Datasheet

:
=
:
=
k9.49R
k98.6R
2UV
1UV
=
V1.10
k98.6 :
( )k9.49k98.6V24.1 :+:x
V
ONTURN
=
-
V
ONTURN
=
-
V24.1 ( )RR
2UV1UV
+x
R
1UV
=
:
=
k06.7
:
x
k9.49V24.1
- V24.1V10
=
R
1UV
x
RV24.1
2UV
-
-
V24.1V
ONTURN
1
2
k9.49A22RV
UVHYS
x:
=
Px
=
V1.A22
=
P
2
==
V
R
HYS
UV
=
:k50
PA22PA22
V1.1
SMC,V100,A3D1o
mA457V
D
x=x mV750 = mW343IP
DD
=
( )
1ID1I
LEDD
-
=
x-
=
¨
¨
©
§
I
LED
x
¸
¸
¹
·
V
IN
Kx
V
O
mA457A97.1
=
x
¸
¸
¹
·
V35
1I
D
-
=
¨
¨
©
§
95.0V48 x
VV
MAXINMAXD
==
--
V75
Q1o 3.8 DPAK,V100,A
LM3409, LM3409HV, LM3409-Q1
SNVS602J –MARCH 2009–REVISED MAY 2013 www.ti.com
The chosen component from step 6 is:
(65)
7. DIODE
Determine minimum D1 voltage rating and current rating:
(66)
(67)
A 100V, 3A diode is chosen with V
D
= 750mV. Determine P
D
:
(68)
The chosen component from step 7 is:
(69)
8. INPUT UVLO
Solve for R
UV2
:
(70)
The closest 1% tolerance resistor is 49.9 kΩ therefore V
HYS
is:
(71)
Solve for R
UV1
:
(72)
The closest 1% tolerance resistor is 6.98 kΩ therefore V
TURN-ON
is:
(73)
The chosen components from step 8 are:
(74)
9. IADJ CONNECTION METHOD
The IADJ pin is left open forcing V
ADJ
= 1.24V.
10. PWM DIMMING METHOD
PWM dimming signal pair is applied to the EN pin and GND at f
DIM
= 1 kHz.
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