9 Jun
2004
9 Jun
'04
9:35 a.m.
hmmm, I thought there were four rows of 18 chips. That sounds like 8 banks of 256K which
seems
like 2meg to me... unless I got something wrong
as for board worth, the 256K chips are worth a bit. I found a site that had them for sale
at $0.75 each.
If you need a PC to run it in, buy one on eBay... LOL (couldn't help that one)
best regards, Steve
-----Original Message-----
From: Doc Shipley <doc(a)mdrconsult.com>
Sent: Jun 9, 2004 1:28 PM
To: General(a)mdrconsult.com, Discussion@mdrconsult.com@null,
On-Topic and Off-Topic Posts <cctalk(a)classiccmp.org>rg>, null@null
Subject: Re: Cheetah Cub board
Steve Thatcher wrote:
that is a Micron part number for a 100ns 256K ram. It
is probably an EXPANDED memory card unless it has two bus connectors then it might be
EXTENDED memory (didn't we just have a conversation about PC memory...)
Ehh? 256K? So either the board provides 16MB of RAM, and the 2Mbyte
silkscreen is just the shipped configuration, or somebody was wasting
87% of their money?
BTW, the 10ns was a brain fart. I dunno much about DRAM, but I can
read the -6, -7, and -etc extensions... :)
So, more relevant to me, is the board itself worth anything outside
the DRAMs? I have a friend doing a 68000 SBC project who could use
some, and I think some of the rest would look fabulous on my Amiga
2000's A2091 board.
One note - It looks like the DIP switches can be set per bank, but
without docs, I'm guessing that taking any of the RAM off will render it
inoperable.
It is more than likely one of the two memory standards
and will work in any OLD PC.
But I don't have any old PCs. :)
BTW, your line wraps seem to be set to 256 or something.
Doc
9 Jun
9 Jun
10:09 a.m.
Steve Thatcher wrote:
hmmm, I thought there were four rows of 18 chips. That
sounds like 8 banks of 256K which seems
like 2meg to me... unless I got something wrong
No, I guess I got it wrong. I thought you meant 256K per chip, or
4MB per bank (16 * 256K plus 2 * 256K for parity).
So are these 256K per bank, or 256K per chip? IOW, if they turn out
to be compatible and I install 16 of them on the Amiga A2091, will that
be a 2MB or 0.25MB upgrade?
I'm a complete retard at memory components. It doesn't make sense to
me at all, even after a fair amount of reading and some explanation from
some Smart People.
as for board worth, the 256K chips are worth a bit. I
found a site that had them for sale at $0.75 each.
Well, that wouldn't suck. Of course that means they're really worth
about $0.20/each. :)
If you need a PC to run it in, buy one on eBay... LOL
(couldn't help that one)
Heh. I'd rather stick my face in the mud.... [family joke]
Doc
10:55 a.m.
hmmm, I
thought there were four rows of 18 chips.
That sounds like 8 banks of 256K which seems
like 2meg to me... unless I got something wrong
No, I guess I got it wrong. I thought you meant 256K
per chip, or 4MB per bank (16 * 256K plus 2 * 256K for parity).
So are these 256K per bank, or 256K per chip? 
2:33 p.m.
The _chips_ are 256K by 1 bit. You feed an 18 bit address into
Confusing correction (for the real hardware hackers here). These DRAMs
take the address in 2 halves of 9 bits each. You give the chip 9 bits of
the address, then the other 9 bits, and finally read/write 1 bit of data.
one chip and you get one data bit out. 9 chips form a
bank with
8 data bits plus a parity bit.
Correct
-tony
10:59 a.m.
Rumor has it that Doc Shipley may have mentioned these words:
Steve Thatcher wrote:
Yes. ;-)
-=-=-=-=-=-=-=-
They're 256K*bit* chips, so it takes 8 (or 9 if you use parity) chips to
make 256K*byte*s of storage. So, just saying "256K" isn't quite enough
information to enumerate exactly what's meant...
hmmm, I thought there were four rows of 18 chips.
