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参数资料
| 型号： | IRU3038CSTR |
| 厂商： | International Rectifier |
| 文件页数： | 9/18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： | 440kHz |
| 占空比： | 95% |
| 电源电压： | 5 V ~ 12 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 ~ 70°C |
| 封装/外壳： | 14-SOIC（0.154"，3.90mm 宽） |
| 包装： | 带卷 (TR) |
| 产品目录页面： | 1384 (CN2011-ZH PDF) |
IRU3038
For a general solution for unconditionally stability for
F P1 = 0
( )
2 p3 R 7 3
ceramic capacitor with very low ESR and any type of
output capacitors, in a wide range of ESR values we
should implement local feedback with a compensation
network. The typically used compensation network for
voltage-mode controller is shown in Figure 7.
F P2 =
F P3 =
1
2 p3 R 8 3 C 10
1
C 12 3 C 11
C 12 +C 11
?
1
2 p3 R 7 3 C 12
2 p3 R 7 3 C 11
F Z2 = 2 p3 C 10 3 (R 6 + R 8 ) ?
Z IN
C 10
R 8
V OUT
R 6
R 7
C 12
C 11
Z f
F Z1 =
1
1
1
2 p3 C 10 3 R 6
F O = R 7 3 C 10 3
R 5
Fb
Vp=V REF
E/A
Comp
Ve
Cross Over Frequency:
V IN
V OSC
3
1
2 p3 Lo 3 Co
---(15)
Gain(dB)
H(s) dB
Where:
V IN = Maximum Input Voltage
V OSC = Oscillator Ramp Voltage
Lo = Output Inductor
Co = Total Output Capacitors
F Z 1
F Z 2
F P 2
F P 3
Frequency
The stability requirement will be satisfied by placing the
poles and zeros of the compensation network according
Figure 7 - Compensation network with local
feedback and its asymptotic gain plot.
to following design rules. The consideration has been
taken to satisfy condition (14) regarding transconduc-
tance error amplifier.
In such configuration, the transfer function is given by:
V e
V OUT
=
1-
1+
g m Z f
g m Z IN
1) Select the crossover frequency:
Fo < F ESR and Fo [ (1/10 ~ 1/6) 3 f S
The error amplifier gain is independent of the transcon-
ductance under the following condition:
2) Select R 7 , so that R 7 >>
2
g m
g m Z f >> 1
and
g m Z IN >>1
---(14)
3) Place first zero before LC’s resonant frequency pole.
F Z1 ? 75% F LC
By replacing Z IN and Z f according to Figure 7, the trans-
former function can be expressed as:
C 11 =
1
2 p 3 F Z1 3 R 7
( )]
1+sR 7 C 12 +C 11
H(s) =
1
sR 6 (C 12 +C 11 )
3
[
(1+sR 7 C 11 ) 3 [1+sC 10 (R 6 +R 8 )]
C 12 C 11
3 (1+sR 8 C 10 )
4) Place third pole at the half of the switching frequency.
f S
F P3 =
2
As known, transconductance amplifier has high imped-
ance (current source) output, therefore, consider should
C 12 =
1
2 p 3 R 7 3 F P3
be taken when loading the E/A output. It may exceed its
source/sink output current capability, so that the ampli-
fier will not be able to swing its output voltage over the
necessary range.
C 12 > 50pF
If not, change R 7 selection.
5) Place R 7 in equation (15) and calculate C 10 :
The compensation network has three poles and two ze-
ros and they are expressed as follows:
C 10 [
2 p 3 Lo 3 F O 3 Co
R 7
3
V OSC
V IN
Rev. 2.0
09/12/02
www.irf.com
9
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