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参数资料
| 型号： | IRU3046CFTR |
| 厂商： | International Rectifier |
| 文件页数： | 9/20页 |
| 文件大小： | 0K |
| 描述： | IC REG CTRLR BUCK PWM VM 24TSSOP |
| 标准包装： | 1 |
| PWM 型： | 电压模式 |
| 输出数： | 2 |
| 频率 - 最大： | 450kHz |
| 占空比： | 90% |
| 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： | 24-TSSOP（0.173"，4.40mm 宽） |
| 包装： | 剪切带 (CT) |
| 产品目录页面： | 1384 (CN2011-ZH PDF) |
| 其它名称： | *IRU3046CFTR IRU3046CFCT |
IRU3046
The master error amplifier is a differential-input transcon-
ductance amplifier. The output is available for DC gain
control or AC phase compensation.
First select the desired zero-crossover frequency (Fo):
F O1 > F ESR and F O1 [ (1/5 ~ 1/10) 3 f S
Use the following equation to calculate R 4 :
The E/A can be compensated with or without the use of
local feedback. When operated without local feedback
the transconductance properties of the E/A become evi-
R 4 =
V OSC F O1 3 F ESR
3
V IN(MASTER) F LC2
3
R 5 + R 6
R 5
3
1
g m
---(13)
dent and can be used to cancel one of the output filter
poles. This will be accomplished with a series RC circuit
from Comp1 pin to ground as shown in Figure 6.
The ESR zero of the LC filter expressed as follows:
Where:
V IN(MASTER) = Maximum Input Voltage
V OSC = Oscillator Ramp Voltage
F O1 = Crossover Frequency for the master E/A
F ESR = Zero Frequency of the Output Capacitor
F ESR =
1
2 p3 ESR 3 Co
V OUT
---(9)
F LC(MASTER) = Resonant Frequency of Output Filter
g m = Error Amplifier Transconductor
R 5 and R 6 = Resistor Dividers for Output Voltage
Programming
For:
R 6
R 5
Fb1
E/A1
Comp1
C 9
Ve
V IN(MASTER) = 5V
V OSC = 1.25V
F O1 = 30KHz
F ESR = 25.26KHz
F LC(MASTER) = 3.57KHz
V REF
Gain(dB)
H(s) dB
F Z
R 4
Frequency
R 5 = 1K
R 6 = 200 V
g m = 600 m mho
This results to: R 4 =29.7K V . Choose: R 4 =29.4K V
To cancel one of the LC filter poles, place the zero be-
fore the LC filter resonant frequency pole:
Figure 6 - Compensation network without local
F Z ? 75%F LC(MASTER)
feedback and its asymptotic gain plot.
The transfer function (Ve / V OUT ) is given by:
F Z ? 0.75 3
For:
2 p
1
L O 3 C O
---(14)
(
) 3 1 + sC sR C
H(s) = g m 3
R 5
R 6 + R 5
9
4 9
---(10)
Lo = 2.2 m H
Co = 900 m F
Fz = 2.67KHz
The (s) indicates that the transfer function varies as a
function of frequency. This configuration introduces a gain
and zero, expressed by:
R 4 = 24.9K V
Using equations (12) and (14) to calculate C 9 , we get:
C 9 = 2003pF
|H(s)| = g m 3
R 5
R 6 3 R 5
3 R 4
---(11)
Choose: C 9 = 2200pF
One more capacitor is sometimes added in parallel with
F Z =
1
2 p 3 R 4 3 C 9
---(12)
C 9 and R 4 . This introduces one more pole which is mainly
used to suppress the switching noise. The additional
pole is given by:
The gain is determined by the voltage divider and E/A's
1
transconductance gain.
Rev. 1.9
09/27/02
F P =
www.irf.com
2 p 3 R 4 3
C 9 3 C POLE
C 9 + C POLE
9
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