HP Laserjet ..again....

From: Tony Duell <ard_at_p850ug1.demon.co.uk>
Date: Sun Sep 19 17:01:15 2004

>
> I didn't understand that you were holding the word width constant. If

I'm not. But I am keeping the size of a 'byte' constant at 8 bits (this
is actually not always the case, but anyway).

> 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

OK....

Let's think of those 16 64K*1 DRAMs again.

If I wire them up as a 16-bit wide memory (to link to a 68000, say), it's
correct that I have 64K _words_ of storage (where one word == 16 bits ==
data bus width). But it's also conventional to call that 128K bytes (the
origianl Mac, for example, had exactly that configuration of memory, it
was called a 128K machine, even though it's really got 64K words of
memory (each being 16 bits or 2 bytes).

My PDP11/45 has 124K words of memory. I have seen this quoted as 248K
bytes (IMHO that is also correct).

> 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.

Usually, but not always The Philips P850 looks to be a 16 bit machine to
the programmer, the front panel has 16 toggles, the registers look to be
16 bits wide, as do the memory locations, etc. At the hardware level it's
an 8 bit machine, 8 bit ALU (a pair of 74181s) 8 bit data bus to memory,
etc.
 
> 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

YEs, becasue a byte is 8 bits, and because the memory chips all store
data (not parity or ECC bits, for example).

It really doesn't matter if the processor accesses that memory 8 bits at
a time or 16 bits at a time. The total number of bits that can be stored
is the same. So, therefore, is the total number of 8 bit bytes.

> 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.

Actually, on the Laserjet 2, the bus is 16 bits wide. It's a 68000
processor.

-tony
Received on Sun Sep 19 2004 - 17:01:15 BST