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Electronics => Beginners => Topic started by: ttyz on September 17, 2013, 03:13:05 pm

Title: Thermal resistance of unknown heatsink
Post by: ttyz on September 17, 2013, 03:13:05 pm
Hello everybody,

I have a small silly question, but I hope somebody could help me. I am now working on my pawer supply project. I have some considerations on thermal design. I have watched Dave's video on this topic, but I still have a question. I found a heatsink but the problem is that I don't know what part number it is, so I cannot look the thermal resistance in the datasheet to do the calculations. Is there any method to calculate or mesure thermal resistance?

Thank you in advance,
Ebgeny
Title: Re: Thermal resistance of unknown heatsink
Post by: Monkeh on September 17, 2013, 03:14:29 pm
Apply known amount of heat, measure temperature..
Title: Re: Thermal resistance of unknown heatsink
Post by: ttyz on September 17, 2013, 03:20:08 pm
Yes, thank you Monkeh. That was my first thought. But I could not really think of the source of this known amount of heat.
Title: Re: Thermal resistance of unknown heatsink
Post by: Balaur on September 17, 2013, 03:25:43 pm
If you want to be rigorous about it:


Or you can use a power resistor, but attaching it to the radiator is a bit more delicate.

PS (sorry for the initial incomplete submit, was a mistake)
Title: Re: Thermal resistance of unknown heatsink
Post by: Dave on September 17, 2013, 03:26:07 pm
You actually need to apply a known amount of power, not heat.
Heat - energy
Power - heat over time

All you need is a power resistor, a power supply and couple of multimeters.
Title: Re: Thermal resistance of unknown heatsink
Post by: ttyz on September 17, 2013, 03:28:32 pm
Thank you, guys, now I think I am getting the idea. I will give it a try.
Title: Re: Thermal resistance of unknown heatsink
Post by: mariush on September 17, 2013, 03:28:40 pm
Get a linear regulator like 7805 or a 7812, connect a lightbulb or a resistor, something that would use a few hundred mA (measure the exact current) ... measure the tab temperature after a minute or so.
Put the linear regulator on the heatsink, repeat and measure temperature.

You have the tab thermal resistance in the datasheet, so now it's just an issue of math to get the heatsink's thermal performance.
Title: Re: Thermal resistance of unknown heatsink
Post by: AG6QR on September 17, 2013, 03:33:15 pm
Yes, thank you Monkeh. That was my first thought. But I could not really think of the source of this known amount of heat.

Attach a resistor or transistor to the heat sink, using the same type of package that it's designed to accept.  Using an adjustable power supply or something else appropriate, dissipate power in the device.  Measure the voltage and current.

It's dead simple with a TO-220 resistor, which they do make.  It's only a bit more complicated with a transistor.  A voltage regulator is pretty easy to throw a known amount of power into.
Title: Re: Thermal resistance of unknown heatsink
Post by: Jay_Diddy_B on September 17, 2013, 03:56:10 pm
Hi,
You can use a little Physics.

First you weigh the heatsink.

Let say it weighs to 200g.

You then multiply this by the specific heat capacity of the material:

Aluminum 0.897 J/g/degrees C

200 x 0.897 = 179 J / degrees C

You can then heat the heatsink  in an oven, or any other way.

Place the heatsink in the orientation that you plan to use it. If you are using a fan, arrange the fan to blow across the heatsink.

Allow the heatsink to cool. While the heatsink is cooling measure and record the temperature changes.

Lets say you raise the heatsink to 60C and the room temperature is 20C

When the temperature has dropped by 63%, (60-20) x 0.64 = 25.6C, Would be 60-25.6 = 34.4C

Record the time, this is the thermal time constant.

Let's say it takes 300 seconds.

Thermal  Resistance = Thermal time constant / Thermal mass = 300 / 179 = 1.67 C/W


I apologize for using physics.

Jay_Diddy_B

Title: Re: Thermal resistance of unknown heatsink
Post by: ttyz on September 17, 2013, 04:08:05 pm
Jay_Diddy_B, cool method, thank you very much, I have to do it this way. Using phisics is always great :)