散热器扩散热阻的计算

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热阻的公式

热阻的公式

热阻的公式热阻这个概念,在物理学中可是相当重要的哟!咱先来说说热阻到底是个啥。

打个比方吧,就像咱们在马路上走路,有时候会遇到一些阻碍,让咱们走得没那么顺畅。

热在传递的过程中也会遇到这样的“阻碍”,这“阻碍”就是热阻。

热阻可以用来衡量热量传递的难易程度。

那热阻的公式是啥呢?热阻(R)等于材料的厚度(d)除以热导率(k)和传热面积(A)的乘积,用公式写出来就是 R = d / (k × A) 。

比如说,有一块厚度为 10 厘米的木板,热导率是 0.1 瓦/(米·开尔文),传热面积是 1 平方米。

那按照公式来算,热阻 R = 0.1 米 /(0.1×1) = 1 开尔文/瓦。

这就意味着,要让 1 瓦的热量通过这块木板,会产生 1 开尔文的温差。

我记得有一次,我在给学生们讲解热阻这个知识点的时候,发生了一件特别有意思的事儿。

当时我拿了两块材质不同、厚度相同的板子,一块是木板,一块是铝板。

我在板子的一边加热,然后让同学们用手去感受另一边的温度变化。

同学们都特别好奇,一个个瞪大眼睛,伸着小手去感受。

结果发现铝板那一边很快就热了,而木板那一边温度变化很小。

这时候我就趁机跟他们说:“同学们,这就是因为木板的热阻大,铝板的热阻小呀,热量在铝板里传递就容易得多,在木板里就困难一些。

”同学们恍然大悟,那场面,可有趣啦!再深入讲讲热阻这个公式哈。

其中的热导率(k),它反映了材料本身导热的能力。

像金属的热导率一般就比较高,比如铜、铝这些,所以它们常常被用来制作散热器,能快速把热量散出去。

而像木头、塑料这些材料,热导率就低,热阻就大。

厚度(d)就很好理解啦,材料越厚,热量要穿过它就越难,热阻也就越大。

传热面积(A)也会影响热阻哦。

如果传热面积大,热量传递就相对容易些,热阻就会小一点。

在实际生活中,热阻的概念用处可大了。

比如说咱们家里的暖气,管道和暖气片的材质选择、厚度设计,都得考虑热阻,这样才能保证咱们的屋子能暖暖和和的。

散热器的扩散热阻计算.docx

散热器的扩散热阻计算.docx

散热器的扩散热阻计算
1、什么是扩散热阻?
当热源与底板的面积相差比较大时,热量从热源中心往边缘扩散所形成热阻叫扩散热阻。

图1扩散热阻示意
2、如何计算?
针对一个底板上贴一个热源的散热状况,如何计算扩散热阻?分两种情况,一个是热源在底板中心,另一个是不在中心。

针对第一种情况的扩散热阻计算如下:
As:热源面积
Ap:底板面积
t:底板厚度
R0:散热器的平均热阻(从地板厚度的一半之处到环境的热阻)k:散热器材料的导热系数
假设其他参数已知,扩散热阻Rc的计算公式如下:
其中,系数λ是一个由Ap和As计算出来的因子。

从上可以看出,总热阻为扩散热阻与散热器的平均热阻的总和。

针对热源不在中心的情况,需要对公示1进行修正。

增加修正系数C:
而C的计算如下:
假设一个正方形的底板,热源也是正方形,由于对称的关系,取四分之一底板来做分析:
图2:2种热源不在中心的状况
如图2所示,左侧的图,热源的坐标为(x,0),右侧的图中热源坐标为(x,x)。

