散热器扩散热阻的计算

合集下载

散热器的计算

散热器的计算

散热器厂的计算金旗舰散热器的计算设I=350mA,Vin=12V,则耗散功率Pd=(12V-5V)*0.35A=2.45W按照TO-220封装的热阻θJA=54℃/W,温升是132℃,设室温25℃,那么将会达到7805的热保护点150℃,7805会断开输出.正确的设计方法是:首先确定最高的环境温度,比如60℃,查出7805的最高结温TJM AX=125℃,那么允许的温升是65℃.要求的热阻是65℃/2.45W=26℃/ W.再查7805的热阻,TO-220封装的热阻θJA=54℃/W,均高于要求值,都不能使用,所以都必须加散热片,资料里讲到加散热片的时候,应该加上4℃/W的壳到散热片的热阻.计算散热片应该具有的热阻也很简单,与电阻的并联一样,即54/ /x=26,x=50℃/W.其实这个值非常大,只要是个散热片即可满足.散热器的计算:总热阻RQj-a=(Tjmax-Ta)/PdTjmax :芯组最大结温150℃Ta :环境温度85℃Pd : 芯组最大功耗Pd=输入功率-输出功率={24×0.75+(-24)×(-0.25)}-9.8×0.25×2=5.5℃/W总热阻由两部分构成,其一是管芯到环境的热阻RQj-a,其中包括结壳热阻RQj-C和管壳到环境的热阻RQC-a.其二是散热器热阻RQd-a,两者并联构成总热阻.管芯到环境的热阻经查手册知 RQj-C=1.0 R QC-a=36 那么散热器热阻RQd-a应<6.4. 散热器热阻RQd-a=[(10/kd) 1/2+650/A]C其中k:导热率铝为2.08d:散热器厚度cmA:散热器面积cm2C:修正因子取1按现有散热器考虑,d=1.0 A=17.6×7+17.6×1×13算得散热器热阻RQd-a=4.1℃/W,散热器选择及散热计算目前的电子产品主要采用贴片式封装器件,但大功率器件及一些功率模块仍然有不少用穿孔式封装,这主要是可方便地安装在散热器上,便于散热。

散热器的散热量计算公式

散热器的散热量计算公式

散热器的散热量计算公式散热器是一种用于降低电子设备或机械设备温度的装置。

它通过将设备产生的热量转移到周围环境中,从而保持设备的正常运行温度。

散热器的散热量计算公式可以帮助我们了解散热器的散热能力和热量传递效率。

散热器的散热量计算公式如下:Q = U * A * ΔT其中,Q表示散热器的散热量,U表示散热器的传热系数,A表示散热器的表面积,ΔT表示散热器与环境之间的温度差。

