Path: utzoo!utgpu!jarvis.csri.toronto.edu!rutgers!sun-barr!ames!amdahl!amdcad!diablo!phil
From: phil@diablo.amd.com (Phil Ngai)
Newsgroups: comp.sys.ibm.pc
Subject: Re: e000=f000 - Summary of responses
Message-ID: <27371@amdcad.AMD.COM>
Date: 19 Sep 89 23:54:47 GMT
References: <403@wjh12.harvard.edu>
Sender: news@amdcad.AMD.COM
Reply-To: phil@diablo.AMD.COM (Phil Ngai)
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Organization: Advanced Micro Devices, Inc. Sunnyvale CA
Lines: 21

In article <403@wjh12.harvard.edu> djb@wjh12.UUCP (David J. Birnbaum) writes:
|>... the sockets on the motherboards will take a 2 64kx8 chips
|>while the BIOS is just 2 32kx8 chips (there is room for 4 roms on my 
|>motherboard)  Of course enough address lines had to be run to the socket
|>to account for the bigger chips, and A15 is unused on a 32kx8 and goes
|>to the Vpp (programing voltage) pin. So A15 is just ignored because the
|>Vpp pin isn't functional unless you have 12.5 volts running to it.
|
|Or is it A16?:

You have to "read between the lines" here.  The A15 mentioned is the
EPROM's A15. Since there are two of them, that's equivalent to the
286's A16.

Of course, this is all a little off the point. What counts is what the
control logic does, but I suppose talking about memory chips is an
easy, if slightly misleading, way to explain why the address map is
the way it is. 
--
Phil Ngai, phil@diablo.amd.com		{uunet,decwrl,ucbvax}!amdcad!phil
"Should the US send assault rifles to Colombia? How about small arms?"