左侧图中的C=1.414(根号2)。

右侧的C=2 ,如何推算需另外讨论。

如图2中的2种热源位置,热源的大小为25*25mm的话,最大的扩散热阻分别计算如下:
总热阻分别为2.259℃/W和3.334℃/W。

对于上述两种状况,平均热阻R0是不变的,即平均热阻与热源的位置无关。

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。

LED散热器各部分热阻及其影响因素

LED散热器各部分热阻及其影响因素
⑵.对流系数的影响:散热器周围环境通风越好,自然对流系数越大,散热器热阻越小
⑶.产热功率的影响:同一散热器,同样环境下,实际产热功率越大,散热器的热阻反而略有减小。
所以散热器的总热阻不仅与散热器的散热面积、几何尺寸、表面材料的辐射系数等自身因素有关,还受LED的产热功率以及周围环境的对流系数等外部因素的影响,并不是一个恒定的数值。但一般来说,在自然对流情况下对流系数变化并不大,正常情况下LED产热功率的变化也不会太大,对热阻的影响应该很小。为便于分析和计算,我们在应用时可近似认为散热器的总热阻是一定的。
Q产=a.W⑵
Q导=b.s.(T1-T2)/L⑶
式中
Q产——LED工作时产生的热量
Q导——散热器本身导出的热量
T1——与铝基板接触点处散热器的温度
T2——散热器外表面平均温度
a——LED产热系数
W——为LED灯实际功率
b——散热器材料综合导热系数
s——散热器平均传热面积
L——散热器热传导平均距离
对于特定散热器b、s、L是一定的,因此公式⑶可简化为Q导=m.(T1-T2),其中m=b.s/L,经推导可知m.(T1-T2)=a.W,因此(T1-T2)=a.W/m,带入公式⑴可知R导=a/m,由此公式可以看出对于特定散热器,在LED灯源一定的情况下,散热器的热阻是一个定值。另外,在热阻计算公式中W代表的是LED的总功率,而LED在工作中一部分功率用于发光,一部分功率转变为热能,因此既然是计算热阻,公式中的W换成产热功率(a.W)更为科学,这样R导=1/m=L/(b.s),就是说散热器本身热阻与电阻一样,是一个仅跟散热器本身参数有关的常数,它与散热器平均传热距离成正比,与散热器平均传热面积、散热器材料导热系数成反比。
LED散热器各部分热阻及其影响因素

散热器散热计算公式:

散热器散热计算公式:

(一)散热器选择通用原则 散热器热阻Rsa 是选择散热器的主要依据。

Rsa=c ajm P TT−-(R jc+R cs)式中:R sa────散热器热阻,℃/W;R jc────半导体器件结壳热阻,℃/W;R cs────接触热阻,℃/W;T jm ────半导体器件最高工作结温,℃;T a────环境温度,℃;P c ────半导体器件耗散功率,W;T jm,P c,R jc可以从器件技术参数表中查到,或计算得到;T a是实际工作环境温度;R cs与接触材料的种类和接触压力有关,可以根据接触材料(如硅脂)的热阻参数估算得到。

所选择的散热器,其热阻值应小于以上的计算值,就可满足散热的要求。

散热器的热阻与材质,结构,表面状态,表面颜色,几何尺寸及冷却条件等有关;应该按照有关的标准用实验的方法测试得到,常用的散热器热阻曲线有3种,(1)热阻——长度曲线,(2)热阻——风速曲线,(3)功耗——温升曲线。

用CFD技术模拟仿真运算可以得到散热器的热阻值,风压及温度分布状况,为散热器选择提供参考依据。

(二)电力半导体用散热器的选择和使用原则 摘自JB/T9684-2000一﹑散热器选择的基本原则电力半导体器件用散热器选择要根据器件的耗散功率,器件结壳热阻,接触热阻,以及器件最高工作结温和冷却介质温度来综合考虑。

选用散热器时要了解散热器的散热能力范围,冷却方式,技术参数和结构特点,一种器件仅从热阻参数看,可能有多种散热器均能满足散热要求,但应结合冷却,安装,通用互换和经济性来综合考虑。

二﹑器件与散热器紧固力的要求为使器件与散热器组装后又良好的热接触,必须采用合适的安装力或安装力矩,其值由器件制造厂或器件标准给出,具有较小的范围,组装时应严格遵守不要超出范围,当器件厂未给出紧固力时,按照器件管壳与散热器接触的面积,可采用1~1.5KN/cm2的紧固力。

为了改善散热器与器件的接触,增加有效接触面积,提高散热效果,在散热器和器件之间可涂一薄层导电导热性物质如硅脂。

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.

