Texas Instruments LM3409 Demonstration Board LM3409EVAL/NOPB LM3409EVAL/NOPB 数据表
产品代码
LM3409EVAL/NOPB
mW
577
m
190
A
74
.
1
R
I
P
2
DSON
2
RMS
T
T
=
:
x
=
x
=
-
1
1.97A
x
+
x
x
=
¨
¨
©
§
V
35
12
1
95
.
0
V
48 x
I
RMS
T-
2
¸
¸
¹
·
¸¸
¹
·
¨¨
©
§
1.027A
1.97A
1.74A
=
I
RMS
T-
I
LED
x
=
1
D
2
x
+
x
¸¸
¸
¹
·
¸
¸
¹
¹
·
¨
¨
©
©
§
¨¨
¨
©
§
12
1
i
L-PP
'
I
LED
I
RMS
T-
=
A
51
.
1
=
A
97
.
1
V
35 x
95
.
0
V
48 x
I
D
I
LED
T
=
x
=
V
IN
K
x
I
V
LED
O
x
V
75
V
V
MAX
IN
MAX
T
=
=
-
-
F
2
.
2
C
C
2
IN
1
IN
P
=
=
I
RMS
IN-
kHz
528
A
97
.
1
x
x
=
mA
831
=
ns
440
s
45
.
1
x
P
t
t
OFF
ON
x
f
I
I
SW
LED
RMS
IN
x
x
=
-
2
C
MIN
IN
=
x
=
-
F
96
.
3
P
C
IN
C
MIN
IN
=
-
=
=
F
98
.
1
P
V
44
.
1
A
97
.
1
x
s
45
.
1
P
v
PP
IN
'
-
t
I
ON
LED
x
1
t
ON
=
1
t
OFF
=
-
-
ns
440
=
s
45
.
1
P
kHz
528
f
SW
=
R
SNS
:
1
.
0
I
LED
=
2
1.027A
-
A
97
.
1
=
099
.
0
5
:
x
V
24
.
1
I
LED
=
R
5
SNS
x
-
2
i
PP
L
'
-
V
ADJ
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
= 190m
Ω
and Q
g
= 20nC. Determine I
T-RMS
and P
T
:
(63)
(64)
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25
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