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Politics : Formerly About Advanced Micro Devices -- Ignore unavailable to you. Want to Upgrade?

To: Maxwell who wrote (35175)	7/29/1998 12:33:00 AM
From: Time Traveler	Read Replies (1) \| Respond to of 1572637

Maxwell, After spending almost three days consulting with your experts, you are still mislead by them. Low K dielectric is still very important, even when everything is copper. The low K means lower capacitance. The lower capacitance means higher speed and lower power consumption. In a text book, the power dissipation is P = I^2 * R where I = DC current and R = DC resistance (Bill, are you taking notes?), but we are talking about AC. This textbook equation thus does not apply. In fact, the most generic equation is P = integral[t1,t2]( i(t) * v(t) * dt ) / (t2 - t1), notice i(t) and v(t) are both functions of time. t1 and t2 are the starting and ending times of a period, or whatever you want to define power as. The low K is thus extremely important at all frequencies. Low K will increase the performance of all the gates in the chain. Time Traveler

To: Maxwell who wrote (35175)	7/29/1998 1:14:00 AM
From: Paul Engel	Read Replies (1) \| Respond to of 1572637

Maxwell - Re: " The barrier layer is TaN. The sidewall coverage is 50A! " I think you'll find 50 A barrier layers are a disaster. Most work today uses barrier films (TiN mostly but some use TaN) of 15 to 30 Nanometers = 150 to 300 Angstroms. Multiplying this by 2 (two sidewalls) and you have 300 to 600 A of an 1800 A (minimum) trench consumed by barrier films. Hence, 16 to 32% of the THEORETICAL copper line width isn't even copper! Bye Bye copper advantage! Paul