|
MAX15112 Datasheet, PDF (18/23 Pages) Maxim Integrated Products – High-Efficiency, 12A, Current-Mode Synchronous Step-Down Regulator with Integrated Switches | |||
|
◁ |
MAX15112
High-Efficiency, 12A, Current-Mode Synchronous
Step-Down Regulator with Integrated Switches
The peak current-mode controllerâs modulator gain is
attenuated by the equivalent divider ratio of the load
resistance and the current-loop gain. GMOD becomes:
GMOD
=
gMC
Ã
1+
RLOAD
fSWx L
Ã
1
K S
Ã
(1-
D)
-
0.5
where RLOAD = VOUT/IOUT(MAX), fSW is the switching
frequency, L is the output inductance, D is the duty cycle
(VOUT/VIN), and KS is the slope compensation factor
calculated as:
KS
=
1+
VSLOPE Ã
VIN
fSW Ã L
- VOUT
Ã
gMC
where VSLOPE = 130mV and gMC = 80A/V.
The power modulatorâs dominant pole is a function of the
parallel effects of the load resistance and the current-
loop gainâs equivalent impedance. Assuming that ESR
of the output capacitor is much smaller than the parallel
combination of the load and the current loop, fPMOD can
be calculated as:
fPMOD
=
1
2Ï Ã COUT Ã RLOAD
+
[K S Ã (1- D) - 0.5]
2Ï Ã fSW Ã L Ã COUT
The power modulator zero is:
fZMOD
=
fZESR
=
2Ï
Ã
1
COUT
Ã
ESR
The total system transfer can be written as:
GAIN(s) = GFF(s) Ã GEA(s) Ã GMOD(DC)
à GFILTER(s) à GSAMPLING(s)
where:
GFF
(s)
=
R2
R1+ R2
Ã
sCFFR1+ 1
sCFF (R1||R2) +
1
GEA
(s)
=
10
AVEA(dB)/20
Ã
sC
sCCRC + 1
C ï£«ï£¬ï£¬ï£ 10
AVEA(dB)/20
gM


+
1
GFILTER(s) = RLOAD
Ã
sC OUT


ï£
2Ï
Ã
sCOUTESR + 1
1
+ K S Ã (1- D) -
RLOAD 2Ï Ã fSW Ã
0.5
L



-1
+
1
GSAMPLING(s) =
s2
(Ï Ã fSW )2
1
+
s
Ï Ã fSW
à QC
+1
where QC
=
1
Ï Ã [K S Ã (1- D) - 0.5]
The dominant poles and zeros of the transfer loop gain
are:
fP1
<<
2Ï
Ã
CC
gM
Ã10 AVEA(dB)/20
fP2
=
2Ï
Ã
C
OUT


ï£
1
RLOAD
1
+
KS
à (1- D) -
fSW Ã L
0.5 -1


fP3
=
fSW
2
fZ1
=
2Ï
Ã
1
C CR C
fZ2
=
2Ï
Ã
1
C OUTESR
The order of pole occurrence is:
fP1 < fP2 < fZ1 < fCO < fP3 < fZ2
���������������������������������������������������������������� Maxim Integrated Productsââ 18
|
▷ |