User manual
700 mA x 0.467
1.95: x 50 mA x 700 kHz
= 4.79 PF
C
O
=
C
O
=
r
D
x 'i
LED-PP
x f
SW
I
LED
x D
L1 = 47 PH
340 mA x 0.533
700 mA
I
L-RMS
=
I
LED
D'
I
L-RMS
=
700 mA
0.533
I
L-RMS
= 1.32A
I
LED
'I
L-PP
x D'
2
¸
¸
¹
·
¨
¨
©
§
¸
¸
¹
·
¨
¨
©
§
1 +
12
1
x
1 +
12
1
x
2
x
x
24V x 0.467
=
47 PH x 700 kHz
= 340 mA
'i
L-PP
=
L1 x f
SW
V
IN
x D
24V x 0.467
=
350 mA x 700 kHz
= 46 PH
L1 =
'i
L-PP
x f
SW
V
IN
x D
R8 = R7 = 1.4 k:
R1 = 12.4 k:
R9 = 0.2:
1.24V x 1.4 k:
=
= 700 mA
I
LED
=
1.24V x R8
R9 x R1
0.2: x 12.4 k:
30107716
700 mA x 12.4 k: x 0.2:
=
1.24V
= 1.4 k:
R8 =
I
LED
x R1 x R9
1.24V
30107715
150 mV
=
700 mA
= 0.214:
R9 =
V
SNS
I
LED
C7 = 1 nF
R10 = 35.7 k:
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Design Procedure
The chosen components from Section 6.3 are:
(9)
6.4 Average LED Current
Solve for R9:
(10)
Assume R1 = 12.4 kΩ and solve for R8:
(11)
The closest standard resistor for R9 is 0.2Ω and the closest for R8 (and R7) is actually 1.4 kΩ therefore
I
LED
is:
(12)
The chosen components from Section 6.4 are:
(13)
6.5 Inductor Ripple Current
Solve for L1:
(14)
The closest standard inductor is 47 µH therefore the actual Δi
L-PP
is:
(15)
Determine minimum allowable RMS current rating:
(16)
The chosen component from Section 6.5 is:
(17)
6.6 Output Capacitance
Solve for C
O
:
(18)
7
SNVA415C–June 2010–Revised May 2013 AN-2010 LM3423 Buck-Boost 2 Layer Evaluation Board
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