I never seen exact schematic for that but I guess that must be something similar to this:
http://www.diyaudio.com/forums/power-supplies/271172-hq-low-noise-4-quadrant-psu-2xlm3886.html
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Very interesting. I have an amplifier that used a high power regulator that seems to have been damaged due to an accidental short at the amplifier outputs...
Hi Braca,
These are good points, and I agree that including the front face in the cooling arrangement can bring down the original 2K/W figure. Not sure that it will drop to 1.3, seems a lot, but that's more gut feeling.
The other thing I thought about is that with the separate block + heatsink arrangement you insert another thermal resistance namely between block and heatsink, that would not be there if the chip was mounted directly on the final heatsink. That may introduce another 0.1K/W.
Finally, the block mounting on the plastic face of the chip will delete any direct heat dissipation by radiation and convection from the front face - again, not sure how much that would be but it is a factor worsening the total arrangement.
I wonder if we can actually measure these effects in some meaningful way?
Jan
I have only just reached your post302 so don't know what others have contributed.
The copper spreader is a good idea, but a number of conditions need to apply to make it really useful.
a.) the copper should be direct to the back of the device using Thermal goop only. This should preferably be by a clamping method rather than the standard offset device hole. 2 holes devices are often better and maybe nearly as good as clamping.
b.) the extra thickness of the copper would negate the advantage of the very good thermal conductivity, if the copper to heatsink interface area is too small. You need a big interface area. I'll guess and suggest the copper/heatsink be at least ~4times the device backplate area. I'll guess again and say that 10times is too big.
c.) the copper must be electrically insulated from the heatsink. Again a clamping method would be better than using just a couple of bolts into a thin piece of copper. The copper must be in intimate contact with the isolator and the isolator must be in intimate contact with the heatsink over as close to 100% of the copper area as possible. That means pulling the copper to the heatsink with sufficient bolts/clamps such that no distortion of the copper or heatsink creates high thermal resistance "gaps" filled with thermal goop.
d.) The copper is now Live and the layout should try to prevent accidents.
It's 5:1 approximately.
Excellent info and work. Thanks.
Mica is a terrible thermal conductor, I do not recommend it.
For maximum transfer, use a graphene sheet to a copper spreader, then isolate the copper to heatsink.
If you don't like graphene, go with a copper loaded paste.
With the isolated part, that is an accurate statement. A such, the heat flux out of the top will be 1/5th of the lower path.
With a non isloated part, the difference will be in excess of an order of magnitude, AND the copper tab will not be isothermal.
I believe you meant to say lower gradient to front of package. Same thermal difference but a longer path.
jn
BTW there is an update to 'update my Dynaco', really shouldn't be an issue unless you like to push the power to the max, then the stock power supply collapses at low frequencies. see here eg switches to the metal tabbed version along with a beefier HS plate.
Andrew,
Thanks for the information concerning post #302. Referring to your (b) comment, I am restricted to the 2"x1/4"x7" pieces of copper. This is a 14 sqin piece of copper which is not bad. The thing I am not sure enough about is "how far from the centre of the chip will the heat will radiate. The most heat will be directly around the chip, but not sure how far down the length of copper the heat will venture.
If I divide the length in half and join the 2 pieces to make a 3.5" x 4" x 1/4" copper spreader, it may give more surface area directly around the chip, but then not sure if the 2 pieces will work better than a single homogeneous piece of copper.
Any ideas on this???
Regards,
Myles
Thanks for the information concerning post #302. Referring to your (b) comment, I am restricted to the 2"x1/4"x7" pieces of copper. This is a 14 sqin piece of copper which is not bad. The thing I am not sure enough about is "how far from the centre of the chip will the heat will radiate. The most heat will be directly around the chip, but not sure how far down the length of copper the heat will venture.
If I divide the length in half and join the 2 pieces to make a 3.5" x 4" x 1/4" copper spreader, it may give more surface area directly around the chip, but then not sure if the 2 pieces will work better than a single homogeneous piece of copper.
Any ideas on this???
Regards,
Myles
For aluminium I have accepted what others posted and found that in practice it works.
The heat conducts away from the heat source very well, out to a radius of about 10 times the thickness of the aluminium. 8mm thick aluminium will be close to isothermal out to ~80mm from the heat source. i.e. a 180mwide heatsink with a 20mm wide source in the middle will be nearly isothermal over the whole backplate, with the corners possibly being just slightly cooler.
Copper is better. I have no data to support this guess, but I reckon a 12times to 15times the thickness will work well for copper.
1/4" thick should be nearly isothermal out to at least 3" from the source.
