参数资料
型号: LM2574N-005
厂商: ON Semiconductor
文件页数: 12/26页
文件大小: 0K
描述: IC REG BUCK 5V 0.5A 8DIP
产品变化通告: Product Discontinuation 27/Jun/2007
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: 500mA
同步整流器:
工作温度: -40°C ~ 125°C
安装类型: 通孔
封装/外壳: 8-DIP(0.300",7.62mm)
包装: 管件
供应商设备封装: 8-PDIP
其它名称: LM2574N-005OS
LM2574, NCV2574
Procedure (Adjustable Output Version: LM2574?ADJ)
V out
x 10
E x T + (V
* V out)
V x m s
E x T + (40 * 24) x 24 x 1000 + 105 V x m s
Procedure
4. Inductor Selection (L1)
A. Use the following formula to calculate the inductor Volt x
microsecond [V x m s] constant:
6
in V F[Hz]
in
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 23. 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 27.
D. From the inductor code, identify the inductor value. Then
select an appropriate inductor from 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 Load . The inductor current
rating can also be determined by calculating the inductor
peak current:
Example
4. Inductor Selection (L1)
A. Calculate E x T V x m s constant :
40 52
B. E x T + 185 V x m s
C. I Load(max) = 0.4 A
Inductance Region = 1000
D. Proper inductor value = 1000 m H
Choose the inductor from Table 2.
I p ( max ) + I Load ( max ) )
V in * V out t on
2L
t on + out x 1.0
where t on is the “on” time of the power switch and
V
V in f osc
For additional information about the inductor, see the inductor
section in the “External Components” section of this data
sheet.
5. Output Capacitor Selection (C out )
5. Output Capacitor Selection (C out )
A. Since the LM2574 is a forward?mode switching regulator with
voltage mode control, its open loop 2?pole?1?zero frequency
characteristic has the dominant pole?pair determined by the
A.
C out w 13, 300 x
40
24 x 1000
+ 22.2 m F
output capacitor and inductor values.
To achieve an acceptable ripple voltage, select
For stable operation, the capacitor must satisfy the following
requirement:
C out = 100 m F electrolytic capacitor.
in ( max )
C out w 13, 300
V
V out x L m H
m 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 8.0 V is appropriate, and a 10 V
or 16V rating is recommended.
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