User manual
1A x 0.467
1.95: x 50 mA x 700 kHz
= 6.84 PF
C
O
=
C
O
=
r
D
x 'i
LED-PP
x f
SW
I
LED
x D
L1 = 33 PH
485 mA x 0.533
1A
I
L-RMS
=
I
LED
D'
I
L-RMS
=
1A
0.533
I
L-RMS
= 1.88A
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
=
33 PH x 700 kHz
= 485 mA
'i
L-PP
=
L1 x f
SW
V
IN
x D
24V x 0.467
=
500 mA x 700 kHz
= 32 PH
L1 =
'i
L-PP
x f
SW
V
IN
x D
R8 = R7 = 1 k:
R1 = 12.4 k:
R9 = 0.1:
1.24V x 1.0 k:
=
= 1.0A
I
LED
=
1.24V x R8
R9 x R1
0.1: x 12.4 k:
1A x 12.4 k: x 0.1:
=
1.24V
= 1.0 k:
R8 =
I
LED
x R1 x R9
1.24V
30100815
100 mV
=
1A
= 0.1:
R9 =
V
SNS
I
LED
C7 = 1 nF
R10 = 35.7 k:
www.ti.com
Design Procedure
(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 0.1Ω and the closest for R8 (and R7) is actually 1 kΩ therefore I
LED
is:
(12)
The chosen components from step 3 are:
(13)
6.5 Inductor Ripple Current
Solve for L1:
(14)
The closest standard inductor is 33 µH therefore the actual Δi
L-PP
is:
(15)
Determine minimum allowable RMS current rating:
(16)
The chosen component from step 4 is:
(17)
6.6 Output Capacitance
Solve for C
O
:
(18)
A total value of 6.6 µF (using 3 2.2 µF X7R ceramic capacitors) is chosen therefore the actual Δi
LED-PP
is:
7
SNVA403C–July 2009–Revised May 2013 AN-1985 LM3429 Buck-Boost Evaluation Board
Submit Documentation Feedback
Copyright © 2009–2013, Texas Instruments Incorporated