参数资料
型号: IR3086AMPBF
厂商: International Rectifier
文件页数: 28/33页
文件大小: 0K
描述: IC PHASE CONTROLLER OVP 20MLPQ
标准包装: 100
系列: XPhase™
应用: 处理器
电流 - 电源: 10mA
电源电压: 8.4 V ~ 14 V
工作温度: 0°C ~ 125°C
安装类型: 表面贴装
封装/外壳: 20-MLPQ
供应商设备封装: 20-MLPQ(4x4)
包装: 剪切带 (CT)
其它名称: *IR3086AMPBF
IR3086APbF
R PHASE41 =10k , R PHASE42 =768 , R PHASE43 =2.80k
Bootstrap Capacitor C BST
Choose C BST =0.1uF
Decoupling Capacitors for Phase IC and Power Stage
Choose C VCC =0.1uF, C VCCL =0.1uF
VOLTAGE LOOP COMPENSATION
Type III compensation is used for the converter with only ceramic output capacitors. The crossover frequency and
phase margin of the voltage loop can be estimated as follows.
f C 1 =
R DRP
2 π ? C E ? G CS ? R FB ? R LE
=
2 π ? ( 62 ? 22 * 10
? 6
576
) ? 34 ? 162 ? ( 0 . 5 * 10 ? 3 / 6 )
= 146 kHz
θ C 1 = 90 ? A tan( 0 . 5 ) ?
180
π
= 63 °
Choose R FB 1 =
2
3
? R FB =
2
3
? 162 = 110 ?
Choose the desired crossover frequency fc (=140kHz) around fc1 estimated above, and calculate
C FB =
1
4 π ? f C ? R FB 1
=
1
4 π ? 140 * 10 3 ? 110
= 5 . 2 nF , choose C FB =5.6nF
C DRP =
( R FB + R FB 1 ) ? C FB
R DRP
=
( 162 + 110 ) ? 5 . 6 * 10 ? 9
576
= 2 . 7 nF
R CP =
( 2 π ? f C ) 2 ? L E ? C E ? R FB ? V RAMP
V O
=
( 2 π ? 140 * 10 3 ) 2 ? (100 * 10 ? 9 / 6) ? ( 22 * 10 ? 6 ? 62 ) ? 162 * 0. 75
1 . 3 ? 20 * 10 ? 3
= 1 . 65 k ?
C CP =
10 ? L E ? C E
R CP
=
10 ? (100 * 10 ? 9 / 6) ? ( 22 * 10 ? 6 * 62 )
1 . 65 ? 10 3
= 27 nF
Choose C CP1 =47pF to reduce high frequency noise.
CURRENT SHARE LOOP COMPENSATION
The crossover frequency of the current share loop f CI should be at least one decade lower than that of the voltage
loop f C . Choose the crossover frequency of current share loop f CI =3.5kHz , and calculate C SCOMP ,
F MI =
R PWMRMP * C PWMRMP * f SW * V PWMRMP
( V I ? V PWMRMP ? V DAC ) * ( V I ? V DAC )
=
18 .2 * 10 3 * 100 * 10 ? 12 * 800 * 10 3 * 0 .75
( 12 ? 0 . 75 ? 1 . 3 ) * ( 12 ? 1 . 3 )
= 0 . 011
C SCOMP =
=
0. 65 * R PWMRMP * V I * I O * G CS _ ROOM * R LE * [1 + 2 π * f CI * C E * ( V O I O )] * F MI
V O ? 2 π ? f CI * 1 . 05 * 10 6
0 . 65 * 18 .2 * 10 3 * 12 * 105 * 34 * ( 0 . 5 * 10 ? 3 6 ) * [1 + 2 π * 3500 * 22 * 10 ? 6 * 62 * (1. 33 ? 105 * 9 . 1 * 10 ? 4 ) 105] * 0 .011
( 1 . 33 ? 105 * 9 . 1 * 10 ? 4 ) ? 2 π ? 3500 * 1 . 05 * 10 6
= 20 . 6 nF
Choose C SCOMP =22nF
Page 28 of 33
May 13, 2009
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