User manual
=
= 0.13 PF
C8 =
1
rad
sec
Z
P2
x 5e
6
:
1
1.596 x 5e
6
:
Z
Z1
=
5 x 4510
= 1.596
Z
P2
=
5 x T
U0
min(Z
P1
, Z
Z1
)
rad
sec
=
5 x 4510
36k
rad
sec
0.533 x 620V
=
1.467 x 1A x 0.05:
= 4510
T
U0
=
(1 + D) x I
LED
x R6
D' x 620V
1.95: x 0.533
2
=
0.467 x 33 PH
= 36k
Z
Z1
=
D x L1
r
D
x D'
2
rad
sec
1.467
=
1.95: x 6.8 PF
= 110k
Z
P1
=
r
D
x C
O
1 + D
rad
sec
R6 = 0.05:
=
0.05:
= 4.9A
I
LIM
=
R6
245 mV 245 mV
=
5A
= 0.049:
R6 =
I
LIM
245 mV 245 mV
C6 = C17 = C19 = 2.2 PF
1- 0.677
= 1.45A
1 - D
MAX
D
MAX
I
CO-RMS
= I
LED
x
= 1A x
0.677
1A x 0.467
1.95: x 6.6 PF x 700 kHz
= 52 mA
'i
LED-PP
=
r
D
x C
O
x f
SW
I
LED
x D
'i
LED-PP
=
Design Procedure
www.ti.com
(19)
Determine minimum allowable RMS current rating:
(20)
The chosen components from step 5 are:
(21)
6.7 Peak Current Limit
Solve for R6:
(22)
The closest standard resistor is 0.05 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
:
8
AN-1985 LM3429 Buck-Boost Evaluation Board SNVA403C–July 2009–Revised May 2013
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