参数资料
型号: ADP1823ACPZ-R7
厂商: Analog Devices Inc
文件页数: 21/32页
文件大小: 0K
描述: IC REG CTRLR BUCK PWM VM 32LFCSP
标准包装: 1
PWM 型: 电压模式
输出数: 2
频率 - 最大: 720kHz
占空比: 90%
电源电压: 3.7 V ~ 20 V
降压:
升压:
回扫:
反相:
倍增器:
除法器:
Cuk:
隔离:
工作温度: -40°C ~ 125°C
封装/外壳: 32-VFQFN 裸露焊盘,CSP
包装: 标准包装
产品目录页面: 791 (CN2011-ZH PDF)
其它名称: ADP1823ACPZ-R7DKR
ADP1823
Type II Compensator
Next choose the high frequency pole f P1 to be half of f SW .
f P 1 =
f P 1 =
G
(dB)
PHASE
–180°
–270°
–1
S
LO
PE
f Z
f P
–1
S
LO
PE
1
f SW
2
Because C HF << C I , Equation 29 is simplified to
1
2 π R Z C HF
(36)
(37)
C HF
Solving for C HF in Equation 36 and Equation 37 yields
C HF =
FROM
V OUT
R TO P
R BOT
R Z
EA
C I
COMP
TO PWM
1
π f SW R Z
Type III Compensator
(38)
VREF
VRAMP
0V
Figure 28. Type II Compensation
If the output capacitor ESR zero frequency is sufficiently low
G
(dB)
–90°
PHASE
–1
S
LO
PE
f Z
+1
S
LO
PE
f P
–1
SL
O
PE
(≤? 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, R Z , with the following
–270°
C HF
equation:
R FF
C FF
R Z
C I
R Z =
R TOP V RAMP f ESR f CO
V IN f LC 2
(31)
FROM
V OUT
R TO P
R BOT
EA
COMP
TO PWM
where:
VREF
VRAMP
f CO is chosen to be 1/10 of f SW .
V RAMP is 1.3 V.
Next choose the compensation capacitor to set the compensa-
tion zero, f Z1 , to the lesser of ? of the crossover frequency or ?
of the LC resonant frequency.
0V
Figure 29. Type III Compensation
If the output capacitor ESR zero frequency is greater than half
of the crossover frequency, use a Type III compensator as
shown in Figure 29. Set the poles and zeros as follows:
= SW =
f Z 1 =
f CO f
4 40
1
2 π R Z C I
(32)
f P 1 = f P 2 =
1
2
f SW
(39)
f Z 1 = LC =
= =
or
f
2
1
2 π R Z C I
(33)
or
f Z 1 = f Z 2 =
f CO f SW
4 40
1
2 π R Z C I
(40)
C I =
f Z 1 = f Z 2 =
=
f LC 1
2 π R Z C I
Solving for C I in Equation 32 yields
20
π R Z f SW
Solving for C I in Equation 33 yields
(34)
(41)
2
Use the lower zero frequency from Equation 40 or Equation 41.
Calculate the compensator resistor, R Z .
C I =
1
π R Z f LC
(35)
R Z =
R TOP V RAMP f Z 1 f CO
V IN f LC 2
(42)
Use the larger value of C I from Equation 34 or Equation 35.
Next calculate C I .
Because of the finite output current drive of the error amplifier,
C I needs to be less than 10 nF. If it is larger than 10 nF, choose a
larger R TOP and recalculate R Z and C I until C I is less than 10 nF.
C I =
1
2 π R Z f Z 1
(43)
Rev. D | Page 21 of 32
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