MAX5083ATE+T Maxim Integrated Products, MAX5083ATE+T Datasheet - Page 13

• Manufacturer Part Number: MAX5083ATE+T
• Description: IC DC-DC CONV 1.5A 16-TQFN
• Manufacturer: Maxim Integrated Products
• Type: Step-Down (Buck)r

Specifications of MAX5083ATE+T

• Internal Switch(s): Yes
• Synchronous Rectifier: No
• Number Of Outputs: 1
• Voltage - Output: 1.23 ~ 32 V
• Current - Output: 1.5A
• Frequency - Switching: 250kHz
• Voltage - Input: 7.5 ~ 40 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

Pick a value for the feedback resistor R5 in Figure 3
(values between 1kΩ and 10kΩ are adequate).
C7 is then calculated as:
f
(G
G
f
Since G
f
is then calculated by:
Figure 3. Error Amplifier Compensation Circuit (Closed-Loop
and Error-Amplifier Gain Plot) for Ceramic Capacitors
C
C
P2
EA(fC)
EA
is:
occurs between f
is set at one-half the switching frequency (f
) at f
GAIN
(dB)
V
EA(fC)
= 2π x f
OUT
C
G
MOD(fC)
is due primarily to C6 and R5. Therefore,
x G
C
C
C6
C
7
6
MOD(fC)
R3
=
x C6 x R5 and the modulator gain at
______________________________________________________________________________________
=
CLOSED-LOOP
GAIN
Z2
2
R4
=
R6
f
Z1
π
f
C
(
R
and f
×
2
f
Z2
× ×
5
π
0 8
REF
)
= 1, C6 is calculated by:
L
×
2
.
G
f
P2
C
×
G
1
×
MOD(DC)
C
L
. The error-amplifier gain
MOD(DC)
R5
f
LC
OUT
f
P2
×
EA
C8
C
×
f
P3
OUT
×
R
1.5A, 40V, MAXPower Step-Down
C7
2
5
π
×
EA
GAIN
f
C
FREQUENCY
COMP
2
SW
). R6
Since R3 >> R6, R3 + R6 can be approximated as R3.
R3 is then calculated as:
f
For larger ESR capacitors such as tantalum and alu-
minum electrolytic ones, f
f
f
equal to f
quency is higher than f
loop crossover frequency. The equations that define
the error amplifier’s poles and zeroes (f
and f
lower than the closed-loop crossover frequency. Figure
4 shows the error amplifier feedback as well as its gain
response for circuits that use higher-ESR output capac-
itors (tantalum or aluminum electrolytic).
Pick a value for the feedback resistor R5 in Figure 4 (val-
ues between 1kΩ and 10kΩ are adequate).
C7 is then calculated as:
The error amplifier gain between f
mately equal to R5/R6 (given that R6 << R3). R6 can
then be calculated as:
C6 is then calculated as:
P3
ZESR
Z2
is set at 5xf
remain the same as before however, f
P3
< f
) are the same as before. However, f
C
ZESR
, then f
DC-DC Converters
C
C
R
C
. The output capacitor’s ESR zero fre-
8
. Therefore, C8 is calculated as:
6
7
C
R
=
R
C
=
=
6
occurs between f
3
6
(
2
≈
2
2
≈
=
π
π
Compensation When f
π
×
×
R
×
2
LC
C
π
5
C
0 8
C
ZESR
OUT
.
×
7
×
6
but lower than the closed-
f
×
10
1
LC
×
f
C
R
×
C
1
1
R
6
7
2
f
×
0 5
LC
can occur before f
×
5
×
.
ESR
P2
f
×
C
LC
×
×
f
6
P3
R
P2
f
and f
2
SW
5
−
and f
Z1
1
)
P3
, f
P2
Z2
P3
is approxi-
is now set
C
P2
, f
. f
> f
P1
Z1
is now
ZESR
, f
C
and
. If
P2
13
,