I noticed several people posted that their CPU uses x watts. How do you guys know that? Are you going by what is on spec sheets? Or is their a (simple) formula you use to calculate the power consumption?
The reason I ask is I know about the spec sheets. But what happens if you change the voltage? The reason I ask is that I am interested in buying a "45 watt" Mobile Athlon. However, my mobo will probably run it at 1.575 volts rather than 1.45 volts. How much hotter will that run? How does that compare to the 1.6 v T-bred B 1700+ I own now.
TIA
How do you calculate how many watts a CPU uses?
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There's a useful utility in the SilentPCReview.com -- Web Links:
CPU Power by Kostik
That will do everything you need. (note that a "mobile Athlon" is just a Barton, running at lower voltage. )
CPU Power by Kostik
That will do everything you need. (note that a "mobile Athlon" is just a Barton, running at lower voltage. )
A decent estimate for power increase:
WattsNew = WattsOld * (SpeedNew/SpeedOld) * (VNew*Vnew/(Vold*Vold))
hence temp goes up as the square of the voltage, but linearly with speed.
The actual watts of a CPU going thru the heatsink seems to be approximately 6/7 of "maximum wattage" when the CPU is loaded, judging from what I've seen. That varies depending on CPU, HS, and so on.
Your mobile Barton at that voltage should well be cooler than your old thoroughbred at the same GHz.
An easy estimate for (1 + X)(1+X) where X is pretty small, is (1+2X).
Thus, moving from 1.45 volts to 1.575 volts would be about 20% more watts - 20% hotter. Your delta-T case to CPU would be ~20% bigger. If previously your CPU was at 40C, case at 30C, then now your CPU would be at 42C.
An 1800 GHz 45 watt Barton @ 1.45 volts going to 2200 GHz @ 1.575v = rated 64 watts -- still a fairly cool CPU. Probably less than your thoroughbred, but pretty similar.
the wesson
WattsNew = WattsOld * (SpeedNew/SpeedOld) * (VNew*Vnew/(Vold*Vold))
hence temp goes up as the square of the voltage, but linearly with speed.
The actual watts of a CPU going thru the heatsink seems to be approximately 6/7 of "maximum wattage" when the CPU is loaded, judging from what I've seen. That varies depending on CPU, HS, and so on.
Your mobile Barton at that voltage should well be cooler than your old thoroughbred at the same GHz.
An easy estimate for (1 + X)(1+X) where X is pretty small, is (1+2X).
Thus, moving from 1.45 volts to 1.575 volts would be about 20% more watts - 20% hotter. Your delta-T case to CPU would be ~20% bigger. If previously your CPU was at 40C, case at 30C, then now your CPU would be at 42C.
An 1800 GHz 45 watt Barton @ 1.45 volts going to 2200 GHz @ 1.575v = rated 64 watts -- still a fairly cool CPU. Probably less than your thoroughbred, but pretty similar.
the wesson
Thanks for the answers guys!
I do have a follow up question though. I've been running some numbers I noticed something strange. For example, if I take a 2500 Barton and overclock it to 2600 speeds, it uses more watts than if I take a 2800 Barton and underclock it to 2600 speeds. This is odd to me since both processors would be running identical speeds and voltage. Would there really be a 1-2 watt difference between them? Or is it simply an "error" in the formula.
I do have a follow up question though. I've been running some numbers I noticed something strange. For example, if I take a 2500 Barton and overclock it to 2600 speeds, it uses more watts than if I take a 2800 Barton and underclock it to 2600 speeds. This is odd to me since both processors would be running identical speeds and voltage. Would there really be a 1-2 watt difference between them? Or is it simply an "error" in the formula.
If I remember correctly, the TDP was the same in the program for the 2500 - 2800. Use the figures for the 2800; I'm pretty sure those are the most accurate.discdude wrote:I do have a follow up question though. I've been running some numbers I noticed something strange. For example, if I take a 2500 Barton and overclock it to 2600 speeds, it uses more watts than if I take a 2800 Barton and underclock it to 2600 speeds. This is odd to me since both processors would be running identical speeds and voltage. Would there really be a 1-2 watt difference between them? Or is it simply an "error" in the formula.
Amd's data-sheets list the same spec's (vcore, amperage, TPD, and MPD) for the 2500-2800, so the program is just regurgitating AMD's bad information.
If the 2500's 68.3 watts is correct, then the 2800 should be 77.6
If the 2800 is correct at 68.3, then the 2500 should be 60.1.
No real way of knowing which is correct, but since there's not enough difference to make a difference in any decision making, so I would probably just go with whatever the higher wattage was. It's all just ballpark guessing anyway.
If the 2500's 68.3 watts is correct, then the 2800 should be 77.6
If the 2800 is correct at 68.3, then the 2500 should be 60.1.
No real way of knowing which is correct, but since there's not enough difference to make a difference in any decision making, so I would probably just go with whatever the higher wattage was. It's all just ballpark guessing anyway.
Rusty075 wrote:It's all just ballpark guessing anyway.
True. In any case, it should give off less heat than my old Palomino 1700+ which wasn't *that* hard to cool.
Still, I would like to lower the overall power consumption of my system to well under 100 watts (~80 watts maybe?). It uses slightly more than 100 watts at idle right now, as measured by my Kill-a-watt. I figure a power supply more efficient than my FSP300-60BT (65% efficient, no PFC) should do most of the work.
Maybe replace my Radeon 8500 with a 9600?
Oh well, thanks for the help again guys.