MAX5080EVKIT Maxim Integrated Products, MAX5080EVKIT Datasheet - Page 14

• Manufacturer Part Number: MAX5080EVKIT
• Description: EVAL KIT FOR MAX5080
• Manufacturer: Maxim Integrated Products

Specifications of MAX5080EVKIT

• Main Purpose: DC/DC, Step Down
• Outputs And Type: 1, Non-Isolated
• Voltage - Output: 3.3V
• Current - Output: 1A
• Voltage - Input: 4.5 ~ 40V
• Regulator Topology: Buck
• Frequency - Switching: 250kHz
• Board Type: Fully Populated
• Utilized Ic / Part: MAX5080
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• Power - Output: -
• 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
then calculated by:
1A, 40V, MAXPower Step-Down
DC-DC Converters
Figure 3. Error Amplifier Compensation Circuit (Closed-Loop
and Error-Amplifier Gain Plot) for Ceramic Capacitors
14
C
C
P2
EA(fC)
EA
is:
occurs between f
is set at 1/2 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
=
R4
2
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
MOD(DC)
L
. The error-amplifier gain
R5
f
OUT
LC
f
×
P2
EA
C8
C
×
f
P3
OUT
×
R
C7
2
5
π
×
EA
GAIN
f
C
FREQUENCY
COMP
2
SW
). R6 is
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
C
R
C
8
C
. The output capacitor’s ESR zero fre-
R
6
. Therefore C8 is calculated as:
7
=
R
C
6
C
=
=
3
occurs between f
≈
(
6
2
≈
2
2
=
π
π
π
Compensation When f
R
2
×
×
×
5
C
π
LC
C
C
0 8
OUT
×
×
ZESR
7
.
6
10
but lower than the closed-
f
f
LC
×
C
C
R
×
1
1
×
1
2
R
7
6
×
0 5
×
f
×
LC
5
can occur before f
.
f
ESR
LC
×
C
P2
×
×
6
f
2
P3
R
f
P2
and f
SW
5
−
and f
1
Z1
)
P3
, f
P2
Z2
P3
is approxi-
is now set
C
P2
, f
. f
> f
P1
Z1
is now
ZESR
, f
C
and
P2
. If
,