Texas Instruments LM3409 Demonstration Board LM3409EVAL/NOPB LM3409EVAL/NOPB 데이터 시트
제품 코드
LM3409EVAL/NOPB
A
1
2
+
=
+
I
I
LED
MAX
L
=
-
2
A
22
.
1
=
mA
445
i
PP
L
'
-
H
22
L1
P
=
ns
x 700
V
14
=
H
22
P
mA
445
=
i
PP
L
=
'
-
t
V
OFF
O
x
L1
t
V
L1
OFF
O
x
=
=
H
8
.
21
P
=
ns
700
V
14 x
mA
450
i
PP
L
'
-
:
=
=
k
4
.
15
R
pF
470
C
OFF
OFF
=
f
SW
=
t
OFF
1-
¸¸
¹
·
¨¨
©
§
V
IN
x
K
V
O
kHz
503
=
ns
700
1-
¸
¹
·
¨
©
§
V
14
V
24
90
.
0
x
t
OFF
=
1
ln
k
4
.
15
pF
490
-
x
:
x
-
¨¨
©
§
ns
700
=
¸¸
¹
·
V
14
V
24
.
1
ln
R
(C
OFF
+ 20 pF)
t
OFF
OFF
x
x
-
=
1
-
¨
¨
©
©
§
¸
¸
¹
¹
·
V
O
V
24
.
1
=
R
OFF
-
- 1
¸¸
¹
·
¨¨
©
§
V
O
x
K
V
IN
-
x
x
1
ln
f
C
OFF
+ 20 pF
SW
¨¨
©
§
¸¸
¹
·
V
24
.
1
V
O
=
:
=
k
5
.
15
x
x
ln
kHz
500
pF
490
¸
¹
·
-
1
¨
©
§
V
24
.
1
V
14
-
- 1
¸
¹
·
¨
©
§
x
V
24
90
.
0
V
14
R
OFF
SNVS602J – MARCH 2009 – REVISED MAY 2013
η
= 0.90
1. NOMINAL SWITCHING FREQUENCY
Assume C
OFF
= 470pF and
η
= 0.90. Solve for R
OFF
:
(75)
The closest 1% tolerance resistor is 15.4 k
Ω
therefore the actual t
OFF
and target f
SW
are:
(76)
(77)
The chosen components from step 1 are:
(78)
2. INDUCTOR RIPPLE CURRENT
Solve for L1:
(79)
The closest standard inductor value is 22 µH therefore the actual
Δ
i
L-PP
is:
(80)
The chosen component from step 2 is:
(81)
3. AVERAGE LED CURRENT
Determine I
L-MAX
:
(82)
Assume V
ADJ
= 1.24V and solve for R
SNS
:
28
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