Newby - [lease be gentle!

My PC has a K7S5A V1.0 motherboard. It has 256M of memory in slot DDR1.

I feel like adding a further 256M. What should I be looking for?

Is a single 512 better then 2 x 256?

--


Regards

John

I'd check with the manufacturer or the motherboard book, but in many boards
ram works better in pairs
"John" wrote in message
...
My PC has a K7S5A V1.0 motherboard. It has 256M of memory in slot DDR1.

I feel like adding a further 256M. What should I be looking for?

Is a single 512 better then 2 x 256?

--


Regards

John

John wrote:
My PC has a K7S5A V1.0 motherboard. It has 256M of memory in slot DDR1.

I feel like adding a further 256M. What should I be looking for?

Is a single 512 better then 2 x 256?

Although I do not have this version of MB (I have K7S5A pro v.5.0), you
might encounter a problem with a single 512 MB stick. I installed a
Kingston Value Ram PC2700 512MB stick and encountered errors even though
the stick passed on a friends computer (different MB) easily. The
advice to seek a size and brand that has been proven to work is good
advice on this MB. I might add that I have two sticks of PC 2700 256MB
installed and it works flawlessly. One is a Kingston VR and the other a
Nanya. Your results may vary.

If you're in the US (I believe they also do business in the UK), visit:

www.crucial.com

(This is the retail operation of Micron.) They guarantee that RAM purchased
by using their memory selector will be compatible with your board. The
highest capacity DIMMs that Crucial sells for your system are 512 MB.

If you prefer to deal with a local merchant, you can probably use generic
184 pin DDR memory, PC2100 or faster (unbuffered, non-ECC). (This is the
commonest sort.) A couple of things:

You mainboard supports both DDR and the older SDRAM. Don't mix them. (DDR
should give better performance, and it's now cheaper than SDRAM, so stay
with it.)

I don't think that the board supports dual-channel operation, so there may
be little performance difference between one DIMM or two. Some mainboards
are fussy about memory when all of the DIMM slots are filled, but I wouldn't
expect that to apply to your system (with two DDR DIMM slots).

Address scrambled. Replace nkbob with bobkn.

"John" wrote in message
...
My PC has a K7S5A V1.0 motherboard. It has 256M of memory in slot DDR1.

I feel like adding a further 256M. What should I be looking for?

Is a single 512 better then 2 x 256?

--


Regards

John

John wrote:

My PC has a K7S5A V1.0 motherboard. It has 256M of memory in slot DDR1.

I feel like adding a further 256M. What should I be looking for?

Is a single 512 better then 2 x 256?


Is there some truly logical reason why you don't submit your question to
ECS? That's most like the best place to get an intelligent, well
informed, response to your question!

John wrote:
My PC has a K7S5A V1.0 motherboard. It has 256M of memory in slot DDR1.

I feel like adding a further 256M. What should I be looking for?

If you're running 266 FSB then 256 meg PC2100, or faster.


Is a single 512 better then 2 x 256?


Not for adding another 256 Meg. It just increases the cost.

Ken wrote:

John wrote:

My PC has a K7S5A V1.0 motherboard. It has 256M of memory in slot DDR1.

I feel like adding a further 256M. What should I be looking for?

Is a single 512 better then 2 x 256?

Although I do not have this version of MB (I have K7S5A pro v.5.0),
you might encounter a problem with a single 512 MB stick. I installed a
Kingston Value Ram PC2700 512MB stick and encountered errors even though
the stick passed on a friends computer (different MB) easily.

You don't say what kind of errors but it sounds like the typical "high
density" memory problem.

Each DIMM socket in the K7S5A supports two ranks with up to 256 Meg in each
rank for 2x256=512 Meg max total per slot (and 2 slots makes the 1 gig
maximum). A "standard" 512 Meg stick is made that way: 256 Meg per rank and
2 ranks.

A "high density" DIMM, however, has the whole 512 Meg in one rank and, so,
exceeds the slot's capacity.


