参数资料
型号: ISL6323BCRZ
厂商: Intersil
文件页数: 31/36页
文件大小: 0K
描述: IC PWM CTRLR SYNC BUCK DL 48QFN
标准包装: 43
应用: 控制器,AMD SVI
输入电压: 5 V ~ 12 V
输出数: 2
输出电压: 最高 2V
工作温度: 0°C ~ 70°C
安装类型: 表面贴装
封装/外壳: 48-VFQFN 裸露焊盘
供应商设备封装: 48-QFN(7x7)
包装: 管件
ISL6323B
.
-------------------------------- > f 0
R 1 = R FB ? --------------------------------------------
R C = R FB ? ----------------------------------------------------------
2 ? π ? V PP ? R FB ? f 0
C 1 = --------------------------------------------
( 2 ? π ) 2 ? f 0 ? f HF ? ( L ? C ) ? R FB ? V P-P
Case 1:
1
2 ? π ? L ? C
2 ? π ? f 0 ? V P-P ? L ? C
0.66 ? V IN
0.66 ? V IN
C C = ----------------------------------------------------
C ? ESR
L ? C – C ? ESR
L ? C – C ? ESR
R FB
0.75 ? V IN
C 2 = -----------------------------------------------------------------------------------------------------
-------------------------------- ≤ f 0 < -------------------------------------
V P-P ? ( 2 ? π ) 2 ? f 02 ? L ? C
R C = R FB ? ------------------------------------------------------------------
V P-P ? ? 2 π ? ? f 0 ? f HF ? L ? C ? R FB
R C = ------------------------------------------------------------------------------------------
0.75 V IN ? ( 2 ? π ? f HF ? L ? C – 1 )
Case 2:
1 1
2 ? π ? L ? C 2 ? π ? C ? ESR
0.66 ? V IN
(EQ. 51)
?
? ?
2
(EQ. 52)
C C = -------------------------------------------------------------------------------------
PP ? R FB ?
( 2 ? π ) 2 ? f 2 ? V L ? C
f 0 > -------------------------------------
( 2 ? π ) 2 ? f 0 ? f HF ? ( L ? C ) ? R FB ? V P-P
Case 3:
0.66 ? V IN
0
1
2 ? π ? C ? ESR
0.75 ? V IN ? ( 2 ? π ? f HF ? L ? C – 1 )
C C = -----------------------------------------------------------------------------------------------------
In the solutions to the compensation equations, there is a
R C = R FB ? ----------------------------------------------
2 ? π ? V P-P ? R FB ? f 0 ? L
-------------------------------- > f 0
2 ? π ? L ? C
R C = R FB ? ----------------------------------------------------------
0.66 ? V
2 ? π ? V PP ? R FB ? f 0
2 ? π ? f 0 ? V P-P ? L
0.66 ? V IN ? ESR
0.66 ? V IN ? ESR ? C
C C = ------------------------------------------------------------------
Compensation Without Loadline Regulation
The non load-line regulated converter is accurately modeled
as a voltage-mode regulator with two poles at the L-C
resonant frequency and a zero at the ESR frequency. A
type III controller, as shown in Figure 23, provides the
necessary compensation.
The first step is to choose the desired bandwidth, f 0 , of the
compensated system. Choose a frequency high enough to
assure adequate transient performance but not higher than 1/3
of the switching frequency. The type-III compensator has an
extra high-frequency pole, f HF . This pole can be used for added
noise rejection or to assure adequate attenuation at the error-
single degree of freedom. For the solutions presented in
Equation 53, R FB is selected arbitrarily. The remaining
compensation components are then selected according to
Equation 53.
In Equation 53, L is the per-channel filter inductance divided
by the number of active channels; C is the sum total of all
output capacitors; ESR is the equivalent-series resistance of
the bulk output-filter capacitance; and V PP is the peak-to-
peak sawtooth signal amplitude as described in Electrical
Specifications on page 9.
1
Case 1:
2 ? π ? f 0 ? V P-P ? L ? C
IN
0.66 ? V IN
C C = ----------------------------------------------------
-------------------------------- ≤ f 0 < -------------------------------------
amplifier high-order pole and zero frequencies. A good general
rule is to choose f HF = 10f 0 , but it can be higher if desired.
Case 2:
1 1
2 ? π ? L ? C 2 ? π ? C ? ESR
R C
V P-P ? ( 2 ? π ) 2 ? f 02 ? L ? C
R C = R FB ? ------------------------------------------------------------------
( 2 ? π ) 2 ? f 02 ? V PP ? R FB ? L ? C
Choosing f HF to be lower than 10f 0 can cause problems with
too much phase shift below the system bandwidth.
C 2
C C
COMP
0.66 ? V IN
0.66 ? V IN
C C = -------------------------------------------------------------------------------------
(EQ. 53)
f 0 > -------------------------------------
FB
Case 3:
1
2 ? π ? C ? ESR
2 ? π ? f 0 ? V P-P ? L
R C = R FB ? ----------------------------------------------
2 ? π ? V P-P ? R FB ? f 0 ? L
C 1
R 1
R FB
VSEN
ISL6323B
0.66 ? V IN ? ESR
0.66 ? V IN ? ESR ? C
C C = ------------------------------------------------------------------
FIGURE 23. COMPENSATION CIRCUIT WITHOUT LOAD-LINE
REGULATION
31
FN6879.1
May 12, 2010
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