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Wattage calculations
Hey! This is my first thread i make here at xtreme. I make it because there have been some discussions on how to calculate the heat radiation of an overclocked cpu.
For example a IRC friend is running a Tbred B 1700+(It’s radiate 59.8W according to AMD’s datasheets) @ 2150mhz 1.85V that would radiate about 117W and he only has an cheap all aluminium cooler master heatsink. That’s impossible and know morphling1 comes with his 1700+@2700mhz 2.2V that’s actually would radiate 208W. So there seams to be something wrong here.
I have calculated it with the same formula the site http://www.benchtest.com/calc.html uses.
Witch is Overclocked Watts = Default Watts * (Overclocked Mhz \ Default Mhz) * (Overclocked Vcore \ Default Vcore)². So for example a 1700+@2700mhz 2.2V calculation would look like this.
59.8*(2700/1467)*(2.2/1.6)² = 208.0847Watts
If it was a 2600+ @ 2700mhz 2.2V it would look like this
68.31*(2700/2133)*(2.2/1.65)² = ~154Watts @ 88% 135W
So what’s up with this what do the experts over here say?
Regards
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I actually dont know, since u dont know how big the current is u cant really calculate it, and the increase of the current isnt linear when u increase vcore.
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hmm isn't the current stated in AMD datasheets?
(59.8W = 1700+ Tbred B CORE)
And I should be 59.8w/1.6v=37.375A and 37.4Amps is in AMD datasheet.. So..
It would also be about R = U / I = > 1.6/37.375 = 0,042809364548494983277591973244147 ohm =)
P= U²/R => 1.6V²/0.042809364548494983277591973244147 = 59.800000000000000000000000000206watts
So i think that u got the right numbers now. Anyone really good for this like an master in electronics or anything that would clear this out what the cpu would radiator at say 2800mhz 2.2V
I can clearly see that it should radiate 113,05937500000000000000000000039W with 2.2V at default frequency.
Then 2800-1467=1333 =) 1333/1467=0,90865712338104976141785957736878 over default clock
113,05937500000000000000000000039Watt*1,9086571233 8104976141785957736878 = 215,79158145875937286980231765556Watts
And this is somthing totalywrong if this isn't right it should be and thats means that.
Overclocked Watts = Default Watts * (Overclocked Mhz \ Default Mhz) * (Overclocked Vcore \ Default Vcore)² allso is right!!
Thats mean that Morphling is watercooling a 215W(100%) cpu :slobber: at 88% it will be 189,89659168370824812542603953684Watts know thats a challange for a prome thats only can cool about 180W :P
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The current is only specified for default voltage, i wanna know what happens to the current when u increase voltage (practically).
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this means i am cooling about 120w on my old coolermaster heatsink @ ~55C FL, impossible!
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m0rm0r
Maybe it is wrong with the AMD datasheets or just that an processor don't follow ohmslaw
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the formula or the datasheet must be wrong ;)
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Becuse of the higher vcore from tbred a i think the wattage is probely right.
If it would radiate about 50W it would be ~100W instead of ~120W. But it can't answer that :(
I can contact AMD but i think the mail will endup at some stupid office that's dont know a §§§§ =)
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That 88% figure used here is for a really big loading. Like running Prime 95 & FAH both at the same time or burn k7.
But from what I've read morphling 1 is mostly running the 3D marks and other bench marks when he's been running his system all out. These programs don't use as high a percentage so the heat output should be lower would it not?
I don't know much about O/Cing but this seems to make sense to me.
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It says that my proc is dissipating 82.1Watts
If you want to be sure, an experience to do (if you're wc) would be to check the temp before and after the wb then calculate the time it takes to the water to go through the wb and then divide the two
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That may actually be right. i.e. Morphling's putting out 189 watts and a prommy can take around 180. But how often do you see people with prommies that are still around ambient, like opps bench rig. i.e his puts out 150 watts. If anything, that might be a little conservative.
