参数资料
型号: IRDC3637
厂商: International Rectifier
文件页数: 9/21页
文件大小: 0K
描述: BOARD EVAL SYNC BUCK REGULATOR
产品变化通告: (EP) Parts Discontinuation 25/May/2012
标准包装: 1
主要目的: DC/DC,步降
输出及类型: 1,非隔离
输出电压: 1.8V
电流 - 输出: 7A
输入电压: 4.5 ~ 5.5 V
稳压器拓扑结构: 降压
频率 - 开关: 400kHz
板类型: 完全填充
已供物品:
已用 IC / 零件: IR3637
产品目录页面: 1383 (CN2011-ZH PDF)
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IR3637SPBF
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 ac-
ceptable 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 = 5.5V
V OSC = 1.25V
R 6 Fb
R 5
V REF
E/A
Comp
C 9
R 4
Ve
C POLE
Fo = 40kHz
F ESR = 26.5kHz
F LC = 7.50kHz
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 = 1.5 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 = 5.6kHz
R 4 = 16K ?
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 = 1.77nF
Choose C 9 = 1.8nF
|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
Rev. 1.1
06/16/05
www.irf.com
9
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