max5073atit Maxim Integrated Products, Inc., max5073atit Datasheet - Page 20

• Manufacturer Part Number: max5073atit
• Description: Max5073 2.2mhz, Dual-output Buck Or Boost Converter With Internal Power Mosfets
• Manufacturer: Maxim Integrated Products, Inc.

2.2MHz, Dual-Output Buck or Boost Converter
with Internal Power MOSFETs
where V
to 1V.
The transconductance error amplifier gain is:
The total loop gain at f
or
4) Place a zero at or below the LC double pole:
5) Place a high-frequency pole at f
Procedure 2 (see Figure 7)
If the output capacitor used is a low-ESR ceramic type,
the ESR frequency is usually far away from the targeted
unity crossover frequency (f
pensation is recommended. Type III compensation pro-
vides two-pole zero pairs. The locations of the zero and
poles should be such that the phase margin peaks at f
The
phase margin at f
the two zeros at or below the double pole to avoid the
conditional stability issue.
1) Select a crossover frequency:
2) Calculate the LC double-pole frequency, f
3) Place a zero
20
G
______________________________________________________________________________________
M
f
f
Z
C
R
=
OSC
F
=
f
V
=
f
Z
f
OSC
C
V
P
IN
V
is a peak-to-peak ramp amplitude equal
=
OSC
= 5
f
LC
2
π
×
C
C
×
(
is a good number to get about 60°
=
ESR
ESR
0 8
F
G
. However, it is important to place
G
.
R
2π
E A
=
M
1
C
F
/
×
2π
f
should be equal to 1
×
+
C
×
+
×
V
=
G
(
2
IN
C
≤
2
C
×
g
E A
π
L
F
π
). In this case, Type III com-
f
m
R
ESR
×
/
OUT
SW
×
20
×
1
F
at
g
×
1
f
C
f
m
=
×
C
R
0 75
1
×
×
F
f
P
×
×
LC
.
L
C
= 0.5 x f
ESR
L
OUT
OUT
OUT
×
)
V
f
)
LC
OUT
×
LC
SW
V
:
0 8
OUT
.
.
C
.
where:
and R
4) Calculate C
5) Place a pole
6) Place a second zero, f
7) Place a second pole
Figure 7. Type III Compensation Network
whichever is lower.
the switching frequency.
F
≥ 10kΩ.
C
I
=
C
C
R
I
2π
I
CF
I
C
for a target unity crossover frequency, f
R
R
×
F
1
(
I
R1
R2
f
=
P
f
=
=
=
C
V
1
(
OUT
2
2π
2
2
=
×
π
π
π
2
×
L
×
×
×
π
OUT
V
0 5
(
REF
f
f
f
0 75
V
ZERO ESR
Z
P
.
×
1
IN
.
2
2
R
Z2
×
1
R
×
×
I
=
F
×
-
+
C
1
f
1
g
SW
×
, at 0.2 x f
CF
×
C
,
C
M
R
2
C
f
LC
C
I
OUT
F
π
F
×
I
−
C
×
)
F
R
×
R
×
R
at f
F
×
I
R
F
1
C
×
F
V
ZERO,ESR
I
×
COMP
OSC
C
C
C
F
CF
)
or at f
−
)
1
at 1/2
.
LC
C
:
,