The
advice to seek a size and brand that has been proven to work is good
advice on this MB. I might add that I have two sticks of PC 2700 256MB
installed and it works flawlessly. One is a Kingston VR and the other a
Nanya. Your results may vary.

"Dee" wrote in message
...
John wrote:

My PC has a K7S5A V1.0 motherboard. It has 256M of memory in slot DDR1.

I feel like adding a further 256M. What should I be looking for?

Is a single 512 better then 2 x 256?


Is there some truly logical reason why you don't submit your question to
ECS? That's most like the best place to get an intelligent, well
informed, response to your question!


Dee - I was concerned that the builder of the PC would have fitted the
existing memory and that I perhaps needed to match to that rather than to
the motherboard.

--


Regards

John

David Maynard wrote:
Ken wrote:

John wrote:

My PC has a K7S5A V1.0 motherboard. It has 256M of memory in slot DDR1.

I feel like adding a further 256M. What should I be looking for?

Is a single 512 better then 2 x 256?

Although I do not have this version of MB (I have K7S5A pro
v.5.0), you might encounter a problem with a single 512 MB stick. I
installed a Kingston Value Ram PC2700 512MB stick and encountered
errors even though the stick passed on a friends computer (different
MB) easily.


You don't say what kind of errors but it sounds like the typical "high
density" memory problem.

Each DIMM socket in the K7S5A supports two ranks with up to 256 Meg in
each rank for 2x256=512 Meg max total per slot (and 2 slots makes the 1
gig maximum). A "standard" 512 Meg stick is made that way: 256 Meg per
rank and 2 ranks.

A "high density" DIMM, however, has the whole 512 Meg in one rank and,
so, exceeds the slot's capacity.

You may well be correct about the "Rank" thing. Personally, I don't
fully understand it. I DO understand that the memory circuitry is
sometimes designed to see the memory in different configurations, such
as low density and high density. One would think however if the density
were the problem, that the full size would not be detected during POST.
I have seen many situations were this was the case, and it made sense
when it detected only half of the memory on the installed stick.
Apparently there is a factor in addition to the density that will cause
memory to not be used effectively?

I don't mean this as a critical remark, but only as a fact: You
apparently know much more about the access circuity of memory on MBs
than I do. Although I am curious about the "Rank" thing, I have read an
explanation (perhaps even written by you) and still do not fully
understand it. I am afraid it is going to take a graphic illustration
for this old fart to understand rank.

I do appreciate your comments, and I am sure that others less dense
than myself will understand them fully. My only purpose in posting my
comments were to alert the OP that not all memory will work on a given
MB. Thanks again for your comments.



The advice to seek a size and brand that has been proven to work is
good advice on this MB. I might add that I have two sticks of PC 2700
256MB installed and it works flawlessly. One is a Kingston VR and the
other a Nanya. Your results may vary.

Ken wrote:

David Maynard wrote:

Ken wrote:

John wrote:

My PC has a K7S5A V1.0 motherboard. It has 256M of memory in slot DDR1.

I feel like adding a further 256M. What should I be looking for?

Is a single 512 better then 2 x 256?

Although I do not have this version of MB (I have K7S5A pro
v.5.0), you might encounter a problem with a single 512 MB stick. I
installed a Kingston Value Ram PC2700 512MB stick and encountered
errors even though the stick passed on a friends computer (different
MB) easily.



You don't say what kind of errors but it sounds like the typical "high
density" memory problem.

Each DIMM socket in the K7S5A supports two ranks with up to 256 Meg in
each rank for 2x256=512 Meg max total per slot (and 2 slots makes the
1 gig maximum). A "standard" 512 Meg stick is made that way: 256 Meg
per rank and 2 ranks.

A "high density" DIMM, however, has the whole 512 Meg in one rank and,
so, exceeds the slot's capacity.


You may well be correct about the "Rank" thing. Personally, I don't
fully understand it. I DO understand that the memory circuitry is
sometimes designed to see the memory in different configurations, such
as low density and high density. One would think however if the density
were the problem, that the full size would not be detected during POST.

That is often what happens, yes.

