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参数资料
| 型号： | IRU3038CSTR |
| 厂商： | International Rectifier |
| 文件页数： | 8/18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
Note that this method requires that the output capacitor
should have enough ESR to satisfy stability requirements.
In general, the output capacitor’s ESR generates a zero
typically at 5KHz to 50KHz which is essential for an
acceptable phase margin.
The ESR zero of the output capacitor expressed as fol-
lows:
Where:
V IN = Maximum Input Voltage
V OSC = Oscillator Ramp Voltage
Fo = Crossover Frequency
F ESR = Zero Frequency of the Output Capacitor
F LC = Resonant Frequency of the Output Filter
R 5 and R 6 = Resistor Dividers for Output Voltage
Programming
F ESR =
1
2 p3 ESR 3 Co
---(8)
g m = Error Amplifier Transconductance
For:
V OUT
V IN = 5V
V OSC = 1.25V
R 6
Fb
Fo = 30KHz
F ESR = 26.5KHz
R 5
Vp=V REF
E/A
Comp
C 9
R 4
Ve
F LC = 5KHz
R 5 = 1K
R 6 = 1K
g m = 600 m mho
Gain(dB)
H(s) dB
This results to R 4 =26.52K V . Choose R 4 =26.1K V
To cancel one of the LC filter poles, place the zero be-
fore the LC filter resonant frequency pole:
F Z ? 75%F LC
F Z
Frequency
Figure 6 - Compensation network without local
feedback and its asymptotic gain plot.
F Z ? 0.75 3
For:
2 p
1
L O 3 C O
---(13)
The transfer function (Ve / V OUT ) is given by:
Lo = 10 m H
Co = 300 m F
(
)
H(s) = g m 3
R 5
R 6 + R 5
3
1 + sR 4 C 9
sC 9
---(9)
F Z = 3.8KHz
R 4 = 26.1K V
Using equations (11) and (13) to calculate C 9 , we get:
|H(s)| = g m 3
3 R 4
---(10)
2 p 3 R 4 3 C 9
C 9 3 C POLE
The (s) indicates that the transfer function varies as a
function of frequency. This configuration introduces a gain
and zero, expressed by:
R 5
R 6 3 R 5
1
F Z = ---(11)
The gain is determined by the voltage divider and E/A's
transconductance gain.
First select the desired zero-crossover frequency (Fo):
C 9 ? 1800pF
One more capacitor is sometimes added in parallel with
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:
1
F P =
2 p 3 R 4 3
C 9 + C POLE
The pole sets to one half of switching frequency which
results in the capacitor C POLE:
Fo 3 F ESR
g m
R 4 = 3 3 3
Fo > F ESR and F O [ (1/5 ~ 1/10) 3 f S
Use the following equation to calculate R 4 :
V OSC R 5 + R 6 1
V IN F LC2 R 5
---(12)
1
C POLE =
p 3 R 4 3 f S - 1
C 9
f S
for F P <<
2
?
1
p 3 R 4 3 f S
8
www.irf.com
Rev. 2.0
09/12/02
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