MAX15021ATI+ Maxim Integrated Products, MAX15021ATI+ Datasheet - Page 19

• Manufacturer Part Number: MAX15021ATI+
• Description: IC REG SYNC DUAL 28-TQFN-EP
• Manufacturer: Maxim Integrated Products
• Type: Step-Down (Buck)r

Specifications of MAX15021ATI+

• Internal Switch(s): Both
• Synchronous Rectifier: Yes
• Number Of Outputs: 2
• Voltage - Output: 0.6 ~ 5.5 V
• Current - Output: 2A, 4A
• Frequency - Switching: 500kHz ~ 4MHz
• Voltage - Input: 2.5 ~ 5.5 V
• Operating Temperature: -40°C ~ 125°C
• Mounting Type: Surface Mount
• Package / Case: 28-TQFN Exposed Pad
• Power - Output: 2.76W
• Topology: Buck
• Output Voltage: 0.6 V to 5.5 V
• Output Current: 2 A, 4 A
• Input Voltage: 2.5 V to 5.5 V
• Duty Cycle (max): 100 %
• Switching Frequency: 500 KHz to 4 MHz
• Maximum Operating Temperature: + 125 C
• Mounting Style: SMD/SMT
• Minimum Operating Temperature: - 40 C
• Operating Supply Voltage: 2.5 V to 5.5 V
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant

The locations of the zeros and poles should be such
that the phase margin peaks around f
Set the ratios of f
er, e.g., f
60° of phase margin at f
important to place the two zeros at or below the double
pole to avoid the conditional stability issue.
The following procedure is recommended:
1) Select a crossover frequency, f
2) Calculate the LC double-pole frequency, f
where C
3) Select the feedback resistor, R
4) Place the compensator’s first
5) The gain of the modulator (Gain
The gain of the error amplifier (Gain
quencies is:
The total loop gain is the product of the modulator gain
and the error amplifier gain at f
as follows:
So:
Regulator with Tracking/Sequencing Capability
Gain
tenth the switching frequency (f
3.3kΩ to 30kΩ.
zero at or below the output filter’s
double-pole, f
the regulator’s pulse-width modulator, LC filter,
feedback divider, and associated circuitry—at the
crossover frequency is:
MOD
Gain
4
CO
OUT
f
Z
C [ F]
×
×
f
LC
F
(2
= f
E/A
2
= ×
is the output capacitor of the regulator.
f
[MHz]
µ
π
π
CO
P
4
×
×
= 5 is a good number to get approximately
CO
=
Gain
=
f
f
f
LC
CO
CO
(2
2
CO
______________________________________________________________________________________
-to-f
2
π
π
≈
π
[kHz])
[kHz] C [ F] R [k ] 1
, as follows:
×
[kHz]
MOD
×
2
×
f
Z
f
CO
π
R [k ] 0.5 f
CO
and f
×
F
CO
[kHz] C [ F] R [k ]
Dual, 4A/2A, 4MHz, Step-Down DC-DC
×
x Gain
[MHz])
2
≤
L[ H]
Ω
. Whichever technique, it is
×
f
1
P
SW
I
µ
C
-to-f
p
×
CO
×
1
OUT
10
E/A
[kHz]
1
2
CO
×
×
I
should be equal to 1,
SW
1
×
CO
MOD
×
C
µ
[ F] L[ H]
E/A
= 1
µ
L[ H] C
F
F
OUT
LC
equal to one anoth-
CO
):
, at or below one-
f
, in the range of
µ
Z1
) in midband fre-
×
)—comprised of
×
[kHz]
Ω
.
µ [
=
×
F
=
F]
2
µ
π
LC
OUT
Ω
×
R
:
1
F
[ F]
µ
×
C
F
Solving for C
6) For those situations where f
If a ceramic capacitor is used, then the capacitor ESR
zero, f
the switching frequency, that is f
f
(f
erode the phase margin at the crossover frequency.
For example, f
tribution to phase loss at the crossover frequency f
only about 11°:
Once f
7) Place the second zero (f
8) Place the third pole (f
9) Calculate R
where V
ESR
P2
as with low-ESR tantalum capacitors, the compen-
sator’s second pole (f
f
system Bode plot, the loop gain maintains its
+20dB/decade slope up to
quency verses flattening out soon after the 0dB
crossover. Then set:
whichever is lower, and calculate R
lowing equation:
quency and calculate C
) should be placed high enough not to significantly
ESR
. In this case, the frequency of the second pole
ESR
C pF]
P2
. This provides additional phase margin. On the
C
FB
I
[
is known, calculate R
, is likely to be located even above one-half of
CF
R [k ] R [k ]
= 0.6V (typ).
2
[ F]
I
R [k ]
R [k ]
n =
:
=
1
I
P2
Ω
2
(
2
as:
Ω =
Ω =
π
=
can be set at 5 x f
(
×
2
f
1
π
CO
f
2
2
P2
×
f
π
π
Ω
P2
0.5 f
[kHz] L[ H] C
×
×
= 5 x f
P2
f
×
= f
f [kHz] C [ F]
P3
P2
4 R [k ]
Z2
×
CF
) should be used to cancel
V
×
) at 1/2 the switching fre-
Z2
ESR
SW
OUT_
[kHz] C [ F]
×
from:
LC
I
) at 0.2 x f
CO
:
F
1
1
1
[MHz] R [k ]
1
µ
/
2
V
< f
[V] V
FB
Ω
LC
×
×
of the switching fre-
CO
×
CO
[V]
−
I
I
< f
×
µ
, so that its con-
µ
OUT
1
FB
< f
CO
F
using the fol-
CO
[V]
ESR
[ F]
µ
Ω
< f
or at f
)
< f
)
SW
SW
CO
/2 <
LC
19
/2,
is
,