Can these hetsinks dissipate 125W ? - diyAudio
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Old 30th June 2001, 06:47 PM   #1
hifi is offline hifi  Sweden
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http://www.webra.se/html/products/w-222.htm

im thinking of using this heatsink in 20cm length for Sides in an ALEPH 4 monoblock will it be able to dissipate the heat?...i think the back is to thin only 5mm but its so damn cheap...one with 10mm back cost like 3times as musch...(in this case) can i use say an 5mm copper or aluminium "coldplate" to mount the mosfets on and then mount it on the heatsinks?

your opinion?

/micke
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Old 30th June 2001, 08:56 PM   #2
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Micke,

The W-222 heatsink (304mm x 200mm x 40mm) has a thermal resistance of between 0.25 and 0.3 degC/W. Using these sinks to dissipate 125W each would mean a dT of between 31 and 37 degC. With an ambient temperature of 25degC the heatsinks could reach 62 degC. This is not ideal as this is probably too hot to touch but is certainly not unacceptable for heatsink temperatures.

I would recommend using Cu/Al backing plates for each mosfet 2-3 times the mounting area of the package. Mount the mosfet directly to the plate (with some thermal grease) and isolate the plate from the sink with a mica sheet and thermal grease on both sides. I did this with my zen and although it doesn't have any affect on the temperature of the heatsink it dramatically reduces the case-heatsink thermal resistance and thus the operating temperature of the device junction. I feel this will greatly enhance the life expectancy of the devices.

Good luck

Dan
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Old 1st July 2001, 11:21 AM   #3
hifi is offline hifi  Sweden
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hmm 62c is to hot.....hmm what about making the heatsiks 300x304x40? that should lower the cw to 0,2-0,25?

/micke
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Old 1st July 2001, 07:59 PM   #4
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micke,

Listen, 125 W dissipation through any kind of heatsink will allways raise the temperature of the heatsink to levels which are dangereous to the touch. Lowering the thermal resistance of the heatsink will lower it's temperature, but your junction temperature (inside the MOSFET) will remain high. Too high, because a quick calculation revealed to me that the junction temperature would be about 169C, which is well above the maximum junction temperature rating given by IRF (150C). Your precious MOSFET's will be ruined pretty quick this way.....(I'm talking days here).

The heatsink temperature will indeed be about 62C, as calculated by Ding, but as I've made clear above, that's only half of the story. Lowering the thermal resistance to 0.2 K/W will lower the heatsink temperature to 50C but your junction temperature will still be 156C...

You should try to share the amount of power which is to be dissipated among several heatsinks. For instance, using 2 heatsinks with a 0.4 K/W thermal resistance will yield a junction temperature of about 103C and a heatsink temperature of 50C.

A friendly word of advice though; before engaging in a project which is highly dependent on good thermal "management", get a good grasp of the underlying fundamentals. Because if you don't, you're bound to run into some serious trouble...

Gordon
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Old 1st July 2001, 09:34 PM   #5
hifi is offline hifi  Sweden
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well the 125W is divided over 6 Mosfets in To247 capsules that will result in 125/6 = 21W per mosfet...

if the thermal resistance betwwen junction and case and case to sink would be like 1.3 for worst case it will only result in an temp rise of 1,3*21 =27,3C and if the heatsink will have and Cw of 0,25 (1,25*0,25=31,25c) and the roomtemp is about 25c the heatsink will not rise beyond 31,5+25 = 56,25c and the worstcase junction of 27,3 will leave the mosfets att 56,25+27,3= 83,5C well within the safty margin.....ok!?!

The junction temp was never the point of this discussion but what i really want to know is how the temperature gradients look in the heatsink when its 30x30cm and only has a 5mm "back" maybe i need to use some kind of heat spreader to fully make use of the larger heatsink?...

/micke



[Edited by hifi on 07-01-2001 at 06:30 PM]
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Old 2nd July 2001, 12:14 AM   #6
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Micke,
Just my two cents worth...I've never been happy with various "heat spreader" schemes because the thermal resistance between the heat spreader and the heat sink tends to be high enough to offset gains.
Case in point: My first version of the water-cooled heat sink (since abandoned) used an aluminum L bracket bolted to a hollow rectangular aluminum girder. The output devices were mounted to the L. The difference in temperature was quite remarkable. The devices themselves (naturally) were hottest. But the L was much, much hotter than the girder. Yes, I used heat sink grease. Yes, it was tightly bolted. It's just that every time you cross a mechanical boundary you're going to have trouble getting the heat across.
The same is true for every amp I've ever prodded around in. I wouldn't think that your main problem is spreading the heat around (5mm should be enough to prevent serious hot spots), it's going to be getting that heat transferred to the surrounding air, which is more a function of the number of fins and the surface area of each fin.
In short, I'd recommend skipping any "heat spreader" and just mounting the devices directly to the heat sink. The fewer mechanical boundaries, the better.
If all else fails, there's always the old standby--fans.
(Now you know the frustration I felt when I started on my Aleph 2s. Up until I had the water-cooled idea, I was ready to tear my hair out.)
Good luck.

