ADP1873ARMZ-0.3-R7 Analog Devices Inc, ADP1873ARMZ-0.3-R7 Datasheet - Page 28

• Manufacturer Part Number: ADP1873ARMZ-0.3-R7
• Description: 300kHz, Light Load Eff Enabled
• Manufacturer: Analog Devices Inc

Specifications of ADP1873ARMZ-0.3-R7

• Pwm Type: Current Mode
• Number Of Outputs: 1
• Frequency - Max: 300kHz
• Duty Cycle: 84%
• Voltage - Supply: 2.75 V ~ 20 V
• Buck: Yes
• Boost: No
• Flyback: No
• Inverting: No
• Doubler: No
• Divider: No
• Cuk: No
• Isolated: No
• Operating Temperature: -40°C ~ 125°C
• Package / Case: 10-MSOP, Micro10™, 10-uMAX, 10-uSOP
• Frequency-max: 300kHz
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• Other names: ADP1873ARMZ-0.3-R7TR

ADP1872/ADP1873
The inductor peak current is approximately
Therefore, an appropriate inductor selection is 1.0 μH with
DCR = 3.3 mΩ (7443552100) from Table 7 with peak current
handling of 20 A.
Current Limit Programming
The valley current is approximately
Assuming a lower side MOSFET R
as the valley current limit from Table 6 and Figure 70 indicates
that a programming resistor (RES) of 100 kΩ corresponds to an
A
Choose a programmable resistor of R
sense gain of 24 V/V.
Output Capacitor
Assume a load step of 15 A occurs at the output and no more
than 5% is allowed for the output to deviate from the steady
state operating point. The ADP1872’s advantage is, because the
frequency is pseudo-fixed, the converter is able to respond
quickly because of the immediate, though temporary, increase
in switching frequency.
Assuming the overall ESR of the output capacitor ranges from
5 mΩ to 10 mΩ,
Therefore, an appropriate inductor selection is five 270 μF
polymer capacitors with a combined ESR of 3.5 mΩ.
Assuming an overshoot of 45 mV, determine if the output
capacitor that was calculated previously is adequate:
CS
of 24 V/V.
15 A + (5 A × 0.5) = 17.5 A
= 0.003 × (15 A)
15 A − (5 A × 0.5) = 12.5 A
ΔV
= 1.11 mF
= 1.4 mF
C
C
=
P
=
DCR
OUT
OUT
2
1 (
DROOP
×
8 .
(
1
LOSS
=
300
=
×
−
(
10
45
2
(
)
= 0.05 × 1.8 V = 90 mV
V
×
=
×
−
OUT
6
mV
10
DCR
f
×
SW
15
(
3
−
15
)
2
×
Δ
×
(
A
Δ
−
×
L
2
(
A
V
(
90
I
. 1 (
= 675 mW
×
I
Δ
OVSHT
)
LOAD
2
2
LOAD
I
V
) 8
mV
2
DROOP
LOAD
2
)
2
)
−
)
(
)
V
OUT
ON
)
2
RES
of 4.5 mΩ, choosing 13 A
)
= 100 kΩ for a current-
Rev. A | Page 28 of 40
Choose five 270 μF polymer capacitors.
The rms current through the output capacitor is
The power loss dissipated through the ESR of the output
capacitor is
Feedback Resistor Network Setup
It is recommended to use R
Compensation Network
To calculate R
parameter and the current-sense gain variable are required. The
transconductance parameter (G
sense loop gain is
where A
(see the Programming Resistor (RES) Detect Circuit and Valley
Current-Limit Setting sections).
The crossover frequency is 1/12
The zero frequency is 1/4
P
300 kHz/12 = 25 kHz
25 kHz/4 = 6.25 kHz
= 100 kΩ
=
= 250 pF
I
=
R
G
R
=
C
COUT
RMS
T
CS
COMP
COMP
2
1
2
25
CS
=
×
×
=
×
. 3
=
15
and R
= (I
1
A
10
=
14
1
2
3
=
kΩ
CS
25
×
COMP
3
(
1
RMS
2
f
×
1
13
R
CROSS
+
μF
π
×
100
ON
ON
1
×
2 .
)
R
3
. 6
10
2
, C
×
f
1 (
COMP
are taken from setting up the current limit
(
CROSS
25
V
× ESR = (1.5 A)
V
300
=
3
×
8 .
+
COMP
1
1
IN
−
10
×
24
V
1
f
0
,
10
MAX
L
×
ZERO
f
8 .
3
6 .
ZERO
×
10
−
×
, and C
th
×
3
V
1
. 0
V
0
. 6
f
of the crossover frequency:
×
3
B
−
)
SW
6 .
005
×
= 15 kΩ. Calculate R
25
×
2
V
V)
2
×
OUT
13
1
×
π
th
M
PAR
. 3
8 .
=
f
2 .
10
) is 500 μA/V, and the current-
G
of the switching frequency:
CROSS
=
141
)
. 8
V
2
, the transconductance
M
V
×
30
3
× 1.4 mΩ = 3.15 mW
33
G
500
V
=
×
C
CS
V
kΩ
IN
A/V
. 1
25
OUT
OUT
,
×
MAX
49
×
10
10
A
×
−
V
6
V
3
×
OUT
REF
×
8
. 1
T
3 .
11
as
×
10
−
3
×
1
0
8 .
6 .