Datasheet

Data Sheet ADP1874/ADP1875
Rev. A | Page 33 of 44
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
Choosing R
B
= 1 kΩ as an example, calculate R
T
as follows:
kΩ2
V6.0
V)6.0V8.1(
kΩ1 =
−
×=
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 section
and the Valley Current-Limit Setting section).
The crossover frequency is 1/12 the switching frequency.
300 kHz/12 = 25 kHz
The zero frequency is 1/4 the crossover frequency.
25 kHz/4 = 6.25 kHz
( )
( )
CS
MREF
OUT
L
OUT
OUT
L
ZERO
CROSS
CROSS
COMP
GGV
V
R
CESRs
CESRRs
ff
f
R
11
1
)(1
2
2
2
2
22
×××
××+
++
×
+
=
( )
( )
8.1
15
3.810500
1
6.0
8.1
0011.00035.02521
0011.0)0035.0)158.1((2521
25.625
25
6
22
2
2
22
×
××
×
×
×××π+
×+××π+
×
+
=
−
k
k
kk
K
R
COMP
= 60.25 kΩ
ZERO
COMP
COMP
fR
C
π
=
2
1
=
33
1025.61025.6014.32
1
×××××
= 423 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
)(
+×××××
++×××××=
+×
++×=
−
−
BIAS
lowerFET
SW
BIAS
DR
upperFET
SW
DR
LOSSDR
IVREGCfVREG
IVCfVP
= 57.12 mW
mW6.55
)002.05103.310300()V5V13(
)()(
93
)(
=
+×××××−=
+×××−=
−
BIAS
total
SW
IN
LDODISS
IVREGCfVREGVP
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