Grey
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Old 2nd July 2001, 09:11 AM   #7
hifi is offline hifi  Sweden
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yes...but i want passive cooling....it feels pretty dumb do have a amplifiere with -98db in distorsion or whatever and then use a noicy fan to cool it.....

ok...well im buying the heatsinks directly from the manufacturer therefore its quite cheap...about 20$ for an 20x30x4cm piece...

I think i will use 30x30x4cm pieces and use them as sides in an monoblock construction.

Would there be any benifit in mounting a smaler heatsink on the front side of the mosfet? it could serv as a "preassure spreader" for the mounting screw aswell for som additional cooling?


/micke
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Old 2nd July 2001, 01:46 PM   #8
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/micke

Your calculation fails to take into account the effect the TOTAL thermal resistance has on junction temperature. This is critical because every other temperature in the thermal system is calculated from the junction temperature. You break down the total thermal resistance in a way that's not realistic.


Basic thermal model;

delta T= P*Rth tot
or
delta T= P*(Rj-c + Rc-h + Rh-a)

This simplified model of heat transfer is widely used throughout the engineering community. You're right when you say that the power which is to be dissipated is spread among several devices. However, for several devices mounted in parallel more complicated models exist that take into account several specific topics. For instance, when used in parallel various thermal resistances are calculated in a different way. Although more realistic, for relatively simple cases the basic model is more then sufficient.

I've been using it for years with excellent results and I've built several amps with high dissipation rates which are still safe to touch. Don't forget that good thermal management increases the life-expectancy of your amp by orders of magnitude as well. I've designed and built a class A amp which is still functioning (and measuring)just fine after 11 years of heavy usage...On the other hand, I've seen people with designs that completely lacked good thermal management that had to change transistors every week. Designs beyond redemption, lett me tell ya...

Temperature gradients are inevitable in any heatsink; use good thermal interface material (low thermal resistance, < 0.35 K/W) and spread out the various devices equally among the surface of your heatsink.

Additional heatsinks on top of the MOSFET will provide only a little amount of extra cooling because of the fact that it's thermal resistance will be fairly high (10K/W or so). They will run extremely hot while only dissipating a few watts.

Gordon
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Old 2nd July 2001, 02:16 PM   #9
hifi is offline hifi  Sweden
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Tj = Tamb + *(Rj-c + Rc-h + Rh-a)

where:

Rj-c = 0,08 cw approx should be wery low cuz of the large contact are of six T0247..

Rc-h = 0,3cw aluoxide isolator

Rh-a = 0,20cw Heatsink 30x30x4

Tamb = roomtemp of say 25c

Tj = 25 + 125*(0,08+0,3+0,20)

Tj = 97,5c

damn that is hot but still within the saifty margins?...
what would your suggestion be to lower the junction temp??
Im already using separate heatsinks since this is one side of an Aleph 4 amp.

/micke

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Old 2nd July 2001, 06:28 PM   #10
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/micke

I agree with your calculations but I don't see how you can come up with a Rj-c of 0.08. The surface area covered by the 6 devices is only one variable in the total calculation (which is made by the manufacturer) of the Rj-c. IRF states that the Rj-c is 0.83 K/W max, and I use 0.5 K/W as an absolute maximum value. This is based on the graph of the transient thermal response in the IRFP 240 datasheet.

A total thermal resistance of 0.58 (which you use) is very low, almost unachievable so. Anyway, a junction temp of 97.5 is well within the safety margins (150C), so no problems there. To lower it even further use an extremely good thermal material (Rc-h <0.25K/W)between MOSFET and heatsink, such as Bergquist Sil-Pad 1500 (0.23K/W)or SilPad 2000 (0.2K/W).

If you use a thermal material of 0.2 K/W the junction temperature would be 85C, so there is still room for improvement. Be careful not to get carried away though; I mean a 10C drop in junction temperature is nice and will extend the life-expectancy of your MOSFET's, but be aware of the fact that the described thermal materials can be expensive....

Gekko
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