欢迎访问ic37.com |
会员登录 免费注册
发布采购

JAW075A 参数 Datasheet PDF下载

JAW075A图片预览
型号: JAW075A
PDF下载: 下载PDF文件 查看货源
内容描述: JAW050A和JAW075A电源模块; DC-DC转换器36伏至75伏直流输入, 5伏直流输出; 50瓦到75瓦 [JAW050A and JAW075A Power Modules; dc-dc Converters 36 Vdc to 75 Vdc Input, 5 Vdc Output; 50 W to 75 W]
分类和应用: 转换器电源电路DC-DC转换器
文件页数/大小: 16 页 / 435 K
品牌: LINEAGEPOWER [ LINEAGE POWER CORPORATION ]
 浏览型号JAW075A的Datasheet PDF文件第8页浏览型号JAW075A的Datasheet PDF文件第9页浏览型号JAW075A的Datasheet PDF文件第10页浏览型号JAW075A的Datasheet PDF文件第11页浏览型号JAW075A的Datasheet PDF文件第12页浏览型号JAW075A的Datasheet PDF文件第14页浏览型号JAW075A的Datasheet PDF文件第15页浏览型号JAW075A的Datasheet PDF文件第16页  
JAW050A and JAW075A Power Modules; dc-dc Converters:  
36 Vdc to 75 Vdc Input, 5 Vdc Output; 50 W to 75 W  
Data Sheet  
April 2008  
Custom Heat Sinks  
Thermal Considerations (continued)  
Heat Transfer with Heat Sinks (continued)  
A more detailed model can be used to determine the  
required thermal resistance of a heat sink to provide  
necessary cooling. The total module resistance can be  
separated into a resistance from case-to-sink (θcs) and  
sink-to-ambient (θsa) as shown in Figure 25.  
These measured resistances are from heat transfer  
from the sides and bottom of the module as well as the  
top side with the attached heat sink; therefore, the  
case-to-ambient thermal resistances shown are gener-  
ally lower than the resistance of the heat sink by itself.  
The module used to collect the data in Figures 23 and  
24 had a thermal-conductive dry pad between the case  
and the heat sink to minimize contact resistance. The  
use of Figure 23 is shown in the following example.  
TC  
TS  
TA  
PD  
θcs  
θsa  
8-1304(F).e  
Figure 25. Resistance from Case-to-Sink and  
Sink-to-Ambient  
Example  
If an 82 °C case temperature is desired, what is the  
minimum airflow necessary? Assume the JAW075A  
module is operating at VI = 55 V, an output current of  
15 A, longitudinal orientation, maximum ambient air  
temperature of 40 °C, and the heat sink is 1/4 inch.  
For a managed interface using thermal grease or foils,  
a value of θcs = 0.1 °C/W to 0.3 °C/W is typical. The  
solution for heat sink resistance is:  
(TC TA)  
θsa = ------------------------ θcs  
PD  
Solution  
Given: VI = 55 V  
IO = 15 A  
This equation assumes that all dissipated power must  
be shed by the heat sink. Depending on the user-  
defined application environment, a more accurate  
model, including heat transfer from the sides and bot-  
tom of the module, can be used. This equation pro-  
vides a conservative estimate for such instances.  
TA = 40 °C  
TC = 82 °C  
Heat sink = 1/4 inch.  
Determine PD by using Figure 20:  
PD = 14 W  
EMC Considerations  
Then solve the following equation:  
For assistance with designing for EMC compliance,  
refer to the FLTR100V10 Filter Module Data Sheet  
(DS99-294EPS).  
(TC TA)  
θca = ------------------------  
PD  
(82 40)  
-----------------------  
θca =  
14  
Layout Considerations  
θca = 3.0 °C/W  
Use Figure 23 to determine air velocity for the 1/4 inch  
heat sink.  
Copper paths must not be routed beneath the power  
module standoffs. For additional layout guidelines,  
refer to the FLTR100V10 Filter Module Data Sheet  
(DS99-294EPS).  
The minimum airflow necessary for this module is  
1.1 m/s (220 ft./min.).  
Lineage Power  
13  
 复制成功!