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ADP1882ARMZ-0.6-R7 датащи(PDF) 29 Page - Analog Devices

номер детали ADP1882ARMZ-0.6-R7
подробное описание детали  Synchronous Current-Mode with Constant On-Time,PWM Buck Controller
PDF  40 Pages
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производитель  AD [Analog Devices]
домашняя страница  http://www.analog.com
Logo AD - Analog Devices

ADP1882ARMZ-0.6-R7 датащи(HTML) 29 Page - Analog Devices

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ADP1882/ADP1883
Rev. 0 | Page 29 of 40
Inductor
Determine the inductor ripple current amplitude as follows:
3
LOAD
L
I
I ≈
Δ
= 5 A
then calculate for the inductor value
V
2
.
13
V
8
.
1
10
300
V
5
)
V
8
.
1
V
2
.
13
(
)
(
3
,
×
×
×
−
=
×
×
Δ
−
=
IN,MAX
OUT
SW
L
OUT
MAX
IN
V
V
f
I
V
V
L
2
L
I
= 1.03 μH
The inductor peak current is approximately
15 A + (5 A × 0.5) = 17.5 A
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.
PDCR(LOSS) = DCR ×
= 0.003 × (15 A)2 = 675 mW
Current Limit Programming
The valley current is approximately
15 A − (5 A × 0.5) = 12.5 A
Assuming a lower-side MOSFET RON of 4.5 mΩ, choosing 13 A
as the valley current limit from Table 7 and Figure 71 indicates
that a programming resistor (RES) of 100 kΩ corresponds to an
ACS of 24 V/V.
Choose a programmable resistor of RRES = 100 kΩ for a current-
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.
ΔVDROOP = 0.05 × 1.8 V = 90 mV
Assuming the overall ESR of the output capacitor ranges from
5 mΩ to 10 mΩ,
)
mV
90
(
10
300
15
2
)
(
3
×
×
×
=
A
V
f
DROOP
SW
()
()
2
Δ
×
Δ
×
=
I
C
LOAD
OUT
= 1.11 mF
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.
2
2
2
6
2
2
2
)
8
.
1
(
)
mV
45
8
.
1
(
)
A
15
(
10
1
)
(
)
(
−
−
×
×
=
−
Δ
−
×
=
−
OUT
OVSHT
OUT
LOAD
OUT
V
V
V
I
L
C
= 1.4 mF
Choose five 270 μF polymer capacitors.
The rms current through the output capacitor is
A
49
.
1
V
2
.
13
V
8
.
1
10
300
μF
1
)
V
8
.
1
V
2
.
13
(
3
1
2
1
)
(
3
1
2
1
3
,
,
=
×
×
×
−
×
=
×
×
−
×
=
MAX
IN
OUT
SW
OUT
MAX
IN
RMS
V
V
f
L
V
V
I
The power loss dissipated through the ESR of the output
capacitor is
PCOUT = (IRMS)2 × ESR = (1.5 A)2 × 1.4 mΩ = 3.15 mW
Feedback Resistor Network Setup
It is recommended that RB = 15 kΩ be used. Calculate RT as
follows:
RT = 15 kΩ ×
V
6
.
0
V)
6
.
0
V
8
.
1
(
−
= 30 kΩ
Compensation Network
To calculate RCOMP, CCOMP, and CPAR, the transconductance
parameter and the current-sense gain variable are required. The
transconductance parameter (GM) is 500 μA/V, and the current-
sense loop gain is
GCS =
A/V
7
.
7
005
.
0
26
1
1
=
×
=
ON
CS R
A
where ACS and RON are taken from setting up the current limit
(see the Programming Resistor (RES) Detect Circuit and Valley
Current-Limit Setting sections).
The crossover frequency is 1/12 of the switching frequency:
300 kHz/12 = 25 kHz
The zero frequency is 1/4 of the crossover frequency:
25 kHz/4 = 6.25 kHz
3
.
8
10
500
10
11
.
1
10
25
141
.
3
2
10
25
.
6
10
25
10
25
2
6
3
3
3
3
3
×
×
×
×
×
×
×
×
×
+
×
×
=
×
π
×
+
=
−
−
REF
OUT
CS
M
OUT
CROSS
ZERO
CROSS
CROSS
COMP
V
V
A
G
C
f
f
f
f
R
×
8
.
0
8
.
1
= 75 kΩ
ZERO
COMP
COMP
f
R
C
π
=
2
1
=
3
3
10
25
.
6
10
75
14
.
3
2
1
×
×
×
×
×
= 340 pF



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