(The following is my original sysopt post. I wanted thought I'd share it here to help jump-start thought by involving more minds and ESPECIALLY because so many around here are extreme overclockers pushing serious wattage into their ubercooling systems.)
I know many of us are wondering about how to calculate the wattage of our GPUs. With no solid numbers released to the public, the public is forced to guess. Well, I decided it'd be time for an EDUCATED guess so that we know what kind of heat these bad boys throw off and design our aftermarket cooling solutions around them.
I may be wrong here, but I think I've found a way to estimate the power consumption of a GPU. It's most likely a little bit off, perhaps far off. All I know is that the power consumption figures seem about right.
I decided it'd be time to figure out just how much heat my R350 puts out at 400MHz. This requires knowing the transistor count (~110 million) and process technology (.15-micron) as well.
The goal is to find the unknowns based on a knowns. To start, I used a reference CPU that I knew the wattage of, a Thoroughbred 2400+ (68.3W Max @ 2.0GHz). I then divided the wattage by the approximate number of transistors (37 million) in the CPU to get watts/transistor.
68.3/37000000
1.8459459459459459459459459459459e-6
This resulted in that number of watts per transistor AT 2GHz operational frequency. To get their frequency @ 400MHz, I had to set up the following equation:
(1.85x10-6 watts/tr)/(2GHz)=(x watts/tr)/(0.4GHz)
answer*.4
7.3837837837837837837837837837838e-7
answer/2
3.6918918918918918918918918918919e-7
this resulted in 3.69x10-7 watts/transistor at 400MHz.
Now, the radeon 9800 GPU does not have 37 million transistors; it has 110 million. So I simply multiplied that value by 110 million to get the total heat output of the GPU @ 130nm:
answer*110000000
40.59
which ends up being 40.59W.
But... the Radeon 9800 GPU is not .13-micron; it is .15-micron. Since the relationship is linear, I set up an equation to find a compensating factor:
.15-.13
_______ = .1333333...
.15
So with this equation we find that on average a .15-micron transistor runs 13.3% hotter than a .13-micron transistor.
Multiply the wattage by 113% and we get the extra heat output factored into the total wattage:
40.59*1.13333333333333333333333333333333333
46.002
which ends up being 46.002 watts.
Now, one final addition. The R350 core does not run at 1.65v, most are set at 1.75v. Again, another ratio must be established, but this is not as simple as the earlier one. Heat and voltage is not a linear relationship; it is an exponential one.
The equation for finding a fudge factor for the additional voltage is as follows:
overvolt^2
_____________
original volt^2
=
(1.75/1.65)^2 = 1.1248852157943067033976124885216
multiplying this value by the wattage at 1.65v and you get a wattage that factors in the additional .1v:
answer*46.002
51.74696969696969696969696969697
So the heat output of our R350 GPU running at 400MHz on a .15-micron process at 1.75v is, in theory, at most 51.75W. People with a sense of this heat output would immediately get alarmed by the anemic size of the R350 heatsink in comparison to the CPU-like thermal output. Undoubtedly, the R350 has been measured to reach nearly 80 degrees celsius under full load.
However, there is no way in hell a heatsink like that would be able to even remotely cool nearly 52W of thermal power. GPUs are deeply pipelined devices and the transistors are almost always under-utilized. And 52W is a theoretical value - the most thermal loading programs have been able to load CPUs is 88% or 7/8 of their maximum power.
Since there is no graphics program dedicated to filling the pipelines to their maximum with predictable values, GPUs can never see even 88% of their theoretical output. Even under "full load" rendering conditions, the number probably hovers between 66% and 75% utilization.
Now these are numbers I've pulled out of nether regions, but they are more close to reality IMHO.
51.74*(2/3)=34.49333333333333333333333333333...
51.74*(3/4)=38.805
arithmetic average:
(34.493+38.805)/2
= an average of 36.65W under "full load" rendering/gaming conditions.
which still does heat up the anemic heatsink quite a bit.
Now, based on those values, let's calculate the degC/W of the stock heatsink. C/W is a measure of how many degrees celsius a cooling apparatus allows a heat source to rise over the ambient (environment) temperature. The lower, the better.
Xbit labs says their R350 heated up to 79.6 degrees Celsius under load. They said this was on a lab bench in the open air, so I'll assume their room temperature was around 22 degrees celsius. This is a ballpark figure, whose basis on a 20MHz clock differential has almost no bearing on the results.
(79.6-22)=57.6 (this is how much the heatsink allowed the core to rise above ambient, ouch)
heat output of R350 @ 400MHz (from my possibly skewed calculations)=36.65 under "gaming" load
57.6 C / 36.65 W -> 1.57 degrees C/W
(Note that a small AMD stock cooler has about .35 degrees C/W in comparison)
No doubt about it, the stock cooler is very poor, but it does keep the GPU below the error-producing heat point. GPUs tend to be more tolerant than CPUs of heat because they run at much lower clock speeds on a given process and are loaded on average far less.
I hope this gives some folks some insight as to how to perform ballpark estimates on GPU thermal output (or output of any chips) when no solid wattage information is released to the public - as for this calculation, I know I'm a bit off, but I think I'm pretty close to reality.
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that CANT be right! if the fx5800 would only produce 40W of heat it would never need the dustbuster at all! dont forget it has a huge cooler and still reaches up to 100+°C under load in a closed case, thats where the termal throttling sets in, at 120°C the gpu shuts itself down.
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