Pravin, "Isn't Intel's operating voltage on any given technology typically lower (by 0.1-0.2 Volts) than AMD's?"
The answer is no. First, there is no such thing as
"any GIVEN technology". There are shrinks, compromises,
process tweaks, and specification fudging, all in the
name of achieving market-driven frequency targets.
All other things are secondary. The worst case is
when they jack up voltage when requesting lower junction
temperatures. That was exactly what Intel did on top
"0.18u" Pentium-IIIs with 1.75V Vcc and 62C die
temperature. This was an indication of a wall.
Then Intel moved to super-long-pipe design of P4, and
the technology requirements were slightly relieved
due to frequency-favoring micro-architecture.
The recent batch of P4 is at the same 1.75V Vcc,
however the total leakage is "only" 8.7A,
or 16-22%.
It is also clear that the mobile variant of 0.13
is heavily optimized for lower voltage. Given their
leakage data (for deep sleep state, P-III-M datasheet):
x y
0.95 1.7
1.05 2.65 **
1.1 3.01 **
1.15 3.99
1.4 8.04
The dependence seems to be exponential,
y=0.0785exp(3.39x), so the leakage at 1.75V
would be at 27A, which would be 2x of the leakage
on "regular" Tulatin parts. (I hope process
engineers could help me here with formulas).
Still, there is a published fact (Intel datasheet
#249657-002, August 2001) that the regular
Tulatin 0.13u design is leaking 10A, or 50%, the best
achievement of new Intel technology so far.
- Ali |
