MAX8650EEG+ Maxim Integrated Products, MAX8650EEG+ Datasheet - Page 22

• Manufacturer Part Number: MAX8650EEG+
• Description: IC CNTRLR STP DWN 24-QSOP
• Manufacturer: Maxim Integrated Products
• Type: Step-Down (Buck)r

Specifications of MAX8650EEG+

• Internal Switch(s): No
• Synchronous Rectifier: No
• Number Of Outputs: 1
• Voltage - Output: 0.7 ~ 5.5 V
• Current - Output: 25A
• Frequency - Switching: 200kHz ~ 1.2MHz
• Voltage - Input: 4.5 ~ 28 V
• Operating Temperature: -40°C ~ 85°C
• Mounting Type: Surface Mount
• Package / Case: 24-QSOP
• Power - Output: 762mW
• Output Voltage: 0.7 V to 5.5 V
• Output Current: 25 A
• Input Voltage: 4.5 V to 28 V
• Mounting Style: SMD/SMT
• Maximum Operating Temperature: + 85 C
• Minimum Operating Temperature: - 40 C
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant

4.5V to 28V Input Current-Mode Step-Down
Controller with Adjustable Frequency
Figure 10. Simplified Gain Plot for the f zMOD > f C Case
The transconductance error amplifier has a DC gain,
G
er transconductance, which is equal to 110µS, R
the output resistance of the error amplifier, which is
30MΩ. A dominant pole is set by the compensation
capacitor (C
and the compensation resistor (R
by the compensation resistor (R
tion capacitor (C
and R
occurs near the crossover frequency (f
The crossover frequency, f
than the power-modulator pole fp
be less than or equal to 1/5 the switching frequency.
Select a value for f
At the crossover frequency, the total loop gain must
equal 1, and is expressed as:
22
POWER
MODULATOR
EA(DC)
______________________________________________________________________________________
C
GAIN
(dB)
0dB
to cancel the output-capacitor ESR zero if it
= g
FB
DIVIDER
G
mEA
C
f
EA fc
pdEA
), the amplifier output resistance (R
( )
C
f
f
x R
pEA
zEA
). There is an optional pole set by C
C
f
=
×
f
pMOD
in the range:
pMOD
O
CLOSE LOOP
G
2π
=
=
, where g
MOD fc
2π
2π
×
C
<<
×
×
( )
C
C
C
C
f
, should be much higher
1
C
1
×
F
C
f
1
C
×
(
mEA
×
≤
R
×
C
f
zMOD
R
V
f
O
R
) and the compensa-
MOD
5
C
S
V
OUT
C
C
FB
), and a zero is set
+
is the error-amplifi-
R
C
. Also, f
C
ERROR
AMP
= 1
). Thus:
)
C
FREQUENCY
should
O
O
is
),
F
For the case where f
Then R
where g
The error-amplifier compensation zero formed by R
and C
Calculate the value of C
If f
C
As the load current decreases, the modulator pole also
decreases; however, the modulator gain increases
accordingly and the crossover frequency remains the
same.
Figure 11. Simplified Gain Plot for the f zMOD < f C Case
POWER
MODULATOR
F
, from COMP to GND. The value of C
zMOD
GAIN
(dB)
C
0dB
C
mEA
should be set at the modulator pole fp
can be calculated as:
is less than 5 x f
G
FB
DIVIDER
= 110µS.
C
R
MOD fc
C
C
=
C
G
=
( )
F
EA fc
R
g
(
f
pMOD
R
mEA
LOAD
=
( )
LOAD
zMOD
=
2π
G
C
CLOSE LOOP
=
MOD dc
×
×
as follows:
×
g
R
V
C
+
is greater than f
f
V
mEA
FB
C
S
, add a second capacitor,
OUT
f
f
zMOD
(
1
S
× ×
×
×
×
L C
f
)
zMOD
G
×
L
×
MOD fc
)
R
f
f
×
C
C
pMOD
OUT
R
f
C
C
( )
F
ERROR
AMP
is:
C
:
FREQUENCY
MOD
C
.