参数资料
型号: LT3080EQ#TRPBF
厂商: Linear Technology
文件页数: 16/28页
文件大小: 0K
描述: IC REG LDO ADJ 1.1A 5-DDPAK
产品培训模块: More Information on LDOs
标准包装: 750
稳压器拓扑结构: 正,可调式
输出电压: 可调至 36V
输入电压: 1.2 V ~ 36 V
电压 - 压降(标准): 1.35V @ 1.1A
稳压器数量: 1
电流 - 输出: 1.1A
电流 - 限制(最小): 1.1A
工作温度: -40°C ~ 125°C
安装类型: 表面贴装
封装/外壳: TO-263-6,D²Pak(5 引线+接片),TO-263BA
供应商设备封装: D2PAK-5
包装: 带卷 (TR)
LT3080
APPLICATIONS INFORMATION
= 0.36A
1A –
R P =
= 3.65 ?
The second technique for reducing power dissipation,
shown in Figure 9, uses a resistor in parallel with the
LT3080. This resistor provides a parallel path for current
flow, reducing the current flowing through the LT3080.
This technique works well if input voltage is reasonably
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) =
5.5V, V OUT = 3.3V, V OUT(MIN) = 3.2V, I OUT(MAX) = 1A and
I OUT(MIN) = 0.7A. Also, assuming that R P carries no more
than 90% of I OUT(MIN) = 630mA.
Calculating R P yields:
5.5V – 3.2V
0.63A
(5% Standard value = 3.6Ω)
The maximum total power dissipation is (5.5V – 3.2V) ?
1A = 2.3W. However the LT3080 supplies only:
5.5V – 3.2V
3.6 ?
Therefore, the LT3080’s power dissipation is only:
P DIS = (5.5V – 3.2V) ? 0.36A = 0.83W
R P dissipates 1.47W of power. As with the first technique,
choose appropriate wattage resistors to handle and dis-
sipate the power properly. With this configuration, the
LT3080 supplies only 0.36A. Therefore, load current can
increase by 0.64A to 1.64A while keeping the LT3080 in
its normal operating range.
C1
V CONTROL
LT3080
+
IN
R P
V IN
SET
R SET
OUT
V OUT
C2
3080 F09
Figure 9. Reducing Power Dissipation Using a Parallel Resistor
3080fc
16
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