MAX5951ETJ+ Maxim Integrated Products, MAX5951ETJ+ Datasheet - Page 18

• Manufacturer Part Number: MAX5951ETJ+
• Description: IC BUCK PWM CTRLR 12V/5V 32-TQFN
• Manufacturer: Maxim Integrated Products
• Type: Step-Down (Buck)r

Specifications of MAX5951ETJ+

• Internal Switch(s): No
• Synchronous Rectifier: No
• Number Of Outputs: 1
• Voltage - Output: 0.8 ~ 5.5 V
• Current - Output: 10A
• Frequency - Switching: 100kHz ~ 1MHz
• Voltage - Input: 8 ~ 16 V
• Operating Temperature: -40°C ~ 85°C
• Mounting Type: Surface Mount
• Package / Case: 32-TQFN Exposed Pad
• Power - Output: 2.76W
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant

12V/5V Input Buck PWM Controller
Figure 5 shows the error-amplifier feedback, as well as
its gain response for circuits that use low-ESR output
capacitors (ceramic). In this case, f
f
f
compensate for the gain-and-phase loss due to the
double pole. Choose the inductor (L) and output
capacitor (C
Selection and Output Capacitor Selection sections.
Pick a value for the feedback resistor R5 in Figure 5
(values between 1kΩ and 10kΩ are adequate). C7 is
then calculated as:
f
mented with C7 >> C8 and R3 >> R6, in which case,
the error amplifier gain (G
C6 and R5. Therefore:
The modulator gain at f
Since G
R3 is then calculated as:
f
then calculated by:
f
18
ZESR
Z1
C
P2
P3
occurs between f
is set at 5 x f
is set to 0.5 x f
is set at 1/2 the switching frequency (f
______________________________________________________________________________________
.
EA(fC)
G
MOD fC
x G
OUT
C
G
C
R
C
. Therefore, C8 is calculated as:
C
EA fc
MOD(fC)
7
6
( )
R
6
8
) as described in the Inductor
=
LC
=
3
( )
=
=
2
Z2
≈
=
2
f
Compensation when f
π
C
π
and f
C
2
2
R
(
=
×
2
π
× ×
π
×
and f
5
is:
2
π
0 5
L C
= 1, C6 is calculated by:
×
×
C
×
π
)
EA
.
2
R
G
f
6 0 5
×
LC
Z2
1
5 5
G
× ×
×
1
) at f
MOD DC
×
f
1
C
1
P2
MOD DC
L C
× ×
OUT
f
×
LC
is set to f
×
.
C
. The circuit is imple-
C
(
×
6
C
×
(
6
OUT
×
f
R
C
f
is due primarily to
SW
×
2
5
)
π
R
C
)
5
×
occurs before
LC
f
C
2
in order to
SW
C
< f
). R6 is
ZESR
For larger ESR capacitors such as tantalum and alu-
minum electrolytic, f
f
remain the same as before; however, f
equal to f
quency is higher than f
loop crossover frequency. The equations that define
the error amplifier’s poles and zeros (f
and f
lower than the closed-loop crossover frequency. Figure
5 shows the error-amplifier feedback, as well as its gain
response for circuits that use higher ESR output capac-
itors (tantalum, aluminum electrolytic, etc.).
Pick a value for feedback resistor R5 in Figure 5 (values
between 1kΩ and 10kΩ are adequate). C7 is then cal-
culated as:
The circuit is implemented with C7 >> C8 and R3 >>
R6, in which case the error-amplifier gain between f
and f
The modulator gain at f
Since G
ed as:
f
R3 is then calculated as:
f
ZESR
P2
P3
is set to f
is set at 5 x f
P3
P3
, then f
) are the same as before. However, f
is approximately equal to:
EA(fC)
ZESR
G
MOD fC
ZESR
C
x G
R
occurs between f
. The output capacitor’s ESR zero fre-
6
C
C
( )
. C6 is then calculated as:
C
≈
. Therefore, C8 is calculated as:
7
MOD(fC)
R
C
8
=
(
2
3
=
6
ZESR
=
) π
2
≈
=
Compensation when f
(
R
π
2
C
2
2
2
5
π
C
×
π
LC
× ×
π
R
R
is:
)
×
OUT
×
0 5
2
= 1, R6 can then be calculat-
L C
5
6
can occur before f
×
G
.
R
but lower than the closed-
× ×
G
f
LC
MOD DC
R
5 5
1
1
×
L C
1
MOD DC
6
×
× ×
f
OUT
LC
×
ESR
P2
(
C
(
×
OUT
6
f
and f
C
×
R
)
f
5
)
C
2
×
Z1
P3
f
C
P2
, f
2
. f
Z2
is now set
Z1
C
P2
C
, f
. If f
> f
and f
P1
is now
ZESR
, f
C
P2
Z2
P2
>
,