Datasheet

D1 o 12A, 100V, DPAK
mW600mV600A1VIP
FDDD
=
x
=
x
=
A1II
LEDMAXD
==
-
V91V21V70VVV
OMAXINMAXRD
=
+
=
+
=
--
Q1 o 32A, 100V, DPAK
mW82m50A28.1RIP
2
DSON
2
RMSTT
=
:x
=
x
=
-
x
I
RMST
=
-
I
LED
D
c
=
x
A28.1
=
0.467
A1
533.0
D
=
A2.1A1
=
x
677.01-
677.0
I
MAXT-
V91V21V70VVV
OMAXINMAXT
=+=+=
--
C
IN
= 4 x 4.7 PF
x
A1
=
I
LED
I
RMSIN-
=
1- 0.677
677.0
1.45A
x
1- D
MAX
D
MAX
=
C
IN
==
kHz504mV100 x
467.0A1 x
F27.9 P=
f
'v
SWPPIN-
x
DI
LED
x
LM3424
www.ti.com SNVS603B –AUGUST 2009–REVISED OCTOBER 2009
10. INPUT CAPACITANCE
Solve for the minimum C
IN
:
(137)
To minimize power supply interaction a 200% larger capacitance of approximately 20 µF is used, therefore the
actual Δv
IN-PP
is much lower. Since high voltage ceramic capacitor selection is limited, four 4.7 µF X7R capacitors
are chosen.
Determine minimum allowable RMS current rating:
(138)
The chosen components from step 10 are:
(139)
11. NFET
Determine minimum Q1 voltage rating and current rating:
(140)
(141)
A 100V NFET is chosen with a current rating of 32A due to the low R
DS-ON
= 50 mΩ. Determine I
T-RMS
and P
T
:
(142)
(143)
The chosen component from step 11 is:
(144)
12. DIODE
Determine minimum D1 voltage rating and current rating:
(145)
(146)
A 100V diode is chosen with a current rating of 12A and V
D
= 600 mV. Determine P
D
:
(147)
The chosen component from step 12 is:
(148)
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