Except that when attached to an good heatsink, one is pulling a lot of heat out of the copper. This will increase the copper temperature gradient along that guessimated limiting radius. By attaching a good heatsink backplate, the limiting radius for isothermal operation of the copper is MUCH lower than for copper radiating out into the air.
The 3886 backplate area is ~232mm²
4times that is 928mm² giving 18.3mm*50.8mm as the minimum copper size for an effective interplate.
10times that is 2320mm² giving 45mm*50.8mm as the maximum (my guessimate) copper size for an effective interplate.
This makes a nonsense of using a 7" long copper plate for a dual 3886. At least 4off 3886 spread out along the 7" length is the absolute minimum number of devices to make effective use of the copper interplate. And may be sufficient for upto 9off 3886 if you could find room to fit in all the clamping bolts between all the PCBs.
I would be tempted to cut the 7" length into 4 equal pieces and use each piece to cool one 3886.
The heat conducts away from the heat source very well, out to a radius of about 10 times the thickness of the aluminium. 8mm thick aluminium will be close to isothermal out to ~80mm from the heat source. i.e. a 180mwide heatsink with a 20mm wide source in the middle will be nearly isothermal over the whole backplate, with the corners possibly being just slightly cooler.
Copper is better. I have no data to support this guess, but I reckon a 12times to 15times the thickness will work well for copper.
1/4" thick should be nearly isothermal out to at least 3" from the source.
Except that when attached to an good heatsink, one is pulling a lot of heat out of the copper. This will increase the copper temperature gradient along that guessimated limiting radius. By attaching a good heatsink backplate, the limiting radius for isothermal operation of the copper is MUCH lower than for copper radiating out into the air.
The 3886 backplate area is ~232mm²
4times that is 928mm² giving 18.3mm*50.8mm as the minimum copper size for an effective interplate.
10times that is 2320mm² giving 45mm*50.8mm as the maximum (my guessimate) copper size for an effective interplate.
This makes a nonsense of using a 7" long copper plate for a dual 3886. At least 4off 3886 spread out along the 7" length is the absolute minimum number of devices to make effective use of the copper interplate. And may be sufficient for upto 9off 3886 if you could find room to fit in all the clamping bolts between all the PCBs.
I would be tempted to cut the 7" length into 4 equal pieces and use each piece to cool one 3886.
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As promised, I'm enclosing a note dealing with the heat transfer issues that arose in conjunction with the experiment involving the cooling of the LM3886TF by means of a copper cooling block that encloses the entire package.
The calculations are as accurate as they can be under the circumstances, not the last of which is the fact that the manufacturer himself is not quite sure which value of thermal resistance is valid for the package.
Nevertheless, the calculation results show that it is indeed possible to achieve much higher cooling rates by enclosing the package in a cooling block. I think that the absolute value of the cooling power thus achiveable is not that important, for other issues will inevitably crop up when pushing it quite far. In that i concur with billshurv as regards his comment on the operating point significance.
Although the theory I employed in analyzing this problem is the simplest possible in the field of Heat Transfer, the phenomena involved are probably not easy to grasp for everyone. This is all the more the reason for granting the benefit of a doubt to a non-mainstream claim. I found this a bit lacking in the previous discussion.
While delving into this matter I also learned a few things, and I'm glad for that.
I also stand corrected for stating that the purpose of the test the colleague from the Serbian forum performed was the one of endurance. It was due to the lack of full information about the experiment.
Please fill free to comment and ask.
Regards,
Braca
The calculations are as accurate as they can be under the circumstances, not the last of which is the fact that the manufacturer himself is not quite sure which value of thermal resistance is valid for the package.
Nevertheless, the calculation results show that it is indeed possible to achieve much higher cooling rates by enclosing the package in a cooling block. I think that the absolute value of the cooling power thus achiveable is not that important, for other issues will inevitably crop up when pushing it quite far. In that i concur with billshurv as regards his comment on the operating point significance.
Although the theory I employed in analyzing this problem is the simplest possible in the field of Heat Transfer, the phenomena involved are probably not easy to grasp for everyone. This is all the more the reason for granting the benefit of a doubt to a non-mainstream claim. I found this a bit lacking in the previous discussion.
While delving into this matter I also learned a few things, and I'm glad for that.
I also stand corrected for stating that the purpose of the test the colleague from the Serbian forum performed was the one of endurance. It was due to the lack of full information about the experiment.
Please fill free to comment and ask.
Regards,
Braca
Hello Jan,
I've just posted my note on the LM3886 thermal issues, and now would like to comment on the points you raised. Actually, they are implicitly dealt with in the note, and I hope that I haven't been "too scientific" at that (I read with pleasure your thread on this subject).
Re. your second comment, fixing the block to the main heat sink does indeed involve an additional thermal resistance, but it will be less than 0.1K/W because the area of the block is larger than the area of the package probably by a factor of at least two, and would thus halve the above value. In any case, the extra cooling rate gained by the block is much more worth.
The convection of heat from the package face is implicity included in the thermal resistance figure quoted. In spite of being dependent upon the difference of the fourth powers of the casing and air temperatures, the radiation of heat is probaly a second-order effect in this case. Again, both of them are more than compensated for in the cooling block configuration.
Depending upon the accuracy required, a measurement of these effects might not be an easy task at all. I did some successful work in a different application (gas-wall heat transfer), but the sensors are expensive, and the experimental set-up non-trivial. There are heat flux sensors in the form of thin foils, but if e.g. heat transfer rate between two adjoining parts is to be measured, even a thin foil would have to be milled into one of the parts and would thus become a part of the heat transfer equation. So i think that the best one can do is to measure the casing temperature (D. Self, Chapter 15), and then use the heat transfer theory.
Regards,
Braca
I have only skimmed the analysis so need to give it a proper read tonight, but did go back to page 1 and compare Tom's results with the spec. I have noted one possible issue with some of his results (or my misreading of the specs) in so much as power dissipation peaks at around half max power for a particular voltage rail, so in theory if you can (say) do 40W out at +/-30V rails you can do 80W. I am missing something there.
Anyway assuming the curves are correct then the difference between T and TF on the small heatsink is about 7W dissipation if my extrapolations are correct. So for 'spec level' output (and leaving aside egg frying heatsinks) for lower rails under 30V you don't need to give the package much help to get a lot more power out. I'm happy with Silpads and T packages, but for those who like the ease of use of TFs interesting to see.
Usual caveats about real music crest factor apply 🙂
Anyway assuming the curves are correct then the difference between T and TF on the small heatsink is about 7W dissipation if my extrapolations are correct. So for 'spec level' output (and leaving aside egg frying heatsinks) for lower rails under 30V you don't need to give the package much help to get a lot more power out. I'm happy with Silpads and T packages, but for those who like the ease of use of TFs interesting to see.
Usual caveats about real music crest factor apply 🙂
Andrew,
Thanks for the information and also to Braca, Tom and others on this thread. Once I get around to building, I will try a 51mm x 51mm x 6mm copper block on each of the 3875 chips with each of the copper blocks mated to a block of 228mm x 152mm x 9mm aluminum, which is one side of the chassis, (building monoblocks).
I will also try a 89mm x 51mm x 6mm copper block on each of the 4780 chips with each copper block mated to a block of 228mm x 152mm x 9mm aluminum. (building a paralled amp).
Since I have 8 pieces of copper bar, I plan to do some thermodynamic calculations to see if an optimum area of copper spreader can be obtained. I do not listen to music at high levels, so calculations can be kept real!!!!. Once the amps are operational, plans are to record some temperatures at various points and possibly reverse engineer the process to obtain some data. May take a while, but in the meantime, thanks for all the information so far.
Two question: Since all chips are TF type, If one does not want the copper block to be electrically conductive with the chip, but wants the heat transfer to happen, what is the best interface between the chip and copper: "Pads, non conductive grease, others?????
Is drilling and countersinking various diameter holes in the aluminum HS block a waste of time, or can it be an aid in the convection process from the HS to the air???.
regards,
Myles
Thanks for the information and also to Braca, Tom and others on this thread. Once I get around to building, I will try a 51mm x 51mm x 6mm copper block on each of the 3875 chips with each of the copper blocks mated to a block of 228mm x 152mm x 9mm aluminum, which is one side of the chassis, (building monoblocks).
I will also try a 89mm x 51mm x 6mm copper block on each of the 4780 chips with each copper block mated to a block of 228mm x 152mm x 9mm aluminum. (building a paralled amp).
Since I have 8 pieces of copper bar, I plan to do some thermodynamic calculations to see if an optimum area of copper spreader can be obtained. I do not listen to music at high levels, so calculations can be kept real!!!!. Once the amps are operational, plans are to record some temperatures at various points and possibly reverse engineer the process to obtain some data. May take a while, but in the meantime, thanks for all the information so far.
Two question: Since all chips are TF type, If one does not want the copper block to be electrically conductive with the chip, but wants the heat transfer to happen, what is the best interface between the chip and copper: "Pads, non conductive grease, others?????