我们来了解一下散热器的传热系数U。

传热系数是指单位时间内通过单位面积的热量传递量与温度差之间的比值。

它决定了散热器传热的效率和能力。

散热器的传热系数受到散热器材料、散热片结构和流体状态等因素的影响。

散热器的表面积A也是计算散热量的重要参数。

表面积越大,散热器与周围环境之间的热交换面积就越大,从而能够更快地将热量散发出去。

温度差ΔT是指散热器表面温度与环境温度之间的差值。

温度差越大,散热器的散热能力越强。

散热器的散热量计算公式可以帮助我们评估散热器的散热性能。

通过调整散热器材料、改进散热片结构和优化流体状态,可以提高散热器的传热系数和表面积,从而提高散热器的散热能力。

除了散热器自身的设计和性能,散热器的散热量还受到其他因素的影响。

例如,周围环境的温度和湿度、设备的功率和工作状态等都会对散热器的散热效果产生影响。

在实际应用中,我们可以根据设备的功率、工作温度和环境温度等参数,计算出散热器所需的散热量。

然后,根据散热器的传热系数和表面积,选择合适的散热器型号和规格。

散热器的散热量计算公式是评估散热器散热性能的重要工具。

通过合理选择散热器和优化散热系统设计,可以有效降低设备温度,提高设备的可靠性和稳定性。

在未来的发展中,我们可以期待散热器技术的进一步创新和提高,以满足不断增长的散热需求。

散热器的计算公式

散热器的计算公式

散热器的计算公式
散热器是一种用来散发热量的设备,广泛应用于各个领域,包
括建筑、工业、汽车等。

计算散热器的散热能力对于确保设备正常
运作非常重要。

以下是一些常用的散热器计算公式。

1. 热功率计算
散热器的主要功能是散发热量,因此计算热功率是散热器设计
的关键。

热功率可根据以下公式计算:
热功率 (W) = 热量传导系数 (U) ×温度差(ΔT) × 表面积 (A)
其中,热量传导系数是指散热器材料的热导率,温度差是散热
器表面的温度与周围环境温度之差,表面积是指散热器的外表面积。

2. 散热器尺寸计算
散热器尺寸的计算涉及到散热片的数量和间距。

以下是一些常
用的散热器尺寸计算公式:
- 散热片数量 (N) = 热功率 (W) / 单个散热片的散热能力 (Q)
其中,单个散热片的散热能力可由散热片的热导率 (K) 和表面积 (A) 计算得出。

- 散热片间距 (D) = 散热器高度 (H) / (散热片数量 (N) - 1)
3. 散热器材料选择
散热器材料的选择是散热器设计中的另一个重要因素。

常用的散热器材料包括铝、铜、不锈钢等。

根据散热需求和成本考虑,选择适当的材料是非常关键的。

4. 其他因素考虑
除了以上的计算公式外,散热器设计还需要考虑其他因素,例如流体流量、风速、散热器的布局等。

这些因素会对散热器的散热能力产生影响,需要进行综合考虑。

综上所述,散热器设计的计算公式涉及热功率、散热器尺寸、材料选择等因素。

根据实际需求合理使用这些公式可以确保散热器的有效运作。

热传导的热阻计算

热传导的热阻计算

热传导的热阻计算热传导是指热量从一个物体传递到另一个物体的过程,通过固体、液体或气体中分子的热振动来实现。

在实际应用中,我们经常需要计算热传导的热阻,以评估材料的热传导性能或设计热传导设备。

本文将介绍热阻的概念以及几种常用的热阻计算方法。

一、热阻的概念热阻是指单位面积上的温度梯度与单位时间内通过该面积的热流量之比,通常用符号R表示。

热阻的单位是K/W(开尔文/瓦特),表示温度梯度每增加1开尔文时,通过单位面积的热流量减少的程度。

热阻与热传导率(热导率)有密切关系,热传导率表示材料单位厚度上单位面积的热流量,其倒数即为热阻。

因此,热阻可以通过热传导率计算得到。

二、热阻的计算方法1. 薄壁材料的热阻计算对于薄壁材料,例如金属板材、塑料薄膜等,可以使用以下公式计算热阻:R = 1 / (k * A)其中,R为热阻,k为材料的热传导率,A为单位面积。

2. 复合层的热阻计算对于由多层材料组成的复合层结构,可以使用以下公式计算总的热阻:R_total = R_1 + R_2 + R_3 + ... + R_n其中,R_total为总的热阻,R_1、R_2、R_3等分别为每个材料层的热阻。

需要注意的是,计算复合层的热阻时,需要根据不同的传热模式(如平行传热、串联传热等)进行相应的修正。

3. 不均匀材料的热阻计算对于不均匀材料,如中空球、多孔材料等,可以使用以下公式计算热阻:R_total = L / (k_1 * A_1 + k_2 * A_2 + k_3 * A_3 + ... + k_n * A_n)其中,R_total为总的热阻,L为传热长度,k_1、k_2、k_3等为不同材料的热传导率,A_1、A_2、A_3等为不同材料的传热面积。

三、热阻计算的应用热阻计算在工程领域有着广泛的应用,例如在隔热材料的选择中,通过计算不同材料的热阻,可以评估不同材料的隔热效果,选择最适合的材料。

另外,在散热系统的设计中,通过计算散热组件的热阻,可以确定散热器的尺寸和工作参数,以确保系统的稳定性和可靠性。

大功率半导体器件用散热器风冷热阻计算公式和应用软件

大功率半导体器件用散热器风冷热阻计算公式和应用软件

大功率半导体器件用散热器风冷热阻计算公式和应用软件2012-03-12 14:17:31 作者:来源:中国电力电子产业网文章概要如下:一、计算公式为了推导风冷散热器热阻计算公式作如下设定:1,散热器是由很多块金属平板组成,平板一端连在一起成一块有一定厚度的基板,平板之间存在间隙,散热器的基本单元是一块平板;2,平板本身具有一定的长度、宽度和厚度(L×l×b)。

平板的横截面积A =L × b;3,由n个平板(齿片)组成的散热器如图一所示,平板(齿片)数为n ;4,由此可见,参数L即为散热器长,或称“截长”;5,设散热器端面周长为“S”。

大功率半导体器件安装在基板上,工作时产生的热通过接触面传到散热器的过程属于固体导热。

散热器平板周围是空气。

风冷条件下平板上的热要传到空气中属于固体与流体间的传热。

所以风冷散热器总热阻等于两部分热阻之和:Rzo(总热阻)= Rth(散热器内固体传热)+ Rthk(散热器与空气间的传热热阻)引用埃克尔特和..德雷克著的“传热与传质”中的基本原理和公式。

推导出如下实用公式:Ks 为散热器金属材料的导热系数。

20℃时,纯铝:KS = 千卡/ 小时米℃;纯铜:Ks = 332 千卡/ 小时米℃;参数L、l、b、S的单位:米;风速us 单位:米/秒如散热器端面的周边长为S 、散热器的长为L,忽略两端面的面积,散热器的总表面积为: A = S L 。

代入上式后,强迫风冷条件下散热器总热阻公式也可写成:对某一型号的散热器来说参数Ks、b、n、S 都是常数。

用此公式即可求出不同长度L、不同风速us条件下的总热阻,并可作出相应曲线。

本公式的精确性受到多种因素的影响存在一定误差。

主要有:ⅰ,受到环境空气的温度、湿度、气压等自然因素的影响。

如散热器金属的热导系数“Ks”与金属成分及散热器工作时温度有关,本文选用的是20℃时的纯铝。

ⅱ,文中所用的“风速”是指“平均风速”。

电子仪器中散热器热阻的计算方法

电子仪器中散热器热阻的计算方法
1)如果守门员在球门内,那么它突出球门线的高度会相对较小。 球从侧面袭来时,守门员的防守区域相对于突出球门线的高度会缩 小:
2)如果守门员离开球门线向前,球从底线袭来时守门员会漏球或 者会导致“乌龙球”。
因此,实践证明守门员在球门线上的位置为最佳位置。 4.结论 本文对Voronoi图功能及足球机器人的运动进行了介绍.提出并 实现一种简单、高效的Voronoi多边形面积计算方法。创造性的提出了 机器人判别动作的算法模型。利用Vomnoi图的几何特性分析并解决 足球机器人的任务规划问题。实践证明,用voronoi图来提出的足球机 器人动作判别算法提高了足球机器人的判别能力,降低了判别的失误 率,并实现了足球机器人的分工协作,在一定程度上增强了足球机器人
科技信息
。机械与电子O
2008年第4期
电子仪器中散热器热阻的计算方法
汪择宏 (中国电子科技集团公司第41研究所 山东青岛266555)
【摘要】在电子仪器中,绝大多数大功率元器件的散热是通过散热器来完成的。因此合理选择散热器尤为重要。本文提出了该类散热器热
阻的计算方法,为整机大功率元器件的热设计提供了必要的依据。
的智能化。《
【参考文献】 [11张颖霞,杨宜民,陈波.多智能体团队合作在机器人足球赛中的应用.微机发 展.2004年第7期. [2]吴丽娟,翟玉人.足球机器人系统众角色分配策略的设计.基础自动化.2000 年第7卷. [3]尚路彦,张小川,李祖枢.关于机器人足球区域射门动作算法的改进.重庆工 学院学报.2∞5年3月. [4]柳长安,刘冈,刘春阳.机器人足球防守算法研究.哈尔滨工业大学学报.2002 年第36卷第7期.

