Datasheet
ADP1870/ADP1871 Data Sheet
Rev. B | Page 30 of 44
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 follows:
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
1011.11025141.32
1025.61025
10
25
2
6
3
3
33
3
×
××
×××××
×
×+×
×
=
×
π
×
+
=
−
−
REF
OUT
CS
M
OUT
CROSS
ZERO
CROSS
CROSS
COMP
V
V
GG
Cf
ff
f
R
= 100 kΩ
ZERO
COMP
COMP
fR
C
π
=
2
1
=
33
1025.61010014.32
1
×××××
= 250 pF
Loss Calculations
Duty cycle = 1.8/12 V = 0.15
R
ON (N2)
= 5.4 mΩ
t
BODY(LOSS)
= 20 ns (body conduction time)
V
F
= 0.84 V (MOSFET forward voltage)
C
IN
= 3.3 nF (MOSFET gate input capacitance)
Q
N1,N2
= 17 nC (total MOSFET gate charge)
R
GATE
= 1.5 Ω (MOSFET gate input resistance)
( )
[ ]
2
1
LOAD
N2(ON)N1(ON)N1,N2(CL)
IRDRDP ××−+×=
= (0.15 × 0.0054 + 0.85 × 0.0054) × (15 A)
2
= 1.215 W
2
)(
)(
×××=
F
LOAD
SW
LOSSBODY
LOSSBODY
VI
t
t
P
= 20 ns × 300 × 10
3
× 15 A × 0.84 × 2
= 151.2 mW
P
SW(LOSS)
= f
SW
× R
GATE
× C
TOTAL
× I
LOAD
× V
IN
× 2
= 300 × 10
3
× 1.5 Ω × 3.3 × 10
−9
× 15 A × 12 × 2
= 534.6 mW
( )
[ ]
( )
[ ]
))002.00.5103.310300(0.5(
))002.062.4103.310300(62.4(
93
93
)(
+×××××+
+×××××=
+×+
+×=
−
−
BIASREG
lowerFET
SWREG
BIAS
DR
upperFET
SW
DR
LOSSDR
IVCfV
IVCfVP
= 57.12 mW
mW6.55
)002.05103.310
300()V5V13(
)()(
93
)(
=
+×××××−=
+×
××−=
−
BIASREG
total
SWREG
IN
LDODISS
IVCfVVP
P
COUT
= (I
RMS
)
2
× ESR = (1.5 A)
2
× 1.4 mΩ = 3.15 mW
2
)(
LOAD
LOSSDCR
IDCRP ×=
= 0.003 × (15 A)
2
= 675 mW
P
CIN
= (I
RMS
)
2
× ESR = (7.5 A)
2
× 1 mΩ = 56.25 mW
P
LOSS
= P
N1,N2
+ P
BODY(LOSS)
+ P
SW
+ P
DCR
+ P
DR
+ P
DISS(LDO)
+ P
COUT
+ P
CIN
= 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