LSI Corporation Message Board

STOCKTALK

We've detected that you're using an ad content blocking browser plug-in or feature. Ads provide a critical source of revenue to the continued operation of Silicon Investor. We ask that you disable ad blocking while on Silicon Investor in the best interests of our community. If you are not using an ad blocker but are still receiving this message, make sure your browser's tracking protection is set to the 'standard' level.

Technology Stocks : LSI Corporation

Public Reply Prvt Reply Mark as Last Read File

Previous 10 Next 10 Previous Next

To: Grand Poobah who wrote (11009)	3/20/1998 5:11:00 AM
From: shane forbes	Read Replies (3) 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.

Report TOU Violation

Share This Post

Public Reply Prvt Reply Mark as Last Read File

Previous 10 Next 10 Previous Next