参数资料
型号: IR3802AMTRPBF
厂商: International Rectifier
文件页数: 14/21页
文件大小: 0K
描述: IC REG BUCK SYNC ADJ 6A 15QFN
产品培训模块: SupIRBuck? Family and POL Design Tools Overview
标准包装: 1
系列: SupIRBuck™
类型: 降压(降压)
输出类型: 可调式
输出数: 1
输出电压: 0.6 V ~ 12 V
输入电压: 2.5 V ~ 21 V
PWM 型: 电压模式
频率 - 开关: 300kHz
电流 - 输出: 6A
同步整流器:
工作温度: -40°C ~ 125°C
安装类型: 表面贴装
封装/外壳: 15-PowerVQFN
包装: 标准包装
供应商设备封装: PQFN(5x6)
产品目录页面: 1384 (CN2011-ZH PDF)
其它名称: IR3802AMTRPBFDKR
PD-60357
IR3802AMPbF
To cancel one of the LC filter poles, place the
zero before the LC filter resonant frequency pole:
Z IN
V OUT
C 3
F z = 0 . 75 *
F z = 75 % F LC
1
2 π L o * C o
- - - (17)
C 7
R 10
R 8
R 3
C 4
Z f
Use equations (15) and (16) to calculate C4.
One more capacitor is sometimes added in
parallel with C4 and R3. This introduces one
R 9
Fb
E/A
Comp
Ve
more pole which is mainly used to suppress the
switching noise.
Gain(dB)
V REF
F P =
2 π * R 3 * 4
The additional pole is given by:
1
C * C POLE
C 4 + C POLE
H(s) dB
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
=
C POLE
For F P <<
1
π * R 3 * F s ?
F s
2
1
C 4
?
1
π * R 3 * F s
feedback and its asymptotic gain plot
As known, a transconductance amplifier has high
impedance (current source) output, therefore,
consideration should be taken when loading the
error amplifier output. It may exceed its
For a general solution for unconditional stability
for any type of output capacitors, in a wide range
of ESR values we should implement local
feedback with a compensation network (type III).
The typically used compensation network for
voltage-mode controller is shown in figure 15.
In such a configuration, the transfer function is
source/sink output current capability, so that the
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
1 ? g m Z f
V e
2 π * R 3 ? ? 4
? C * C 3 ?
C 4 + C 3 ? ?
given by:
=
V o 1 + g m Z IN
The error amplifier gain is independent of the
transconductance under the following condition:
g m * Z f >> 1 and g m * Z in >> 1
- - - (18)
F P 2 =
F P 3 =
F z 1 =
1
2 π * R 10 * C 7
1
?
1
2 π * R 3 * C 4
?
?
1
2 π * R 3 * C 3
By replacing Z in and Z f according to figure 15, the
transformer function can be expressed as:
F z 2 =
1
2 π * C 7 * ( R 8 + R 10 )
?
1
2 π * C 7 * R 8
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 )
F o = R 3 * C 7 *
V in 1
V osc 2 π * L o * C o
H ( s ) =
*
? ? C 4 + C 3 ? ?
Cross over frequency is expressed as:
*
11/04/08
14
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