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Technology Stocks : Intel Corporation (INTC) -- Ignore unavailable to you. Want to Upgrade?

To: Proud_Infidel who wrote (58705)	6/24/1998 7:34:00 PM
From: Paul Engel	Read Replies (1) \| Respond to of 186894

Brian - AMAT, DIe Shrinks, and Fab Capacity This question is valid but has some confusion based on SEMANTICS. IF a device is "shrunk" from one size to a SMALLER SIZE, using finer lithography feature sizes, more interconnect, etc., the die size is reduced and MORE DIE SITES/WAFER are available. The essence of the argument is "What device does the new SHRUNKEN DEVICE replace?". If the SHRUNKEN DIE replaces the "UNSHRUNKEN" DIE and equipment/capacity issues are not a constraint (which is rarely the case) AND YIELDS (as defined as percent of good die/possible die) ARE NOT DEGRADED - then productiion capacity will be increased. But let me give you a counter example. Intel's 0.25 micron Pentium II/Deschutes is 131 sq. mm. in size (area). It directly replaces the 0.35 micron Pntium II which had a die size of 203 sq. mm. HOWEVER, Intel is replacing all of its 0.35 micron Pentium MMX devices which were 128 to 140 sq. mm in size with the new 0.25 micron Pentium II. The "net-net" capacity increase is marginal, since since MOST of the 0.25 micron Pentium II devices replace Pentium MMX chips of comparable size. There is an incremental capacity increase because the 0,25 micron Pentium II replacesd the 0.35 micron Pentium II but that production level was relatively small ( a few million/quarter) vs the approximately 18 million/quarter Pentium MMX (0.35 micron) production levels. Also, the 0.25 micron Pentium II process requires an extra metal layer - which actually requires several other layers (Interlevel oxides) as well as via mask and etching and the additional metal mask and etching. This ultimately reduces capacity somewhat since more equipment is required and longer cycle times result. So, the answer is not always SIMPLE! Paul