MAX5072ETJ+ Maxim Integrated Products, MAX5072ETJ+ Datasheet - Page 23

• Manufacturer Part Number: MAX5072ETJ+
• Description: IC CONV BUCK/BOOST 32-TQFN
• Manufacturer: Maxim Integrated Products
• Type: Step-Down (Buck), Step-Up (Boost)r

Specifications of MAX5072ETJ+

• Internal Switch(s): Yes
• Synchronous Rectifier: No
• Number Of Outputs: 2
• Voltage - Output: 0.8 ~ 28 V
• Current - Output: 1A, 2A
• Frequency - Switching: 200kHz ~ 2.2MHz
• Voltage - Input: 4.5 ~ 23 V
• Operating Temperature: -40°C ~ 85°C
• Mounting Type: Surface Mount
• Package / Case: 32-TQFN Exposed Pad
• Power - Output: 2.76W
• Output Voltage: 5.2 V
• Output Current: 2 A
• Input Voltage: 5.5 V to 23 V
• Supply Current: 2.2 mA
• Switching Frequency: 1250 KHz
• Mounting Style: SMD/SMT
• Maximum Operating Temperature: + 85 C
• Minimum Operating Temperature: - 40 C
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant

The transconductance error-amplifier gain is:
The total loop gain at f
or
Place a zero at or below the LC double pole:
Place a high-frequency pole at f
Procedure 2 (See Figure 8):
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.
Select a crossover frequency:
Calculate the LC double-pole frequency, f
Place a zero
where:
and R
f
f
F
Z
C
≥ 10kΩ.
=
R
f
F
f
Z
f
C
P
=
Converter with POR and Power-Fail Output
=
f
= 5
LC
V
C
2
OSC
π
F
=
C
______________________________________________________________________________________
C
×
=
is a good number to get about 60°
F
2π
. However, it is important to place
G
G
R
(
2
ESR
0 8
=
E A
1
M
C
f
F
π
.
C
×
/
2π
should be equal to 1
×
×
×
≤
×
=
C
G
L
0 75
+
C
×
f
V
.
OUT
SW
F
g
E A
20
2
IN
). In this case, Type III com-
R
m
/
π
1
1
F
at
2.2MHz, Dual-Output Buck or Boost
1
×
×
×
×
P
=
×
×
g
f
R
f
= 0.5 x f
LC
0 75
C
m
1
f
C
F
LC
.
×
OUT
×
×
L
ESR
OUT
R
×
F
SW
LC
f
LC
)
V
:
.
OUT
C
.
Calculate C
Place a pole
Place a second zero, f
is lower.
Place a second pole
switching frequency.
The boost converter compensation gets complicated
due to the presence of a right-half-plane zero
F
phase while adding positive (+1) slope to the gain curve.
It is important to drop the gain significantly below unity
before the RHP frequency. Use the following procedure
to calculate the compensation components.
Calculate the LC double-pole frequency, f
right half plane zero frequency.
where:
Target the unity-gain crossover frequency for:
ZERO,RHP
C
I
=
. The right-half-plane zero causes a drop in-
C
I
CF
for a target unity crossover frequency, f
2π
f
f
(f
LC
ZERO RHP
P
R
R
1
=
×
1
D
I
R
=
=
f
(
=
C
=
2
(
,
= −
MIN
2
π
2
2
×
2π
π
1
π
Boost Converter Compensation
π
(
×
Z2
f
×
)
L
P
×
×
0 5
×
2
OUT
=
, at 0.2f
V
=
.
R
V
1
f
V
OUT
f
1
L
ZERO ESR
I
I
=
IN
Z
1
IN
OUT MAX
(
OUT OUT
×
−
1
2
×
2
V
−
2
×
×
f
D
C
×
π
SW
OUT
π
1
(
D
C
R
I
C
,
C
C
C
×
×
)
OUT
F
at f
I
F
2
R
or at f
×
L
−
F
R
1
OUT
)
R
ZERO,ESR
R
(
×
×
F
MIN
×
I
C
C
LC
×
V
I
OSC
CF
)
C
, whichever
F
LC
)
)
.
−
, and the
at 1/2 the
1
C
23
: