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A704 参数 Datasheet PDF下载

A704图片预览
型号: A704
PDF下载: 下载PDF文件 查看货源
内容描述: 开关模式LED驱动器 [switch mode LED driver]
分类和应用: 驱动器开关
文件页数/大小: 14 页 / 408 K
品牌: ADDTEK [ ADDTEK CORP ]
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A704  
In this design, the Vbridge is 425 V and IFSM is 30A. So the ESR of the chock should > (425/30)=14.1. If the  
power loss of such high ESR chock is too high, the designer can replace the high ESR chock with a low ESR chock  
and add a negative-temperature-coefficient (NTC) resistor to limit the inrush current. As long as the (NTC (hot)  
resistance + low ESR Chock resistance) is lower than the resistance of the high ESR chock, the efficiency of the A704  
DC-DC can be improved.  
The hold-up and input filter capacitor required at the diode bridge output have to be calculated at the minimum  
AC input voltage. The minimum capacitor value can be calculated as:  
Vo,max × Io,max ×(12× freq×tc )  
(2×Vm2in,ac Vm2in,dc )×η × freq  
C1≥  
In this design,  
freq  
freq  
Where,  
is the AC input frequency, as a rule,  
is 50 ~ 60Hz.  
is the conduction angle of the AC input, use 45°  
tc  
η is efficiency of the system , and  
conduction angle if unknown.  
12×0.35×(10.5)  
5.4uF  
C1≥  
(2×902 902 )×0.8×60  
The voltage rating of the capacitor should be at least 1.15 times greater than the peak input voltage for a safety  
margin. For example, the input 220Vac input, the input capacitor voltage should exceed than as:  
V
1.15 × 2 ×V  
V  
360V  
max,ac  
max,cap  
max,ac  
Choose a 400V/ 10µF electrolytic capacitor in the example.  
Such electrolytic capacitor has sizable ESR component. The large ESR of these capacitors makes it inappropriate  
to absorb the high frequency ripple current generated by the buck converter. Thus, adding a small MLCC capacitor in  
parallel with the electrolytic capacitor is recommended.  
4. Choose the power choke  
The inductor selection should make the Buck converter work in CCM; the inductor value depends on the ripple  
current in the LEDs. For example, assume a +/- 30% ripple current in the LEDs. Then, the inductor can be calculated  
as:  
V (1D)T  
o
s
L =  
0.6I ×η  
o
In this design example, Ts is 16.6µ s, D=0.12, Vo=12Vdc , η =0.8, Io=0.35  
12× (10.12)×16.6µ  
L =  
= 1.05mH  
0.6× 0.35× 0.8  
Choose the power choke inductance is 1mH.  
Contack Info:Samsun Zhang 13556850583 IC9898@163.com  
Copyright © 2008 ADDtek Corp.  
9
A704_V0.6 -- AUGUST 2008