Ed Casas (edc@ece.ubc.ca) Sun, 10 Dec 2000 16:40:56 -0800
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Date: Sun, 10 Dec 2000 16:40:56 -0800 From: Ed Casas <edc@ece.ubc.ca> Subject: Re: Asgn 4 Solution> Some questions about solution#1. I don't quite understand the way you > calculate the total memory especially for EPROM. Since we have 4 chips in > each bank and each chip is 16k bytes, Each chip is "16k x 4" or 16k nybbles, not bytes. > then we would have 64k bytes for each bank. There are 16 k words per bank, but each word is 2 bytes, thus 32k bytes. > Don't we have 128k bytes of total memory? No. 64k Bytes total. > But the solution given is 64k bytes? In the calculation, where > is the 2 bytes/word come from? 1 byte is 8 bits. A 16-bit word is 2 bytes. -- Ed Casas edc@ece.ubc.ca http://casas.ece.ubc.ca +1 604 822-2592
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