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| 型号： | LM2575T-5G |
| 厂商： | ON Semiconductor |
| 文件页数： | 12/28页 |
| 文件大小： | 0K |
| 描述： | IC REG BUCK 5V 1A TO220-5 |
| 标准包装： | 50 |
| 类型： | 降压（降压） |
| 输出类型： | 固定 |
| 输出数： | 1 |
| 输出电压： | 5V |
| 输入电压： | 4.75 V ~ 40 V |
| PWM 型： | 电压模式 |
| 频率 - 开关： | 52kHz |
| 电流 - 输出： | 1A |
| 同步整流器： | 无 |
| 工作温度： | -40°C ~ 125°C |
| 安装类型： | 通孔 |
| 封装/外壳： | TO-220-5 |
| 包装： | 管件 |
| 供应商设备封装： | TO-220-5 |
| 产品目录页面： | 1116 (CN2011-ZH PDF) |
| 其它名称： | LM2575T-5GOS |
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LM2575, NCV2575
Procedure (Adjustable Output Version: LM2575 ? Adj) (continued)
Procedure
4. Inductor Selection (L1)
A. Use the following formula to calculate the inductor Volt x
microsecond [V x m s] constant:
Example
4. Inductor Selection (L1)
A. Calculate E x T [V x m s] constant:
x 10
E x T + 12 8.0 x 8.0 x 1000 + 51 [V x m s]
ExT + V
in
V out
V out
V on
6
F[Hz]
[V x m s]
12 52
B. Match the calculated E x T value with the corresponding
number on the vertical axis of the Inductor Value Selection
Guide shown in Figure 21. This E x T constant is a measure
of the energy handling capability of an inductor and is
dependent upon the type of core, the core area, the number
of turns, and the duty cycle.
C. Next step is to identify the inductance region intersected by
the E x T value and the maximum load current value on the
horizontal axis shown in Figure 21.
D. From the inductor code, identify the inductor value. Then
select an appropriate inductor from the Table 1 or Table 2.
The inductor chosen must be rated for a switching
frequency of 52 kHz and for a current rating of 1.15 x I Ioad .
The inductor current rating can also be determined by
calculating the inductor peak current :
B. E x T = 51 [V x m s]
C. I Load(max) = 1.0 A
Inductance Region = L220
D. Proper inductor value = 220 m H
Choose the inductor from the Table 1 or Table 2.
I
p(max)
+ I
Load(max)
)
V
in
V out t on
2L
V out 1
t on +
V
f osc
where t on is the “on” time of the power switch and
x
in
For additional information about the inductor, see the
inductor section in the “External Components” section of
this data sheet.
12
Cout w 7.785
+ 53 μ F
5. Output Capacitor Selection (C out )
A. Since the LM2575 is a forward ? mode switching regulator
with voltage mode control, its open loop 2 ? pole ? 2 ? zero
frequency characteristic has the dominant pole ? pair
determined by the output capacitor and inductor values .
For stable operation, the capacitor must satisfy the
5. Output Capacitor Selection (C out )
A.
8.220
To achieve an acceptable ripple voltage, select
C out = 100 m F electrolytic capacitor.
following requirement:
Cout w 7.785
V in(max)
V out x L [ μ H]
[ μ F]
B. Capacitor values between 10 m F and 2000 m F will satisfy
the loop requirements for stable operation. To achieve an
acceptable output ripple voltage and transient response, the
output capacitor may need to be several times larger than
the above formula yields.
C. Due to the fact that the higher voltage electrolytic capacitors
generally have lower ESR (Equivalent Series Resistance)
numbers, the output capacitor’s voltage rating should be at
least 1.5 times greater than the output voltage. For a 5.0 V
regulator, a rating of at least 8V is appropriate, and a 10 V
or 16 V rating is recommended.
http://onsemi.com
12
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相关代理商/技术参数 |
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|---|---|
| LM2575T-5G | 制造商:ON Semiconductor 功能描述:IC STEP-DOWN REGULATOR TO-220-5 制造商:ON Semiconductor 功能描述:IC, STEP-DOWN REGULATOR, TO-220-5 |
| LM2575T-ADJ | 功能描述:直流/直流开关转换器 RoHS:否 制造商:STMicroelectronics 最大输入电压:4.5 V 开关频率:1.5 MHz 输出电压:4.6 V 输出电流:250 mA 输出端数量:2 最大工作温度:+ 85 C 安装风格:SMD/SMT |
| LM2575T-ADJ | 制造商:Texas Instruments 功能描述:SWITCHING REG 1A ADJ 2575 TO2205 |
| LM2575T-ADJ/LB02 | 制造商:Texas Instruments 功能描述:SIMPLE SWITCHER 1A -40/+125 #01202511 |
| LM2575T-ADJ/LB03 | 功能描述:直流/直流开关转换器 RoHS:否 制造商:STMicroelectronics 最大输入电压:4.5 V 开关频率:1.5 MHz 输出电压:4.6 V 输出电流:250 mA 输出端数量:2 最大工作温度:+ 85 C 安装风格:SMD/SMT |
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