Rainbow Electronics MAX15022 User Manual
Page 21
MAX15022
Dual, 4A/2A, 4MHz, Step-Down DC-DC
Regulator with Dual LDO Controllers
______________________________________________________________________________________
21
The locations of the zeros and poles should be such
that the phase margin peaks around f
CO
.
Set the ratios of f
CO
-to-f
Z
and f
P
-to-f
CO
equal to one anoth-
er, e.g., f
CO
= f
P
= 5 is a good number to get approximately
f
Z
f
CO
60° of phase margin at f
CO
. Whichever technique, it is
important to place the two zeros at or below the double
pole to avoid the conditional stability issue.
The following procedure is recommended:
1) Select a crossover frequency, f
CO
, at or below one-
tenth the switching frequency (f
SW
):
2) Calculate the LC double-pole frequency, f
LC
:
where C
OUT
is the output capacitor of the regulator.
3) Select the feedback resistor, R
F
, in the range of
3.3k
Ω to 30kΩ.
4) Place the compensator’s first zero
at or below the output filter’s dou-
ble-pole, f
LC
, as follows:
5) The gain of the modulator (Gain
MOD
)—comprised of
the regulator’s PWM, LC filter, feedback divider, and
associated circuitry—at the crossover frequency is:
The gain of the error amplifier (Gain
E/A
) in midband fre-
quencies is:
The total loop gain is the product of the modulator gain
and the error amplifier gain at f
CO
should be equal to 1,
as follows:
Gain
MOD
x Gain
E/A
= 1
So:
Solving for C
I
:
6) For those situations where f
LC
< f
CO
< f
ESR
< f
SW
/2,
as with low-ESR tantalum capacitors, the compen-
sator’s second pole (f
P2
) should be used to cancel
f
ESR
. This provides additional phase margin. On the
system Bode plot, the loop gain maintains its
+20dB/decade slope up to 1/2 of the switching fre-
quency verses flattening out soon after the 0dB
crossover. Then set:
f
P2
= f
ESR
If a ceramic capacitor is used, then the capacitor ESR
zero, f
ESR
, is likely to be located even above 1/2 of the
switching frequency, that is f
LC
< f
CO
< f
SW
/2 < f
ESR
. In
this case, the frequency of the second pole (f
P2
) should
be placed high enough not to significantly erode the
phase margin at the crossover frequency. For example,
f
P2
can be set at 5 x f
CO
, so that its contribution to phase
loss at the crossover frequency f
CO
is only about 11°:
f
P2
= 5 x f
CO
Once f
P2
is known, calculate R
I
:
7) Place the second zero (f
Z2
) at 0.2 x f
CO
or at f
LC
,
whichever is lower, and calculate R
1
using the fol-
lowing equation:
8) Place the third pole (f
P3
) at 1/2 the switching fre-
quency and calculate C
CF
from:
9) Calculate R
2
as:
where V
FB
= 0.6V (typ).
R [k ] R [k ]
V
[V]
V
[V] V
[V]
2
1
FB
OUT_
FB
Ω
Ω
=
×
−
C
[ F]
1
2
0.5 f
[MHz] R [k ]
CF
SW
F
n
=
Ч
Ч
Ч
(
)
π
Ω
R [k ]
1
2
f
[kHz] C [ F]
1
Z2
I
Ω =
Ч
Ч
π
μ
R [k ]
1
2
f
[kHz] C [ F]
I
P2
I
Ω =
Ч
Ч
π
μ
C pF]
2
f
[kHz] L[ H] C
[ F]
4 R [k ]
I
CO
OUT
F
[
=
Ч
Ч
Ч
(
)
Ч
π
μ
μ
Ω
4
1
(2
f
[kHz])
C
[ F] L[ H]
2
f
[kHz] C [ F] R [k ]
1
CO
2
OUT
CO
I
F
Ч
Ч
Ч
Ч
Ч
Ч
Ч
Ч
=
π
μ
μ
π
p
Ω
Gain
2
f
[kHz] C [ F] R [k ]
E/A
CO
I
F
=
Ч
Ч
Ч
π
μ
Ω
Gain
4
1
(2
f
[MHz])
L[ H] C
[ F]
MOD
CO
2
OUT
= Ч
Ч
Ч
Ч
π
μ
μ
C [ F]
1
2
R [k ] 0.5 f
[kHz]
F
F
LC
μ
π
=
Ч
Ч
Ч
Ω
f
[MHz]
1
2
L[ H] C
F]
LC
OUT
≈
Ч
Ч
π
μ
μ
[
f
[kHz]
f
[kHz]
10
CO
SW
≤
f
1
2
R
C
Z1
F
F
=
Ч
Ч
π