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| 型号： | SC402BMLTRT |
| 厂商： | Semtech |
| 文件页数： | 23/32页 |
| 文件大小： | 1425K |
| 描述： | IC REG DL BCK/LINEAR SYNC 32MLPQ |
| 标准包装： | 1 |
| 系列： | EcoSpeed®, SmartDrive™ |
| 拓扑： | 降压（降压）同步（1），线性（LDO）（1） |
| 功能： | 任何功能 |
| 输出数： | 2 |
| 频率 - 开关： | 1MHz |
| 电压/电流 - 输出 1： | 可调至 0.6V，10A |
| 电压/电流 - 输出 2： | 可调至 0.75V，200mA |
| 带 LED 驱动器： | 无 |
| 带监控器： | 无 |
| 带序列发生器： | 是 |
| 电源电压： | 3 V ~ 28 V |
| 工作温度： | -40°C ~ 85°C |
| 安装类型： | 表面贴装 |
| 封装/外壳： | 32-VFQFN 裸露焊盘 |
| 供应商设备封装： | 32-MLPQ（5x5） |
| 包装： | 标准包装 |
| 其它名称： | SC402BMLDKR |
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SC402B
23
Applications Information (continued)
There are two values of load current to evaluate con-
tinuous load current and peak load current. Continuous
load current relates to thermal stresses which drive the
selection of the inductor and input capacitors. Peak load
current determines instantaneous component stresses
and filtering requirements such as inductor saturation,
output capacitors, and design of the current limit circuit.
The following values are used in this design.
V
IN
= 12V + 10%
V
OUT
= 1.5V + 4%
f
SW
= 300kHz
Load = 10A maximum
Frequency Selection
Selection of the switching frequency requires making a
trade-off between the size and cost of the external filter
components (inductor and output capacitor) and the
power conversion efficiency.
The desired switching frequency is 300kHz which results
from using components selected for optimum size and
cost.
A resistor (R
TON
) is used to program the on-time (indirectly
setting the frequency) using the following equation.
SW
TON
f
pF
25
k
R
To select R
TON
, use the maximum value for V
IN
, and for T
ON
use the value associated with maximum V
IN
.
SW
INMAX
OUT
ON
f
V
V
T
T
ON
= 379 ns at 13.2V
IN
, 1.5V
OUT
, 300kHz
Substituting for R
TON
results in the following solution.
R
TON
= 133.3k&, use R
TON
= 130k&
Inductor Selection
In order to determine the inductance, the ripple current
must first be defined. Low inductor values result in smaller
size but create higher ripple current which can reduce
efficiency. Higher inductor values will reduce the ripple
"
"
"
"
current/voltage and for a given DC resistance are more
efficient. However, larger inductance translates directly
into larger packages and higher cost. Cost, size, output
ripple, and efficiency are all used in the selection process.
The ripple current will also set the boundary for PSAVE
operation. The switching will typically enter PSAVE mode
when the load current decreases to 1/2 of the ripple
current. For example, if ripple current is 4A then PSAVE
operation will typically start for loads less than 2A. If ripple
current is set at 40% of maximum load current, then PSAVE
will start for loads less than 20% of maximum current.
The inductor value is typically selected to provide a ripple
current that is between 25% to 50% of the maximum load
current. This provides an optimal trade-off between cost,
efficiency, and transient performance.
During the on-time, voltage across the inductor is
(V
IN
- V
OUT
). The equation for determining inductance is
shown next.
RIPPLE
ON
OUT
IN
I
T
)
V
V
(
L
Example
In this example, the inductor ripple current is set equal to
45% of the maximum load current. Therefore ripple
current will be 45% x 10A or 4.5A. To find the minimum
inductance needed, use the V
IN
and T
ON
values that corre-
spond to V
INMAX
.
H
99
.
0
A
5
.
4
ns
379
)
5
.
1
2
.
13
(
L
A slightly larger value of 1礖 is selected. This will decrease
the maximum I
RIPPLE
to 4.43A.
Note that the inductor must be rated for the maximum DC
load current plus 1/2 of the ripple current.
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