User manual

rad
sec
Z
P3
= 25k x 10 = 250k
Z
P3
= max (Z
P1
, Z
Z1
) x 10 = Z
Z1
x 10
rad
sec
=
= 0.28 PF
C8 =
1
rad
sec
Z
P2
x 5e
6
:
1
0.709 x 5e
6
:
Z
P1
=
5 x 5360
= 0.709
Z
P2
=
5 x T
U0
min(Z
P1
, Z
Z1
)
rad
sec
=
5 x 5360
19k
rad
sec
0.533 x 620V
=
1.467 x 700 mA x 0.06:
= 5360
T
U0
=
(1 + D) x I
LED
x R6
D' x 620V
1.95: x 0.533
2
=
0.467 x 47 PH
= 25k
Z
Z1
=
D x L1
r
D
x D'
2
rad
sec
1.467
=
1.95: x 40 PF
= 19k
Z
P1
=
r
D
x C
O
1 + D
rad
sec
R6 = 0.06:
=
0.06:
= 4.1A
I
LIM
=
R6
245 mV 245 mV
=
4A
= 0.061:
R6 =
I
LIM
245 mV 245 mV
C6 = 4 x 10 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
1.95: x 40 PF x 700 kHz
= 6 mA
'i
LED-PP
=
r
D
x C
O
x f
SW
I
LED
x D
'i
LED-PP
=
Design Procedure www.ti.com
A total value of 40 µF (using 4 10 µF X7R ceramic capacitors) is chosen therefore the actual Δi
LED-PP
is:
(19)
Determine minimum allowable RMS current rating:
(20)
The chosen components from Section 6.6 are:
(21)
6.7 Peak Current lmit
Solve for R6:
(22)
The closest standard resistor is 0.06 Ω therefore I
LIM
is:
(23)
The chosen component from Section 6.7 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)
To attenuate switching noise, calculate ω
P3
:
(30)
8
AN-2010 LM3423 Buck-Boost 2 Layer Evaluation Board SNVA415C–June 2010–Revised May 2013
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