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散热器扩散热阻的计算

散热器扩散热阻的计算

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.。

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,℃/W;Rtj───半导体器件内热阻,℃/W;Rtc───半导体器件与散热器界面间的界面热阻,℃/W;Rtf───散热器热阻,℃/W;Tj───半导体器件结温,℃;Tc───半导体器件壳温,℃;Tf───散热器温度,℃;Ta───环境温度,℃;Pc───半导体器件使用功率,W;ΔTfa ───散热器温升,℃;散热计算公式:Rtf =(Tj-Ta)/Pc - Rtj –Rtc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,Pc是设计要求的参数,Rtc 可从热设计专业书籍中查表。

(1)计算总热阻Rt:Rt= (Tjmax-Ta)/Pc(2)计算散热器热阻Rtf 或温升ΔTfa:Rtf = Rt-Rtj-Rtc ΔTfa=Rtf×Pc(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ΔTfa和Pc 选择散热器,查所选散热器的散热曲线(Rtf 曲线或ΔTfa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ΔTfa,然后计算散热器的综合换热系数α:α=7.2ψ1ψ2ψ3{√√[(Tf-Ta)/20]}式中:ψ1───描写散热器L/b对α的影响,(L为散热器的长度,b为两肋片的间距);ψ2───描写散热器h/b对α的影响,(h为散热器肋片的高度);ψ3───描写散热器宽度尺寸W增加时对α的影响;√√[(Tf-Ta)/20]───描写散热器表面最高温度对周围环境的温升对α的影响;以上参数可以查表得到。

散热器的计算

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。

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。

热阻计算[详解]

热阻计算[详解]

热阻计算热阻计算2008-01-13 22:21一般,热阻公式中,Tcmax =Tj - P*Rjc 的公式是在假设散热片足够大而且接触足够良好的情况下才成立的,否则还应该写成Tcmax =Tj - P*(Rjc+Rcs+Rsa)。

Rjc表示芯片内部至外壳的热阻,Rcs表示外壳至散热片的热阻,Rsa表示散热片的热阻。

没有散热片时,Tcmax =Tj - P*(Rjc+Rca)。

Rca表示外壳至空气的热阻。

一般使用条件用Tc =Tj - P*Rjc的公式近似。

厂家规格书一般会给出,Rjc,P等参数。

一般P是在25度时的功耗。

当温度大于25度时,会有一个降额指标。

举个实例:一、三级管2N5551规格书中给出25度(Tc)时的功率是1.5W(P),Rjc是83.3度/W。

此代入公式有:25=Tj-1.5*83.3可以从中推出Tj为150度。

芯片最高温度一般是不变的。

所以有Tc=150-Ptc*83.3,其中Ptc表示温度为Tc时的功耗。

假设管子的功耗为1W,那么,Tc=150-1*83.3=66.7度。

注意,此管子25度(Tc)时的功率是1.5W,如果壳温高于25度,功率就要降额使用。

规格书中给出的降额为12mW/度(0.012W/度)。

我们可以用公式来验证这个结论。

假设温度为Tc,那么,功率降额为0.012*(Tc-25)。

则此时最大总功耗为1.5-0.012*(Tc-25)。

把此时的条件代入公式得出:Tc=150-(1.5-0.012*(Tc-25))×83.3,公式成立。

一般情况下没办法测Tj,可以经过测Tc的方法来估算Ttj。

公式变为:Tj=Tc+P*Rjc同样与2N5551为例。

假设实际使用功率为1.2W,测得壳温为60度,那么:Tj=60+1.2*83.3=159.96此时已经超出了管子的最高结温150度了!按照降额0.012W/度的原则,60度时的降额为(60-25)×0.012=0.42W,1.5-0.42=1.08W。

