Datasheet

F27.0C
FS
P=
F0.33C
CMP
P=
:10R
FS
=
1
=
F28.0
1
P==C
FS
10:
x
sec
rad
k360
10:
3P
Zx
1010(max
1Z
xZ
=
x
=
,
1Z1P
ZZ
=
3P
Z
3P
Z
)
sec
rad
k36
sec
rad
k36010
=
x
F30.0
1
1
C
CMP
P===
675.0
e5
sec
rad
6
x :
e5
6
2P
xZ
:
= =
sec
rad
675.0=
sec
rad
k19
56305 x56305 x
1P
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
k36===
533.095.1
2
x:
H33467.0 Px
Dr
2
D
c
x
L1D x
1Z
Z
sec
rad
k19===
1.467
F40
1.95:
Px
C
O
r
D
x
D1+
1P
Z
:
16.5R
SLP
k
=
=
R
SLP
x 1Le5.1
13
xx RRV
SNSTO
= =
:k5.16
Px H33e5.1
13
x:x k3.14V21 :1.0
R
SLP
LM3424
SNVS603B –AUGUST 2009–REVISED OCTOBER 2009 www.ti.com
8. SLOPE COMPENSATION
Solve for R
SLP
:
(127)
The chosen component from step 8 is:
(128)
9. LOOP COMPENSATION
ω
P1
is approximated:
(129)
ω
Z1
is approximated:
(130)
T
U0
is approximated:
(131)
To ensure stability, calculate ω
P2
:
(132)
Solve for C
CMP
:
(133)
To attenuate switching noise, calculate ω
P3
:
(134)
Assume R
FS
= 10Ω and solve for C
FS
:
(135)
The chosen components from step 9 are:
(136)
42 Submit Documentation Feedback Copyright © 2009, Texas Instruments Incorporated
Product Folder Links: LM3424