Texas Instruments LM3423 Evaluation Board LM3423BBLSCSEV/NOPB LM3423BBLSCSEV/NOPB Datenbogen
Produktcode
LM3423BBLSCSEV/NOPB
A
89
.
1
12
1
1
I
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
678
2
x
¸¸
¹
·
¨¨
©
§
A
1
x
533
.
0
A
1
PP
-
=
=
L
D
V
IN
x
f
1
L
SW
x
kHz
01
5
H
33
x
P
467
.
0
V
4
2
x
mA
678
=
'
i
=
=
D
V
IN
x
f
SW
x
467
.
0
V
4
2
x
P
H
32
=
1
L
PP
-
'
i
L
kHz
01
5
700 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
= 49.9 k
:
25
25
=
49.9 k
:
x 1 nF
f
SW
=
R
T
x C
T
= 501 kHz
25
25
=
500 kHz x 1 nF
= 50 k
:
R
T
=
f
SW
x C
T
SNVS574E – JULY 2008 – REVISED MAY 2013
2. SWITCHING FREQUENCY
Assume C
T
= 1 nF and solve for R
T
:
(101)
The closest standard resistor is 49.9 k
Ω
therefore f
SW
is:
(102)
The chosen component from step 2 is:
(103)
3. AVERAGE LED CURRENT
Solve for R
SNS
:
(104)
Assume R
CSH
= 12.4 k
Ω
and solve for R
HSP
:
(105)
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:
(106)
The chosen components from step 3 are:
(107)
4. INDUCTOR RIPPLE CURRENT
Solve for L1:
(108)
The closest standard inductor is 33 µH therefore
Δ
i
L-PP
is:
(109)
Determine minimum allowable RMS current rating:
(110)
The chosen component from step 4 is:
Copyright © 2008–2013, Texas Instruments Incorporated
41
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