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ADP1828YRQZ-R7 数据表(PDF) 22 Page - Analog Devices

部件名 ADP1828YRQZ-R7
功能描述  Synchronous Buck PWM, Step-Down, DC-to-DC Controller
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

ADP1828YRQZ-R7 数据表(HTML) 22 Page - Analog Devices

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ADP1828
Rev. 0 | Page 22 of 32
Use the following guidelines for selecting between Type II and
Type III compensators:
If
2
CO
ESRZ
f
f
≤
, use Type II compensation.
If
2
CO
ESRZ
f
f
>
, use Type III compensation.
GAIN
FREQUENCY
PHASE
PHASE CONTRIBUTION AT CROSSOVER
OF VARIOUS ESR ZERO CORNERS
fSW
fCO
fESR3
fESR2
fESR1
0dB
fLC
–40dB/dec
–20dB/dec
0°
–90°
–180°
Φ1
Φ2
Φ3
Figure 37. LC Filter Bode Plot
The following equations are used for the calculation of the
compensation components as shown in Figure 38 and Figure 39:
I
Z
Z1
C
R
f
π
=
2
1
(24)
)
(
2
1
FF
TOP
FF
Z2
R
R
C
f
+
π
=
(25)
HF
I
HF
I
Z
P1
C
C
C
C
R
f
+
π
=
2
1
(26)
FF
FF
P2
C
R
f
π
=
2
1
(27)
where:
fZ1 is the zero produced in the Type II compensation.
fZ2 is the zero produced in the Type III compensation.
fP1 is the pole produced in the Type II compensation.
fP2 in the pole produced in the Type III compensation.
Type II Compensator
fZ
fP
CHF
CI
FB
RZ
RBOT
RTOP
VOUT
INTERNAL
VREF
EA
COMP
–1
SLO
PE
–1
SLO
PE
PHASE
–180°
–270°
G
(dB)
Figure 38. Type II Compensation
If the output capacitor ESR zero frequency is sufficiently low
(≤½ of the crossover frequency), use the ESR to stabilize the
regulator. In this case, use the circuit shown in Figure 38.
Calculate the compensation resistor, RZ, with the following
equation:
2
LC
IN
CO
ESR
RAMP
TOP
Z
f
V
f
f
V
R
R =
(28)
where:
fCO is chosen to be 1/10 of fSW.
VRAMP is 1.0 V.
Next, choose the compensation capacitor to set the compensa-
tion zero, fZ1, to the lesser of ¼ of the crossover frequency or ½
of the LC resonant frequency
I
Z
SW
CO
Z1
C
R
f
f
f
π
=
=
=
2
1
40
4
(29)
or
I
Z
LC
Z1
C
R
f
f
π
=
=
2
1
2
(30)
Solving for CI in Equation 29 yields
SW
Z
I
f
R
C
π
=
20
(31)
Solving for CI in Equation 30 yields
LC
Z
I
f
R
C
π
=
1
(32)



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