参数资料
型号: IR3504MTRPBF
厂商: International Rectifier
文件页数: 33/42页
文件大小: 0K
描述: IC CTRL XPHASE3 SVID 32-MLPQ
标准包装: 3,000
系列: XPhase3™
应用: 处理器
电流 - 电源: 10mA
电源电压: 4.75 V ~ 7.5 V
工作温度: 0°C ~ 100°C
安装类型: 表面贴装
封装/外壳: 32-VFQFN 裸露焊盘
供应商设备封装: 32-MLPQ(5x5)
包装: 带卷 (TR)
IR3504
Type III Compensation for AVP Applications
Determine the compensation at no load, the worst case condition. Assume the time constant of the resistor and
capacitor across the output inductors matches that of the inductor, the crossover frequency and phase margin of
the voltage loop can be estimated by (25) and (26), where R LE is the equivalent resistance of inductor DCR.
f C 1 =
R DRP
2 π * C E ? G CS * R FB ? R LE
(25)
θ C 1 = 90 ? A tan( 0 . 5 ) ?
180
π
(26)
Choose the desired crossover frequency fc around fc1 estimated by (25) or choose fc between 1/10 and 1/5 of
the switching frequency per phase, and select the components to ensure the slope of close loop gain is -20dB
/Dec around the crossover frequency. Choose resistor R FB1 according to (27), and determine C FB and C DRP from
(28) and (29).
R FB 1 =
1
2
R FB
to
R FB 1 =
2
3
R FB
(27)
C FB =
1
4 π ? f C ? R FB 1
(28)
C DRP =
( R FB + R FB 1 ) ? C FB
R DRP
(29)
R CP and C CP have limited effect on the crossover frequency, and are used only to fine tune the crossover
frequency and transient load response. Determine R CP and C CP from (30) and (31).
R CP =
C CP =
( 2 π ? f C ) 2 ? L E ? C E ? R FB ? 5
V I
10 ? L E ? C E
R CP
(30)
(31)
C CP1 is optional and may be needed in some applications to reduce the jitter caused by the high frequency noise.
A ceramic capacitor between 10pF and 220pF is usually enough.
Type III Compensation for Non-AVP Applications
Resistor R DRP and capacitor C DRP are not needed. Choose the crossover frequency fc between 1/10 and 1/5 of
the switching frequency per phase and select the desired phase margin θ c. Calculate K factor from (32), and
determine the component values based on (33) to (37),
K = tan[ ? ( C + 1 . 5 )]
4 180
π θ
(32)
R CP = R FB ?
( 2 π ? L E ? C E ? f C ) 2 ? 5
V I ? K
(33)
Page 33
C CP =
C CP 1 =
K
2 π ? f C ? R CP
1
2 π ? f C ? K ? R CP
(34)
(35)
July 28, 2009
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