参数资料
型号: ADP1828ACPZ-R7
厂商: Analog Devices Inc
文件页数: 24/36页
文件大小: 0K
描述: IC REG CTRLR BUCK PWM VM 20LFCSP
标准包装: 1,500
PWM 型: 电压模式
输出数: 1
频率 - 最大: 720kHz
占空比: 93%
电源电压: 3 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: 20-WFQFN 裸露焊盘,CSP
包装: 带卷 (TR)
ADP1828
Use the larger value of C I from Equation 31 or Equation 32.
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 Z1
(40)
f P1 =
Next, choose the high frequency pole, f P1 , to be ? of f SW .
1
f SW
2
Because C HF << C I , Equation 26 is simplified to
(33)
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.
Because C HF << C I , combining Equation 26 and Equation 36
yields
f P1 =
1
2 π R Z C HF
(34)
C HF =
1
π f SW R Z
(41)
Combine Equation 33 and Equation 34, and solve for C HF ,
Next, calculate the feedforward capacitor C FF . Assuming R FF <<
C HF =
f Z2 =
1
π f SW R Z
Type III Compensator
(35)
R TOP , then Equation 25 is simp lified to
1
2 π C FF R TOP
(42)
C FF =
G
(dB)
–90°
–1
SL
O
PE
f Z
+1
SL
O
PE
f P
–1
SL
O
PE
Solving C FF in Equation 42 yields
1
2 π R TOP f Z2
(43)
PHASE
–270°
C HF
where f Z2 is obtained from Equation 37 or Equation 38.
The feedforward resistor, R FF , can be calculated by combining
Equation 27 and Equation 36
R FF =
π C FF f SW
V OUT
R FF
R TO P
R BOT
C FF
FB
R Z
EA
C I
COMP
1
(44)
Check that the calculated component values are reasonable. For
instance, capacitors smaller than about 10 pF should be avoided.
INTERNAL
VREF
Figure 40. Type III Compensation
If the output capacitor ESR zero frequency is greater than ? of
the crossover frequency, use the Type III compensator as shown
in Figure 40. Set the poles and zeros as follows:
In addition, the ADP1828 error amplifier has a finite output
current drive, so R Z values less than 3 kΩ and C I values greater
than 10 nF should be avoided. If necessary, recalculate the compen-
sation network with a different starting value of R TOP . If R Z is too
small or C I is too big, start with a larger value of R TOP . This com-
pensation technique should yield a good working solution.
= =
f P1 = f P2 =
f Z 1 = f Z2 =
1
f SW
2
f CO f SW
4 40
1
2 π R Z C I
(36)
(37)
In general, aluminum electr olytic capacitors have high ESR, and
Type II compensation is adeq uate. However, if several aluminum
electrolytic capacitors are connected in parallel, and produce a
low effecti ve ESR, then Type III compensation is needed. In
addition, ceramic capacitors have very low ESR (only a few
or
f Z 1 = f Z2 =
f LC
2
=
1
2 π R Z C I
(38)
milliohms) making Type III compensation a better choice.
Type III compensation offers better performance than Type II
in terms of more low frequency gain and more phase margin
and less high frequency gain at the crossover frequency.
Use the lower zero frequency from Equation 37 or Equation 38.
Calculate the compensator resistor, R Z
R Z =
R TOP V RAMP f Z1 f CO
V IN f LC 2
(39)
Rev. C | Page 24 of 36
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