(完整版)散热器设计的基本计算

(完整版)散热器设计的基本计算

? 条件
Rthjc——器件手册查询
Rthcs——材料热阻:
R =L /( K ·S ) th 绝缘垫
绝缘垫厚度
绝缘垫
绝缘垫接触 c 的面积
Rthsa——散热器热阻曲线图查询
T 结温——器件手册查询(待计算数值)
T 环温——任务指标中的工作环境要求
P ——电路设计计算
? 计算
T 结温 =( Rthjc+Rthcs+Rthsa)· P+ T 环温 <手册推荐结温
电流 VabI (A) 电压 Vab=Va- Vb (V) 电阻 R=Vab/I (Ω) 电阻串联 R=R1+R2+… 电阻并联 1/R=1/R1+ 1/ R2+…
2、 热阻:在热路中,各种介质及接触状态,对热量的传递表现出的不同阻碍作用—— 在热路中产生温度差 , 形成对热路中两点间指标性的评价。
绝缘垫厚度
绝缘垫
绝缘垫接触 c 的面积
Rthsa——散热器热阻曲线图查询
T 环境 ——任务指标中的工作环境要求
? 计算
J1 的最大结温: Tjmax1=( Rthjc1+Rthcs1)·Pj1+Rthsa·( Pj1+Pj2)+ T 环境
J2的最大结温: Tjmax2=(Rthjc2+ Rthcs2)·Pj2+Rthsa·( Pj1+Pj2)+ T 环境 ? 注: 判定计算出的最大结温,是否小于手册推荐结温;
符号—— Rth
单位——℃ /W。
? 稳态热传递的热阻计算 : Rth= (T1-T2)/P
T1——热源温度(无其他热源) (℃ ) T2——导热系统端点温度 (℃ ) ? 热路中材料热阻的计算 : Rth=L/(K·S)
L——材料厚度 (m) S——传热接触面积 ( m2)

热阻计算

热阻计算

热阻计算一般,热阻公式中,Tcmax =Tj - P*Rjc的公式是在假设散热片足够大而且接触足够良好的情况下才成立的,否则还应该写成Tcmax =Tj - P*(Rjc+Rcs+Rsa)。

Rjc表示芯片内部至外壳的热阻,Rcs表示外壳至散热片的热阻,Rsa表示散热片的热阻。

没有散热片时,Tcmax =Tj - P*(Rjc+Rca)。

Rca 表示外壳至空气的热阻。

一般使用条件用Tc =Tj - P*Rjc的公式近似。

厂家规格书一般会给出,Rjc,P等参数。

一般P是在25度时的功耗。

当温度大于25度时,会有一个降额指标。

一、可以把半导体器件分为功率器件和小功率器件。

1、大功率器件的额定功率一般是指带散热器时的功率,散热器足够大时且散热良好时,可以认为其表面到环境之间的热阻为0,所以理想状态时壳温即等于环境温度。

功率器件由于采用了特殊的工艺,所以其最高允许结温有的可以达到175度。

但是为了保险起见,一律可以按150度来计算。

适用公式:Tc =Tj -P*Rjc。

设计时,Tj最大值为150,Rjc已知,假设环境温度也确定,根据壳温即等于环境温度,那么此时允许的P也就随之确定。

2、小功率半导体器件,比如小晶体管,IC,一般使用时是不带散热器的。

所以这时就要考虑器件壳体到空气之间的热阻了。

一般厂家规格书中会给出Rja,即结到环境之间的热阻。

(Rja=Rjc+Rca)。

同样以三级管2N5551为例,其最大使用功率1.5W是在其壳温25度时取得的。

假设此时环境温度恰好是25度,又要消耗1.5W的功率,还要保证结温也是25度,唯一的可能就是它得到足够良好的散热!但是一般像2N5551这样TO-92封装的三极管,是不可能带散热器使用的。

所以此时,小功率半导体器件要用到的公式是:Tc =Tj - P*RjaRja:结到环境之间的热阻。

一般小功率半导体器件的厂家会在规格书中给出这个参数。

2N5551的Rja,厂家给的值是200度/W。

汽车散热器选择的计算方法

汽车散热器选择的计算方法

散热器选择的计算方法一,各热参数定义:Rja———总热阻,℃/W;Rjc———器件的内热阻,℃/W;Rcs———器件与散热器界面间的界面热阻,℃/W;Rsa———散热器热阻,℃/W;Tj———发热源器件内结温度,℃;Tc———发热源器件表面壳温度,℃;Ts———散热器温度,℃;Ta———环境温度,℃;Pc———器件使用功率,W;ΔTsa ———散热器温升,℃;二,散热器选择:Rsa =(Tj-Ta)/Pc - Rjc -Rcs式中:Rsa(散热器热阻)是选择散热器的主要依据。