鼍P事p


碰壤曩鞲
2●
5 lO 15 20

大功率半导体器件用散热器风冷热阻计算公式和应用软件

大功率半导体器件用散热器风冷热阻计算公式和应用软件

大功率半导体器件用散热器风冷热阻计算公式和应用软件-CAL-FENGHAI.-(YICAI)-Company One1大功率半导体器件用散热器风冷热阻计算公式和应用软件2012-03-12 14:17:31作者:来源:中国电力电子产业网文章概要如下:一、计算公式为了推导风冷散热器热阻计算公式作如下设定:1,散热器是由很多块金属平板组成,平板一端连在一起成一块有一定厚度的基板,平板之间存在间隙,散热器的基本单元是一块平板;2,平板本身具有一定的长度、宽度和厚度(L×l×b)。

平板的横截面积A =L × b;3,由n个平板(齿片)组成的散热器如图一所示,平板(齿片)数为n ;4,由此可见,参数L即为散热器长,或称“截长”;5,设散热器端面周长为“S”。

大功率半导体器件安装在基板上,工作时产生的热通过接触面传到散热器的过程属于固体导热。

散热器平板周围是空气。

风冷条件下平板上的热要传到空气中属于固体与流体间的传热。

所以风冷散热器总热阻等于两部分热阻之和:Rzo(总热阻)= Rth(散热器内固体传热)+ Rthk(散热器与空气间的传热热阻)引用埃克尔特和..德雷克着的“传热与传质”中的基本原理和公式。

推导出如下实用公式:Ks 为散热器金属材料的导热系数。

20℃时,纯铝:KS = 千卡/ 小时米℃;纯铜:Ks = 332 千卡/ 小时米℃;参数L、l、b、S的单位:米;风速us 单位:米/秒如散热器端面的周边长为S 、散热器的长为L,忽略两端面的面积,散热器的总表面积为: A = S L 。

代入上式后,强迫风冷条件下散热器总热阻公式也可写成:对某一型号的散热器来说参数 Ks、b、n、S 都是常数。

用此公式即可求出不同长度L、不同风速us条件下的总热阻,并可作出相应曲线。

本公式的精确性受到多种因素的影响存在一定误差。

主要有:ⅰ,受到环境空气的温度、湿度、气压等自然因素的影响。

如散热器金属的热导系数“Ks”与金属成分及散热器工作时温度有关,本文选用的是20℃时的纯铝。

热阻的计算方法

热阻的计算方法

热阻的计算方法首先确定要散热的电子元器件,明确其工作参数,工作条件,尺寸大小,安装方式,选择散热器的底板大小比元器件安装面略大一些即可,因为安装空间的限制,散热器主要依靠与空气对流来散热,超出与元器件接触面的散热器,其散热效果随与元器件距离的增加而递减。

对于单肋散热器,如果所需散热器的宽度在表中空缺,可选择两倍或三倍宽度的散热器截断即可。

关于散热器选择的计算方法参数定义:Rt -------- 总内阻,C /W ;Rtj -------- 半导体器件内热阻,C /W;Rte ------- 半导体器件与散热器界面间的界面热阻,C /W;Rtf -------- 散热器热阻,C /W;Tj --------- 半导体器件结温,C;Te -------- 半导体器件壳温,C;Tf --------- 散热器温度,C;Ta -------- 环境温度,C;Pe -------- 半导体器件使用功率,W ;△Tfa -------- 散热器温升,C;散热计算公式:Rtf =(Tj-Ta) / Pe - Rtj -Rte散热器热阻Rff是选择散热器的主要依据。

Tj和Rtj是半导体器件提供的参数,Pe是设计要求的参数,Rte可从热设计专业书籍中查表。

(1)计算总热阻Rt: Rt= (Tjmax-Ta) / Pe(2)计算散热器热阻Rtf 或温升△ Tfa : Rtf = Rt —Rtj —Rte △ Tfa = Rtf x Pe(3)确定散热器:按照散热器的工作条件(自然冷却或强迫风冷),根据Rtf或厶Tfa和Pe选择散热器,查所选散热器的散热曲线(Rtf曲线或△ Tfa线),曲线上查出的值小于计算值时,就找到了合适的散热器。

