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P
D
= I
D
x V
FD
= 700 mA x 600 mV = 420 mW
I
D-MAX
= I
LED
= 700 mA
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
= 897 mA
2
x 50 m: = 40 mW
x
x
700 mA
D =
0.533
I
T-RMS
=
I
LED
D'
0.467 = 897 mA
0.677
1 - 0.677
x 700 mA = 1.46A
I
T-MAX
=
V
T-MAX
= V
IN-MAX
+ V
O
= 70V + 21V = 91V
C3 = 68 PF
1- 0.677
= 1.01A
1 - D
MAX
D
MAX
I
CO-RMS
= I
LED
x
= 700 mA x
0.677
700 mA x 0.467
=
100 mV x 700 kHz
= 4.67 PF
C
IN
=
'V
IN-PP
x f
SW
I
LED
x D
C8 = 1.0 PF
R20 = 10:
C12 = 0.1 PF
=
= 0.4 PF
C12 =
1
rad
sec
1
10: x 250k
10: x Z
P3
www.ti.com Design Procedure
Assume R20 = 10Ω and solve for C12:
(31)
Since PWM dimming can be evaluated with this board, a much larger compensation capacitor C8 = 1.0 µF
is chosen and a smaller high frequency capacitor C12 = 0.1 µF is chosen.
The chosen components from Section 6.8 are:
(32)
6.9 Input Capacitance
Solve for the minimum C
IN
:
(33)
To minimize power supply interaction a much larger capacitance of 68 µF is used, therefore the actual
Δv
IN-PP
is much lower.
Determine minimum allowable RMS current rating:
(34)
The chosen components from Section 6.9 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 Section 6.10 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
:
(43)
9
SNVA415C–June 2010–Revised May 2013 AN-2010 LM3423 Buck-Boost 2 Layer Evaluation Board
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