On Wed, 27 Oct 2004 03:34:01 GMT,
(The little lost angel) put
finger to keyboard and composed:

So both the dynamic power and static power rises exponentially to the
voltage applied? Thus if I take a 3Ghz Prescott, changes the voltage
from say 1.4V to 1.5V and measure an increase of 5W power while idle,
this would imply that the increase in leakage current is 500A!?? Since
the current increases with the square, so 5W = 0.1*0.1 * A therefore 5
/ 0.01 = A = 500A? At cube, it becomes even crazier

Ok, I did that before I saw the formula you gave later as (VDD1/VDD2),
thus is it (1.4/1.5)^3 * A = 5, ...

VDD1/VDD2 is a dimensionless quantity. Hence the units on both sides
of your equation do not balance, ie on the left you have amps, while
on the right you have watts. Then there's your mistake in equating
ratiometric changes with absolute ones ...

... hence A= 6.15 which sounds a WHOLE lot
more credible. Would this also mean that the actual leakage can be
1.5V x 6.15A = 9W or would it be 1.5V x (6.15A + X) where X was the
original leakage current at 1.4V?

Please don't laugh too hard if the maths is hopelessly naive & wrong
PPPP


- Franc Zabkar
--
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