User manual

1
=
F097.0
1
P==C12
10:
x
sec
rad
M04.1
10:
3P
Zx
1010max(
1P
xZ
=
x
=
,
1Z1P
ZZ
sec
rad
k104
=
sec
rad
M04.110
=
x
3P
Z
3P
Z
)
F11.0
1
1
C8 P===
76.1
e5
sec
rad
6
x :
e5
6
2P
xZ
:
= =
sec
rad
76.1=
sec
rad
k52
59005 x59005 x
1Z
Z
2P
=Z
),min(
1Z1P
ZZ
T5
0U
x
=T
0U
= 5900=
04.0A1 :
x
V310762.0
x
RI
LIMLED
x
V310D
x
c
sec
rad
k52===
762.0925.2
2
x:
H33P
Dr
2
D
c
x
L1
1Z
Z
sec
rad
k104===
2
F6.6
2.925:
Px
C
O
r
D
x
2
1P
Z
0.04:R6 =
:04.0
===
6.1A
mV245mV245
I
LIM
R6
:=== 041.0
6A
R6
mV245mV245
I
LIM
F2.2C6 = C17 = C19 P=
x
A1
=
I
LED
I
RMSCO-
=
1- 0.683
683.0
1.47A
x
1- D
MAX
D
MAX
=
Design Procedure www.ti.com
(20)
The chosen components from step 5 are:
(21)
6.7 Peak Current Limit
Solve for R6:
(22)
The closest standard resistor is 0.04 Ω therefore I
LIM
is:
(23)
The chosen component from step 6 is:
(24)
6.8 Loop Compensation
ω
P1
is approximated:
(25)
ω
Z1
is approximated:
(26)
T
U0
is approximated:
(27)
To ensure stability, calculate ω
P2
:
(28)
Solve for C8:
(29)
Since PWM dimming can be evaluated with this board, a much larger compensation capacitor C8 = 1.0 µF
is chosen.
To attenuate switching noise, calculate ω
P3
:
(30)
Assume R20 = 10Ω and solve for C12:
(31)
The chosen components from step 7 are:
8
AN-1986 LM3429 Boost Evaluation Board SNVA404B–July 2009–Revised May 2013
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