That sounds like 8
banks of 256K which seems
like 2meg to me... unless I got something wrong
No, I guess I got it wrong. I thought you meant 256K per chip, or 4MB
per bank (16 * 256K plus 2 * 256K for parity).
So are these 256K per bank, or 256K per chip? IOW, if they turn out to be compatible and I install
16 of them on the
Amiga A2091, will that be a 2MB or 0.25MB upgrade?
Neither -- 16 would give you a 0.5MegaByte upgrade. 8 would give you 0.25MB.[1]
Heh. I'd rather stick my face in the mud....
[family joke]
My brain's *way* too frelled up right now to get into that...
Vioxx & Beer -- *bad* combination. My Doc (not you, Doc ;-) says I gotta
keep taking the Vioxx, so no beer. :-((~~
HTH,
Roger "Merch" Merchberger
[1] I figured I shouldn't mention until later that the upgrade you mention
would give you 256K*words* of memory, as the Amiga's a 16-bit machine... 4
bits = nybble, 8 bits = byte, CPU data width = word. (The last part is a
bit simplistic, but...)
--
Roger "Merch" Merchberger | "Profile, don't speculate."
sysadmin, Iceberg Computers | Daniel J. Bernstein
zmerch(a)30below.com |
2:31 p.m.
So are these 256K per bank, or 256K per chip? IOW,
if they turn out
to be compatible and I install 16 of them on the Amiga A2091, will that
be a 2MB or 0.25MB upgrade?
Most DRAM chips are 1 bit wide. A 256K DRAM stores 256K _bits_ (not bytes
or words), arranged as 262144 locations of 1 bit each. 8 such chips store
256K btes, 256K locations of 8 bits each (bit 0 is stored in the
locations of one of the chips, bit 1 in another chip, and so on). 9 chips
were used to store a byte + a parity bit (very common on PCs)
16 of them, used conventionally (which I assume the A2091 does) would
store 512K bytes or 256K 16-bit words. It would be a 0.5M upgrade I guess
-tony
3:13 p.m.
Tony Duell wrote:
So I talked to the memory guy here, and he says that the MT part
number is bogus. He says that there was a company doing counterfeits in
the late 80s and these are some of them.
Back to the DRAM nomenclature, the 1 bit vs 1 byte explanation helps
a lot. I'm still confused about the multiplier.
The Amiga docs show 16 256Kx4 DRAMs for a total of 2MB, so does that
mean that these chips are 256Kx1?
Doc
So are these
256K per bank, or 256K per chip? IOW, if they turn out
to be compatible and I install 16 of them on the Amiga A2091, will that
be a 2MB or 0.25MB upgrade?
Most DRAM chips are 1 bit wide. A 256K DRAM stores 256K _bits_ (not bytes
or words), arranged as 262144 locations of 1 bit each. 8 such chips store
256K btes, 256K locations of 8 bits each (bit 0 is stored in the
locations of one of the chips, bit 1 in another chip, and so on). 9 chips
were used to store a byte + a parity bit (very common on PCs)
16 of them, used conventionally (which I assume the A2091 does) would
store 512K bytes or 256K 16-bit words. It would be a 0.5M upgrade I guess
7:28 p.m.
Back to the DRAM nomenclature, the 1 bit vs 1 byte
explanation helps
a lot. I'm still confused about the multiplier.
The Amiga docs show 16 256Kx4 DRAMs for a total of 2MB, so does that
mean that these chips are 256Kx1?
256*4 DRAMs are what the numbers imply. 256K (262144) locations of 4 bits
(1 nybble) each. So 2 of _those_ chips store 256K bytes, 16 of them store
8*256K bytes == 2 Mbytes
The 256K *1 and 256K *4 chips have totally different pinouts (in fact I
think the former is a 16 pin chip, the latter a 20 pin chip), you
certainly can't interchange them.
-tony
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