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large vdroop.. am I calculating it correctly?

post #1 of 4
Thread Starter 
Hello. I have a new G0 q6600 with a vid of 1.325. Right now it's at the stock clock rate, and the bios is set to 1.325 vcore.

Here are the numbers from CPUz:

idle: 1.120v
load: 1.264v

That's .12v difference. Whats going on? Also, since my vid is 1.325, does that mean I shouldn't set the vcore to anything lower in the bios? It seems like I have read too many conflicting definitions of vid.

Thanks
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post #2 of 4
if you are running stock just run the auto voltage setting...

so therefore vcore should be 1.325

the drop from BIOS--->OS can't be fix. so its normal to see 1.264v on load.

the reason why the chip is reading 1.120v on idle is because you have speedstep enabled. It lowers your voltage and speed of the chip at idle...then ramps it up when you work it.

EDIT- if you would like to see what your vdroop really is...you need to disable C1E and EIST in your bios...then your chip is running full speed all the time and you'll see your voltage at idle increase. That (number) - 1.264v is your vdroop.
Edited by maddux - 12/12/08 at 12:27pm
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post #3 of 4
Thread Starter 
Maddux, yea it is actually set to auto voltage. Should speedstep be disabled when figuring out vdroop?

Edit: ok I will disable those and check again
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post #4 of 4
Thread Starter 
Well this makes more sense. It drops from 1.28 at idle to 1.264 at load with the bios settings disabled. Thanks for the help.
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