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Old 12th March 2005, 07:40 AM   #1
djdamix is offline djdamix  France
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Default Heatsink size ???

I'm developping a amp powered with +/- 65V DC, and I'm using 3 IRFP9240 & 3xIFP240 for the output stage.

I've set the current bias to about 150mA. I'm using the heatsink on the photo attached and put a fan in front of it (PC power supply fan).

However when I test it on a resistive load (8 ohm) at full power (musical signal), the amp become quite hot (about the limit where you can touch it). On the scope the voltage goes from about +55 to -55V. All the signal looks clean.

Any comment ? Does the heatsink appears to small for you ? (I haven't the datasheet of it)

Thanks.
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Old 12th March 2005, 07:45 AM   #2
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if its not too hot to touch, then its fine. My car amp gets too hot to keep your hand on... hmm.

also we don't know how big your roll of solder is. lol
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Old 12th March 2005, 07:52 AM   #3
djdamix is offline djdamix  France
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lol the heatsink is 12cm x 10 cm x 6cm

;-)
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Old 12th March 2005, 07:53 AM   #4
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what was the ambient temperature like though in comparison to how hot it can get?
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Old 12th March 2005, 08:54 AM   #5
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With 65vdc and 6 irfp's you will need a very large heatsink.
Worst case dissipation will be well over 200 watts.
That would require a 0.25 heatsink.
imo, you may consider adding a couple of irfp's.
With 8 irfp's the heatsink should be around 0.30

From the dimensions you posted your heatsink is somewhere around 0.70.
Putting a vent on it can make it drop to ~ 0.25 , with a big high air volume vent.
For a vent with known volume displacement rate, you can divide this by the prop cross sectional area, which will give you the air speed.
On the web you can find graphs for heatsink thermal resistance at different air speed levels.

You can get the data of your heatsink at
www.fischerelektronik.de
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Old 12th March 2005, 11:48 AM   #6
Giaime is offline Giaime  Italy
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You can also try something that a friend of mine has done, that is to put each complementary pair of transistors on a salvaged CPU heatsink, each one with their own fan. That would be noisy but most of them are rated for 0.30 °C/W, so are enought for a single couple of IRF's.
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Old 12th March 2005, 01:32 PM   #7
Mr Evil is offline Mr Evil  United Kingdom
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Wth six output devices and those rails, I calculate approximately 1.1C/W as bare minimum. How does the size of that heatsink compare to others of that thermal resistance? Note that worst case dissipation occurs at around half output power, not full.
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Old 12th March 2005, 02:07 PM   #8
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Evil one,

i suppose the 1.1 C/W you calculated is for an optimal scenario ?
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Old 12th March 2005, 05:58 PM   #9
Mr Evil is offline Mr Evil  United Kingdom
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That's assuming an ambient temp of 50C, a max junction temp of 150C, worst case power dissipation at about half of +/-65V (sine wave output) of approximately 66W with an 8 ohm load, or 11W per transistor, a junction-heatsink thermal resistance of 2C/W.

Thus temperature rise junction-heatsink of 11 * 2 = 22C per transistor, then the maximum allowed temperature rise heatsink-ambient is 78C, implying a minimum thermal resistance of 78/66 = 1.18C/W, or 1.1 to allow a bit of headroom.

I hope there's no mistakes in there!

There are a number of assumptions in there: Ambient temp of 50C is pretty high unless the heatsink is in an unventilated case or you live in a hot country. 2C/W junction-heatsink thermal resistance depends on how good the transistor package is and what mounting method you use; it could be anywhere from 1-4C/W. Worst case power dissipation occurs with a sine wave, but real music will be a lot easier than this. So overall it's quite conservative, but then again it's not a good idea to run very close to 150C junction temp (unless you're using one of the nice newer 175-200C junction temp devices perhaps).
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Old 12th March 2005, 07:52 PM   #10
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Default Multiply each heatsink fin by its side.... and add the partial results

Each 9 squared inches represents 10 watts.

Your amplifier may dissipate 200 watts, producing 100W rms or a little more.

All you fins added, must reach 180 squared inches for one channel....or 360 squared inches for a stereo unit.

This is practice result formula...where 10 watts of heat need 9 square inch of metal surface (side A counted....side b not counted).... is length multiplied by width.

So, both sides of a 9 square inch aluminium plate, measuring 3 inches by 3 inches will dissipate 10 watts of heat.

Imagine your amplifier dissipating twice the output power...half of power over speaker and half of heat....so, if you deduce will produce 100 watts, believe in 200 watts heat.

This is a good size if heatsink is external, and positioned in the correct position for convection air movement...were hot air have the tendency to go up, sucking fresh air from under the heatsink.

If you can isolate heatsinks, each one related the other, and both heatsinks insulated from chassis, using wood or another kind of insulator, you will have better heat transfer, avoiding insulators, using only silicone thermal grease compound.

But remember that plus 65 and minuw 65 represents 130 volts that can be very dangerous...so..... use damn insulators!

When you feel the heatsink alike a man with fever..... temperature is OK....but if you cannot keep you finger touching it with strengh...while you count "one thousand and one, one thousand and two, one thousand and three"...if you cannot keep you finger....no good!...increase it, or use some fan blower to keep it cool.....as when you have outside case, or even heatsink with 50 degrees temperature....inside the case...in the solid state waffer, temperature is not so big distance from "melt" temperature.

Good luck

Carlos
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