参数资料
型号: IRU3037CFPBF
厂商: International Rectifier
文件页数: 8/21页
文件大小: 0K
描述: IC REG CTRLR BUCK PWM VM 8-TSSOP
标准包装: 100
PWM 型: 电压模式
输出数: 1
频率 - 最大: 220kHz
占空比: 95%
电源电压: 4.2 V ~ 25 V
降压:
升压:
回扫:
反相:
倍增器:
除法器:
Cuk:
隔离:
工作温度: 0°C ~ 70°C
封装/外壳: 8-TSSOP(0.173",4.40mm 宽)
包装: 管件
IRU3037/IRU3037A & (PbF)
2 π × ESR × Co
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
F ESR = ---(8)
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 = 5V
V OSC = 1.25V
R 6
R 5
Fb
E/A
Comp
C 9
Ve
Fo = 30KHz
F ESR = 26.52KHz
F LC = 2.9KHz
R 5 = 1K
R 6 = 1.65K
V REF
Gain(dB)
H(s) dB
R 4
g m = 600 μ mho
This results to R 4 =104.4K ? . Choose R 4 =105K ?
To cancel one of the LC filter poles, place the zero be-
fore the LC filter resonant frequency pole:
F Z
Frequency
Figure 6 - 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 = 10 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 = 2.17KHz
R 4 = 86.6K ?
Using equations (11) and (13) to calculate C 9 , we get:
The (s) indicates that the transfer function varies as a
function of frequency. This configuration introduces a gain
and zero, expressed by:
C 9 = 698pF
Choose C 9 = 680pF
|H(s)| = g m ×
R 5
R 6 × R 5
× R 4
---(10)
One more capacitor is sometimes added in parallel with
C 9 and R 4 . This introduces one more pole which is mainly
F Z =
1
2 π× R 4 × C 9
---(11)
used to supress the switching noise. The additional pole
is given by:
1
The gain is determined by the voltage divider and E/A's
transconductance gain.
F P =
2 π × R 4 ×
C 9 × C POLE
C 9 + C POLE
First select the desired zero-crossover frequency (Fo):
Fo > F ESR and F O ≤ (1/5 ~ 1/10) × f S
The pole sets to one half of switching frequency which
results in the capacitor C POLE:
× × g m
R 4 = ×
Use the following equation to calculate R 4 :
V OSC Fo × F ESR R 5 + R 6 1
V IN F LC2 R 5
---(12)
1
C POLE =
π× R 4 × f S - 1
C 9
f S
for F P <<
2
?
1
π× R 4 × f S
8
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