MAX8543 MAXIM [Maxim Integrated Products], MAX8543 Datasheet - Page 24

• Manufacturer Part Number: MAX8543
• Description: Step-Down Controllers with Prebias Startup, Lossless Sensing, Synchronization, and OVP
• Manufacturer: MAXIM [Maxim Integrated Products]

Step-Down Controllers with Prebias Startup,
Lossless Sensing, Synchronization, and OVP
The crossover frequency, f
than the power-modulator pole f
be less than or equal to 1/5th the switching frequency.
Select a value for f
At the crossover frequency, the total loop gain must
equal 1, and is expressed as:
For the case where f
then R
where g
The error-amplifier compensation zero formed by R
and C
is calculated by:
If f
from COMP to GND. The value of C
follows:
As the load current decreases, the modulator pole also
decreases; however, the modulator gain increases
accordingly and the crossover frequency remains
the same.
24
zMOD
______________________________________________________________________________________
C
C
mEA
should be set at the modulator pole f
can be calculated as:
is less than 5 x f
G
G
R
= 110µS.
C
MOD fc
EA fc
C
C
( )
=
C
=
G
f
f
zEA
pEA
( )
F
C
g
R
EA fc
f
(
pMOD
R
mEA
LOAD
=
×
in the range:
LOAD
( )
zMOD
G
=
2π
=
=
MOD fc
G
2π
2π
×
×
=
C
MOD dc
<<
×
V
, add a second capacitor C
+
R
g
×
×
f
V
is greater than f
FB
( )
S
C
C
(
mEA
OUT
C
f
C
f
C
S
, should be much higher
1
× ×
(
1
×
1
C
F
×
L C
×
≤
×
f
zMOD
×
×
G
L
)
PMOD
×
f
V
5
)
R
S
R
×
MOD fc
V
)
OUT
R
C
FB
C
×
f
OUT
C
pMOD
R
F
f
. Also, f
( )
C
C
is calculated as
= 1
C
PMOD
:
C
should
. C
C
C
F
For the case where f
The power-modulator gain at f
The error-amplifier gain at f
R
where g
C
C
Below is a numerical example to calculate R
values of the typical operating circuit of Figure 1
(MAX8544):
A
R
g
V
I
R
C
ESR = 5mΩ
OUT(MAX)
mc
OUT
C
VCS
DC
LOAD
G
C
F
OUT
MOD dc
is calculated from:
is calculated as:
is calculated from:
= 1 / (A
= 2.5mΩ
= 11 (for ILIM1 = GND)
= 2.5V
= 360µF
(
= V
R
mEA
C
)
= 15A
OUT
=
=
=
G
VCS
G
g
36 36
= 110µS.
C
V
mc
MOD fc
EA fc
V
OUT
C
.
FB
/ I
C
x R
×
( )
=
OUT(MAX)
( )
F
R
0 167
0 167
R
(
R
.
.
×
R
LOAD
DC
LOAD
=
LOAD
=
LOAD
g
zMOD
=
2π
g
) = 1 / (11 x 0.0025) = 36.7S
mEA
×
+
G
mEA
+
×
(
(
MOD dc
×
600 10
600 10
×
(
+
R
C
f
f
S
S
f
= 2.5 / 15 = 0.167Ω
is less than f
×
S
C
(
×
is:
f
×
×
G
S
×
×
1
× ×
(
C
×
R
L
L
MOD fc
L C
×
)
f
is:
C
zMOD
L
)
f
3
3
C
)
×
×
)
)
)
×
×
( )
×
f
f
OUT
f
zMOD
pMOD
(
(
zMOD
R
0 8 10
0 8 10
f
.
.
C
C
×
×
×
f
C
zMOD
:
−
−
6
6
C
)
)
=
and C
4 50
.
C