参数资料
型号: IR3637ASTRPBF
厂商: International Rectifier
文件页数: 9/19页
文件大小: 0K
描述: IC REG CTRLR BUCK PWM VM 8-SOIC
标准包装: 1
PWM 型: 电压模式
输出数: 1
频率 - 最大: 660kHz
占空比: 76%
电源电压: 4.5 V ~ 5.5 V
降压:
升压:
回扫:
反相:
倍增器:
除法器:
Cuk:
隔离:
工作温度: 0°C ~ 125°C
封装/外壳: 8-SOIC(0.154",3.90mm 宽)
包装: 标准包装
其它名称: IR3637ASTRPBFDKR
IR3637ASPbF
F ESR = ---(8)
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:
1
2 π × ESR × Co
V OUT
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
g m = Error Amplifier Transconductance
For:
V IN = 5.5V
V OSC = 1.25V
R 6 Fb
R 5
V REF
E/A
Comp
C 9
R 4
Ve
C POLE
Fo = 60KHz
F ESR = 26.5KHz
F LC = 9.20KHz
R 5 = 1K
R 6 = 1.25K
g m = 600 μ mho
Gain(dB)
H(s) dB
This results to R 4 =16.06K ? . Choose R 4 =16K ?
To cancel one of the LC filter poles, place the zero be-
fore the LC filter resonant frequency pole:
F Z
Frequency
Figure 9 - Compensation network without local
feedback and its asymptotic gain plot.
F Z ? 75%F LC
F Z ? 0.75 ×
For:
2 π
1
L O × C O
---(13)
The transfer function (Ve / V OUT ) is given by:
Lo = 1.0 μ H
Co = 300 μ F
(
) × 1 + sC sR C
H(s) = g m ×
R 5
R 6 + R 5
4 9
9
---(9)
F Z = 6.9KHz
R 4 = 16K ?
Using equations (11) and (13) to calculate C 9 , we get:
|H(s)| = g m ×
× R 4
---(10)
F Z =
---(11)
C 9 × C POLE
2 π × R 4 ×
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 × R 5
1
2 π× R 4 × C 9
The gain is determined by the voltage divider and E/A's
transconductance gain.
First select the desired zero-crossover frequency (Fo):
Fo > F ESR and F O ≤ (1/5 ~ 1/10) × f S
C 9 = 1.44nF
Choose C 9 = 1.5nF
One more capacitor is sometimes added in parallel with
C 9 and R 4 . This introduces one more pole which is mainly
used to supress the switching noise. The additional pole
is given by:
1
F P =
C 9 + C POLE
The pole sets to one half of switching frequency which
results in the capacitor C POLE:
× × g m
V IN
F LC2
R 5
π× R 4 × f S - 1
for F P <<
Use the following equation to calculate R 4 :
V OSC Fo × F ESR R 5 + R 6 1
R 4 = ×
---(12)
C POLE =
f S
2
1
C 9
?
1
π× R 4 × f S
www.irf.com
9
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