MAX5089 Maxim Integrated Products, MAX5089 Datasheet - Page 17
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MAX5089
Manufacturer Part Number
MAX5089
Description
2.2MHz - 2A Buck Converters with an Integrated High-Side Switch
Manufacturer
Maxim Integrated Products
Datasheet
1.MAX5089.pdf
(24 pages)
Available stocks
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Part Number
Manufacturer
Quantity
Price
Company:
Part Number:
MAX5089ATE+
Manufacturer:
Maxim Integrated Products
Quantity:
135
Calculate the modulator gain (G
quency.
where V
The transconductance error amplifier gain at f
The total loop gain at f
or
Place a zero at or below the LC double pole:
Place a high-frequency pole at f
C
Procedure 2 (see Figure 4)
When using a low-ESR ceramic-type capacitor as the
output capacitor, the ESR frequency is much higher
than the targeted unity-gain crossover frequency (f
In this case, Type III compensation is recommended.
Type III compensation provides a low-frequency pole
(≈DC) and two pole-zero pairs. The locations of the
zero and poles should be such that the phase margin
peaks at f
The
ly 60° of phase margin at f
place the two zeros at or below the double pole to
avoid conditional stability.
CF
is:
f
f
C
Z
G
OSC
M
=
R
C
F
f
f
.
C
P
=
is the 1V
=
=
V OSC
5
V IN
V
C
is a good number to get approximate-
OSC
C
______________________________________________________________________________________
CF
F
G
G
P-P
×
E/A
V
=
C
M
(ESR
=
FB
ESR 2
ramp amplitude and V
should be equal to 1:
= G
2
= g
π
×
π
C
2.2MHz, 2A Buck Converters with an
×
+
V
+
. However, it is important to
×
E/A
IN
m
R
( π
R
ESR
2
F
1
F
1
x R
×
π
= 1
P
×
M
×
g
×
×
= 0.5 x f
f
) at the crossover fre-
m
SW
f c L)
F
f
LC
f
C
×
×
×
ESR
L)V
×
SW
V OUT
OUT
V FB
. Therefore
FB
C
Integrated High-Side Switch
is:
= 0.6V.
C
).
First, select the crossover frequency so that:
Calculate the LC double-pole frequency, f
where:
with R
Calculate C
Place a second zero, f
er is lower.
Place a pole f
Place a second pole (f
at 1/2 the switching frequency.
F
Place a zero f
≥ 10kΩ.
C
C
A
CF
A
for a target unity crossover frequency, f
C
(
R
F
=
=
P
1
R
f
=
1
LC
A
=
(2
2
=
Z
π
2
=
π
f
=
2
C
×
2
π
=
π
×
2π f
Z2
2
π f
×
f
P2
0.5 f
≤
×
π
2
C
×
, at 0.2 x f
π
0.75 f
R
×
×
f
× ×
SW
20
Z2
V
×
1
A
×
L C
1
=
IN
ZESR
R
L C
×
1
SW
1
×
2
1
F
1
×
C
×
×
C
π
C
×
OUT
F
LC
A
R
×
A
×
C
×
OUT
F
)
R
R
www.DataSheet4U.com
F
C
C
×
−
F
F
1
at f
×
A
R
R
or at f
at 0.75 f
×
×
V
F
A
C
OSC
C ) -1
ZESR
CF
F
LC
LC
)
:
×
.
, whichev-
LC
C
17
: