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?ViewItem&Item=200191071834&Category=162&_trksid=p3907.m29
Has anyone mounted two lm4780's on the same heatsink and ran into trouble?
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?ViewItem&Item=200191071834&Category=162&_trksid=p3907.m29
Has anyone mounted two lm4780's on the same heatsink and ran into trouble?
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.
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.
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/audio/files/Overture_Design_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.
http://www.national.com/appinfo/audio/files/Overture_Design_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.
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.
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.
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??
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??
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
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
Where did you get the 31W? I thought you got it from the Overture Design Guide linked to in an earlier post in this thread. So I thought you were looking at the 31W of Pd that shows up on that spread sheet with the LM4780 selected, 8 ohms and a +/-35V supply. But then you say 45W and 100W into the sub. So +/-35V is going to do much more power than 45W into 8 ohms. 45W is more like +/-30V supply results. So put your numbers in here where appropriate.
Supply: +/-30V
Max Pd is at 22.80W/channel.
Total Pd per IC = 48.74W (Design Guide takes into account Q current of IC).
Total for your design = 97.48W
150C-40C = 110C
110C/97.48W = 1.13C/W
1.13 - 0.8 - 0.2 = 0.13C/W
Another way to look at it is what is the temperature rise of a heat sink with ~ 97W of power dissipation in it. if 1C/W then it is 97C. For one LM4780 the heat sink needed for these conditions is 1.26C/W. But you get a lower JC resistance because the 0.8C/W is for one package. Basically, you have two packages so they act like parallel resistors (2 paths for the power to travel). So I think the more realistic number is somewhere in the area of 0.5C/W. For example, 0.5C/W * 100W = 50C over ambient or 90C. Now you have a total of 1C/W for the package and washer but the package is only dissipating ~ 49W so another 49C over 90C giving ~140C. Because you are using two packages you have to reduce the package thermal numbers to be 0.5C/W. Now this gives a total for the heat sink of 0.63C/W, sounds reasonable. Work the numbers backwards again to see what you get.
-SL
Supply: +/-30V
Max Pd is at 22.80W/channel.
Total Pd per IC = 48.74W (Design Guide takes into account Q current of IC).
Total for your design = 97.48W
150C-40C = 110C
110C/97.48W = 1.13C/W
1.13 - 0.8 - 0.2 = 0.13C/W
Another way to look at it is what is the temperature rise of a heat sink with ~ 97W of power dissipation in it. if 1C/W then it is 97C. For one LM4780 the heat sink needed for these conditions is 1.26C/W. But you get a lower JC resistance because the 0.8C/W is for one package. Basically, you have two packages so they act like parallel resistors (2 paths for the power to travel). So I think the more realistic number is somewhere in the area of 0.5C/W. For example, 0.5C/W * 100W = 50C over ambient or 90C. Now you have a total of 1C/W for the package and washer but the package is only dissipating ~ 49W so another 49C over 90C giving ~140C. Because you are using two packages you have to reduce the package thermal numbers to be 0.5C/W. Now this gives a total for the heat sink of 0.63C/W, sounds reasonable. Work the numbers backwards again to see what you get.
-SL
Hi,
the Spike protection is scaled with the chip temperature.
The good looking limits are quoted for chip temperatures of 25degC.
If the chip runs hotter then Spike automatically reduces the limits.
At transient peak junction temperature I would not be surprised to learn that Spike has reduced the current limits to near zero amps.
The hotter you run these protected chips the earlier the protection mechanisms will trigger.
I recommend that the sinks quoted in the National design guide be doubled to keep chip temperature down and thus maximise the output currents before triggering Spike.
the Spike protection is scaled with the chip temperature.
The good looking limits are quoted for chip temperatures of 25degC.
If the chip runs hotter then Spike automatically reduces the limits.
At transient peak junction temperature I would not be surprised to learn that Spike has reduced the current limits to near zero amps.
The hotter you run these protected chips the earlier the protection mechanisms will trigger.
I recommend that the sinks quoted in the National design guide be doubled to keep chip temperature down and thus maximise the output currents before triggering Spike.
AndrewT said:
andrew, that's completely unreasonable. to halve the thermal impedance you have to increase the surface area of the heat sink more than 3X
i've run paralleled, bridged and stereo LM4780's for hours at their maximum rating with the heat sinks specified in the National spreadsheet without thermal intervention.
Hi Jack,
the chart on page 14 (of lm3886.pdf) suggests Rths-a of 2.5C/W for 70V PSU and 40degC ambient.
I suggest 1.25C/W. What's so unreasonable about that?
For a dual inside the 4780 that would equate to 0.53C/W
National assumes Rthc-s=0.2C/W so 1.25 for a single chip becomes an effective Rthc-a=1.45C/W and twice the dissipation requires 1.45/2 - 0.2=0.525C/W.
the chart on page 14 (of lm3886.pdf) suggests Rths-a of 2.5C/W for 70V PSU and 40degC ambient.
I suggest 1.25C/W. What's so unreasonable about that?
For a dual inside the 4780 that would equate to 0.53C/W
National assumes Rthc-s=0.2C/W so 1.25 for a single chip becomes an effective Rthc-a=1.45C/W and twice the dissipation requires 1.45/2 - 0.2=0.525C/W.
If you want to do the hardest thermal testing then run the amps at the peak power dissipation point (this is at 50% efficiency so Pout = Pd, this way set your Pout to equal the calculated Pd) with a 20Hz sine wave. If all is well there is no musical signal that will ever equal this level of thermal testing. BTW, setting Pout equal to the calculated max Pd is, in fact, very close to the peak. I have done it with a thermal couple connected to the heat sink. I let it fully stabilize then would move the output power just a little each direction and see the heat sink cool a tad. Of course, you have to let it stabilize first so very time consuming. If you see Spike protection (see AN-898) or thermal shut down activating then it still may be ok for real world signals, depends on how fast the protections come on and to what extent. I built an amp with two LM3886 in BTL and it would not do sine waves on the bench but has never has issues with music.
Remember, trust the llama.
-SL
Remember, trust the llama.
-SL
Update:
Ran my lm4780 parallel on my heat sink today, reached 153 Watts on a 4 ohm load. To my surprise it didn't' get scorching hot like I expected. Next I will attach the stereo lm4780 to the other side of the heat sink and attempt stability without a fan.
I attached my recorded testing numbers in an excel file if anyone is interested.
Tested with an 8ohm and 4ohm load and using +-36V rails
Ran my lm4780 parallel on my heat sink today, reached 153 Watts on a 4 ohm load. To my surprise it didn't' get scorching hot like I expected. Next I will attach the stereo lm4780 to the other side of the heat sink and attempt stability without a fan.
I attached my recorded testing numbers in an excel file if anyone is interested.
Tested with an 8ohm and 4ohm load and using +-36V rails
If you can't get a thermocouple try an electronic kitchen thermometer on the heat sink -- one of the metal ones sold at Bed n Bath or Williams Sonoma are actually quite good.
Same spread sheet idea but record temperature over time -- you will note that at some point the "sink" and it's ability to transfer energy into the air becomes limited -- so you will have a graph in which the X-Axis is time, the Y-Axis is temperature and the Series are the power levels into chip.
So instead of making us download an Excel file -- use an Excel graph, paste it to MSPaint and attach as a GIF.
I use this method to calculate the thermal impedance of heat sinks -- instead of a chip I use a 10 ohm, 50 Watt Dale metal power resistor which can be attached to the heat sink with 6-32 screws and some thermal paste.
Same spread sheet idea but record temperature over time -- you will note that at some point the "sink" and it's ability to transfer energy into the air becomes limited -- so you will have a graph in which the X-Axis is time, the Y-Axis is temperature and the Series are the power levels into chip.
So instead of making us download an Excel file -- use an Excel graph, paste it to MSPaint and attach as a GIF.
I use this method to calculate the thermal impedance of heat sinks -- instead of a chip I use a 10 ohm, 50 Watt Dale metal power resistor which can be attached to the heat sink with 6-32 screws and some thermal paste.
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