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I
D-MAX
= I
LED
= 1A
V
RD-MAX
= V
IN-MAX
+ V
O
= 70V + 21V = 91V
Q1 o 40A, 100V, DPAK
P
T
= I
T-RMS
2
x R
DSON
= 1.28A
2
x 50 m: = 82 mW
x x
1A
D =
0.533
I
T-RMS
=
I
LED
D'
0.467 = 1.28A
0.677
1 - 0.677
x 1A = 2.1A
I
T-MAX
=
V
T-MAX
= V
IN-MAX
+ V
O
= 70V + 21V = 91V
C2 = C3 = C16 = C18 = 4.7 PF
1- 0.677
= 1.45A
1 - D
MAX
D
MAX
I
IN-RMS
= I
LED
x
= 1A x
0.677
1A x 0.467
=
100 mV x 700 kHz
= 6.66 PF
C
IN
=
'V
IN-PP
x f
SW
I
LED
x D
C8 = 1.0 PF
R20 = 10:
C12 = 0.1 PF
=
= 0.091 PF
C12 =
1
rad
sec
1
10: x 1.1M
10: x Z
P3
rad
sec
Z
P3
= 110k x 10 = 1.1M
Z
P3
= max (Z
P1
, Z
Z1
) x 10 = Z
P1
x 10
rad
sec
www.ti.com Design Procedure
(30)
Assume R20 = 10Ω and solve for C12:
(31)
The chosen components from step 7 are:
(32)
6.9 Input Capacitance
Solve for the minimum C
IN
:
(33)
To minimize power supply interaction a 3x 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:
(34)
The chosen components from step 8 are:
(35)
6.10 NFET
Determine minimum Q1 voltage rating and current rating:
(36)
(37)
A 100V NFET is chosen with a current rating of 40A due to the low R
DS-ON
= 50 mΩ. Determine I
T-RMS
and
P
T
:
(38)
(39)
The chosen component from step 9 is:
(40)
6.11 Diode
Determine minimum D1 voltage rating and current rating:
(41)
(42)
A 100V diode is chosen with a current rating of 12A and V
D
= 600 mV. Determine P
D
:
9
SNVA403C–July 2009–Revised May 2013 AN-1985 LM3429 Buck-Boost Evaluation Board
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