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C2 = C3 = C16 = C18 = 4.7 PF
x
A1
=
I
LED
I
RMSIN-
=
1- 0.677
677.0
1.45A
x
1- D
MAX
D
MAX
=
C
IN
==
kHz504mV100 x
467.0A1 x
F27.9 P=
f
'v
SWPPIN-
x
DI
LED
x
C12 = 0.22 PF
C10 = 1 PF
R20 = 10:
C12 = =
= 0.28 PF
1
10:xZ
P3
1
rad
sec
10:x 360k
1010(max
1P
xZ
=
x
=
,
1Z1P
ZZ
sec
rad
k36
sec
rad
k36010
=
x
=
3P
Z
3P
Z
)
C10 = =
= 0.30 PF
1
Z
P2
x 5e
6
:
1
0.675 x 5e
6
:
rad
sec
= =
sec
rad
675.0=
sec
rad
k19
56305 x56305 x
1Z
Z
2P
=Z
),min(
1Z1P
ZZ
T5
0U
x
T
U0
=
= = 5630
D' x 620V
(1 + D) x I
LED
x R6
0.533 x 620V
1.467 x 1A x 0.04:
sec
rad
k36===
533.095.1
2
x:
H33467.0 Px
Dr
2
D
c
x
L1Dx
1Z
Z
sec
rad
k19===
1.467
F40
1.95:
Px
C
O
r
D
x
D1+
1P
Z
R
SLP
= 10 k:
www.ti.com Design Procedure
(28)
8.10 Loop Compensation
ω
P1
is approximated:
(29)
ω
Z1
is approximated:
(30)
T
U0
is approximated:
(31)
To ensure stability, calculate ω
P2
:
(32)
Solve for C
CMP
:
(33)
To attenuate switching noise, calculate ω
P3
:
(34)
Assume R
FS
= 10Ω and solve for C
FS
:
(35)
The chosen components from step 9 are:
(36)
8.11 Input Capacitance
Solve for the minimum C
IN
:
(37)
To minimize power supply interaction a 200% larger capacitance of approximately 20 µF is used, therefore
the actual Δv
IN-PP
is much lower. Since high voltage ceramic capacitor selection is limited, four 4.7 µF X7R
capacitors are chosen.
Determine minimum allowable RMS current rating:
(38)
The chosen components from step 10 are:
(39)
11
SNVA397A–August 2009–Revised May 2013 AN-1967 LM3424 Buck-Boost Evaluation Board
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