Tj 和Rjc 是发热源器件提供的参数,Pc 是设计要求的参数,Rcs 可从热设计专业书籍中查表,或采用Rcs=截面接触材料厚度/(接触面积X 接触材料导热系数)。

(1)计算总热阻Rja:Rja= (Tjmax-Ta)/Pc(2)计算散热器热阻Rsa 或温升ΔTsa:Rsa = Rja-Rtj-RtcΔTsa=Rsa×Pc(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Rsa 或ΔTsa 和Pc 选择散热器,查所选散热器的散热曲线(Rsa 曲线或ΔTsa 线),曲线上查出的值小于计算值时,就找到了合适的热阻散热器及其对应的风速,根据风速流经散热器截面核算流量及根据散热器流阻曲线上风速对应的阻力压降,选择满足流量和压力工作点的风扇。

散热器热阻曲线三,散热器尺寸设计:对于散热器,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ΔTsa,然后计算散热器的综合换热系数α:α=7.2ψ1ψ2ψ3{√√ [(Tf-Ta)/20]}式中:ψ1———描写散热器L/b 对α的影响,(L 为散热器的长度,b 为两肋片的间距);ψ2———描写散热器h/b 对α的影响,(h 为散热器肋片的高度);ψ3———描写散热器宽度尺寸W 增加时对α的影响;√√ [(Tf-Ta)/20]———描写散热器表面最高温度对周围环境的温升对α的影响;以上参数可以查表得到。

散热片热阻计算

散热片热阻计算

散热片热阻计算散热片热阻是指散热片在散热过程中阻碍热量传递的程度。

散热片是一种用于散热的设备,通常由金属制成,具有较好的导热性能。

在电子设备、汽车发动机、空调等各种应用中,散热片起着重要的散热作用。

散热片的热阻是指单位面积上热量通过散热片的难度,其计算公式为:热阻 = 温度差 / 热流率热阻越小,热量传递越顺畅,散热效果越好。

散热片的热阻主要由以下几个因素决定:1. 散热片材料的导热性能:散热片通常采用导热性能较好的金属材料,如铝、铜等。

这些金属具有较高的热导率,能够快速传导热量,从而降低热阻。

2. 散热片的结构形式:散热片的结构形式也会影响其热阻。

散热片通常采用片状或翅片状的结构,增加了散热面积,提高了热量的散发能力。

同时,翅片的设计也会影响热阻的大小,合理的翅片结构能够增加热量的传导效率。

3. 散热片与散热介质之间的接触热阻:散热片通常需要与散热介质(如风扇、散热鳍片等)接触,将热量传递给散热介质。

接触热阻取决于接触面的平整度、接触面积、接触介质的导热性能等因素。

为了减小接触热阻,通常需要采取一些措施,如增加接触面积、使用导热硅脂等。

4. 散热片的尺寸和形状:散热片的尺寸和形状也会影响热阻。

一般来说,散热片的尺寸越大,散热面积越大,热量传递能力越强,热阻越小。

同时,散热片的形状也会影响热量的传导效率,如翅片的形状和密度等。

在实际应用中,为了降低散热片的热阻,可以采取以下措施:1. 选择导热性能好的材料:选择导热性能好的金属材料,如铝、铜等,能够提高散热片的热传导能力,降低热阻。

2. 设计合理的翅片结构:合理设计翅片的形状和密度,增加散热面积,提高热量的散发能力。

3. 优化散热片与散热介质的接触:采取一些措施,如增加接触面积、使用导热硅脂等,减小散热片与散热介质之间的接触热阻。

4. 增大散热片的尺寸:增大散热片的尺寸,增加散热面积,提高热量的传导效率。

散热片的热阻是影响散热效果的重要指标。

通过选择合适的材料、合理设计翅片结构、优化散热片与散热介质的接触方式以及增大散热片的尺寸等措施,可以有效降低散热片的热阻,提高散热效果,确保设备的正常运行和稳定性。