对于型材散热器,当无法找到热阻曲线或温升曲线时,可以按以下方法确定:按上述公式求出散热器温升△ Tfa,然后计算散热器的综合换热系数 a :a = 7.2 2 1 2 2 2 3{ W [(Tf-Ta) / 20]} 式中:2 1 ------- 描写散热器L/b对a的影响,(L为散热器的长度,b为两肋片的间距);2 2 ------- 描写散热器h/b对a的影响,(h为散热器肋片的高度);2 3 ------- 描写散热器宽度尺寸W增加时对a的影响;W [(Tf-Ta) /20] -------------- 描写散热器表面最高温度对周围环境的温升对a的影响;以上参数可以查表得到。

  1. 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
  2. 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
  3. 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。

散热器扩散热阻的计算Accident?Consider the scenario where a designer wishes to incorporate a newly developed device into a system and soon learns that a heat sink is needed to cool the device. The designer finds a rather large heat sink in a catalog which marginally satisfies the required thermal criteria. Due to other considerations, such as fan noise and cost constraints, an attempt to use a smaller heat sink proved futile, and so the larger heat sink was accepted into the design. A prototype was made which, unfortunately, burned-out during the initial validation test, the product missed the narrow introduction time, and the project was canceled. What went wrong?The reasons could have been multi-fold. But, under this scenario, the main culprit could have been the spreading resistance that was overlooked during the design process. It is very important for heat sink users to realize that, unless the heat sink is custom developed for a specific application, thermal performance values provided in vendor's catalogs rarely account for the additional resistances coming from the size and location considerations of a heat source. It is understandable that the vendors themselves could not possibly know what kind of devices the users will be cooling with their products.Figure 1 - Normalized local temperature rise with heat sources of different size; from L to R, source area =100%, 56%, 25%, 6%, of heat-sink areaIntroductionSpreading or constriction resistances exist whenever heat flows from one region to another in different cross sectional area. In the case of heat sink applications, the spreading resistance occurs in the base-plate when a heat source of a smaller footprin footprint area is mounted on a heat sink with a larger base-plate area. This results in a higher local temperature at the location where the heat source is placed. Figure 1 illustrates how the surface temperature of a heat sink base-plate would respond as the size of the heat source is progressively reduced from left to right with all other conditions unchanged: the smaller the heat source, the more spreading has to take place, resulting in a greater temperature rise at the center. In this example, the effect of the edge surfaces of the heat sink is ignored and the heat source is assumed to be generating uniform heat flux. In cases where the footprint of a heat sink need not be much larger than the size of the heat source, the contribution of the spreading resistance to the overall device temperaturerise may be insignificant and usually falls within the design margin. However, in an attempt to remove more heat from today's high performance devices, a larger heat sink is often used and, consequently, the impact of spreading resistance on the performance of a heat sink is becoming an important factor that must not be ignored in the design process. It is not uncommon to find in many high performance, high power applications that more than half the total temperature rise of a heat sink is attributed to the spreading resistance in the base-plate.The objectives of this article are:1) to understand the physics and parameters associated with spreading resistance2) to provide a simple design correlation for accurate prediction of the resistance3) to discuss and clarify the concept of spreading resistance with an emphasis on the practical use of the correlation in heat sink applicationsThe correlation provided herein was originally developed in references 1 and 2. This article is an extension of the earlier presentation.Spreading ResistanceBefore we proceed with the analysis, let us attend to what the temperature distributions shown in Fig. 1 are telling us. The first obvious one, as noted earlier, is that the maximum temperature at the center increases as the heat source becomes smaller. Another important observation is that, as the temperature rises in the center, the temperatures along the edges of the heat sink decrease simultaneously. It can be shown that this happens in such a way that the area-averaged surface temperature of the heat sink base-plate has remained the same. In other words, the average heat sink thermal performance is independent of the size of a heat source. In fact, as will be seen later, it is also independent of the location of the heat source.The spreading resistance can be determined from the following set of parameters: ∙footprint or contact area of the heat source, A s∙footprint area of the heat sink base-plate, A p∙thickness of the heat sink base-plate, t∙thermal