参数资料
型号: LTC1703IG#TRPBF
厂商: Linear Technology
文件页数: 29/36页
文件大小: 0K
描述: IC REG SW DUAL SYNC VID 28SSOP
标准包装: 2,000
应用: 控制器,移动式 Intel Pentium? III
输入电压: 3 V ~ 7 V
输出数: 2
输出电压: 0.9 V ~ 2 V
工作温度: -40°C ~ 85°C
安装类型: 表面贴装
封装/外壳: 28-SSOP(0.209",5.30mm 宽)
供应商设备封装: 28-SSOP
包装: 带卷 (TR)
LTC1703
APPLICATIO S I FOR ATIO
( )
L ≤ V IN – V OUT ? C ?
Veryquickly,thefeedbackloopwillrealizethatsomething
has changed and will move at the bandwidth allowed by
the external compensation network towards a new duty
cycle. If the bandwidth is set to 50kHz, the COMP pin will
get to 60% of the way to 90% duty cycle in 3 μ s. Now the
inductor is seeing 3.5V across itself for a large portion of
the cycle, and its current will increase from 1A at a rate set
by di/dt = V/L. If the inductor value is 0.5 μ H, the di/dt will
be 3.5V/0.5 μ H or 7A/ μ s. Sometime in the next few micro-
seconds after the switch cycle begins, the inductor current
will have risen to the 5A level of the load current and the
output capacitor will stop losing charge.
Note that the output voltage will stop dropping before the
inductor current reaches this new output current level.
Recall that any practical output capacitor looks like a pure
capacitance in series with some amount of ESR. When a
load transient hits, virtually all of the initial voltage drop at
the output is due to IR drop across the ESR. The output
capacitance begins to discharge at the same time and
continues until the inductor current rises to match the new
output current level.
The output voltage, however, will turn around and start
heading the right way before this happens. The next time
the top MOSFET turns on, the inductor current will begin
increasing linearly. This increasing current flows almost
entirely into the capacitor, going through the ESR as it
does so (Figure 16). Positive di/dt in the inductor causes
positive dv/dt in the ESR, regardless of what the “pure”
capacitance is doing. The output voltage will turn around
when the positive dv/dt across the ESR exceeds the
negative dv/dt across the pure capacitance. If the expected
load step ( ? I) is known, an optimum inductor value can be
chosen:
ESR
? I
Making L smaller than this optimum value yields little or no
improvement in transient response. As the output voltage
recovers, the inductor current will briefly rise above the
level of the output current to replenish the charge lost from
the output capacitor. With a properly compensated loop,
the entire recovery time will be inside of 10 μ s.
Most loads care only about the maximum deviation from
ideal, which occurs somewhere in the first two cycles after
the load step hits. During this time, the output capacitor
does all the work until the inductor and control loop regain
control. The initial drop (or rise if the load steps down) is
entirely controlled by the ESR of the capacitor and amounts
to most of the total voltage drop. To minimize this drop,
reduce the ESR as much as possible by choosing low ESR
I L
I OUT
I L
I OUT
V ESR
V ESR
V CAP
V CAP
I L
V OUT
V SW
L
C OUT
+
V ESR
+
V CAP
V OUT
I OUT
V OUT
TRANSIENT
I L > I OUT
V OUT(NOMINAL)
TIME
1703 F16b
1703 F16a
HITS
V OUT
TURNS
Figure 16a. Capacitor Parasitics
Affecting Transient Recovery
AROUND
Figure 16b. Transient Recovery Curves
1703fa
29
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