参数资料
| 型号： | SC4524SETRT |
| 厂商： | Semtech |
| 文件页数： | 18/21页 |
| 文件大小： | 0K |
| 描述： | IC REG BUCK 2A 8SOIC |
| 标准包装： | 1 |
| 类型： | 降压（降压） |
| 输出数： | 1 |
| 输入电压： | 2.8 V ~ 30 V |
| PWM 型： | 电流模式 |
| 频率 - 开关： | 1.4MHz |
| 电流 - 输出： | 2A |
| 同步整流器： | 无 |
| 工作温度： | -40°C ~ 85°C |
| 安装类型： | 表面贴装 |
| 封装/外壳： | 8-SOIC（0.154"，3.90mm 宽） |
| 包装： | 标准包装 |
| 供应商设备封装： | 8-SOIC-EP |
| 产品目录页面： | 1358 (CN2011-ZH PDF) |
| 其它名称： | SC4524SEDKR |
SC4524
POWER MANAGEMENT
Applications Information
w
=
=
gives a first-order estimate of C 5 :
w
= -
5 &
w z1 is shown to be less than w p2 in Figure 11. Making
w w
w
= -
=
?
is also shown in Figure 11. Its mid-
band gain (between z 1 and p 3 ) is * 5 ì ì
ü ü . The
& ?
(12)
loop gain at high frequency. Placing p 3 at about
w
& ?
(13)
( w =
=
w p I
=
=
In addition C 5 and R 5 form a zero with angular frequency:
5 &
The output-to-control transfer function
Y Y Y
Y Y Y
? 5 ?
í 5 + 5 Y
overall loop gain T(s) is the product of the control-to-
output and the output-to-control transfer functions. To
simplify 7 M w Bode plot, the feedback network is
assumed to be resistive. If the overall loop gain is to
cross 0dB at one tenth of the switching frequency
) at ?20dB/decade, then its mid-band gain
(between z 1 and p 2 ) will be
w
w w & 5
w Q Q
& 5
w 5
Notice that R 5 determines the mid-band loop gain of the
converter. Increasing R 5 increases the mid-band gain and
the crossover frequency. However it reduces the phase
margin. C 6 is a small ceramic capacitor to roll off the
gives:
p I5
Computed R 5 , C 5 and C 6 can indeed result in near optimal
load transient responses in over half of the applications.
However in other cases empirically determined
compensation networks based on optimized load
transient responses may differ from those calculated by
a factor of 3. Therefore checking the transient response
of the converter is imperative. Starting with calculated
R 5 , C 5 and C 6 (using n=1 in Equations (11)-(13)), apply
the largest expected load step to the converter at the
maximum operating V IN . Observe the load transient
response of the converter while adjusting R 5 , C 5 and C 6 .
Choose the largest R 5 , the smallest C 5 and C 6 so that
the inductor current waveform does not show excessive
ringing or overshoot.
* 5 ì ì
This is also equal to * 5
* 5 ì ì
ü ü =
* 5
í 5 + 5 Y
Q
ü ü . Therefore
Re-arranging,
? 5 ? w & 5
? 5
í 5 + 5
.
?
Y
Feedforward capacitor C 11 boosts phase margin over a
limited frequency range and is sometimes used to
improve loop response. C 11 will be more effective if
5 >> 5 ?? 5 .
Example: Determine the compensation components for
the 550kHz 12V to 3.3V converter in Figure 13(a).
5 = ì ì +
ü ü
?
í
5
5
? w &
Y Q* *
(11)
For the converter, w = 0UDGV , , = $ and
& = m ) . n is assumed to be 1 in (11) and (12).
? 2006 Semtech Corp.
18
www.semtech.com
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