参数资料
型号: LT3085EMS8E#PBF
厂商: Linear Technology
文件页数: 17/28页
文件大小: 0K
描述: IC REG LDO ADJ .5A 8-MSOP
产品培训模块: LT3085 Adjustable Low Dropout Regulator
More Information on LDOs
标准包装: 50
稳压器拓扑结构: 正,可调式
输出电压: 可调
输入电压: 1.2 V ~ 36 V
电压 - 压降(标准): 1.35V @ 500mA
稳压器数量: 1
电流 - 输出: 500mA
电流 - 限制(最小): 500mA
工作温度: -40°C ~ 125°C
安装类型: 表面贴装
封装/外壳: 8-TSSOP,8-MSOP(0.118",3.00mm 宽)裸露焊盘
供应商设备封装: 8-MSOP-EP
包装: 管件
产品目录页面: 1331 (CN2011-ZH PDF)
LT3085
APPLICATIONS INFORMATION
Without series resistor R S , power dissipation in the LT3085
Calculating R P yields:
R P = = 7.30 Ω
equals:
P TOTAL = ( 5V – 3.3V ) ?
0.5A
60
+ ( 5V – 3.3V ) ? 0.5A
5.5V – 3.2V
315mA
(5% Standard value = 7.Ω)
R S = = 2.4 Ω
=0.86W
If the voltage differential (V DIFF ) across the NPN pass
transistor is chosen as 0.5V, then R S equals:
5V – 3.3V ? 0.5V
0.5A
Power dissipation in the LT3085 now equals:
The maximum total power dissipation is (5.5V – 3.2V) ?
0.5A = 1.2W. However the LT3085 supplies only:
5.5V – 3.2V
0.5A – = 0.193A
7.5 Ω
Therefore, the LT3085’s power dissipation is only:
P DIS = (5.5V – 3.2V) ? 0.193A = 0.44W
P TOTAL = ( 5V – 3.3V ) ?
0.5A
60
+ ( 0.5V ) ? 0.5A = 0.26W
R P dissipates 0.71W of power. As with the ?rst technique,
choose appropriate wattage resistors to handle and dis-
sipate the power properly. With this con?guration, the
The LT3085’s power dissipation is now only 30% compared
to no series resistor. R S dissipates 0.6W of power. Choose
appropriate wattage resistors to handle and dissipate the
power properly.
LT3085 supplies only 0.36A. Therefore, load current can
increase by 0.3A to 0.143A while keeping the LT3085 in
its normal operating range.
The second technique for reducing power dissipation,
shown in Figure 9, uses a resistor in parallel with the
C1
V CONTROL
LT3085
IN
V IN
LT3085. This resistor provides a parallel path for current
?ow, reducing the current ?owing through the LT3085.
This technique works well if input voltage is reasonably
+
R P
constant and output load current changes are small. This
technique also increases the maximum available output
current at the expense of minimum load requirements.
As an example, assume: V IN = V CONTROL = 5V, V IN(MAX) =
SET
R SET
OUT
V OUT
C2
3085 F09
5.5V, V OUT = 3.3V, V OUT(MIN) = 3.2V, I OUT(MAX) = 0.5A and
I OUT(MIN) = 0.35A. Also, assuming that R P carries no more
Figure 9. Reducing Power Dissipation Using a Parallel Resistor
than 90% of I OUT(MIN) = 630mA.
3085fb
17
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