Author Topic: TO-263S heat dissipation  (Read 425 times)

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Offline reyntjensm

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TO-263S heat dissipation
« on: November 26, 2020, 04:53:34 pm »
I have a project with two mosfets( https://fscdn.rohm.com/en/products/databook/datasheet/discrete/transistor/mosfet/r6020pnjfratl-e.pdf ) in a TO-263S package close to each other. According to LTspice the mosfet's would dissipate 40W. How can i get the heat out of such package? I can't find any heatsink for such a package who is able to dissipate 20W. Should this be done via a very large copper pour? Max PD in datasheet is 305W, how can they ever get the heat out of such a package?
 

Online wraper

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Re: TO-263S heat dissipation
« Reply #1 on: November 26, 2020, 05:04:49 pm »
Quote
According to LTspice the mosfet's would dissipate 40W. How can i get the heat out of such package?
Forget about it. It's cheaper and way easier to use more MOSFETs.
Quote
Max PD in datasheet is 305W, how can they ever get the heat out of such a package?
They don't (under reasonable conditions). If you want to dissipate even 20W from single MOSFET, you will need to use a through hole part mounted on a heatsink.
 

Offline TimFox

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Re: TO-263S heat dissipation
« Reply #2 on: November 26, 2020, 05:18:23 pm »
This is a common (and commonly misunderstood) way to specify thermal power limits on power semiconductors.
There are two important lines: 
Maximum junction temperature Tj = 150 C
Maximum power at Tcase = 25 C,  304 W.
Those two terms determine the relevant parameter of thermal resistance from junction to case:
RT jc = (150 - 25) K / (304 W) = 0.411 K/W
This thermal resistance is in series with the interface resistance between case and heatsink, and the resistance between heatsink and ambient air.  That external resistance is generally much larger than the internal resistance.
I find the “304 W” value suspicious, since it may not be determined to 1% accuracy.
Note the spec of 80 K/W for total thermal resistance when mounted on 25 x 27 x 0.8 mm FR4 pub (note 5).  Mounted thusly in 25 C air limits the power to (125 K)/(80 K/W) = 1.56 W.
« Last Edit: November 26, 2020, 05:23:27 pm by TimFox »
 

Offline Doctorandus_P

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Re: TO-263S heat dissipation
« Reply #3 on: November 26, 2020, 05:55:41 pm »
You can probably dissipate 20W from a TO263 if you solder it directly to a PCB with aluminimum substrate. I don't know the name of that, but it's commonly used for LED panels with COB.

Another option is to make a big hole in the PCB and then solder in a spacial stud for mounting a heatsink on, but that is also a pretty specialized solution.

The most common solution to satisfy a need to dissipate 20Watt would be to use another package such as TO220.

A more modern way would be to use a solution to avoid having to waste 20W as heat.
 

Offline Siwastaja

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Re: TO-263S heat dissipation
« Reply #4 on: November 26, 2020, 06:00:35 pm »
Today, silicon is cheap, aluminium is expensive.

This means, buy more semiconductors. If nothing else, Rth_junction-to-case parallel, making heatsinking much easier, meaning you can run the heatsink at higher temperature while maintaining low enough Tj.
 


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