参数资料
型号: LT4351IMS#TRPBF
厂商: Linear Technology
文件页数: 16/20页
文件大小: 0K
描述: IC CTRLR MOSFET DIODE-OR 10MSOP
标准包装: 2,500
应用: 并行/冗余电源
FET 型: N 沟道
输出数: 1
内部开关:
延迟时间 - 关闭: 600ns
电源电压: 1.2 V ~ 18 V
电流 - 电源: 1.4mA
工作温度: -40°C ~ 85°C
安装类型: 表面贴装
封装/外壳: 10-TFSOP,10-MSOP(0.118",3.00mm 宽)
供应商设备封装: 10-MSOP
包装: 带卷 (TR)
LT4351
APPLICATIONS INFORMATION
R DS <
= 1.5m ?
DesignExample
The following demonstrates the calculations involved for
setting design components for a 5V system that requires
5A. Two supplies are used to do this. The V IN supply will
be deemed in spec when it is within ±5% of nominal. Allow
5% of hysteresis for UV.
So,
UV FAULT = 4.75V, UV HYST = 0.25V
OV FAULT = 5.5
Two separate resistive dividers are used.
For the UV divider:
For regulation, the MOSFETs must have:
15mV
2 ? 5A
This very low value cannot be accomplished with a single
set of MOSFETs so a decision must be made whether to
use multiple MOSFETs or to live with an unregulated off-
set. Since low mΩ R DS(ON) is available, the IR drop using
a single MOSFET would still be acceptable. For R DS(ON)
= 4mΩ the drop is 2 ? 5A ? 4mΩ = 40mV. The finished
schematic is shown in Figure 15.
Layout Considerations
UV HYST 0.25V
10 μ A
R2 ? V UV
R2 =
R1 =
=
I UVHYST
UV FAULT – V UV
=
= 25k ( Use 24.9k )
24.9k ? 0.3V
4.75V – 0.3V
There are two considerations for board layout. The first
is that V IN and V DD bypass capacitors should be as close
to the part as possible. The GND pin should represent the
common tie point. The resistive dividers for UV and OV
should tie here as well.
R A =
= 1.5k use 1.47k (1%)
R1=1.68k.Theclosest1%valueis1.69k
The OV resistors are set as a straight resistive divider.
If the current in the R A , R B divider is 200μA, then:
0.3V
200mA
then
Take care that current flow to the load (both through V IN
and GND), does not inadvertently produce errors due to
IR drops in PCB traces.
Keep the traces to the MOSFETs wide and short and close
to the part. The PCB traces associated with the power path
through the MOSFETs should have low resistance.
R B = ?
– 1 ? R A = ?
– 1 ? 1.47k
? OV FAULT
? V OV
? ? 5.5
? ? 0.3
?
?
R B = 25.48, use 25.5k
Si4838DY
V IN
5V
1
10 μ F
OUT
4.7 μ H
R B
25.5k
1%
R2
24.9k
1%
R1
1.69k
1%
10 μ F
0.1 μ F
7
6
4
3
V IN
UV
OV
SW
1
GATE
LT4351
10
OUT
STATUS
9
2k
5V
2k
10 μ F
220 μ F
R A
1.47k
1%
MBR0530
MBR0530
2
1 μ F
V DD
GND
5
FAULT
8
4351 F15
Figure 15. 5V/5A Design Example
4351fd
16
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