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参数资料
| 型号： | IRU3039CHTR |
| 厂商： | International Rectifier |
| 文件页数： | 11/22页 |
| 文件大小： | 0K |
| 描述： | IC REG CTRLR BUCK PWM VM 20-MLPQ |
| 标准包装： | 1 |
| PWM 型： | 电压模式 |
| 输出数： | 1 |
| 频率 - 最大： | 400kHz |
| 占空比： | 88% |
| 电源电压： | 4.75 V ~ 20 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： | 20-MLPQ |
| 包装： | 剪切带 (CT) |
| 其它名称： | *IRU3039CHTR IRU3039CHTRCT |
IRU3039(PbF)
The IRU3039’s error amplifier is a differential-input
transconductance amplifier. The output is available for
DC gain control or AC phase compensation.
First select the desired zero-crossover frequency (Fo):
Fo > F ESR and F O ≤ (1/5 ~ 1/10) × 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
V IN
Fo × F ESR R 5 + R 6 1
× F LC 2 × R 5 × g m
---(18)
F ESR = ---(14)
dent and can be used to cancel one of the output filter
poles. This will be accomplished with a series RC circuit
from Comp pin to ground as shown in Figure 12.
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 = 18V F LC = 2.8KHz
V OSC = 3.3V R 5 = 1K
Fo = 20KHz R 6 = 3.16K
F ESR = 12KHz g m = 700 μ mho
This results to R 4 =12.08K
R 6
Fb
Choose R 4 =14K
F Z ? 0.75 ×
---(19)
R 5
Vp=V REF
Gain(dB)
H(s) dB
E/A
Comp
C 9
R 4
Ve
To cancel one of the LC filter poles, place the zero be-
fore the LC filter resonant frequency pole:
F Z ? 75%F LC
1
2 π L O × C O
For:
Lo = 4.7 μ H
F Z = 2.1KHz
Co = 660 μ F
R 4 = 14K
( )
H(s) = g m ×
---(15)
R 5 1 + sR 4 C 9
R 6 + R 5
sC 9
C 9 × C POLE
2 π× R 4 ×
F Z Frequency
Figure 12 - Compensation network without local
feedback and its asymptotic gain plot.
The transfer function (Ve / V OUT ) is given by:
×
The (s) indicates that the transfer function varies as a
function of frequency. This configuration introduces a gain
and zero, expressed by:
Using equations (17) and (19) to calculate C 9 , we get:
C 9 ? 5300pF; Choose C 9 =5600pF
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 =
C 9 + C POLE
|H(s=j × 2 π× F O )| = g m ×
R 5
R 6 × R 5
× R 4
---(16)
The pole sets to one half of switching frequency which
results in the capacitor C POLE:
F Z =
---(17)
for F P <<
1
2 π× R 4 × C 9
|H(s)| is the gain at zero cross frequency.
C POLE =
www.irf.com
1
π× R 4 × f S - 1
C 9
f S
2
?
1
π× R 4 × f S
11
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