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To: Grand Poobah who wrote (11009)	3/20/1998 1:41:00 AM
From: Jock Hutchinson	Respond to of 25814

Dear Mr Grand Poobah, I am amazed at the quality , variety and integrity of the posts on this board. You are a tremendous addition.

To: Grand Poobah who wrote (11009)	3/20/1998 5:11:00 AM
From: shane forbes	Read Replies (3) \| Respond to of 25814

GP: That was truly outstanding! ASCII MOSFETs who would have thunk it! RE: Does anyone know what process technology the Gresham fab is supposed to run (e.g. 8", 0.35 micron)? It's actually going to be 8", 0.25 u (0.18 L-effective). RE: I don't have any idea of what percentage of their products are manufactured with the leading edge processes or what their yields are like. I suspect their yields are quite good. Indirectly their Gross Margin has been going up (fell last q - still around 45% which is not bad) and though the Gross Margin is a product of several things I nevertheless suspect part of the reason it is quite good is because of the favorable yields. No clue what percentage of products are manufactured using 0.35 or below. RE: The second thing that I like about LSI is that they seem to have gotten the infrastructure where they can run the large designs (>100k, 200k gates) LSI's average designs now have 400k gates. Not too shabby huh! ---- Finally on the Phsyics thingie I had a feeling 1000 times the "diameter" of an atom was not the crux of the problem. Heck they are doing 0.25u and that would imply 2500x. If no problem at 2500x can't see much of a quantum problem at 1000x. ---- //ot (Question: I see we have an Engineering Physics major here! Your 10 angstroms for the wavelength of an electron took me a good hour to figure out since I'm so very very rusty! Just for fun - Is this right? E = (p)^2 / 2m with lambda = h/p, this gives lambda = h / (2mE)^(.5) The toughie for me was trying to figure out what value to give E! Settled on Rydberg value for E = 13.6 eV (ionization energy of H) Thus with: E = 13.6 * 1.609 * 10^(-19) J m = 9.109 * 10^(-31) kg h = 6.626 * 10^(-34) J sec Eureka: lambda = 3 angstroms so I'm close enough to your 10 and that makes me happy!) ---- Shane.

To: Grand Poobah who wrote (11009)	3/20/1998 12:04:00 PM
From: Duane L. Olson	Respond to of 25814

G.P..Gee I picked a good time to drop back in on "The Pleasant Thread".. It has always been educational, with the Wizard finding more facts to bemuse one than should be allowed to complicate one's investment decisions.. Now I get to factor in the fact that Below the wavelength of the electrons, we would start seeing even more quantum effects as we got close to the spacing of the silicon atoms in the crystallattice (~3 angstroms) and the size of the atoms themselves (~1 angstrom). Life keeps getting more and more difficult! <G>. However, hope you won't mind if I "borrow" your post and drop it on another thread where I know some folks will appreciate it. Welcome to the LSI, thread, G. P.... we have had Knights, and Kings, Wizards, Mystics and Saints here...about time we had a Poobah! TSO ("The Saintly One")