MAX5089ATE+ Maxim Integrated Products, MAX5089ATE+ Datasheet - Page 17

• Manufacturer Part Number: MAX5089ATE+
• Description: IC DC-DC CONV BUCK 16TQFN
• Manufacturer: Maxim Integrated Products
• Type: Step-Down (Buck)r

Specifications of MAX5089ATE+

• Internal Switch(s): Yes
• Synchronous Rectifier: Yes
• Number Of Outputs: 1
• Voltage - Output: 0.6 ~ 20 V
• Current - Output: 2A
• Frequency - Switching: 200kHz ~ 2.2MHz
• Voltage - Input: 4.5 ~ 23 V
• Operating Temperature: -40°C ~ 125°C
• Mounting Type: Surface Mount
• Package / Case: 16-TQFN Exposed Pad
• Power - Output: 2.67W
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant

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
) at the crossover fre-
f
m
SW
F
f c L)
f
f
LC
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
×
Z2
2π f
2
π f
×
f
≤
×
P2
0.5 f
2
π
C
×
, at 0.2 x f
0.75 f
π
R
×
×
f
× ×
SW
20
V
Z2
1
×
A
L C
×
1
=
IN
ZESR
R
L C
×
1
SW
1
×
2
1
F
1
×
C
×
×
C
π
C
×
OUT
F
A
LC
R
×
A
×
C
×
OUT
F
)
R
R
F
C
C
×
−
F
F
1
at f
×
R
A
R
at 0.75 f
or at f
×
×
V
F
A
C
OSC
C ) -1
ZESR
CF
F
LC
LC
)
:
×
.
, whichev-
LC
C
17
: