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ADP1829ACPZ-R7 数据表(PDF) 21 Page - Analog Devices

部件名 ADP1829ACPZ-R7
功能描述  Dual, Interleaved, Step-Down DC-to-DC Controller with Tracking
PDF  32 Pages
Scroll/Zoom Zoom In 100%  Zoom Out
制造商  AD [Analog Devices]
网页  http://www.analog.com
标志 AD - Analog Devices

ADP1829ACPZ-R7 数据表(HTML) 21 Page - Analog Devices

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ADP1829
Rev. 0 | Page 21 of 32
Type II Compensator
G
(dB)
PHASE
–180°
–270°
fZ
fP
0V
VRAMP
CHF
CI
RZ
RTOP
RBOT
FROM
VOUT
VREF
EA
COMP
TO PWM
–1
SLO
PE
–1
SLO
PE
Figure 28. Type II Compensation
If the output capacitor ESR zero frequency is sufficiently low (≤1/2
of the crossover frequency), use the ESR to stabilize the
regulator. In this case, use the circuit shown in Figure 28.
Calculate the compensation resistor, RZ, with the following
equation:
2
LC
IN
CO
ESR
RAMP
TOP
Z
f
V
f
f
V
R
R
=
(31)
where:
fCO is chosen to be 1/10 of fSW.
VRAMP is 1.3 V.
Next choose the compensation capacitor to set the
compensation zero, fZ1, to the lesser of 1/4 of the crossover
frequency or 1/2 of the LC resonant frequency:
I
Z
SW
CO
Z
C
R
f
f
f
π
=
=
=
2
1
40
4
1
(32)
or
I
Z
LC
Z
C
R
f
f
π
=
=
2
1
2
1
(33)
Solving for CI in Equation 32 yields
SW
Z
I
f
R
C
π
=
20
(34)
Solving for CI in Equation 33 yields
LC
Z
I
f
R
C
π
=
1
(35)
Use the larger value of CI from Equation 34 or Equation 35.
Because of the finite output current drive of the error amplifier,
CI needs to be less than 10 nF. If it is larger than 10 nF, choose a
larger RTOP and recalculate RZ and CI until CI is less than 10 nF.
Next choose the high frequency pole, fP1, to be 1/2 of fSW.
SW
P1
f
f
2
1
=
(36)
Because CHF << CI, Equation 29 is simplified to
HF
Z
P1
C
R
f
π
=
2
1
(37)
Solving for CHF in Equation 36 and Equation 37 yields
Z
SW
HF
R
f
C
π
=
1
(38)
Type III Compensator
0V
VRAMP
G
(dB)
PHASE
–270°
–90°
fZ
fP
CHF
CI
RZ
CFF
RTOP
RBOT
FROM
VOUT
VREF
EA
COMP
TO PWM
RFF
–1
SLO
PE
–1
SLO
PE
+1
SL
OP
E
Figure 29. Type III Compensation
If the output capacitor ESR zero frequency is greater than 1/2 of
the crossover frequency, use Type III compensator as shown in
Figure 29. Set the poles and zeros as follows:
SW
P2
P1
f
f
f
2
1
=
=
(39)
I
Z
SW
CO
Z2
Z1
C
R
f
f
f
f
π
=
=
=
=
2
1
40
4
(40)
or
I
Z
LC
Z
Z
C
R
f
f
f
π
=
=
=
2
1
2
2
1
(41)
Use the lower zero frequency from Equation 40 or Equation 41.
Calculate the compensator resistor, RZ.
2
LC
IN
CO
Z1
RAMP
TOP
Z
f
V
f
f
V
R
R
=
(42)
Next calculate CI.
Z1
Z
I
f
R
C
π
2
1
=
(43)



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