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The Truth about Temperatures and Voltages - Page 16

post #151 of 599
Thread Starter 
Quote:
Originally Posted by blkrnbw View Post
Quote:
Originally Posted by ChickenInferno View Post
I'm beginning to think the same thing, but yet in the literature it keeps showing the voltages drops for different amperages. This may be getting to the point where Intel is keeping this as proprietary knowledge (How they power their CPUs and don't fry them). If they gave out the exact in's and out's of how they powered it; they give out good ideas but leave out so much that it makes it impossible to know what they are really doing.

The one point that keeps coming to mind for me is that if there really is a "Maximum Power Usage Limit" that I think Intel would have published it instead of the vcore limits. Either Vcc and Icc are inherently linked (not in the P=IV sense, but in a if Icc=Vcc*100 or something like that.)

Icc Max is designated by the CPU, whether Icc Max is dynamic is a question that if answered can answer alot of others. If Icc Max is also Icc typical, IE when booted the computer brings Amps up to Icc Max and that is wherr it stays. Then the chart with the different drops with different Amps means for different CPU's that are set at these different Amps this is the drop for max, min, and typical load. It should not be read (as I have been) as the Amps raise then the drop is this. We are getting those charts from ATX PSU spec which do have to included all the family of processors so different Amp levels for different CPU's. But that doesnt make sense because if I remember correctly those charts are examples of specific CPU's IE 775_CPU_5AB or what not, so that screws that idea up. Ok so since Intel wrrote this thing I am going to email them and just ask about the load line and how it should be read. Wish me luck on a real answer.

Brian
Good Luck getting an answer. I'm betting that if you do it won't be for a while as they will send the email off to ten different people before they get an answer.

Quote:
Originally Posted by PizzaMan View Post
OK, to test the power consumption limit we need someone who has a mobo that shows watts and amp in everest. Have them OC as much as possible with stock cooler. Monitor wattage during load. Then put a large heatsink on and OC again as much as possible. Then monitor wattage during load. If the theory is correct then each OC should use roughly the same amount of watts.

Any volunteers?
You could also use the cooling you have now and just turn the fans onto about 25% power and then compare that to 100% power. You would have to absolutely push it to the brink though with temperatures/stability and you would need the cap to be temps.

Quote:
Originally Posted by TwoCables View Post
I honestly do not see any reason why the wattage would be different. So, I'm saying that I am personally 100% confident/sure that the wattage will always be the same no matter what the cooling solution is.
I think so too, but I also think it's possible that thermal resistance from high temperatures could cause higher power consumption. P=IV=(I^2)*R and if the temperatures really do play a role in power that could be one avenue of finding a maximum power consumption if there is one.
Edited by ChickenInferno - 8/22/09 at 7:18am
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post #152 of 599
Quote:
Originally Posted by ChickenInferno View Post
You could also use the cooling you have now and just turn the fans onto about 25% power and then compare that to 100% power. You would have to absolutely push it to the brink though with temperatures/stability and you would need the cap to be temps.
If only Everest told me watts, but it doesn't with my board.


Quote:
Originally Posted by ChickenInferno View Post
I think so too, but I also think it's possible that thermal resistance from high temperatures could cause higher power consumption. P=IV=(I^2)*R and if the temperatures really do play a role in power that could be one avenue of finding a maximum power consumption if there is one.
ummm, can you brake down the variables in P=IV=(I^2)*R for me?
    
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post #153 of 599
Thread Starter 
Power (P) in Watts=Current (I) in amps * Voltage (V) in volts

and this is also equal to Power=Current^2 * Resistance (R) in ohms

so

P=IV=I^2 * R
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post #154 of 599
I have a kill-a-watt plugged in between machine and wall. Watt from wall should be = to watt from PSU, well after efficiency is applied, but that really wont matter here (I dont think). I should be able to just read the watts at idle, with EIST on and wait for the CPU to drop to 6X. Then run prime and kick it all the way in. volts should remain the same so all there is to figure out is Amps (since we will have watt's). If volts stays where it is set to and Amps are same under both loads we should have an answer, although I remember watts climbing last time I checked this. So if CPUID shows volt at same level, and watts increase that must mean that Amp is dynamic. Am I correct?

Brian
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post #155 of 599
Watts will increase under load.
    
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post #156 of 599
Thread Starter 
The voltage does drop when switching into the EIST, C1E, and Speedstep as well as dropping down the multiplier. With all forms of thermal/electrical restrictions turned off, we would need to run check the Kill-a-Watt and then run it under prime or LinX. Then with the same Heatsink fan plugged in but turned away (so that is can't cool and the CPU should heat up more) run the same test. If the Kill-a-Watt reads more power even though the everything is still plugged in, then we have an answer.
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post #157 of 599
Quote:
Originally Posted by ChickenInferno View Post
The voltage does drop when switching into the EIST, C1E, and Speedstep as well as dropping down the multiplier. With all forms of thermal/electrical restrictions turned off, we would need to run check the Kill-a-Watt and then run it under prime or LinX. Then with the same Heatsink fan plugged in but turned away (so that is can't cool and the CPU should heat up more) run the same test. If the Kill-a-Watt reads more power even though the everything is still plugged in, then we have an answer.
That's not going to be an accurate enough of a test. We need to have two differenct OCs using two different voltages and both OD'd to the temp/volts limit. We already know that components use more wattage with higher temps. That's all that test will prove. We need to prove that Two different OCs with two very different thermal resistances use roughly the same amount of watts.
    
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post #158 of 599
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Yes, but if we can prove that the hotter cores force more power to be used, then that's half the battle and it also tells us whether or not we even need to do anything else. If the cores run hotter but use the same amount of power, then obviously heat doesn't play a role in the power used and the proposition that running a hotter but lower OC uses the same maximum power consumption as a cooler higher OC is wrong.
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post #159 of 599
I have to disagree with you. W/o decent change in thermal resistance and a change in OC based on new temperatures the test will not be accurate. If he turns his CPU fan off and re-does his OC based on new temps and ends up with a different OC using different voltage to the CPU then maybe. But then you also have the factor that he's monitoring total system power. If his other components temperatures change then their power usage will also change making the test even more inaccurate.
    
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post #160 of 599
Quote:
Originally Posted by PizzaMan View Post
Watts will increase under load.
Exactly. And this is why I don't see how the temperature will affect the wattage. It has always been the amount of load that affects the wattage. Not only that, but it's the overclock and it's the amount of voltage being put through the CPU.

So, I sincerely doubt that the wattage would be any different with a cold CPU under load vs. a hot one. I mean, it doesn't add up for me.
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