Is drilling and countersinking various diameter holes in the aluminum HS block a waste of time, or can it be an aid in the convection process from the HS to the air???.
regards,
Myles
Hi all,
Very great explanations !
One question : where did you buy copper bars, shop, links ?
In Europe, I only found bar of 1 meter or more, as it's very expensive
Phil.
Very great explanations !
One question : where did you buy copper bars, shop, links ?
In Europe, I only found bar of 1 meter or more, as it's very expensive
Phil.
One can easily be fooled by the datasheet numbers, so be careful.
The manu never publishes the actual nominal or measured thermal resistance value. So as an end user, you absolutely cannot extrapolate based on the numbers. Sometimes the manu will simply offset the value by adding a buffer, sometimes they will add a gain number, sometimes both.
Also, be very careful in the use of the die size for calculations. The die size and the dissipating area are not the same thing.
This is also true for pure bipolars and pure hexfets. The actual silicon being used can be as little as 25% of the effective surface.
Another caveat for thermal transport above the die towards the top.. The epoxy may not even be touching the silicon top surface, as there can be a coating to prevent differential shears against the chip aluminum traces, and there may not be a good adhesion between the epoxy and the aluminum. Where the epoxy touches the die attach solder that has flowed horizontally from under the chip, there will always be a poor attach and thermal transfer capability, so that real estate will be unuseable for topside thermal flow.
jn
The manu never publishes the actual nominal or measured thermal resistance value. So as an end user, you absolutely cannot extrapolate based on the numbers. Sometimes the manu will simply offset the value by adding a buffer, sometimes they will add a gain number, sometimes both.
Also, be very careful in the use of the die size for calculations. The die size and the dissipating area are not the same thing.
This is also true for pure bipolars and pure hexfets. The actual silicon being used can be as little as 25% of the effective surface.
Another caveat for thermal transport above the die towards the top.. The epoxy may not even be touching the silicon top surface, as there can be a coating to prevent differential shears against the chip aluminum traces, and there may not be a good adhesion between the epoxy and the aluminum. Where the epoxy touches the die attach solder that has flowed horizontally from under the chip, there will always be a poor attach and thermal transfer capability, so that real estate will be unuseable for topside thermal flow.
jn
I really appreciate you sharing your notes. Thank you.
You are making the assumption that the tab is filled with solid copper. This is not the case, as I clearly shown in my cross-sections. You can have an LM3886TF X-rayed if you want to make sure, but I would not count on any meaningful thermal conduction through the tab.
I think your calculation of the heat transfer through the sides of the package is assuming that the side of the package is isothermal, no?
I've never seen a plastic package from National/TI that did not have some amount of mold flash. Also, the packages tend to have slanted edges, presumably for mold release. You will have to sand or mill down the packages to make them flat so you can have good heat transfer.
Using your values for the thermal resistance through the top of the package and the back of the package, I arrive at Rth = Rth0 || Rth2 = (2*9)/(2+9) = 1.64 K/W.
I'm taking your area calculations at face value. The calculations do assume that the copper is solid without divots, voids, etc. which from my pictures appears to be a bit of a stretch.
As jneutron points out, sometimes, the IC die is encapsulated in some sort of goop to prevent delamination between the IC and the package material. Furthermore, the top side of the IC is normally covered by an insulator (spin-on glass or similar) of a somewhat uncontrolled thickness to prevent moisture from getting into the IC. The bottom of the IC would be the starting material (probably a doped silicon substrate). This means the thermal path through the bottom of the IC will conduct heat better than that through the top of the die. This means the area of the die should be subtracted from the copper area used for calculating Rth2, or at least, the increase in thermal resistance from the copper slug to the top of the package should be factored into the calculation of Rth2.
I have one burning question, though: WHY? I understand that sometimes it's fun to geek out and completely over-engineer something, but why would I machine blocks of copper to tight tolerances to achieve 1.64 K/W, when an LM3886T with a layer of Bergquist SilPad 1500ST (available from Mouser, et al) would have a thermal resistance around 1.1 K/W?
To get to the 1.64 K/W, you will need a tight (press) fit around the LM3886. Unless you plan to machine each copper block to fit each individual LM3886, you will either not get a good thermal contact or crack the IC package when you tighten the bolts. Neither is desirable.
~Tom
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Also, as the system heats up, the epoxy will expand about twice as fast as the copper. Given that it will be hotter than the copper by design as well, the internal compression will raise forces normal to the chip.
If there is no physical buffer between the chip surface and the epoxy, this will add normal forces to the fray. Normal forces on transistors can change their characteristics as well as cause them to fail.
But I do agree with you...geek city..
jn
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