Texas Instruments LM3409 Demonstration Board LM3409EVAL/NOPB LM3409EVAL/NOPB Datenbogen
Produktcode
LM3409EVAL/NOPB
=
:
=
k
06
.
7
:
x
k
9
.
49
V
24
.
1
-
V
24
.
1
V
10
=
R
1
UV
x
R
V
24
.
1
2
UV
-
-
V
24
.
1
V
ON
TURN
1
2
k
9
.
49
A
22
R
V
UV
HYS
x
:
=
P
x
=
V
1
.
A
22
=
P
2
=
=
V
R
HYS
UV
=
:
k
50
P
A
22
P
A
22
V
1
.
1
SMC
,
V
60
,
A
5
1
D
o
mW
268
mV
750
mA
358
V
I
P
D
D
D
=
x
=
x
=
1-
¸¸
¹
·
¨¨
©
§
x
K
V
O
x I
LED
I
D
= (1- D) x I
LED
=
V
IN
I
D
= 1-
¸
¹
·
¨
©
§
V
14
V
24
90
.
0
x
x 1.02A = 358 mA
V
42
V
V
MAX
IN
MAX
D
=
=
-
-
DPAK
,
V
70
,
A
5.7
Q1
o
mW
129
m
190
mA
830
R
I
P
2
DSON
2
RMS
T
T
=
:
x
=
x
=
-
mA
830
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
02
.
1
I
RMS
T
x
=
-
12
1
1
2
x
+
x
¸
¸
¹
·
¸¸
¹
·
¨¨
©
§
¨
¨
©
§
A
02
.
1
mA
445
V
14
90
.
0
V
24 x
=
mA
660
=
I
D
I
LED
T
=
x
=
I
V
LED
O
x
V
IN
K
x
x
A
02
.
1
V
14
x
V
24
90
.
0
V
V
MAX
IN
MAX
T
=
=
-
-
V
42
SNVS602J – MARCH 2009 – REVISED MAY 2013
6. PFET
Determine minimum Q1 voltage rating and current rating:
(95)
(96)
A 70V, 5.7A PFET is chosen with R
DS-ON
= 190m
Ω
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 k
Ω
therefore V
HYS
is:
(105)
Solve for R
UV1
:
(106)
30
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