参数资料
| 型号： | LT3080IT#PBF |
| 厂商： | Linear Technology |
| 文件页数： | 15/28页 |
| 文件大小： | 0K |
| 描述： | IC REG LDO ADJ 1.1A TO-220-5 |
| 标准包装： | 50 |
| 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： | 通孔 |
| 封装/外壳： | TO-220-5 成形引线 |
| 供应商设备封装： | TO-220-5 |
| 包装： | 管件 |
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LT3080
APPLICATIONS INFORMATION
I CONTROL =
I OUT 1A
= = 17mA
P TOTAL = ( 5V – 3.3V ) ? ?
+ ( 5V – 3.3V ) ? 1A
Thepowerinthedrivecircuitequals:
P DRIVE = (V CONTROL – V OUT )(I CONTROL )
where I CONTROL is equal to I OUT /60. I CONTROL is a function
of output current. A curve of I CONTROL vs I OUT can be found
in the Typical Performance Characteristics curves.
The power in the output transistor equals:
P OUTPUT = (V IN – V OUT )(I OUT )
The total power equals:
P TOTAL = P DRIVE + P OUTPUT
The current delivered to the SET pin is negligible and can
be ignored.
V CONTROL(MAX CONTINUOUS) = 3.630V (3.3V + 10%)
V IN(MAX CONTINUOUS) = 1.575V (1.5V + 5%)
V OUT = 0.9V, I OUT = 1A, T A = 50°C
Power dissipation under these conditions is equal to:
PDRIVE = (V CONTROL – V OUT )(I CONTROL )
60 60
P DRIVE = (3.630V – 0.9V)(17mA) = 46mW
P OUTPUT = (V IN – V OUT )(I OUT )
P OUTPUT = (1.575V – 0.9V)(1A) = 675mW
Total Power Dissipation = 721mW
Junction Temperature will be equal to:
T J = T A + P TOTAL ? θ JA (approximated using tables)
T J = 50°C + 721mW ? 64°C/W = 96°C
In this case, the junction temperature is below the maxi-
mum rating, ensuring reliable operation.
Reducing Power Dissipation
In some applications it may be necessary to reduce
the power dissipation in the LT3080 package without
sacrificing output current capability. Two techniques are
available. The first technique, illustrated in Figure 8, em-
ploys a resistor in series with the regulator’s input. The
voltage drop across R S decreases the LT3080’s IN-to-OUT
differential voltage and correspondingly decreases the
LT3080’s power dissipation.
As an example, assume: V IN = V CONTROL = 5V, V OUT = 3.3V
and I OUT(MAX) = 1A. Use the formulas from the Calculating
Junction Temperature section previously discussed.
Without series resistor R S , power dissipation in the LT3080
equals:
? 1A ?
? 60 ? ?
= 1.73W
If the voltage differential (V DIFF ) across the NPN pass
transistor is chosen as 0.5V, then R S equals:
R S =
5V – 3.3V ? 0.5V
1A
= 1.2 ?
P TOTAL = ( 5V – 3.3V ) ? ?
+ ( 0.5V ) ? 1A = 0.53W
C1
V CONTROL
LT3080
+
–
SET
R SET
IN
OUT
R S
C2
3080 F08
V IN
V IN ?
V OUT
Power dissipation in the LT3080 now equals:
? 1A ?
? 60 ? ?
The LT3080’s power dissipation is now only 30% compared
to no series resistor. R S dissipates 1.2W of power. Choose
appropriate wattage resistors to handle and dissipate the
power properly.
Figure 8. Reducing Power Dissipation Using a Series Resistor
3080fc
15
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