Rainbow Electronics MAX15022 - 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