I have seen many situations were this was the case, and it made sense
when it detected only half of the memory on the installed stick.
Apparently there is a factor in addition to the density that will cause
memory to not be used effectively?

Yes. Another is loading of the memory bus. The 'high density' modules put
more chips on the bus and that can cause problems.


I don't mean this as a critical remark, but only as a fact: You
apparently know much more about the access circuity of memory on MBs
than I do. Although I am curious about the "Rank" thing, I have read an
explanation (perhaps even written by you) and still do not fully
understand it. I am afraid it is going to take a graphic illustration
for this old fart to understand rank.

I understand. It isn't easy to visualize if you're unfamiliar with it.

You have a memory address and how much memory can be addressed is
determined by how many bits are in the address. It's binary, of course, so
it goes as the power of 2. 1 bit has two states (2^1). 2 bits have 4 states
(2^2), 3 have 8 (2^3), 4 have 16 (2^4), 5 have 32 (2^5), 6 have 64 (2^6)
and so on (2^x). Call this how 'deep' the memory can be. (remember 2^5 and
2^6 as we'll use those below)

Now, memory also has data 'width'. For a P4 memory bus that's 64 bits wide
which, a byte being 8 bits, comes to 8 bytes.

So think of it as an XY 'stack' of memory cells. Call Y the memory
addressing and call it the 'depth' of the stack. Call X the memory data bus
and call it 'width'. The address bits are decoded to an address # on the
chip, so, for 5 bits of addressing (32) the stack looks like

address # 31 -- ========
....
address # 3 -- ========
address # 2 -- ========
address # 1 -- ========
address # 0 -- ========
01234567 memory width X (8 bytes, each byte[=] is 8 bits)

^^^^^^^^
||||||||
vvvvvvvv

The memory controller asks for memory location Y and gets, all together in
one stroke, 8 bytes of data on the memory bus (or sends data to it).

Lets use a small address space to make the numbers simpler. Say we have 5
address lines so we can address up to 32 (2^5) memory locations (in binary
we number from 0 to 31) and they're 64 bits wide so that totals 32 x 64 for
2048 bits. Since bytes are 8 bits, in bytes that's 32 memory locations deep
by 8 bytes (64/8) wide so our total memory capacity is 256 bytes. This is
going to be our "rank" (just accept that for the moment)

So, we have a system with a certain addressing capacity, 256 bytes, and now
we're going to put memory chips into it. And you may have guessed that our
capacity of 256 bytes resembles 256 meg if you multiply by a binary meg.

Memory chips are organized the same way, a depth and width, and the
'standard' width is 8 bits so it takes 8 chips, side by side, to fill our
64 bit wide memory data bus (rank). Since we have a maximum of 5 address
bits it's clear that the largest chip we can use is 32 bits deep and 8 of
them makes 256 bytes. The combined width always needs to be 64 bits,
because that's the data bus width, but we can use chips with less memory
depth, like 16 bits deep for 128 bytes, but not deeper ones because we
don't have more than 5 address lines.

Now, to describe the memory chip's organization we'll say it's "32x8" since
that's it's depth (32 bits) and width (8 bits). It's a "256 bit, 32x8,
memory chip" and 8 of them, 'side-by-side', make up a rank that totals 256
bytes and everything matches our system's maximum capacity. (btw, 32x8 is a
standard chip size in real life as the 32 is in megabytes)

Now we'll clarify 'rank' a bit more. We can fit more than 8 chips on a
stick so lets mount 16 on it, 8 chips on each side. We need some way to
select which 'side' of 8 chips is being addressed, because we still only
have 5 address lines, so we design a select signal. It need be just 1 bit
so we can select one or the other side of chips, 0 or 1, which allows us to
use both 'sides' of the memory stick we've made. Note that adding an
address line would not do us any good since the chips are still only 32
bits deep and the added address line would have nothing to address. We need
a 'third dimension', so to speak, to select which side of 8 chips we want
to use. And, in fact, 'side' used to be how it was described. I.E. a
"single sided" or "double sided" stick. But, as we'll see below, using
"side" no longer works so a new name was devised and that name is "rank".
Rank is the chips that fill up one 64 bit wide memory bus with whatever
depth they have. Ours are 32 bits deep and our new memory stick has two
ranks (what we began by calling two sides) for 2 times 256 bytes, 512 bytes
total.

