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参数资料
| 型号： | IR3651STRPBF |
| 厂商： | International Rectifier |
| 文件页数： | 16/19页 |
| 文件大小： | 0K |
| 描述： | IC REG CTRLR BUCK PWM VM 14-SOIC |
| 标准包装： | 2,500 |
| PWM 型： | 电压模式 |
| 输出数： | 1 |
| 频率 - 最大： | 460kHz |
| 占空比： | 80% |
| 电源电压： | 4.5 V ~ 13.2 V |
| 降压： | 是 |
| 升压： | 无 |
| 回扫： | 无 |
| 反相： | 无 |
| 倍增器： | 无 |
| 除法器： | 无 |
| Cuk： | 无 |
| 隔离： | 无 |
| 工作温度： | 0°C ~ 125°C |
| 封装/外壳： | 14-SOIC（0.154"，3.90mm 宽） |
| 包装： | 带卷 (TR) |
| 配用： | IRDC3651-ND - KIT REF DESIGN SYNCH BUCK REG |
IR3651SPBF
To cancel one of the LC filter poles, place the
zero before the LC filter resonant frequency pole:
Z IN
V O
C 3
F z = 0 . 75 *
F z = 75 % F LC
1
2 π L o * C o
- -(16)
C 7
R 10
R 8
R 3
C 4
Z f
Using equations (15) and (16) to calculate C 4 .
One more capacitor is sometimes added in
R 9
Fb
E/A
Comp
Ve
parallel with C 4 and R 3 . This introduces one more
pole which is mainly used to suppress the
switching noise.
The additional pole is given by:
Gain(dB)
H(s) dB
V REF
2 π * R 3 * 4
F P =
1
C * C POLE
C 4 + C POLE
F Z 1
F Z 2
F P 2
F P 3
Frequency
The pole sets to one half of switching frequency
which results in the capacitor C POLE :
Fig.15: Compensation network with local
feedback and its asymptotic gain plot
C POLE =
For F P <<
1
π * R 3 * F s ?
F s
2
1
C 4
?
1
π * R 3 * F s
As known, transconductance amplifier has high
impedance (current source) output, therefore,
consideration should be taken when loading the
error amplifier output. It may exceed its
source/sink output current capability, so that the
For a general solution for unconditionally stability
for any type of output capacitors, in a wide range
of ESR values we should implement local
feedback with a compensation network (typeIII).
The typically used compensation network for
voltage-mode controller is shown in figure 15.
amplifier will not be able to swing its output
voltage over the necessary range.
The compensation network has three poles and
two zeros and they are expressed as follows:
F P 1 = 0
V e 1 ? g m Z f
2 π * R 3 ? ? 4
? C * C 3 ?
C 4 + C 3 ? ?
In such configuration, the transfer function is
given by:
=
V o 1 + g m Z IN
The error amplifier gain is independent of the
F P 2 =
F P 3 =
1
2 π * R 10 * C 7
1
?
?
?
1
2 π * R 3 * C 3
transconductance under the following condition:
g m * Z f >> 1 and g m * Z in >> 1 - -(17)
By replacing Z in and Z f according to figure 15, the
F z 1 =
F z 2 =
1
2 π * R 3 * C 4
1
2 π * C 7 * ( R 8 + R 10 )
?
1
2 π * C 7 * R 8
transformer function can be expressed as:
Cross over frequency is expressed as:
1 ( 1 + sR 3 C 4 ) * 1 + sC 7 ( R 8 + R 10 ) ]
sR 8 ( C 4 + C 3 ) ?
? C 4 * C 3 ? ?
? 1 + sR 3 ? ? ? * ( 1 + sR 10 C 7 )
V in 1
V osc 2 π * L o * C o
H ( s ) =
*
? ? C 4 + C 3 ? ?
F o = R 3 * C 7 *
*
06/08/2011
16
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