Datasheet

=
R
SNS
V
ADJ
x
-
I5
MAXL
= =
:099.0
V24.1
x5 A51.2
II
LEDMAXL
+
=
-
2
A2
+
=
A51.2
=
2
i
PPL
'
-
1.027A
L1
=
H15P
= =
H15 P
1.027A
=
440nsV35 x
L1
tV
OFFO
x
i
PPL
'
-
=
H4.15 P
=
L1
=
i
PPL
'
-
tV
OFFO
x
ns440V35 x
A1
=
pF470C
OFF
:
=
k9.24R
OFF
t
OFF
f
SW
= =
ns440
=
kHz528
1
-
¸
¹
·
¨
©
§
V4895.0 x
V35
¸
¸
¹
·
¨
¨
©
§
V
O
V
IN
xK
1
-
1lnk9.24pF490t
OFF
-x:x
-
=
¨
¨
©
§
ns440
=
¸
¸
¹
·
V24.1
V35
(C
OFF
+ 20 pF)t
OFF
-
=
x R
OFF
x
1
1ln -
¸
¸
¹
·
¨
¨
©
§
V
O
V24.
=
R
OFF
--
1
¸
¸
¹
·
¨
¨
©
§
V
O
xK V
IN
=
R
OFF
:
=
k1.25
x lnpF490 xkHz525
¨
©
§
-
1
V35
¸
¹
·
V24.1
--
1
¨
©
§
¸
¹
·
V35
x V4895.0
-
xx 1lnf(C
OFF
+ 20 pF)
SW
¸
¸
¹
·
¨
¨
©
§
V
O
V24.1
LM3409, LM3409HV, LM3409-Q1
SNVS602J –MARCH 2009–REVISED MAY 2013 www.ti.com
(45)
The closest 1% tolerance resistor is 24.9 kΩ therefore the actual t
OFF
and target f
SW
are:
(46)
(47)
The chosen components from step 1 are:
(48)
2. INDUCTOR RIPPLE CURRENT
Solve for L1:
(49)
The closest standard inductor value is 15 µH therefore the actual Δi
L-PP
is:
(50)
The chosen component from step 2 is:
(51)
3. AVERAGE LED CURRENT
Determine I
L-MAX
:
(52)
Assume V
ADJ
= 1.24V and solve for R
SNS
:
(53)
The closest 1% tolerance resistor is 0.1 Ω therefore the I
LED
is:
24 Submit Documentation Feedback Copyright © 2009–2013, Texas Instruments Incorporated
Product Folder Links: LM3409 LM3409HV LM3409-Q1