This is our 512 memory stick. 2 ranks with 256 per rank and, as it turns
out, that's 8 chips on one side of the stick and 8 on the other. 2 ranks, 2
sides. It means the same thing here.

SIDE 0 SIDE 1
RANK 0 RANK 1
select select
| |
V V
address # 31 -- ======== ========
....
address # 3 -- ======== ========
address # 2 -- ======== ========
address # 1 -- ======== ========
address # 0 -- ======== ========
01234567 01234567
|||||||| ||||||||
------------------- the rank selected feeds through
^^^^^^^^
||||||||
vvvvvvvv
01234567 memory width 8 bytes (64 bits)

Now comes along 'high density' memory chips, but what makes them 'high
density? Well, they change the internal organization of the chip to be 64x4
(instead of the standard 32x8). I.E. 64 bits deep but only 4 bits wide.
They have the same total number of bits (32x8 = 64x4) but it now takes 16
of them to fill up that 64 bit wide memory bus (64/4= 16). So more of them
can be packed into one rank, exactly twice as many, and since they're the
same total capacity as the "standard" chips that means more memory in the
one rank, which is how they come up with the not-so-clear term 'high
density'. It's still 16 chips, just like our 2 rank stick, with a memory
capacity of 512, just like our two rank stick. Everything sounds pretty
much the same...

**BUT NOTICE*** the depth is now more than the number of address lines we
have. Remember our binary table at the very beginning. 64 bits deep takes
*6* address lines and we only have 5! And that means we can only address
HALF the memory in that 'high density' module even though the total number
of bytes is exactly the same as our two rank stick. They're both a "512
memory stick" but, the problem is, the whole 512 is all 'high density'
packed into ONE rank.

Here's our 512 'high density' memory stick

Both sides used There is no second rank
RANK 0 RANK 1
select select
| |
V ---
address # 63 -- ================
....
address # 35 -- ================
address # 34 -- ================
address # 33 -- ================
address # 31 -- ================ ---The maximum we can address
....
address # 3 -- ================
address # 2 -- ================
address # 1 -- ================
address # 0 -- ================ ---Two 4 bit chips per byte
0 1 2 3 4 5 6 7
| | | | | | | |
---------------------------------------------
^^^^^^^^
||||||||
vvvvvvvv
01234567 memory width 8 bytes (64 bits)

Also note that it still takes 16 chips with 8 on each side. It's "double
sided," physically, but there's only one rank. And this is why "double
sided" no longer works for the purpose. It used to be equivalent to two
ranks but now, which is it? One rank or two? It depends on what chips are
used and it's rank that tells the correct story, or deducing it by the chip
organization. Remember, standard chips are x8 and 'High Density' are x4.

Also look at our rank select line. With standard density chips it only has
to drive 8 chips but with 'high density' it has to drive twice as many, 16,
and electronic signals tend to slow down as the number of driven chips
increases so, if the select line circuitry is marginal (or intended to
drive 8 x8 chips), doubling the memory chips on it can cause erratic
operation if it doesn't quite 'get there' fast enough.

I do appreciate your comments, and I am sure that others less dense
than myself will understand them fully. My only purpose in posting my
comments were to alert the OP that not all memory will work on a given
MB. Thanks again for your comments.

You're quite right. I was just trying to help in determining which kinds of
memory are potentially problematic and why. In general it's safer to stay
away from 'high density' even if it fits within the rank capacity, because
of the drive problem, unless the motherboard specifically states it
supports them (in which case we hope they've increased the drive capability
to accommodate them).


The advice to seek a size and brand that has been proven to work is
good advice on this MB. I might add that I have two sticks of PC
2700 256MB installed and it works flawlessly. One is a Kingston VR
and the other a Nanya. Your results may vary.