Datasheet

sec
rad
k36===
533.095.1
2
x:
H33467.0 Px
Dr
2
D
c
x
L1Dx
1Z
Z
sec
rad
k19===
1.467
F40
1.95:
Px
C
O
r
D
x
D1+
1P
Z
0.04:R
LIM
=
:04.0
===
6.13A
mV245mV245
I
LIM
R
LIM
:=== 041.0
6A
R
LIM
mV245mV245
I
LIM
C
O
= 4 x 10 PF
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
2
= =
kHz01595.1 xx:
1 mA
467.0A1 x
F40 P
'i
PP-LED
f
'i
r
SWPP-LEDD
xx
DI
LED
x
C
O
=
2
= F39.8 P=
kHz015mA195.1 xx:
467.0A1 x
C
O
H331L P
=
LM3421, LM3421-Q1
LM3423, LM3423-Q1
SNVS574E –JULY 2008–REVISED MAY 2013 www.ti.com
(111)
5. OUTPUT CAPACITANCE
Solve for C
O
:
(112)
The closest capacitance totals 40 µF therefore Δi
LED-PP
is:
(113)
Determine minimum allowable RMS current rating:
(114)
The chosen components from step 5 are:
(115)
6. PEAK CURRENT LIMIT
Solve for R
LIM
:
(116)
The closest standard resistor is 0.04 Ω therefore I
LIM
is:
(117)
The chosen component from step 6 is:
(118)
7. LOOP COMPENSATION
ω
P1
is approximated:
(119)
ω
Z1
is approximated:
(120)
T
U0
is approximated:
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