Datasheet

k:301R
2UV
=
k:18.2R
1UV
=
R
UV1
(
)
RR1.24V
UV2UV1
+x
V
ONTURN
=
-
= V10.1=
( )
k130k18.21.24V :+:x
k18.2 :
V
ONTURN-
:== k4.18
-
-
1.24VV
ONTURN
x R1.24V
UV2
=R
UV1
:x k1301.24V
-1.24V10V
2.99VA23k130A23RV
2UVHYS
=
x:
=
x
=
PP
===
A
23
3V
R
UV2
P
:k130
V
HYS
A23
P
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-
LM3421, LM3421-Q1
LM3423, LM3423-Q1
SNVS574E –JULY 2008–REVISED MAY 2013 www.ti.com
(131)
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
:
(132)
(133)
The chosen component from step 9 is:
(134)
10. DIODE
Determine minimum D1 voltage rating and current rating:
(135)
(136)
A 100V diode is chosen with a current rating of 12A and V
D
= 600 mV. Determine P
D
:
(137)
The chosen component from step 10 is:
(138)
11. INPUT UVLO
Solve for R
UV2
:
(139)
The closest standard resistor is 130 kΩ therefore V
HYS
is:
(140)
Solve for R
UV1
:
(141)
The closest standard resistor is 18.2 kΩ making V
TURN-ON
:
(142)
The chosen components from step 11 are:
(143)
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