The closest 1% tolerance resistor is 0.15 Ω therefore the I
LED
is:
(10)
The chosen component from step 3 is:
(11)
6.5 Output Capacitance
No output capacitance is necessary.
6.6 Input Capacitance
Determine t
ON
:
(12)
Solve for C
IN-MIN
:
(13)
Choose C
IN
:
(14)
Determine I
IN-RMS
:
(15)
The chosen components from step 5 are:
(16)
(17)
6 AN-1953 LM3409HV Evaluation Board SNVA390D – May 2009 – Revised May 2013
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