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Old 27th November 2004, 09:59 PM   #1
k1jroth is offline k1jroth  Finland
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Default heat sink problem

hello

http://sound.westhost.com/project68.htm i`m buildin this amplifier with extra tarnsistors and +-56V and possible bridge connection.

Transistors are http://www.onsemi.com/pub/Collateral/MJL4281A-D.PDF

Problem is that i can not calculate the size of heatsink.

could someone please help.
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Old 28th November 2004, 05:03 AM   #2
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http://sound.westhost.com/heatsinks.htm

Prosit
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Old 28th November 2004, 07:38 AM   #3
glen65 is offline glen65  United States
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Check these out
http://sound.westhost.com/heatsinks.htm#3

http://home.planet.nl/~heuve345/elec.../lesson11.html

http://homepages.which.net/~paul.hil...sinksBody.html


Hope this helps
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Old 29th November 2004, 02:21 AM   #4
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Default Here is my suggestion, based on practice, but never fails

Your Peak power is around 370 Watts, depending the supply and the distortion you can support.... efficiency can be high as 65 percent, but use 50 percent to be more safe...so, your dissipation will be 740 Watts each channel..... if bridged, 2 X 740 Watts dissipation, better in different heat sinks, if not the amplifier will sink as the "titanic", too much heavy in one side, or other, or front of back....have to put transformer to compensate the other side weigth.

I had an old trick, working for more than 30 years without deception, and my environment has higher temperature than most countries (29 fixed degrees centigrades, from winter to summer, with no more than 20 percent variations, when variates, normally, go less than 29 celsius).

Each 10 watts dissipation needs a 3 inches by 3 inches aluminium with 3 to 4 milimeters thickness, positioned to normal convection refrigeration, without pressurized air passing, and not helped with any other aid...no black anodize, or any helping aid, as corrugated surface or things alike.

So, to be simple, thinking in one very simple heatsink, long and not to high, 74 fins, of 9 squared inches, in paralell, making a "in line" shape heatsink, the most traditional shape will hold your power.... as you see, using that one will be no good, as you will have a entire amplifier surround heatsink i suppose.

But 74 fins, each one with 9 square inches, will result in a total of 666 square inches.

So, measure the heatsink you find, each fin side multiplied by side, sorry to be so obvious, but there are strangers like me that need things explained in those stupid details.....this way, 9 squared inches... is what you will use as reference to calculate.... 9 square inches can dissipate 10 watts, both faces working....in the true, you have 18 square inches of surface in contact with air...but use the 9 square to 10 Watts and you will do not need fan.

Using fan, you reduce to one third, and depending your home temperature or your fan (how many units) you can reduce to 1/4 the heatsink size with safe...remember to use at leat 2 fans...never use only one...because this single one, if fail, silently will send your transistors directly to hell, one way ticket.

To avoid wind noise, reduce voltage to 9 volts or even less, use those LM7808 if input voltage is less than 24 Volts.

If your chassis is made of aluminium, take it in your calculations, but use it as half the efficiency, 4 to 5 squared inches each 10 watts, and those watts are not power output, those watts are dissipation power, that can reach, sometimes, twice the output power in the non optimistic aproach... a safe margin. So, having bad heat transfer, because small metal to metal contact from heatsinks (enormous units) related to the mounting (small metal contacts), you must reduce the heat transference expectation to a half. And also under the amplifier, air circulation is hard, and those surface do not work very well to dissipate heat...beeing almost 20 percent of the other sides efficiency... air circulation likes more than 3 inches of free air under and over the heatsink, if more will be great.

Of course, there are precision calculations, that will probably reduce your size..... but that inform i am giving you, are not theorical, are practical and proved in day by day using, sometimes under sun ligth, and sometimes when working hard, someone put something over the amplifier things by accident, and it will hold the over charge.... reason why i did not made any effort to learn the Mathematics formula, as this was proven, under hard conditions, to work great.

regards,

Carlos
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Old 29th November 2004, 03:43 AM   #5
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Default For one of your channels, and using at least one fan each sink

You will be safe.

The fan is because this heatsink you see has no damping effect, as it is thin, heat transference is immediate.... you have not a cold period waiting the "whole thing" turns hot as the thicker ones...this one is better as it uses less material and the important is surface in contact with air....not thickness.

Also the fan is needed, when you use internal heat sinks alike the one shown..... as cannot have enougth convection because of the air flow resistances caused by down and up grille...the holes over and under the amplifier are not enougth (normally, because if enought, you kid can put a coin and booommm!).

I was trying to evaluate dimensions and i see this one PERFECT to one amplifier channel, if inside (must be, this heat sink is fragile) need a fan....i am exagerated...in my idea, two fans, working over the cabinet, exausting hot air.

Good luck

Carlos
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File Type: jpg perfect with fan.jpg (87.0 KB, 447 views)
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Old 29th November 2004, 04:13 AM   #6
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Default This one, i used the cable size to obtain the high and heatsink deep.

And i calculate it to 560 Watts heatsinking.

This can be good to a 300 or 350 constant watts amplifiers....sinusoidal continuous Rms power over resistive load.

To check i find that this amplifier is used as a mono unit, and put out 300 Watts RMS when start clipping.

So, this heatsink is adequated to your amplifier channel...only one.

regards,

Carlos
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File Type: jpg 560w sinks.jpg (97.9 KB, 398 views)
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Old 29th November 2004, 04:33 AM   #7
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Default This one can hold 700 watts heat easy, even inside enclosure

So, 350 to 400 watts RMS can be used here.... 200 watts channel amplifier can be used.

Carlos
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File Type: jpg 700 w heat sink b.jpg (95.2 KB, 399 views)
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Old 29th November 2004, 05:09 AM   #8
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Default I made a search about that image, and now i know the amplifier is rated to 75W X 2

In my idea, or the heatsinks are made to only 8 ohms...or the amplifier is over rated.

I think i cooperate deeply, beeing attention, you will be have a reasonable "tool" in your mind.

regards,

Carlos
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File Type: jpg power evaluation.jpg (82.9 KB, 360 views)
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Old 29th November 2004, 03:08 PM   #9
AndrewT is offline AndrewT  Scotland
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Hi,
destroyer is about right with the maximum power bridged into 8 ohms.
If you intend running at maximum power continuously then your heatsinks should dissipate about two thirds of your maximum power. ie. about 120watts per channel (or 240watts total when bridged ).
Decide on the maximum Tj you want to run at & remember the rule of thumb for each 10degrees C reduction you can double the life of your transistors.
If you are using the amp domestically then you can probably half your heatsink requirement, and if you expect gentle use then you could half again but expect Tj to be much higher each time you host a loud party.
regards Andrew T.
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Old 29th November 2004, 07:56 PM   #10
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Default Thank you AndrewT, by the support and complementary informations

That was good....and my computer microprocessor...i am having not sucess with fan blowers..... temperature is reaching 55 Celsius.

regards,

Carlos
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