Two lm4780's on same heatsink?

[synthius]

Hey guys,

I am planning on running a lm4780 in stereo config to output 45Watts x 2 channels and also a lm4780 in parallel configuration to output 100Watts into a subwoofer. I will be using sil-pad 400 to isolate from the heatsink. The question I have is whether it is a good idea to mount both lm4780's on the same heatsink. The heatsink is 9 inches long with 21 fins. It can be seen here. http://cgi.ebay.com/ws/eBayISAPI.dll...ksid=p3907.m29

Has anyone mounted two lm4780's on the same heatsink and ran into trouble?

[okapi]

i'm using a single LM4780 running in parallel on a similar sized heat sink and it gets fairly warm so i would caution against mounting two on the same heat sink.

i also have a reasonable amount of heat dissipation into other parts of the chassis along with a copper heat spreader.

i have made other LMXXX amps clip in the past. it happened when the heat sink right next to the chip reached 85 degrees C (measured with a thermocouple). It may in fact start to clip subtly earlier. it certainly was not subtle at 85.

[jackinnj]

That heat sink should have a thermal impedance of about 1.1 degrees C per watt -- the important factor is the absolute value of the power supply rails -- look at the National Semi website and it will tell you the thermal impedance you need.

http://www.national.com/appinfo/audi...gn_Guide15.xls

You will need one 1.1 C/W heat sink per chip with +/- 35VDC rails. If you use a heat sink with higher impedance it will work for a while, but the chip's thermal circuitry will shut it down when the heat sink reaches the critical temperature.

[TheMG]

You'd have find out what the thermal coefficient θsa for that heatsink is. The LM4780 has a whole section on how to calculate how much of a heatsink is required, you may want to have a look at it.

By doing the calculations you'll be able to pick a heatsink that is neither too big nor insufficient, which is usually what you want to aim for in an amplifier design.

[jackinnj]

Originally posted by TheMG
You'd have find out what the thermal coefficient θsa for that heatsink is.

9 inches long, 2.25 wide, 21 fins each about 1 inch high => 1.1 C/W in still air.

[synthius]

Thanks for the calculation jackinnj.

Stereo lm4780
I have done the calculation to determine what the heatsink sa needs to be and it came out to 1.58 C/W
TJmax=150C
Ambient temp=70C
PDmax=31
Jc=0.8
Cs=.2

=((150-70)-31*(0.8+0.2))/31 =1.58

on a side note, i am using sil pad k-10 as the insulator which has a cs of .2 C/W. I'm also driving 8ohm speakers.


Parrallel lm4780
The parrallel lm4780 will be driving a 4 ohm load which will actually be seen as a 8ohm load to each amplfier within the IC. I believe the heatsink sa required for this lm4780 will come out to 1.58 also.
So what I have is two lm4780's which need a heatsink with sa below 1.58 in order not overheat.

Is this correct??

[TheMG]

That means each chip needs a 1.58 coefficient heatsink. If you have 2 chips you need twice the heat dissipation ability therefore 1.28/2 or one heatsink for each chip, whichever is best for you.

[SpittinLLama]

The calculations give the worst case power dissipation with a sine wave. Once you go to music the real power dissipation is less. Something to keep in mind. Making it thermally solid such that your amp can run sine waves all day long at the peak power dissipation point is over kill. Not a bad thing to do but if you are planning on using it in a home environment and you are the planner user then you can back down some.

But something is wrong with your numbers. The Pd you list is per channel. You need to multiply by 4 to get the correct total Pd for both parts on the heat sink. Then use this number to calculate. Also, an ambient of 70C is pretty high, is this really correct? I'd think 35C is plenty. Check the spreadsheet, it shows Pd/Ic just below Max Pd.

-SL

[synthius]

You are right 70C is kind of high, i was thinking worst case within the enclosure. I'll use 40C instead.

[synthius]

Spittin lama, here are the new calculations according to your numbers.


TJmax=150C
Ambient temp=40C
PDmax=31*4=124
Jc=0.8
Cs=.2

=((150-40)-124*(0.8+0.2))/124 = -0.1129 C/W


I would need a heatsink with -0.1129 C/W thermal resistance....this can't be right.