MAX8513 MAXIM [Maxim Integrated Products], MAX8513 Datasheet - Page 24

• Manufacturer Part Number: MAX8513
• Description: Wide-Input, High-Frequency, Triple-Output Supplies with Voltage Monitor and Power-On Reset
• Manufacturer: MAXIM [Maxim Integrated Products]

Wide-Input, High-Frequency, Triple-Output Supplies
with Voltage Monitor and Power-On Reset
This gain is also set by the ratio of R3/R1 where R1 is
calculated in the OUT1 Voltage Setting section. Thus:
Due to the underdamped (Q > 1) nature of the output
LC double pole, the error-amplifier zero frequencies
must be set less than the LC double-pole frequency to
provide adequate phase boost. Set the error-amplifier
first zero, f
the second zero, f
Hence:
Set the error-amplifier f
switching frequency, if f
then set f
The gain of the error amplifier between f
set by the ratio of R3/R
where R
is equal to:
Therefore:
C11 can then be calculated as:
and C12 as:
24
______________________________________________________________________________________
I
P2
is the parallel combination of R1 and R4 and
Z1
C
at 1/2 f
, at 1/4th the LC double-pole frequency and
R
12
I
R
C
R
C
3
=
I
5
R
=
11
3
Z2
S
=
=
f
P
(
R
R
2
, at the LC double-pole frequency.
and f
=
2
=
G
I
4
R
π
π
I
EA fZ fZ
×
3
=
f
P2
and is equal to:
2
ZESR
×
C
=
×
π
×
(
G
P3
R
R
R
C
×
R
at f
R
EA fZ fZ
×
R
1
1
1
1
f
5
1
3
PMOD
at f
-
1
G
+
< 1/2 f
×
×
(
R
2
×
C
ZESR
×
MOD fc
-
×
1
2
4
ZESR
R
)
R
1
5
f
R
Z
R
-
R
4
f
f
4
×
PMOD
PMOD
I
2
I
3
( )
f
2
, and f
P
f
S
)
P
.
×
2
. If f
2
f
and
P
3
-1
ZESR
P2
P3
)
and f
to 1/2 the
> 1/2 f
P3
S
is
,
Below is a numerical example to calculate the error-
amplifier compensation values used in the Typical
Applications Circuit of Figure 5:
V
V
V
V
L1A = 1.8µH
C4 = 47µF/ 6.3V ceramic, with R
f
The LC double-pole frequency is calculated as:
Pick R2 = 8.06kΩ.
The modulator gain at DC is:
Pick f
S
IN
RAMP
OUT1
FB1
= 1.4MHz
= 12V (nomimal input voltage)
= 1.25V
C
G
= 3.3V
= 1V
= 100kHz.
EA fZ fZ
G
MOD fc
R
(
f
f
2
2
PMOD
1
ZESR
π
π
1
=
−
×
( )
G
R
1 8 10
8 06
0 008 47 10
.
MOD DC
3
.
2
.
)
=
×
=
=
=
=
=
=
k
2
Ω
1
(
12
13 3
2
R
×
100
π
f G
-
C MOD fC
1
π
1
6
×
×
.
)
f
×
×
PMOD
×
R
kHz
L A
k



17 4
×
47 10
ESR
=
1
Ω
1 25
G
1



3 3
1
.
17 4
.
100
EA fZ fZ
×
.
V
×
( )
-
×
kHz
×
6
RAMP
×
V
.
V
V
(
0 479
C
IN
kHz
C
kHz
0 363
ESR
-
-
.
6
=
1
4
.
1
4
−

 =

423
=

 =

=
=
= 0.008Ω
2
= 12
2
17 3
)
13 3
=
kHz
=
.
.
6 37
0 363
kHz
0 479
.
k
.
.
Ω
k
Ω