dc器件热阻

dc器件热阻

dc器件热阻
热阻是指热量从器件的晶片上向外传导时受到的阻力,其单位是℃/W。

DC器件的热阻因封装和安装PCB板的条件而异,通常在各IC的技术规格书中会给出标准值。

以散热片为例,散热片的用途是增大散热片面积,以便将DC-DC转换器产生的热量转移到空气中,这会降低热阻,但增加了DC-DC转换器的体积。

当给DC-DC转换器添加散热片时,应考虑散热片装配表面与DC-DC转换器外壳之间的热阻,计算方式如下:R外壳-环境=R外壳-散热片+R散热片-环境,可使用热表面材料(导热硅脂等)将表面热阻减少到最小,使用这种热表面材料,R外壳-散热片值可以达到1'C/W以下。

在实际应用中,应根据具体情况选择合适的散热方式,以降低DC器件的热阻,提高器件的可靠性。

关于热阻

关于热阻

什么是热阻?
所谓“热阻”(thermal resistance),是指反映阻止热量传递的能力的综合参量。

热阻的概念与电阻非常类似,单位也与之相仿——℃/W,即物体持续传热功率为1W时,导热路径两端的温差。

对散热器而言,导热路径的两端分别是发热物体(CPU)与环境空气。

热阻的计算有一套完整的公式,通过下面的公式,我们可以得出一款散热器的热阻值。

Rca=(Tc-Ta)/P
其中Tc为CPU表面温度,Intel和AMD都规定其为CPU铜盖表面正中心位置的温度。

Ta为环境温度,但Ta不同于我们通常理解的“机箱温度”,而是指散热器风扇的进风口温度。

Intel和AMD都严格的规定,Ta是在风扇进风口上方,非常靠近风扇的位置取得的一组测试值的算数平均值。

P代表CPU功率,是Intel和AMD给出的散热设计功率(TDP),也是某款CPU能达到的最大功率。

对于CPU散热器而言,其热阻的计算即:散热器热阻=(CPU表面温度-环境温度)÷导热功率。

实际上,为了更好地理解热阻的含义,我们可以把“热阻”理解为电路中的电阻,两个温度的差值相当于电压,而CPU的功率相当于电流。

在一定的环境温度和CPU功率下,热阻小的散热器会让CPU表面温度更低,其性能也更好。

在弄清了热阻的含义以及它对于评价一款散热器性能的重要性后,我们需要寻找一套科学、适用的工具和方法,借以实测ICE WING 6的热阻,验证它的散热能力.。

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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 acceptedinto the design. A prototype was made which, unfortunately, burned-outduring the initial validation test, the product missed the narrowintroduction 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 heatsink 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 ofthe heat source is progressively reduced from left to right with all otherconditions 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 temperature rise may be insignificant and usually fallswithin 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 sfootprint area of the heat sink base-plate, A pthickness of the heat sink base-plate, tthermal conductivity of the heat sink base-plate, kaverage 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 withheat-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 surfacewhich is, again, independent of the heat 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, whichaccounts 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 aheat sink shown 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 thatthis 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 notexceed . 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 mm thick. According to the catalog, the thermal resistance ofthis heat sink under a given set of conditions is °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 °C/W represents the average heat sink performance. From the problem statement, we summarize:A s = x = m2A p = x = m2t = mk = 200 W/mKR0= °C/WTherefore,Hence, the maximum resistance, R total , is:R total = R o + R c= + = °C/WReaders should note the far right temperature distribution in Fig. 1 whichis 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 ,0) and ,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 resi stances of and °C/W, or the total resistances of and °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 ,0) and , 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 . 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 thinnerboundary 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 = mm corresponds to the front most leading edge location of the heat source and x = 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 theeffect 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 islarge, 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, USAReferences1.S. Lee, S. Song, V. Au, and . Moran, Constriction/Spreading Resistance Model forElectronic Packaging, Proceedings of the 4th ASME/JSME Thermal Engineering JointConference, Vol. 4, 1995, pp. 199-206.2.S. Song, S. Lee, and V. Au, Closed Form Equation for Thermal Constriction/SpreadingResistances with 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 ofApplied Physics, Vol. 31, 1960, pp. 1490-1497.。

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