Datasheet

=
:
=
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,V60,A51D o
mW268mV750mA358VIP
DDD
=
x
=
x
=
1-
¸
¸
¹
·
¨
¨
©
§
xK
V
O
x I
LED
I
D
= (1- D) x I
LED
=
V
IN
I
D
=
1
-
¸
¹
·
¨
©
§
V14
V24 90.0x
x 1.02A = 358 mA
V42VV
MAXINMAXD
==
--
DPAK,V70,A5.7Q1o
mW129m190mA
830
RIP
2
DSON
2
RMSTT
=
:x
=
x
=
-
mA830I
RMST
=
-
I
LED
x
=
1D
2
x
+
x
¸
¸
¸
¹
·
¸
¸
¹
·
¨
¨
©
§
¨
¨
¨
©
§
12
1
i
L-PP
'
I
LED
I
RMST-
A02.1I
RMST
x
=
-
12
1
1
2
x
+
x
¸
¸
¹
·
¸
¸
¹
·
¨
¨
©
§
¨
¨
©
§
A02.1
mA445
V14
90.0V24 x
=
mA660
=
IDI
LEDT
=
x
=
IV
LEDO
x
V
IN
Kx
x A02.1V14
xV24 90.0
VV
MAXINMAXT
==
--
V42
LM3409, LM3409HV, LM3409-Q1
SNVS602J –MARCH 2009–REVISED MAY 2013 www.ti.com
6. PFET
Determine minimum Q1 voltage rating and current rating:
(95)
(96)
A 70V, 5.7A PFET is chosen with R
DS-ON
= 190mΩ and Q
g
= 20nC. Determine I
T-RMS
and P
T
:
(97)
(98)
The chosen component from step 6 is:
(99)
7. DIODE
Determine minimum D1 voltage rating and current rating:
(100)
(101)
A 60V, 5A diode is chosen with V
D
= 750mV. Determine P
D
:
(102)
The chosen component from step 7 is:
(103)
8. INPUT UVLO
Solve for R
UV2
:
(104)
The closest 1% tolerance resistor is 49.9 kΩ therefore V
HYS
is:
(105)
Solve for R
UV1
:
(106)
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