Datasheet

mW577m190A74.1RIP
2
DSON
2
RMSTT
=:
x
=
x
=
-
1
1.97A x
+
xx
=
¨
¨
©
§
V35
12
1
95.0V48 x
I
RMST-
2
¸
¸
¹
·
¸
¸
¹
·
¨
¨
©
§
1.027A
1.97A
1.74A
=
I
RMST-
I
LED
x
=
1D
2
x
+
x
¸
¸
¸
¹
·
¸
¸
¹
·
¨
¨
©
§
¨
¨
¨
©
§
12
1
i
L-PP
'
I
LED
I
RMST-
=
A51.1
=
A97.1V35 x
95.0V48 x
IDI
LEDT
=
x
=
V
IN
Kx
IV
LEDO
x
V75VV
MAXINMAXT
==
--
F2.2CC
2IN1IN
P
==
I
RMSIN-
kHz528A97.1 xx
=
mA831
=
ns440s45.1 xP
tt
OFFON
xfII
SWLEDRMSIN
xx
=
-
2C
MININ
=
x
=
-
F96.3 PC
IN
C
MININ
=
-
= =
F98.1 P
V44.1
A97.1 x s45.1 P
v
PPIN
'
-
tI
ONLED
x
1
t
ON
=
1
t
OFF
=
- - ns440
=
s45.1 P
kHz528
f
SW
=
R
SNS
:1.0
I
LED
=
2
1.027A
-
A97.1
=
099.05 :x
V24.1
I
LED
=
R5
SNS
x
-
2
i
PPL
'
-
V
ADJ
LM3409, LM3409HV, LM3409-Q1
www.ti.com SNVS602J –MARCH 2009–REVISED MAY 2013
(54)
The chosen component from step 3 is:
(55)
4. OUTPUT CAPACITANCE
No output capacitance is necessary.
5. INPUT CAPACITANCE
Determine t
ON
:
(56)
Solve for C
IN-MIN
:
(57)
Choose C
IN
:
(58)
Determine I
IN-RMS
:
(59)
The chosen components from step 5 are:
(60)
6. PFET
Determine minimum Q1 voltage rating and current rating:
(61)
(62)
A 100V, 3.8A PFET is chosen with R
DS-ON
= 190mΩ and Q
g
= 20nC. Determine I
T-RMS
and P
T
:
(63)
(64)
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