ADP1871ACPZ-1.0-R7 Analog Devices Inc, ADP1871ACPZ-1.0-R7 Datasheet - Page 30

• Manufacturer Part Number: ADP1871ACPZ-1.0-R7
• Description: 1.0MHz, Light Load Eff Enabled
• Manufacturer: Analog Devices Inc

Specifications of ADP1871ACPZ-1.0-R7

• Frequency - Max: 1MHz
• Pwm Type: Current Mode
• Number Of Outputs: 1
• Duty Cycle: 45%
• Voltage - Supply: 2.95 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-WFDFN, CSP Exposed Pad
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• Other names: ADP1871ACPZ-1.0-R7; ADP1871ACPZ-1.0-R7TR

ADP1870/ADP1871
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:
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
= 100 kΩ
R
=
COMP
25
= 1.4 mF
P
300 kHz/12 = 25 kHz
25 kHz/4 = 6.25 kHz
C
G
=
I
=
R
×
COUT
RMS
T
OUT
CS
10
1 (
1
2
=
CS
25
=
×
8 .
3
=
f
1
=
15
and R
= (I
CROSS
=
+
×
−
×
1
A
1
2
3
10
10
. 6
45
(
kΩ
CS
(
COMP
f
×
RMS
V
(
CROSS
25
1
1
−
13
R
3
mV
OUT
+
6
ON
μF
×
1
)
ON
×
×
3
2 .
2
, C
1 (
f
are taken from setting up the current limit
× ESR = (1.5 A)
10
(
×
ZERO
)
15
(
−
V
8 .
V
=
2
300
COMP
3
Δ
−
(
−
IN
) A
V
24
0
L
×
V
. 1 (
1
,
6 .
×
MAX
L
×
×
−
, and C
OVSHT
8 .
th
2
×
2
10
) 8
×
2
I
0
V
×
1
. 0
of the crossover frequency:
B
V
π
2
6 .
2
f
3
= 15 kΩ. Calculate R
. 3
LOAD
005
−
f
)
SW
G
CROSS
V)
×
141
)
V
M
th
2
M
PAR
13
OUT
1
) is 500 μA/V, and the current-
G
−
=
of the switching frequency:
500
)
=
8 .
×
C
2 .
, the transconductance
CS
(
2
. 8
V
30
25
)
× 1.4 mΩ = 3.15 mW
OUT
V
33
V
OUT
×
×
kΩ
×
10
V
=
A/V
10
V
×
)
IN
. 1
−
2
OUT
6
V
,
V
)
3
49
MAX
×
OUT
REF
×
8
. 1
A
3 .
11
T
×
as follows:
10
−
3
×
1
0
Rev. A | Page 30 of 44
8 .
6 .
Loss Calculations
Duty cycle = 1.8/12 V = 0.15
R
t
V
C
Q
R
BODY(LOSS)
ON (N2)
GATE
F
IN
N1,N2
= 0.84 V (MOSFET forward voltage)
= 3.3 nF (MOSFET gate input capacitance)
=
= 250 pF
= (0.15 × 0.0054 + 0.85 × 0.0054) × (15 A)
= 1.215 W
= 20 ns × 300 × 10
= 151.2 mW
P
= 300 × 10
= 534.6 mW
= 57.12 mW
P
P
P
+ P
= 1.215 W + 151.2 mW + 534.6 mW + 57.12 mW + 55.6
+ 3.15 mW + 675 mW + 56.25 mW
= 2.655 W
C
P
P
P
+
=
+
=
=
P
P
= 1.5 Ω (MOSFET gate input resistance)
SW(LOSS)
COUT
CIN
LOSS
= 17 nC (total MOSFET gate charge)
DR
DCR
N1,N2(CL)
BODY
DISS
[
5 (
COMP
2
(
(
55
V
= 5.4 mΩ
. 4
13
COUT
(
×
0 .
LOSS
REG
62
= (I
(
= 20 ns (body conduction time)
6 .
(
. 3
LOSS
= P
×
LDO
V
= (I
(
LOSS
×
(
14
×
mW
)
=
300
= f
−
RMS
+ P
(
)
=
(
N1,N2
)
300
RMS
2
f
×
=
5
=
)
[
=
3
V
SW
π
SW
)
×
100
V
CIN
[
DCR
=
× 1.5 Ω × 3.3 × 10
2
DR
(
R
D
)
×
10
C
V
× ESR = (7.5 A)
+ P
2
)
× R
10
COMP
t
×
lowerFET
×
× ESR = (1.5 A)
×
IN
3
BODY
×
(
R
(
3
×
1
1
×
BODY(LOSS)
10
300
GATE
f
−
t
×
3
N1(ON)
SW
I
SW
3
3 .
3
V
f
LOAD
3
(
2
× 15 A × 0.84 × 2
3 .
ZERO
LOSS
V
C
×
REG
×
×
× C
REG
upperFET
×
10
10
. 6
10
)
+
)
25
−
= 0.003 × (15 A)
3
×
TOTAL
+
9
×
−
+ P
(
1
9
×
×
(
I
I
×
×
−
V
BIAS
5
3
LOAD
f
10
0 .
SW
2
SW
. 4
3 .
DR
D
× I
−9
× 1 mΩ = 56.25 mW
2
62
+
3
×
)
)
]
+ P
+
× 1.4 mΩ = 3.15 mW
×
× 15 A × 12 × 2
×
. 0
10
×
LOAD
+
I
C
002
R
V
BIAS
. 0
−
DCR
total
N2(ON)
F
9
002
))
× V
×
×
)
]
+ P
×
5
2
))
V
+
2
IN
]
REG
DR
= 675 mW
×
. 0
2
× 2
I
002
+ P
2
LOAD
+
I
)
DISS(LDO)
BIAS
)