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Old 13th July 2012, 04:03 PM   #1
tsiros is offline tsiros  Greece
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Default improve heatsink

Hello!

I am planning on assembling a chipamp based on the lm4780

So far i have the pcb, the transformer (+-20V ), i have ordered the IC itself and i can get the various thingies easily (rectifiers, caps, cables, fuses, resistors)

the thing i am stuck with is the heatsink.

something tells me it has to be... quite substantial...

now, i have a heatsink that was used for a ~100W amplifier

this one:Click the image to open in full size.

The amp itself is getting the boot, so i am using its heatsink.

If i use it in vertical instead of horizontal orientation i think it can be a bit cooler. I mean, literally cooler, not "dude this is sooo coool".

instead of this:
________
________
________

________
________
________

put it in this orientation:

||||| |||||

Also, i thought of an improvement: if i do several horizontal cuts, will the C/W ratio be improved?
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Old 13th July 2012, 04:07 PM   #2
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Try to run the amp as normally. If it runs too hot, add a fan.
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Old 13th July 2012, 04:20 PM   #3
tsiros is offline tsiros  Greece
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I really really do not want to add a fan

I'd rather get a bigger heatsink...
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Old 15th July 2012, 01:09 PM   #4
jitter is offline jitter  Netherlands
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Quote:
Originally Posted by tsiros View Post
Hello!

I am planning on assembling a chipamp based on the lm4780

So far i have the pcb, the transformer (+-20V ), i have ordered the IC itself and i can get the various thingies easily (rectifiers, caps, cables, fuses, resistors)

the thing i am stuck with is the heatsink.

something tells me it has to be... quite substantial...
The size of the heatsink you need depends on several variables, like the max. power the IC is expected to dissipate and the ambient temperature.

Total power dissipation depends on the load you're planning to drive and the power supply voltage.
On page 11 (12 according to the pdf-reader) of the datasheet, there are some graphs of total dissipated power vs output power per channel and PSU voltage for 4,6 and 8 ohm loads. Look up the graph that corresponds most to your setup.
You can also calculate PDmax with equation No. 2 on page 16 and multiply the result by 2.

Next go to page 17 (or 18) of the datasheet and read the section called "Determining the correct heat sink". In it you will find the equation for determining the size of the heatsink (No. 4). This may seem jibberish, but you know most variables already, they're given in the datasheet:
Tjmax = 150 degr. C;
Tamb = typically 25 degr. C, but given hot Greek summers, you might want to calculate with a slightly higher value for the ambient temp;
Pdmax = the value you calculated or looked up in the graph on page 11;
Өjc = typically 0.8 degr. C/W
Өcs = typically 0.2 degr. C/W with the use of thermal compound;
Өsa = this is value you need to calculate. The result will be in degr. C/W and is the thermal resistance of the heatsink to the air. You'll need a heatsink of this value or less.

E.g. with a PSU of +/- 30 V and speakers of 8 ohms, PDmax is 45.6 W (calculated, graph shows slightly higher). Lets assume a Tamb of 25 degr. C. That gives:
Өsa = [(150-25)-45.6*(0.8+0.2)]/45.6=
[125-45.6]/45.6=1.74 degr. C/W.
The heatsink in this setup needs to have a thermal resistance of 1.74 degr. C/W or less.

Quote:
now, i have a heatsink that was used for a ~100W amplifier

this one:Click the image to open in full size.

The amp itself is getting the boot, so i am using its heatsink.
Look up the thermal resistance of your heatsink. Or if you don't know the brand and type of the heatsink, take measurements and look up the thermal resistance of a heatsink that resembles your's as close as possible. This will give you an estimate of the thermal resistance of your heatsink.

Quote:
If i use it in vertical instead of horizontal orientation i think it can be a bit cooler. I mean, literally cooler, not "dude this is sooo coool".
Actually, you should always use a heatsink in vertical orientation since the natural convection of warm air is upwards, not sideways. This is especially important when there's no forced airflow.

Quote:
Also, i thought of an improvement: if i do several horizontal cuts, will the C/W ratio be improved?
No, it will not be improved since you're taking away surface area. A smaller surface leads to a higher thermal resistance. Horizontal slits might actually interfere with the natural convection.
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Old 15th July 2012, 01:20 PM   #5
jitter is offline jitter  Netherlands
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I just realized that if you assume Өjc+Өcs is typically 0.8 + 0.2=1 degr. C/W, Өja (which is Өjs+Өcs+Өsa) becomes Өsa +1.
The use of equation No.3, but with Өja substituted with Өsa+1 gives PDmax = (Tjmax - Tamb)/(Өsa+1).
This can be rewritten to Өsa = [(Tjmax-Tamb)/PDmax] - 1 to give the easiest calculation of the needed heatsink.

In above example:
Өsa = [(150-25)/45.6] -1 = 1.74 degr. C/W

Last edited by jitter; 15th July 2012 at 01:26 PM.
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Old 15th July 2012, 01:30 PM   #6
tsiros is offline tsiros  Greece
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hey, jitter

yeah I had already done all that you did, including the simplification of the formula

The figure you give, 1.74 C/W is for one channel or two channels?

with my calculations i need a heatsink of <0.7C/W

last, if i cut horizontal, thin, slits, the area is, certainly, increased.
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Old 15th July 2012, 01:37 PM   #7
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Rod Elliot did a nice article which will help you ascertain the thermal characteristics of your heatsink.

DIY Heatsink

All you need to do is calculate the surface area of your heatsink. If in doubt UNDERESTIMATE.
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Old 15th July 2012, 01:54 PM   #8
jitter is offline jitter  Netherlands
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Quote:
Originally Posted by tsiros View Post
hey, jitter

yeah I had already done all that you did, including the simplification of the formula
Then I don't understand the reason for you question in post #1. All you need to do now is find out the thermal resistance of your heatsink.

Quote:
The figure you give, 1.74 C/W is for one channel or two channels?
Two channels. The formula for given for PDmax is for one amp (channel). Since there are two in one package, I multiplied the result by two to get 45.6 W.
Graph 200586a8 (PSU = +/-30 V and RL = 8 ohms) shows the max. total dissipation peaks at 49 W at around 22.5 W per channel power output. At lower or higher power ouputs, the dissipation in the IC is less.

Quote:
with my calculations i need a heatsink of <0.7C/W
If you're using 4 ohm speakers, I'm not surprised you come up with a rather low required thermal resistance (huge heatsink). The graphs don't show anything over +/- 25 V PSU with 4 ohm speakers. This is probably why.

Quote:
last, if i cut horizontal, thin, slits, the area is, certainly, increased.
Depends on how much material you remove while making the slits. But even if the area is increased, it won't be by much and an upset airflow might actually make things worse.

Last edited by jitter; 15th July 2012 at 02:00 PM.
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Old 15th July 2012, 01:59 PM   #9
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Base heatsink calcs on MAXIMUM dissipation.
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Old 15th July 2012, 02:03 PM   #10
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Lo-Fi manufacturers base their heatsink calculations on about 70% of maximum dissipation. This is perfectly valid as most people rarely use their systems at full volume.

This is fine until you decide to run the system at full volume.

For safety and longetivity - use the worse case calculations. For compactness you can use fans at full volume - no-one is going to hear a fan if the system is cranking out 100dB of music.
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