Datasheet

17Maxim Integrated
High-Efficiency, 4A, Step-Down DC-DC
Regulators with Internal Power Switches
MAX15066/MAX15166
As previously mentioned, the power modulator’s dominate
pole is a function of the parallel effects of the load resis-
tance and the current-loop gain’s equivalent impedance:
( )
PMOD
1
S
OUT
LOAD SW
1
f
K 1 D 0.5
1
2 C ESR
R fL
−
=



×− −


π× × + +



×



Knowing that the ESR is typically much smaller than the
parallel combination of the load and the current loop,
e.g.,:
( )
1
S
LOAD SW
K 1 D 0.5
1
ESR
R fL
−


×− −


<< +

×

( )
PMOD
1
S
OUT
LOAD SW
1
f
K 1 D 0.5
1
2C
R fL
−
≈


×− −

π× × +


×

This can be expressed as:
( )
S
PMOD
OUT LOAD SW OUT
K 1 D 0.5
1
f
2 C R 2 f LC

×− −

≈+
π× × π× × ×
Note: Depending on the application’s specifics, the
amplitude of the slope compensation ramp could have a
significant impact on the modulator’s dominate pole. For
low duty-cycle applications, it provides additional damp-
ing (phase lag) at/near the crossover frequency. See the
Closing the Loop: Designing the Compensation Circuitry
section. There is no equivalent effect on the power
modulator zero:
ZMOD ZESR
OUT
1
ff
2 C ESR
= =
π× ×
The effect of the inner current loop at higher frequencies
is modeled as a double-pole (complex conjugate)
frequency term, G
SAMPLING
(s), as shown:
( )
( )
SAMPLING
2
2
SW C
SW
1
Gs
ss
1
fQ
f
=
++
π× ×
π×
where the sampling effect quality factor, Q
C
, is:
( )
C
S
1
Q
K 1 D 0.5
=

π× × − −

and the resonant frequency is:
w
SAMPLING
(s) = π × f
SW
or:
SW
SAMPLING
f
f
2
=
Having defined the power modulator’s transfer function,
the total system transfer can be written as follows
(Figure 3):
Gain(s) = G
FF
(s) × G
EA
(s) × G
MOD
(DC) ×
G
FILTER
(s) × G
SAMPLING
(s)
where:
( )
( )
( )
FF
FF
FF
sC R1 1
R2
Gs
R1 R2
sC R1 | | R2 1
+
= ×
+

+

Leaving C
FF
empty, G
FF
(s) becomes:
( )
FF
R2
Gs
R1 R2
=
+
Also:
( )
( )
CC
AVEA(dB)/20
EA
AVEA(dB)/20
CC
MV
sC R 1
G s 10
10
sC R 1
g
+
= ×



 + +




If R
C
<<
AVEA(dB)/20
MV
10
, the equation simplifies to:
( )
( )
CC
AVEA(dB)/20
EA
AVEA(dB)/20
C
MV
sC R 1
G s 10
10
sC 1
g
+
= ×




+




( )
( )
( )
OUT
FILTER LOAD
1
S
OUT
LOAD SW
sC ESR 1
G sR
K 1 D 0.5
1
sC 1
2R 2f L
−
+
= ×



×− −



++


π× π× ×



