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参数资料
| 型号： | MAX1531ETJ/V+T |
| 厂商： | Maxim Integrated Products |
| 文件页数： | 26/33页 |
| 文件大小： | 0K |
| 描述： | IC PS CTRLR MULTI-OUTPUT 32WQFN |
| 标准包装： | 2,500 |
| 应用： | 五路电源监控器 |
| 电源电压： | 4.5 V ~ 28 V |
| 电流 - 电源： | 1.7mA |
| 工作温度： | -40°C ~ 85°C |
| 安装类型： | 表面贴装 |
| 封装/外壳： | 32-WFQFN 裸露焊盘 |
| 供应商设备封装： | 32-TQFN-EP（5x5） |
| 包装： | 带卷 (TR) |
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Multiple-Output Power-Supply
Controllers for LCD Monitors
f POLE ( LOW ) =
C 10 ≈
Current-mode control has the effect of splitting the
complex pole pair of the output LC filter into a single
low-frequency pole and a single high-frequency pole.
The low-frequency current-mode pole depends on out-
put capacitor C OUT and the equivalent load resistance
R LE , given by the following:
1
2 π × R LE × C OUT
The high-frequency current-mode pole is given by:
Unnecessarily high bandwidth can increase noise
sensitivity while providing little benefit. Good tran-
sient response with low amounts of output capaci-
tance is achieved with a crossover frequency
between 20kHz and 100kHz. The series compensa-
tion capacitor (C10) generates a dominant pole that
sets the desired crossover frequency. Determine
C10 using the following expression:
g m × A DC
2 π × f CROSSOVER × A VEA
f POLE ( HIGH ) =
f SW
2 π × n × D '
where g m is the error amplifier’s transconductance
(100μS typ).
2) The compensation resistor R11, together with capac-
The COMP pin, which is the output of the IC’s internal
transconductance error amplifier, is used to stabilize
the control loop. A series resistor (R11) and capacitor
(C10) are connected between COMP and AGND to
itor C10, provides a zero that is used to cancel the
low-frequency current-mode pole. Determine R11
using the following expression:
form a pole-zero pair. Another pole-zero pair can be
added by connecting a feed-forward capacitor (C23) in
parallel with feedback resistor R1. The compensation
R 11 ≈
1
2 π × f POLE ( LOW ) × C 10
resistor and capacitors are selected to optimize the
loop stability.
The compensation capacitor (C10) creates a dominant
pole at very low frequency (a few hertz). The zero
formed by R11 and C10 cancels the low-frequency cur-
rent-mode pole. The zero formed by R1 and C23 can-
cels the high-frequency current-mode pole and
introduces a preferable higher frequency pole. In appli-
3) Because the error amplifier has limited output cur-
rent (16μA typ), small values of R11 can prevent the
error amplifier from providing an immediate COMP
voltage change required for good transient response
with minimal output capacitance. If the calculated
R11 value is less than 100k Ω , use 100k Ω and recal-
culate C10 using the following formula:
cations where ceramic capacitors are used, the ESR
zero is usually not a concern because the ESR zero
occurs at very high frequency. If the ESR zero does not
C 10 ≈
1
2 π × f POLE ( LOW ) × 100 k Ω
occur at a frequency at least one decade above the
crossover, connect a second parallel capacitor (C2)
between COMP and AGND to cancel the ESR zero. The
Changing C10 also changes the crossover frequen-
cy; the new crossover frequency is:
component values shown in the standard application
circuits (Figure 1 and 2) yield stable operation and fast
transient response over a broad range of input-to-out-
f CROSSOVER =
g m × A DC
2 π × C 10 × A VEA
f CROSSOVER ≤ SW
put voltages.
To design a compensation network for other compo-
nents or applications, use the following procedure to
achieve stable operation:
1) Select the crossover frequency f CROSSOVER
(bandwidth) to be 1/5th the switching frequency
f SW or less:
f
5
The calculated crossover frequency should be less
than 1/5th the switching frequency. There are two
ways to lower the crossover frequency if the calculat-
ed value is greater than 1/5th the switching frequen-
cy: increase the high-side MOSFET R DS(ON ), or
increase the output capacitance. Increasing R DS(ON)
reduces the DC loop gain, which results in lower
crossover frequency. Increasing output capacitance
reduces the frequency of the lower low-frequency
current-mode pole, which also results in lower
crossover frequency. The following formula gives the
26
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