conductivity of the heat sink base-plate, k∙average heat sink thermal resistance, R0We will assume, for the time being, that the heat source is centrally mounted on the base-plate, and the heat sink is cooled uniformly over the exposed finned surface. These two assumptions will be examined in further detail. Figure 2 shows a two-dimensional side view of the heat sink with heat-flow lines schematically drawn in the base-plate whose thickness is greatly exaggerated. At the top, the corresponding surface temperature variation across the center line of the base-plate is shown by the solid line. The dotted line represents the average temperature of the surface which is, again, independent of theheat source size and can be easily determined by multiplying R0 with the total amount of heat dissipation, denoted as Q.As indicated in Fig. 2, the maximum constriction resistance R c, which accounts for the local temperature rise over the average surface temperature, is the only additional quantity that is needed for determining the maximum heat sink temperature. It can be accurately determined from the following correlation.Figure 2 - Two dimensional schematic view of local resistance or temperature variation of a heat sinkshown with heat flow LinesNote that the correlation addresses neither the shape of the heat source nor that of the heat sink base-plate. It was found in the earlier study that this correlation typically results in an accuracy of approximately 5% over a wide range of applications with many combinations of different source/sink shapes, provided that the aspect ratio of the shapes involved does not exceed 2.5. See references 1 and 2 for further discussions.Example ProblemConsider an aluminum heat sink (k = 200W/mK) with base-plate dimensions of 100 x 100 x 1.3 mm thick. According to the catalog, the thermal resistance of this heat sink under a given set of conditions is 1.0 °C/W. Find the maximum resistance of the heat sink if used to cool a 25 x 25 mm device.SolutionsWith no other specific descriptions, it is assumed that the heat source is centrally mounted, and the given thermal resistance of 1.0 °C/W represents the average heat sink performance. From the problem statement, we summarize:∙A s = 0.025 x 0.025 = 0.000625 m2∙A p = 0.1 x 0.1 = 0.01 m2∙t = 0.0013 m∙k = 200 W/mK∙R0 = 1.0 °C/WTherefore,Hence, the maximum resistance, R total , is:R total = R o + R c = 1.0 + 0.66 = 1.66 °C/WReaders should note the far right temperature distribution in Fig. 1 which is the result of a numerical simulation for the present problem in rectangular coordinates.Effect of Source LocationIn the following two sections, we will limit our examination to the current example problem. As we shall see, the result of this limited case study will allow us to draw some general yet useful conclusions. Suppose the same heat source in the above example was not centrally located, but mounted a distance away from the center. Obviously, the maximum temperature would further rise as compared to that found in the above example. Figure 3 shows the local resistances corresponding to two such cases:Figure 3 - Heat-sink local resistance showing the effect of source location:from L to R, heat source at (37.5,0) and (37.5,37.5)the first one is for the case where the heat source is mounted midway along the edge, and the other, where it is mounted on one corner of the heat sink. For these two special cases,the maximum spreading resistance can be calculated by using Eq. (1) for R c with input parameters t and R0 modified as shown below:R c = C x R c (A p,A s,k,t/C,R0/C)(3) with for the first case, and C = 2 for the second case. It is to be noted that this expression is independent of the source size. Numerically, for the current problem with a 25 x 25 mm heat source, it results in the maximum spreading resistances of 1.29 and2.38 °C/W, or the total resistances of 2.29 and3.38 °C/W for the first and second cases, respectively. For both cases, it can be shown that the average surface resistance has not changed from unity.For other intermediate source locations, numerical simulations were carried out and a plot is provided in Fig. 4 for the correction factor C f which can be used to compute the total resistance asR total = R0 + C f R c(4) where R c is determined from Eq. (1), given for the case with the heat source placed at the center.Figure 4 - Correction factor as a function of source locationThe coordinates in Fig. 4 indicate the location of the center of the heat source measured from the center of the base-plate in mm: the case with a centrally located heat source corresponds to (0,0), and the cases shown in Fig. 3 correspond to (37.5,0) and (37.5,37.5) for the first and second cases, respectively. Only one quadrant is shown in Fig. 4 as they would be, owing to the assumption of uniform cooling, symmetrical about (0,0). As can be seen from the figure, the correction factor increases from 1 as the heat source is placed away from