Datasheet

= =
sec
rad
173.1=
sec
rad
k37
56305 x56305 x
1Z
Z
2P
=Z
),min(
1Z1P
ZZ
T5
0U
x
=T
0U
= 5630=
04.0A1467.1 :
xx
V620533.0
x
( )
D1
+
RI
LIMLED
xx
V620D
x
c
sec
rad
k37===
533.095.1
2
x:
H33467.0 Px
Dr
2
D
c
x
L1Dx
1Z
Z
=
sec
rad
= 110k
1 + D
Z
P1
=
1.467
1.95: x 6.8 PF
r
D
x C
O
0.04:R
LIM
=
:04.0
===
6.13A
mV245mV245
I
LIM
R
LIM
:=== 041.0
6A
R
LIM
mV245mV245
I
LIM
F8.6C
O
P=
x
A1
=
I
LED
I
RMSCO-
=
1- 0.677
677.0
1.45A
x
1- D
MAX
D
MAX
=
DI
LED
x
=
'i
PP-LED
SW
fxr
D
x
C
O
0
= =
kHz00795.1 xx:
5 mA
467.0A1 x
F8.6 P
'i
PP-LED
f
'i
r
SWPP-LEDD
xx
DI
LED
x
C
O
=
0
= F84.6 P=
kHz007mA595.1 xx:
467.0A1 x
C
O
LM3429, LM3429-Q1
SNVS616G –APRIL 2009–REVISED MAY 2013 www.ti.com
(105)
The closest standard capacitor is 6.8 µF therefore the actual Δi
LED-PP
is:
(106)
Determine minimum allowable RMS current rating:
(107)
The chosen components from step 5 are:
(108)
6. PEAK CURRENT LIMIT
Solve for R
LIM
:
(109)
The closest standard resistor is 0.04 Ω therefore I
LIM
is:
(110)
The chosen component from step 6 is:
(111)
7. LOOP COMPENSATION
ω
P1
is approximated:
(112)
ω
Z1
is approximated:
(113)
T
U0
is approximated:
(114)
To ensure stability, calculate ω
P2
:
(115)
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