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I have to dissipate \$2W\$ from a voltage regulator. It's a 7805 in a TO-220 package. The datasheet is here.

It's the first time I have to pick those so I would like to have a review of the following decision because I'm afraid of missing something as this sound really complicated for me. So I will put you here my entire reasoning.

\$R_{thJC}\$ is 5 C°/W for the TO-220 package and \$R_{thJA}\$ is 50 C°/W (table 3, page 7). As I will need to dissipate 2W I will have without heatsink 100° of heat on the chip. The room is around 21°. \$T_{op}\$ is 0° to 125° so to be safe I definitely need a heatsink. In that case it will just go around 31° based on that formula \$Max_{ambiant}+ R_{thJC} \times W_{dissipated}\$ or 21 + 50 * 2

But now I'm blocked. For the example I will take this heatsink. He's rated as 40 K/W. I assume K is for kelvin degrees. In that case does it means he's rated as 233°C/W ? I've found that formula :

\$Max_{JunctionTemp} >= Max_{AmbientTemp} + (W_{Dissipated} \times (R_{thJC_{C°/W}} + R_{thHeatSink_{C°/W}}))\$

Which give me :

\$ 21 + (2.5 \times (5 + 233) = 595°C\$

So, there's something wrong as this would mean that the junction between chip and heatsink will be 600° hot ... What have I missed ?

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The K is for Kelvin, but it's a relative change per watt so it's the same thing. But nice question and hopefully someone posts a more formal explanation of it all. –  PeterJ Dec 27 '13 at 11:22
    
As Wouter points out below, be sure to include the resistance between the TO220 case and the heatsink. You should also do the calculations using the maximum ambient temperature and maximum power dissipation (rather than "around 21 degrees"). –  Joe Hass Dec 27 '13 at 14:39
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Degree K = degree C - just with offset zero points. 40K/W = 40 C/W so with 2W = 2 x 40 = 80K = 80C rise over ambient. lways set ambient at highest possible real world value you will experience including hot day inside no aircon etc. Then add a bit :-). If on car firewall = much hotter. | Down to 10C/W is not too hard. | ANY blown air or draft = great improvement. –  Russell McMahon Dec 28 '13 at 20:00
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2 Answers

up vote 2 down vote accepted

Refer to wikipedia

The SI units of thermal resistance are kelvins per watt or the equivalent degrees Celsius per watt (the two are the same since as intervals Δ1 K = Δ1 °C).

K/W is the same as C/W and that's because they represent a temperature difference per watt rather than an absolute temperature.

The result for your calculation using a 40K/W heatsink is:
\$21 + (2.5 \times (5 + 40)) = 112.5°C \$


There seems to be some misconception regarding the meaning of the K/W rating and the cooling ability of a given heatsink.
When you compare two heatsinks, the lower the K/W rating the better the heatsink, a lower K/W rating means it can dissipate more power with less temperature increase.

As an example:
A 40K/W heatsink increases the temperature 40 degree Celsius (above the ambient temperature) for every Watt. A more efficient heatsink (regarding cooling ability) is a model that has lower K/W rating like for example 20K/W because the temperature will increase only by 20 degree Celsius for each dissipated watt.

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Ok, but there's something I don't understand. If I need to dissipate 2Watt with RthJC of 5, in that case does it mean I need a heatsink able at least of 5K/W ? So a 40K/W should do the job ? (So does the 112.5° means the temp of the heatsink or what he could dissipate ?) –  Emmanuel Istace Dec 27 '13 at 11:32
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@EmmanuelIstace A 40K/W heatsink increases the temperature 40 degree Celsius for every Watt. A more efficient heatsink like for example 20K/W increases the temperature 20 degree Celsius for every Watt. A 40K/W heatsink is worse than 5K/W –  alexan_e Dec 27 '13 at 11:35
    
@EmmanuelIstace I can't really reply on the economics. A better heatsink has more metal and/or more fins which should be more expensive to manufacture –  alexan_e Dec 27 '13 at 11:46
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You can solve a heat problem in the same way you would solve a current-through-a-resistor problem. Current is equivalent to heat, resistance is thermal resistance, and voltage is temperature.

You have 2W of heat current through a series of heat resistors: Rj-c (5K/W), add 1K/W for the imperfect contact between the case and your heatsink, and the heatsink-to-air (40K/W). The total is 46K/W. With a heat flow of 2W this will cause a temperature gradient of 98 K: the junction will be 98K hotter than the ambient air.

A difficult question in such calculations is how low you can guarantee the ambient air to be. Let's assume (maximum) 40C. Then the (maximum) junction temperature is 40 + 98 = 138C.

The (Fairchild) LM7805 lists 125C as maximum operating temperature under the 'absolute maxima'. Note that in principle the absolute maxima can NOT be used for design calculations, but the graphs later on have curves up to 125C, so on THAT ground the figure of 125C is OK to use.

125 < 138, so with an ambient temperature of 40C and 2A your heatsink might not be enough. (I say might because I used the worse case figures. But as a designer, you should!)

I suggest you find yourself a heatsink that is a bit bigger, aim for 20K/W. This will also make the heatsink less hot to the tough (but still far too hot for touching comfortably! calculate for yourself how hot it will be.).

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+1 for a nice explanation. The distinction that the ratings for heatsinks are the temperature increase from ambient per watt of energy dumped into it was always un-intuitve to me when I first started working with them. –  Adam Head Dec 27 '13 at 14:58
    
Simply think of the figure as a resistance. The higher C/W, the higher the 'voltage' drop will be for a given 'current'. –  Wouter van Ooijen Dec 27 '13 at 15:35
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