HP Laserjet ..again....

From: Jim Isbell <millenniumfalcon_at_cableone.net>
Date: Sun Sep 19 16:14:38 2004

I didn't understand that you were holding the word width constant. If
you hold the word width constant, yes, you are right. But that is not
what I was talking about. In many early computers, the data buss and
the word width were the same. It was not till they began to make
processors with larger word widths that there was a difference.
 Actually, not till they started making processors. The early computers
didn't have processors in the same sense as we now know them as a single
component.. The buss normally matched the word width.

In this particular case the memory I was counting, on the board in the
printer HP printer, I was counting the memory AS IT IS NORMALLY COUNTED
for an HP laserjet Series II, and I was right, to count it as 8 bits
wide regardless of the word width, I have two 4 meg boards and one 2 meg
board. That has been confirmed. My reason for saying the meg count was
dependent on buss width was because in this case, IT IS. The buss is 8
bits wide and that is how you count the memory on a HP laserjet card.

Tony Duell wrote:

>>Sure the buss width matters if you are measuring memory as it is NORMALY
>>measured, in Bytes, not Bits.
>No it doesn't.
>Suppose I take 16 of the 64K * 1 DRAM chips (4164s). I wire them up to an
>8 bit databus with the apporporate control electronics so I get 128K 8
>bit locations. That stores 128K bytes.
>I now wire them up differently to a 16 bit data bus system. I haev 64K 16
>bit locations. That's _still_ 128K bytes of storage.
Received on Sun Sep 19 2004 - 16:14:38 BST

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