the center. It is worthwhile noting that the increase is, however, very minimal over a wide region near the center, and most increases occur closer to the edges.Unlike C in the earlier expression, C f is case dependent (i.e. it depends on the heat-source size). However, it was found that the plots of C f obtained for many other cases exhibit essentially the same profile as that shown in Fig. 4, with magnitudes at the corners determined from Eq. (3), and the domain of the plot defined by the maximum displacement of the heat source. Based on this observation, a general conclusion can be made: for all practical purposes, as long as the heat source is placed closer to the center than to the edges of the heat sink, the correctional increase in the spreading resistance may be ignored, and C f =1 may be used. As noted above, this would introduce a small error of no greater than 5-10% in the spreading resistance which, in turn, is a fraction of the total resistance.So far, we have assumed a uniform cooling over the entire finned-surface area of the base-plate. Although this is a useful assumption, it is seldom realized in actual situations. It is well known that, due to the thinner boundary layer and the less down-stream heating effect, a device would be cooled more effectively if it is mounted toward the air inlet side. Again, a numerical simulation is carried out using our example problem with the boundary layer effect included.Figure 5 shows the resulting modified correction factor as a function of the distance from the center of the heat sink to the heat source placed along the center line at y=0: x = -37.5 mm corresponds to the front most leading edge location of the heat source and x = 37.5 mm the rear most trailing edge placement.Figure 5 - Correction factor modified for boundary layer effect at y=0As can be seen from the figure, it is possible to realize a small improvement by placing the heat source forward of the center location where C f < 1. However, it was experienced in practice that accommodating a heat source away from the center and ensuring its mounting orientation often cause additional problems during manufacturing and assembly processes.Summary and DiscussionA simple correlation equation is presented for determining spreading resistances in heat sink applications. A sample calculation is carried out for a case with a heat source placed at the center of the heat sink base-plate and a means to estimate the correction factor to account for the effect of changing the heat-source location is provided. It is to be noted that the correlation provided herein is a general solution which reduces to the well known Kennedy's solution3 when R0 approaches 0: the mathematical equivalent of isothermal boundary condition. Kennedy's solution is valid only when R0 is sufficiently small such that the fin-side of the heat sink base-plate is close to isothermal. Otherwise, Kennedy's solution, representing the lower boundary of the spreading resistance, may result in gross underestimation of the resistance.The earlier study revealed that, depending on the relative magnitude of the average heat sink resistance, the spreading resistance may either increase or decrease with the base-plate thickness. If the heat sink resistance is sufficiently small, as in liquid cooled heat sink applications, the spreading resistance always increases with the thickness, and an optimum thickness does not exist. On the other hand, if the heat sink resistance is large, as experienced in most air-cooled applications, the spreading resistance decreases with the thickness and a finite optimum thickness exists.It is to be noted that the present correlation calculates the spreading resistance only in the base-plate and does not account for the effect of additional spreading that may exist in other places, such as the fins in a planar heat sink. This additional spreading in the fins usually affects the spreading resistance in a similar way to a thicker base-plate. The current author found that an increase of 20% in the base-plate thickness during the calculation roughly accounts for the effect of this additional spreading in the fins of the same material for most planar heat sinks under air cooling. No modification is required for pin-fin heat sinks.Seri LeeAmkor Electronics, Inc.1900 South Price RoadChandler, AZ 85248, USATel: +1 (602) 821-2408 x 5459Fax: +1 (602) 821-6730Email: lees@References1.S. Lee, S. Song, V. Au, and K.P. Moran, Constriction/Spreading Resistance Model for ElectronicPackaging, Proceedings of the 4th ASME/JSME Thermal Engineering Joint Conference, Vol. 4, 1995, pp. 199-206.2.S. Song, S. Lee, and V. Au, Closed Form Equation for Thermal Constriction/Spreading Resistanceswith Variable Resistance Boundary Condition, Proceedings of the 1994 IEPS Conference, 1994, pp.111-121.3. D. P. Kennedy, Spreading Resistance in Cylindrical Semiconductor Devices, Journal of AppliedPhysics, Vol. 31, 1960, pp. 1490-1497.。

相关文档
最新文档