Texas Instruments LM3409 Demonstration Board LM3409EVAL/NOPB LM3409EVAL/NOPB 데이터 시트
제품 코드
LM3409EVAL/NOPB
=
R
SNS
V
ADJ
x
-
I
5
MAX
L
=
=
:
099
.
0
V
24
.
1
x
5
A
51
.
2
I
I
LED
MAX
L
+
=
-
2
A
2 +
=
A
51
.
2
=
2
i
PP
L
'
-
1.027A
L1 =
H
15
P
=
=
H
15
P
1.027A
=
440 ns
V
35 x
L1
t
V
OFF
O
x
i
PP
L
'
-
=
H
4
.
15
P
=
L1
=
i
PP
L
'
-
t
V
OFF
O
x
ns
440
V
35 x
A
1
=
pF
470
C
OFF
:
=
k
9
.
24
R
OFF
t
OFF
f
SW
=
=
ns
440
=
kHz
528
1
-
¸
¹
·
¨
©
§
V
48
95
.
0
x
V
35
¸¸
¹
·
¨¨
©
§
V
O
V
IN
x
K
1
-
1
ln
k
9
.
24
pF
490
t
OFF
-
x
:
x
-
=
¨¨
©
§
ns
440
=
¸¸
¹
·
V
24
.
1
V
35
(C
OFF
+ 20 pF)
t
OFF
-
=
x R
OFF
x
1
1
ln -
¸
¸
¹
¹
·
¨
¨
©
©
§
V
O
V
24
.
=
R
OFF
-
-
1
¸¸
¹
·
¨¨
©
§
V
O
x
K
V
IN
=
R
OFF
:
=
k
1
.
25
x
ln
pF
490
x
kHz
525
¨
©
§
-
1
V
35
¸
¹
·
V
24
.
1
-
-
1
¨
©
§
¸
¹
·
V
35
x
V
48
95
.
0
-
x
x
1
ln
f
(C
OFF
+ 20 pF)
SW
¸¸
¹
·
¨¨
©
§
V
O
V
24
.
1
SNVS602J – MARCH 2009 – REVISED MAY 2013
(45)
The closest 1% tolerance resistor is 24.9 k
Ω
therefore the actual t
OFF
and target f
SW
are:
(46)
(47)
The chosen components from step 1 are:
(48)
2. INDUCTOR RIPPLE CURRENT
Solve for L1:
(49)
The closest standard inductor value is 15 µH therefore the actual
Δ
i
L-PP
is:
(50)
The chosen component from step 2 is:
(51)
3. AVERAGE LED CURRENT
Determine I
L-MAX
:
(52)
Assume V
ADJ
= 1.24V and solve for R
SNS
:
(53)
The closest 1% tolerance resistor is 0.1
Ω
therefore the I
LED
is:
24
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