Texas Instruments LM3429 Evaluation Boards LM3429BKBSTEVAL/NOPB LM3429BKBSTEVAL/NOPB Datenbogen
Produktcode
LM3429BKBSTEVAL/NOPB
H
33
1
L
P
=
A
88
.
1
I
12
1
1
I
RMS
L
RMS
L
=
+
x
=
-
-
I
I
LED
RMS
L
x
=
-
12
1
1
2
x
+
¸
¸
¹
¹
·
¨
¨
©
©
§
D
i
PP
L
c
x
'
-
I
LED
D
c
533
.
0
mA
485
2
x
x
¸¸
¹
·
¨¨
©
§
A
1
533
.
0
A
1
PP
-
=
=
L
D
V
IN
x
f
1
L
SW
x
kHz
00
7
H
33
x
P
467
.
0
V
4
2
x
mA
485
=
'
i
=
=
D
V
IN
x
f
SW
x
467
.
0
V
4
2
x
P
H
32
=
1
L
PP
-
'
i
L
kHz
00
7
500 mA x
:
k
4
.
2
1
R
CSH
=
:
k
1
R
R
HSN
=
=
HSP
0.1
R
NS
S
=
:
I
LED
=
=
k
0
.
1
1.24V
:
x
A
0
.
1
=
k
4
.
12
1
.
0
:
x
:
R
R
CSH
SNS
x
R
1.24V
HSP
x
=
1.24V
1.24V
=
R
HSP
:
=
k
0
.
1
:
x
:
x
0.1
k
12.4
A
1
x
x
R
R
I
SNS
CSH
LED
:
=
=
=
1
.
0
1A
R
SNS
I
LED
mV
100
V
SNS
C
T
= 1 nF
R
T
= 35.7 k
:
SNVS616G – APRIL 2009 – REVISED MAY 2013
The chosen components from step 2 are:
(96)
3. AVERAGE LED CURRENT
Solve for R
SNS
:
(97)
Assume R
CSH
= 12.4 k
Ω
and solve for R
HSP
:
(98)
The closest standard resistor for R
SNS
is actually 0.1
Ω
and for R
HSP
is actually 1 k
Ω
therefore I
LED
is:
(99)
The chosen components from step 3 are:
(100)
4. INDUCTOR RIPPLE CURRENT
Solve for L1:
(101)
The closest standard inductor is 33 µH therefore the actual
Δ
i
L-PP
is:
(102)
Determine minimum allowable RMS current rating:
(103)
The chosen component from step 4 is:
(104)
5. OUTPUT CAPACITANCE
Solve for C
O
:
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33
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