Datasheet

A1
2
+=+II
LEDMAXL
=
-
2
A22.1=
mA445
i
PPL
'
-
H22L1 P
=
ns
x700
V14
=
H22P
mA445
=
i
PPL
=
'
-
tV
OFFO
x
L1
tV
L1
OFFO
x
= =
H8.21 P
=
ns700V14 x
mA450i
PPL
'
-
:
=
=
k4.15R
pF470C
OFF
OFF
=
f
SW
=
t
OFF
1
-
¸
¸
¹
·
¨
¨
©
§
V
IN
xK
V
O
kHz503
=
ns700
1-
¸
¹
·
¨
©
§
V14
V2490.0 x
t
OFF
=
1lnk4.15pF490
-
x:x
-
¨
¨
©
§
ns700
=
¸
¸
¹
·
V14
V24.1
lnR(C
OFF
+ 20 pF)t
OFFOFF
xx
-
=
1
-
¨
¨
©
§
¸
¸
¹
·
V
O
V24.1
=
R
OFF
-- 1
¸
¸
¹
·
¨
¨
©
§
V
O
xK V
IN
-xx 1lnfC
OFF
+ 20 pF
SW
¨
¨
©
§
¸
¸
¹
·
V24.1
V
O
=
:
=
k5.15
xx lnkHz500pF490
¸
¹
·
-1
¨
©
§
V24.1
V14
-- 1
¸
¹
·
¨
©
§
x V2490.0
V14
R
OFF
LM3409, LM3409HV, LM3409-Q1
SNVS602J –MARCH 2009–REVISED MAY 2013 www.ti.com
η = 0.90
1. NOMINAL SWITCHING FREQUENCY
Assume C
OFF
= 470pF and η = 0.90. Solve for R
OFF
:
(75)
The closest 1% tolerance resistor is 15.4 kΩ therefore the actual t
OFF
and target f
SW
are:
(76)
(77)
The chosen components from step 1 are:
(78)
2. INDUCTOR RIPPLE CURRENT
Solve for L1:
(79)
The closest standard inductor value is 22 µH therefore the actual Δi
L-PP
is:
(80)
The chosen component from step 2 is:
(81)
3. AVERAGE LED CURRENT
Determine I
L-MAX
:
(82)
Assume V
ADJ
= 1.24V and solve for R
SNS
:
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