Datasheet

ADP1872/ADP1873 Data Sheet
Rev. B | Page 28 of 40
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 7 with peak current
handling of 20 A.
2
)(
LOAD
LOSSDCR
IDCRP ×=
= 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 mΩ, choosing 13 A
as the valley current limit from Table 6 and Figure 70 indicates
that a programming resistor (RES) of 100 kΩ 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. The ADP1872’s advantage is, because the
frequency is pseudo-fixed, the converter is able to respond
quickly because of the immediate, though temporary, increase
in switching frequency.
ΔV
DROOP
= 0.05 × 1.8 V = 90 mV
Assuming the overall ESR of the output capacitor ranges from
5 mΩ to 10 mΩ,
)mV90(10300
15
2
)(
2
3
××
×=
∆×
∆
×=
A
Vf
I
C
DROOPSW
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:
( )
( )
22
26
2
2
2
)8.1()mV458.1(
)A15(101
)(
)(
−−
××
=
−∆−
×
=
−
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
10300μF1
)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 mΩ = 3.15 mW
Feedback Resistor Network Setup
It is recommended to use R
B
= 15 kΩ. Calculate R
T
as
kΩ30
V6.0
V)6.0V8.1(
kΩ15 =
−
×=
T
R
Compensation Network
To calculate 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
A/V33.8
005.024
11
=
×
==
ONCS
CS
RA
G
where A
CS
and R
ON
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
th
of the switching frequency:
300 kHz/12 = 25 kHz
The zero frequency is 1/4
th
of the crossover frequency:
25 kHz/4 = 6.25 kHz
6.0
8.1
3.810500
1011.11025141.3
2
1025.61025
1025
2
6
33
33
3
×
××
×××××
×
×+
×
×
=
×
π
×
+
=
−
−
REF
OUT
CS
M
OUT
CROSS
ZERO
CROSS
CROSS
COMP
V
V
GG
Cf
f
f
f
R
= 100 kΩ
ZERO
COMP
COMP
fR
C
π
=
2
1
=
33
1025.61010014.32
1
×××××
= 250 pF