ADP1882 Analog Devices, ADP1882 Datasheet - Page 29

• Manufacturer Part Number: ADP1882
• Description: Synchronous Current-Mode Buck Controller with Constant On-time and 0.8 V Reference Voltage
• Manufacturer: Analog Devices

Inductor
Determine the inductor ripple current amplitude as follows:
then calculate for the inductor value
The inductor peak current is approximately
Therefore, an appropriate inductor selection is 1.0 μH with
DCR = 3.3 mΩ (7443552100) from Table 8, 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 7 and Figure 71 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. Because the frequency is pseudo-fixed,
the advantage of the ADP1882 is that 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.
CS
of 24 V/V.
= 1.03 μH
15 A + (5 A × 0.5) = 17.5 A
P
15 A − (5 A × 0.5) = 12.5 A
ΔV
= 1.11 mF
Δ
L
=
C
=
DCR(LOSS)
I ≈
=
OUT
(
L
2
5
DROOP
13
(
×
V
V
2 .
I
×
300
IN
LOAD
=
Δ
V
= 0.05 × 1.8 V = 90 mV
= DCR ×
3
300
,
MAX
I
2
−
L
×
×
1
×
= 5 A
×
10
8 .
−
10
f
f
SW
SW
V
V
15
3
3
OUT
)
I
×
2
L
A
×
×
Δ
(
90
)
13
= 0.003 × (15 A)
(
I
1
Δ
×
LOAD
8 .
2 .
V
mV
V
V
V
V
DROOP
IN,MAX
OUT
ON
)
RES
of 4.5 mΩ, choosing 13 A
)
= 100 kΩ for a current-
2
= 675 mW
Rev. 0 | Page 29 of 40
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 that R
follows:
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 of the switching frequency:
The zero frequency is 1/4 of the crossover frequency:
= 1.4 mF
P
R
G
300 kHz/12 = 25 kHz
25 kHz/4 = 6.25 kHz
×
=
= 340 pF
C
C
=
I
=
R
=
COUT
T
RMS
OUT
CS
COMP
COMP
2
1 (
1
2
25
1
0
= 15 kΩ ×
×
CS
8 .
=
8 .
8 .
×
1
. 3
×
=
=
and R
= (I
×
−
A
14
10
1
= 75 kΩ
=
(
10
=
1
2
45
3
(
CS
25
V
COMP
×
×
1
3
RMS
2
−
(
1
R
OUT
f
mV
13
6
+
π
75
CROSS
ON
×
μF
ON
×
1
R
)
. 6
10
2 .
3
(
2
×
, C
−
15
COMP
)
f
×
are taken from setting up the current limit
1 (
=
× ESR = (1.5 A)
25
2
10
CROSS
1
(
V
3
Δ
(
V
300
−
8 .
COMP
26
L
+
1
A
V
×
3
IN
−
. 1 (
×
OVSHT
V
)
0
×
f
10
×
1
2
f
,
I
6 .
ZERO
×
) 8
MAX
L
ZERO
. 6
8 .
1
−
2
, and C
. 0
B
10
3
LOAD
×
V
2
25
0
005
= 15 kΩ be used. Calculate R
V
×
)
6 .
f
3
2
−
)
×
×
SW
2
V)
)
−
×
V
10
=
×
2
(
V
13
M
OUT
π
PAR
7
. 3
1
3
) is 500 μA/V, and the current-
OUT
= 30 kΩ
7 .
f
8 .
G
ADP1882/ADP1883
141
2 .
CROSS
, the transconductance
2
)
A/V
M
V
× 1.4 mΩ = 3.15 mW
)
V
×
500
2
A
×
)
V
C
CS
=
25
V
IN
OUT
×
. 1
OUT
×
,
10
MAX
49
10
×
−
A
6
3
V
V
×
×
OUT
REF
8
. 1
3 .
11
×
10
T
as
−
3