参数资料
| 型号： | SC410MLTRT |
| 厂商： | Semtech |
| 文件页数： | 17/26页 |
| 文件大小： | 0K |
| 描述： | IC REG BUCK SYNC ADJ 3A 10MLPD |
| 标准包装： | 1 |
| 系列： | EcoSpeed® |
| 类型： | 降压（降压） |
| 输出类型： | 可调式 |
| 输出数： | 1 |
| 输出电压： | 0.75 V ~ 7.5 V |
| 输入电压： | 5.5 V ~ 24 V |
| 频率 - 开关： | 200kHz ~ 1MHz |
| 电流 - 输出： | 3A |
| 同步整流器： | 是 |
| 工作温度： | -40°C ~ 85°C |
| 安装类型： | 表面贴装 |
| 封装/外壳： | 10-VFDFN 裸露焊盘 |
| 包装： | 标准包装 |
| 供应商设备封装： | 10-MLPD（3x3） |
| 其它名称： | SC410MLDKR |
SC410
Applications Information (continued)
The following values are used in this design example.
save mode when the load current decreases to 1/2 of the
?
?
?
?
V IN = 12V + 10%
V OUT = 3.3V + 4%
f SW = 500kHz
Load = 3A maximum
ripple current. For example, if ripple current is 2A then
power-save operation will typically start for loads less than
1A. If ripple current is set at 40% of maximum load current,
then power-save will start for loads less than 20% of
maximum current.
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
During the DH on-time, voltage across the inductor is
(V IN - V OUT ). The equation for determining inductance is
shown next.
power conversion efficiency.
The desired switching frequency is 500kHz which results
L
( V IN
V OUT ) t ON
I RIPPLE
1 V IN
25pF f SW
V OUT
V OUT
V INMAX
f SW
L
2 . 204 H
fromusing componentsselectedforoptimumsizeand
cost.
A resistor (R TON ) is used to program the on-time (indirectly
setting the frequency) using the following equation.
R TON 400 1 . 2
To select R TON , use the maximum value for V IN , and for T ON
use the value associated with maximum V IN .
t ON
t ON = 500 ns at 13.2V IN , 3.3V OUT , 500kHz
Substituting for R TON results in the following solution.
R TON = 78.5k?, use R TON = 78.7k?
Now, t ON = 501ns given that R TON = 78.7k ?.
Inductor Selection
In order to determine the inductance, the ripple current
Example
In this example, the inductor ripple current is set equal to
75% of the maximum load current. Therefore ripple
current will be 75% x 3A or 2.25A. To find the minimum
inductance needed, use the V IN and T ON values that corre-
spond to V INMAX .
(13 . 2 V 3 . 3 V ) 501ns
2 . 25 A
A standard value of 2.2μH is selected. This gives a
maximum I RIPPLE of 2.53A. The peak ripple can be calculated
by the equation, below where L TOL is assumed to be an
inductor tolerance of 20%.
I RIPPLE_PEAK = I RIPPLE_MAX x (1 + L TOL ) = 2.705A
Note that the inductor must be rated for the maximum DC
load current plus 1/2 of the ripple current.
I LSAT_MIN = I RIPPLE_PEAK X 0.5 + I OUT = 4.353A
The ripple current under minimum V IN conditions is also
checked using the following equations.
( V IN
mustfirstbedefined. Lowinductorvaluesresultinsmaller
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
t ON _ VINMIN
I RIPPLE
25pF R TON V OUT
V INMIN
V OUT ) t ON
L
10 ns
611 ns
ripple, and efficiency are all used in the selection process.
The ripple current will also set the boundary for power-
I RIPPLE _ VINMIN
(10 . 8 V 3 .3 V ) 612ns
2 . 2 H
2 . 08 A
save operation. The switching will typically enter power-
17
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