熱阻

熱阻(Thermal resistance)是熱傳計算與產品開發很重要的性質與參數,指單位時間內兩點之間溫度差與熱傳量的關係。熱阻愈大,熱傳性能愈差;反之,熱阻愈小,熱傳越好。

熱傳分為熱傳導,熱對流和熱輻射三種機制。其中熱傳導的熱阻,與材料本身的特性和幾何尺寸息息相關

Thermal Resistance

 

熱對流的計算方式為

Convection

 

同理,熱對流的熱阻為

Thermal Resistance

 

其中h會隨對流條件而變動,流體的速度愈快,h愈小,系統熱傳的能力愈好。

Conduction

 

Convection

 

我們常用散熱器底部到環境溫度的熱阻 Rth 或是 Rb-a 來表示其性能,單位是K/W 或是°C/W

  • Rth = (Tb-Ta)/Pd
  • Rth or Rb-a: Thermal resistance
  • Pd: Thermal power
  • Tb: Heat sink base temperature
  • Ta: Ambient temperature

以下圖為例,熱源發熱功率是Pd,設熱平衡之後,散熱器底部的溫度為Tb,環境溫度為T amb,則散熱器的熱阻為 Rth = (Tb-T amb)/Pd 整個散熱器的熱阻包含了先前說的傳導熱阻和對流熱阻,若改用熱傳導係數較高的材料,或是加入熱導管設計,可有效降低熱傳導的熱阻;若是將鳍片設計最佳化,或是加入風扇強制對流,則對流熱阻會下降,同樣改善整體散熱器的熱阻。

T-ambient

實際LED 燈具散熱器計算與分析請參考 Mechatronix 範例:

As an example we take a LED COB model, which has a nominal forward current If of 450mA and a maximal forward current of 900mA.
We will drive the module at a forward current of 500mA with a forward voltage Vf of 35.5V.
The maximal case temperature Tc is 105°C but in our design we aim at a life time case temperature of 75°C.
The ambient temperature for our application is 35°C. The electrical power Pe = Vf x If or 35.5V x 0.5A = 17.75W.
The dissipated power Pd = Pe x efficiency where the efficiency of the COB is around 32% or 17.75W x 0.68 = 12.07W.
This is the amount of energy which need to be cooled down.
dT is the temperature difference between the case temperature Tc we want to acquire and the ambient temperature Ta
dT = Tc = Ta or 75°C – 35°C = 40°C
The required maximal thermal resistance Rth of the LED cooler + the thermal interface material Rth = dT / Pd = 40°C / 12.07W = 3.31°C/W
The thermal interface you use has a major impact on the performance.
We recommend to use either arctic silver, a good thermal grease or a thin 0.1 to 0.15mm phase change or graphite thermal pad.
In this case the thermal resistance from the interface material will be between 0.1 and 0.2°C/W.
This interface resistance you substract from the calculated interface resistance to determine the cooling performance your heat sink needs to be.
So Rth heatsink = 3.31°C/W – 0.2°C/W = 3.11°C/W maximal.
Any led cooler which does better than a thermal resistance (lower value) of 3.11°C/W in free air conditions would make that our LED case temperature Tc will remain below the required 75°C.
Keep in mind that an enclosure around the LED cooler, tilting and other variations will affect the performance of the LED cooler.

Thermal-Calculation

 

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