Datasheet

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
V2.13
V8.1
10300V5
)V8.1V2.13(
)(
3
,
×
××
−
=
×
×Δ
−
=
IN,MAX
OUT
SW
L
OUT
MAXIN
V
V
fI
VV
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 m (7443552100) from Table 8, with peak current
handling of 20 A.
P
DCR(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 R
ON
of 4.5 m, choosing 13 A
as the valley current limit from Table 7 and Figure 71 indicates
that a programming resistor (RES) of 100 k corresponds to an
A
CS
of 24 V/V.
Choose a programmable resistor of R
RES
= 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.
V
DROOP
= 0.05 × 1.8 V = 90 mV
Assuming the overall ESR of the output capacitor ranges from
5 m to 10 m,
)mV90(10300
15
2
)(
3
××
×=
A
Vf
DROOPSW
()
()
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 m.
Assuming an overshoot of 45 mV, determine if the output
capacitor that was calculated previously is adequate.
22
26
2
2
2
)8.1()mV458.1(
)A15(101
)(
)(
−−
××
=
−Δ−
×
=
−
OUTOVSHTOUT
LOAD
OUT
VVV
IL
C
= 1.4 mF
Choose five 270 µF polymer capacitors.
The rms current through the output capacitor is
A49.1
V2.13
V8.1
10300F1
)V8.1V2.13(
3
1
2
1
)(
3
1
2
1
3
,
,
=×
××
−
×=
×
×
−
×=
MAXIN
OUT
SW
OUT
MAXIN
RMS
V
V
fL
VV
I
The power loss dissipated through the ESR of the output
capacitor is
P
COUT
= (I
RMS
)
2
× ESR = (1.5 A)
2
× 1.4 m = 3.15 mW
Feedback Resistor Network Setup
It is recommended that R
B
= 15 k be used. Calculate R
T
as
follows:
R
T
= 15 kΩ ×
V6.0
V)6.0V8.1(
−
= 30 kΩ
Compensation Network
To calc u l ate R
COMP
, C
COMP
, and C
PAR
, the transconductance
parameter and the current-sense gain variable are required. The
transconductance parameter (G
M
) is 500 µA/V, and the current-
sense loop gain is
G
CS
=
A/V7.7
005.026
11
=
×
=
ONCS
RA
where A
CS
and R
ON
are taken from setting up the current limit
(see the Programming Resistor (RES) Detect Circuit and Val le y
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
3.810500
1011.11025141.32
1025.61025
1025
2
6
33
33
3
××
×××××
×
×+×
×
=
×
π
×
+
=
−
−
REF
OUT
CS
M
OUT
CROSS
ZERO
CROSS
CROSS
COMP
V
V
AG
Cf
ff
f
R
×
8.0
8.1
= 75 k
ZERO
COMP
COMP
fR
C
π
=
2
1
=
33
1025.6107514.32
1
×××××
= 340 pF