Datasheet

pole, which also results in lower crossover frequency.
The following formula gives the crossover frequency
as a function the MOSFET R
DS(ON)
and the output
capacitance:
m FB
CROSSOVER
VCS OUT(SET) OUT DS(ON)
g V R11
f
2µ A V C R
××
=
×× ××
Change one or both of these circuit parameters to
obtain the desired crossover. Recalculate ADC and
repeat steps 1 to 3 after making the changes.
4) If f
POLE(HIGH)
is less than the crossover frequency,
cancel the pole with a feed-forward zero. Determine
the value of C23 (feedback capacitor) using the follow-
ing:
POLE(HIGH)
1
C23
2µ f R1
≈
××
C23 also forms a secondary pole with R1 and R2
given by the following:
( )
POLE_SEC
1
f
2 µ R1 | | R 2 C 2 3
=
××
The frequency of this pole should be above the cross-
over frequency for loop stability. The position of this
pole is related to the high-frequency currentmode pole,
which is determined by the inductor current ramp sig-
nal. The inductor current ramp signal must satisfy the
following condition to ensure the pole occurs above
the crossover frequency:
( )
CROSSOVER C
SW CROSSOVER
2 D' R2 f m
m1
R1 R2 f - 2 D' R2 f
π× × × ×
>
+ × π× × ×
If the frequency of the secondary pole is below the
crossover frequency, the frequency of the second-
ary pole must be moved higher, or the crossover
frequency must be moved lower. There are two ways
to increase the frequency of the secondary pole:
increase the high-side MOSFET R
DS(ON)
, or reduce
the step-down inductance, L. As explained before, for
given input and output voltages, the current ramp sig-
nal is proportional to the high-side MOSFET R
DS(ON)
,
and inversely proportional to the inductance. If the
pole occurs below the crossover frequency, the cur-
rent feedback signal is too small. Increasing R
DS(ON)
or reducing the inductance can increase the current
feedback signal. To lower the crossover frequency, use
the methods described in step 3. Repeat steps 1 to 4
after making the changes.
5) For most applications using tantalum or polymer
capacitors, the output capacitor’s ESR forms a second
zero that occurs either below or close to the crossover
frequency. The zero must be cancelled with a pole.
Verify the frequency of the output capacitor’s ESR
zero, which is:
ZERO(ESR)
OUT ESR
1
f
2µ C R
=
××
where R
ESR
is the ESR of the output capacitor C
OUT
.
If the output capacitor’s ESR zero does not occur well
after the crossover, add the parallel compensation
capacitor (C2) to form another pole to cancel the ESR
zero. Calculate the value of C2 using:
ZERO(ESR)
C10
C2
2µ f R11 C10-1
≈
× ××
Applications using ceramic capacitors usually have ESR
zeros that occur at least one decade above the crossover.
Since the ESR zero of ceramic capacitors has little effect
on the loop stability, it does not need to be cancelled.
The following is an example. In the circuit of Figure 1, the
input voltage is 12V, the output voltage is set to 3.3V, the
maximum load current is 1.5A, the typical onresistance
of the high-side MOSFET is 100mΩ, and the inductor is
10μH. The calculated equivalent load resistance is 1.67Ω.
The DC loop gain is:
DC
1.238V 1.67 2000
A 4180
3.3V 100m 3.5
× Ω×
≈=
× Ω×
If the chosen crossover frequency is 20kHz (step 1):
100 S 4180
C10 1.7nF
2µ 20kHz 2000
µ×
≈≈
××
With a 22μF output capacitor, the output pole of the step-
down regulator is (step 2):
POLE(OUT)
1
f 4.3kHz
2µ 22µF 1.67
= =
× ×Ω
Calculate R11 using:
1
R11 22k
2µ 4.3kHz 1.7nF
≈=Ω
××
MAX1530/MAX1531 Multiple-Output Power-Supply
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