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參數(shù)資料
| 型號(hào)： | MAX15003 |
| 廠商： | Maxim Integrated Products, Inc. |
| 英文描述： | Triple-Output Buck Controller with Tracking/Sequencing |
| 中文描述： | 三輸出buck控制器，提供跟蹤/排序功能 |
| 文件頁數(shù)： | 24/32頁 |
| 文件大?。?/td> | 945K |
| 代理商： | MAX15003 |
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M
Triple-Output Buck Controller with
Tracking/Sequencing
24
______________________________________________________________________________________
Type III: Compensation When f
CO
< f
ZERO, 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
(f
ZERO, ESR
) is usually much higher than unity
crossover frequency (f
CO
). In this case, a Type III com-
pensation network is recommended (see Figure 7a).
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 f
CO
.
Set the ratios of f
CO
-to-f
Z
and f
P
-to-f
CO
equal to one
another, e.g., f
CO
= f
P
=5is a good number to get about
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.
Use the following procedure to calculate the compen-
sation network components.
1) Select a crossover frequency, f
CO
:
2) Calculate the LC double-pole frequency, f
LC
:
f
LC
=
2
π×
3) Select R
F
≥
10k
.
4) Place a zero f
Z1
= 1
at 0.75 x f
LC
where
2
π
x R
F
x C
F
5) Calculate C
I
for a target unity-gain crossover fre-
quency, f
C
:
fCOL COUT VRAMP
VINRF
Note:
C
I
is derived by setting the total loop gain at
crossover frequency to unity, e.g., G
EA
(f
CO
) x
G
MOD
(f
CO
) = 1V/V. The total loop gain can be
expressed logarithmically as follows:
6) Place a second zero, f
Z2
, at or below f
LC
thereby
determining R
1
.
7) Place a pole (f
P1
=
1
), at or below f
ZERO,ESR
.
(2
π
x R
1
x C
I
)
8) Place a second pole (f
P2
= 1 ) at or below
2
π
x R
F
x C
CF
one-half the switching frequency.
9) Calculate R2 using the following equation:
where V
FB
= 0.6V.
R
R
VFB
VOUTVFB
2
1
=
×
C
CF
fSWRF
=
×
×
1
π
R
fZEROESRCI
,
1
1
2
=
×
×
π
R
fZ
CI
1
1
2
2
=
×
×
π
20
2
20
2
0
10
10
2
×
×
×
×
[
]
+
×
×
(
)
×
×
log
log
(
)
π
π
f
R
C
G
f
L
C
dB
CO
F
I
MOD DC
CO
OUT
C
I
=
×
× ×
×
×
2
π
C
F
RF
fLC
=
×
×
×
1
0 75
.
2
π
L COUT
×
1
f
CO
fSW
10
≤
R1
R
F
COMP
V
OUT
V
REF
R2
R
I
C
I
C
F
C
CF
-
+
g
M
Figure 7a. Type III Compensation Network
Figure 7b. Type III Compensation Network Response
GAIN
(dB)
1ST ASYMPTOTE
(
ω
R
I
C
F
)
-1
3RD ASYMPTOTE
ω
R
F
C
I
5TH ASYMPTOTE
(
ω
R
I
C
CF
)
4TH ASYMPTOTE
R
F
R
I-1
ω
(rad/sec)
2ND ASYMPTOTE
(R
F
R
I
)
-1
1ST POLE
(AT ORIGIN)
2ND POLE
(R
I
C
I
)
-1
3RD POLE
(R
F
C
CF
)
-1
1ST ZERO
(R
F
C
F
)
-1
2ND ZERO
(R
I
C
I
)
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