参数资料
型号: IRDC3637
厂商: International Rectifier
文件页数: 10/21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
For a general solution for unconditionally stability for any
F P1 = 0
Z IN
( )
2 π× R 7 ×
type of output capacitors, in a wide range of ESR values
we should implement local feedback with a compensa-
tion network. The typically used compensation network
for voltage-mode controller is shown in Figure 10.
V OUT
C 12
F P2 =
F P3 =
1
2 π× R 8 × C 10
1
C 12 × C 11
C 12 +C 11
?
1
2 π× R 7 × C 12
C 10
R 7
C 11
F Z1 =
1
2 π× R 7 × C 11
F Z2 = 2 π× C 10 × (R 6 + R 8 ) ?
R 8
R 6
Z f
1
1
2 π× C 10 × R 6
R 5
Fb
E/A
Comp
Ve
Cross Over Frequency:
×
Gain(dB)
V REF
F O = R 7 × C 10 ×
V IN
V OSC
1
2 π× Lo × Co
---(15)
Where:
H(s) dB
V IN = Maximum Input Voltage
V OSC = Oscillator Ramp Voltage
Lo = Output Inductor
Co = Total Output Capacitors
F Z 1
F Z 2
F P 2
F P 3
Frequency
The stability requirement will be satisfied by placing the
Figure 10 - Compensation network with local
feedback and its asymptotic gain plot.
In such configuration, the transfer function is given by:
poles and zeros of the compensation network according
to following design rules. The consideration has been
taken to satisfy condition (14) regarding transconduc-
tance error amplifier.
V e
V OUT
=
1 - g m Z f
1 + g m Z IN
1) Select the crossover frequency:
Fo < F ESR and Fo ≤ (1/10 ~ 1/6) × f S
g m
[ ( )]
× (1+sR 8 C 10 )
sR 6 (C 12 +C 11 ) C 12 × C 11
2 π × F Z1 × R 7
2 π × R 7 × F P3
The error amplifier gain is independent of the transcon-
ductance under the following condition:
g m Z f >> 1 and g m Z IN >>1 ---(14)
By replacing Z IN and Z f according to Figure 7, the trans-
former function can be expressed as:
1 (1+sR 7 C 11 ) × [1+sC 10 (R 6 +R 8 )]
H(s)= ×
1+sR 7
C 12 +C 11
As known, transconductance amplifier has high imped-
ance (current source) output, therefore, consider should
be taken when loading the E/A output. It may exceed its
source/sink output current capability, so that the ampli-
fier will not be able to swing its output voltage over the
2
2) Select R 7 , so that R 7 >>
3) Place first zero before LC’s resonant frequency pole.
F Z1 ? 75% F LC
1
C 11 =
4) Place third pole at the half of the switching frequency.
f S
F P3 =
2
1
C 12 =
C 12 > 50pF
If not, change R 7 selection.
necessary range.
5) Place R 7 in (15) and calculate C 10 :
The compensation network has three poles and two ze-
ros and they are expressed as follows:
C 10 ≤
2 π × Lo × Fo × Co
R 7
×
V OSC
V IN
10
www.irf.com
Rev.1.1
06/16/05
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