ADP1875 Analog Devices, ADP1875 Datasheet - Page 33

• Manufacturer Part Number: ADP1875
• Description: Synchronous Buck Controller with Constant On-Time, Valley Current Mode, and Power Saving Mode
• Manufacturer: Analog Devices

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
Choosing 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 section
and the Valley Current-Limit Setting section).
The crossover frequency is 1/12 the switching frequency.
The zero frequency is 1/4 the crossover frequency.
= 60.25 kΩ
R
R
1
0
COMP
8 .
6 .
COMP
f
CROSS
×
= 1.4 mF
P
300 kHz/12 = 25 kHz
25 kHz/4 = 6.25 kHz
C
G
=
I
=
R
500
COUT
RMS
f
=
T
OUT
CS
=
1 (
CROSS
1
2
2
CS
=
×
8 .
×
+
=
25
1
1
10
=
= (I
and R
=
−
f
×
1
B
kΩ
1
A
k
ZERO
−
1
2
3
10
2
(
45
= 1 kΩ as an example, calculate R
25
6
CS
(
COMP
×
RMS
+
×
V
(
1
×
K
−
13
. 6
8
mV
×
2
OUT
1
ON
6
μF
3 .
1
)
1 (
25
×
3
2 .
R
×
2
, C
8 .
are taken from setting up the current limit
×
×
× ESR = (1.5 A)
(
ON
k
)
15
(
V
−
1
V
2
15
2
300
COMP
1
V
8 .
0
Δ
2
×
−
−
1
(
IN
6 .
) A
=
L
2
V
−
+
. 1 (
1
,
×
MAX
L
+
×
, and C
24
OVSHT
8 .
V
0
(
1
2
s
10
2
) 8
×
6 .
(
I
(
s
V
+
×
R
2
×
f
2
3
LOAD
V)
−
L
(
SW
)
1
. 0
2
ESR
1
×
+
)
V
π
005
2
2
m
PAR
×
OUT
13
=
ESR
+
1
) is 500 μA/V, and the current-
−
)
25
(
8 .
×
2
2
2 .
2
, the transconductance
(
V
=
C
k
π
kΩ
)
× 1.4 mΩ = 3.15 mW
)
V
C
V
OUT
OUT
×
×
×
. 8
OUT
((
25
V
33
=
1
V
)
k
)
IN
8 .
. 1
2
2
)
OUT
A/V
×
2
,
)
15
MAX
49
. 0
×
)
0035
A
R
+
T
1
L
. 0
as follows:
×
0035
×
V
. 0
V
OUT
0011
REF
)
×
. 0
×
)
2
0011
G
M
Rev. 0 | Page 33 of 44
1
G
)
2
CS
×
Loss Calculations
Duty cycle = 1.8/12 V = 0.15
R
t
V
C
Q
R
BODY(LOSS)
ON (N2)
GATE
IN
F
N1,N2
= 0.84 V (MOSFET forward voltage)
= 3.3 nF (MOSFET gate input capacitance)
=
= 423 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
[
5 (
P
P
P
=
P
=
=
P
VREG
= 1.5 Ω (MOSFET gate input resistance)
SW(LOSS)
COUT
CIN
LOSS
COUT
= 17 nC (total MOSFET gate charge)
N1,N2(CL)
BODY
DR
DISS
DCR
COMP
(
0 .
(
55
2
= 5.4 mΩ
13
. 4
(
×
×
= (I
LOSS
(
= 20 ns (body conduction time)
6 .
62
(
LOSS
= P
LDO
. 3
= (I
(
V
+ P
(
LOSS
300
mW
=
×
14
×
= f
)
−
RMS
)
N1,N2
)
(
(
2
CIN
=
=
RMS
5
=
f
300
)
=
×
π
3
×
SW
)
SW
[
[
V
=
DCR
2
R
× 1.5 Ω × 3.3 × 10
V
60
D
10
(
)
V
× ESR = (7.5 A)
2
+ P
COMP
× R
)
DR
C
t
×
×
× ESR = (1.5 A)
.
×
3
IN
BODY
25
lowerFET
10
1
R
×
(
×
×
BODY(LOSS)
GATE
300
t
−
N1(ON)
1
×
3
(
3
SW
I
f
3
VREG
ZERO
f
3 .
10
(
2
LOAD
×
× 15 A × 0.84 × 2
LOSS
SW
×
×
× C
3
3
VREG
10
3 .
C
10
×
+
)
upperFET
×
= 0.003 × (15 A)
. 6
TOTAL
×
3
(
−
)
+ P
1
10
9
×
×
25
I
−
×
LOAD
(
3
+
−
SW
2
D
5
f
3 .
ADP1874/ADP1875
×
9
× I
−9
SW
× 1 mΩ = 56.25 mW
2
I
0 .
10
V
×
)
×
+ P
× 1.4 mΩ = 3.15 mW
BIAS
×
× 15 A × 12 × 2
×
DR
. 4
+
10
LOAD
3
×
R
V
62
. 0
C
DCR
−
N2(ON)
)
+
F
]
002
9
total
× V
+
×
I
×
+ P
BIAS
. 0
2
5
))
×
002
IN
+
2
]
VREG
×
= 675 mW
DR
2
)
. 0
]
× 2
I
+
))
002
+ P
LOAD
2
+
DISS(LDO)
)
+
I
BIAS
)
+