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MAX15003 Datasheet, PDF (24/32 Pages) Maxim Integrated Products – Triple-Output Buck Controller with Tracking/Sequencing
Triple-Output Buck Controller with
Tracking/Sequencing
Type III: Compensation When fCO < fZERO, ESR
As indicated above, the position of the output capaci-
tor’s inherent ESR zero is critical in designing an appro-
priate compensation network. When low-ESR ceramic
output capacitors are used, the ESR zero frequency
(fZERO, ESR) is usually much higher than unity
crossover frequency (fCO). In this case, a Type III com-
pensation network is recommended (see Figure 7a).
VOUT
CCF
RI R1
CI
R2
RF
CF
-
gM
VREF +
COMP
Figure 7a. Type III Compensation Network
Use the following procedure to calculate the compen-
sation network components.
1) Select a crossover frequency, fCO:
fCO
≤
fSW
10
2) Calculate the LC double-pole frequency, fLC :
fLC
=
2π×
1
L × COUT
3) Select RF ≥ 10kΩ.
4) Place a zero fZ1 = 1
at 0.75 x fLC where
2π x RF x CF
CF
=
1
2π×RF ×0.75×fLC
5) Calculate CI for a target unity-gain crossover fre-
quency, fC:
CI
=
2π×fCO ×L×COUT ×VRAMP
VIN ×RF
Note: CI is derived by setting the total loop gain at
GAIN
(dB)
crossover frequency to unity, e.g., GEA(fCO) x
GMOD(fCO) = 1V/V. The total loop gain can be
expressed logarithmically as follows:
3RD ASYMPTOTE
ωRFCI
1ST ASYMPTOTE
(ωRICF)-1
2ND ASYMPTOTE
(RFRI)-1
4TH ASYMPTOTE
RFRI-1
5TH ASYMPTOTE
(ωRICCF)-1
1ST POLE
(AT ORIGIN)
1ST ZERO
(RFCF)-1
2ND POLE
(RICI)-1
2ND ZERO
(RICI)-1
3RD POLE ω(rad/sec)
(RFCCF)-1
Figure 7b. Type III Compensation Network Response
As shown in Figure 7b, a Type III compensation net-
work introduces two zeros and three poles into the con-
trol loop. The error amplifier has a low-frequency pole
at the origin, two zeros, and higher frequency poles.
The locations of the zeros and poles should be such
that the phase margin peaks at fCO.
Set the ratios of fCO-to-fZ and fP-to-fCO equal to one
another,
e.g.,
fCO
fZ
=fCfPO=
5
is
a
good
number
to
get
about
60° of phase margin at fCO. Whichever technique, it is
important to place the two zeros at or below the double
pole to avoid the conditional stability issue.
[ ] 20 × log10 2π × fCO × RF × CI +
( ) ⎡
20
×
log10
⎢
⎢
⎣
GMOD(DC)
2π × fCO 2 × L × COUT
⎤
⎥
⎥
⎦
=
0dB
6) Place a second zero, fZ2, at or below fLC thereby
determining R1.
R1
=
1
2π×fZ2 ×CI
7) Place a pole (fP1 = 1
), at or below fZERO,ESR.
(2π x R1 x CI)
R1
=
1
2π×fZERO,ESR ×CI
8) Place a second pole (fP2 = 1
) at or below
2π x RF x CCF
one-half the switching frequency.
CCF
=
1
π×fSW ×RF
9) Calculate R2 using the following equation:
R2
=
R1 ×
VFB
VOUT −VFB
where VFB = 0.6V.
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