GA-8VM800PMD-775-RH 1gb + 512mb = 1gb ????

GA-8VM800PMD-775-RH

I'm trying to mix a 1gb stick, with a 512mb stick. But I only get a total
of 1gb, instead of 1.5gb, as expected.

Can't find anything about "all sticks must be the same size" on the Gigabyte
site.

http://www.gigabyte.us/Products/Moth...ProductID=2333

Same problem with BIOS version F5... and BIOS F6.
(I also tried swapping the stick positions.)

Any idea on how to get 1.5gb total?

I really can't afford to keep buying more and more memory... and throwing
existing memory in the garbage.
I *ONLY* want/need 1.5gb total.

Do I need to change any BIOS settings? Move any jumper pins?

(I'm using Windows-XP with all the latest updates.)

Thanks.

"A_Michigan_User" wrote:
GA-8VM800PMD-775-RH

I'm trying to mix a 1gb stick, with a 512mb stick. But I only get a total
of 1gb, instead of 1.5gb, as expected.

Can't find anything about "all sticks must be the same size" on the Gigabyte
site.

http://www.gigabyte.us/Products/Moth...ProductID=2333

Same problem with BIOS version F5... and BIOS F6.
(I also tried swapping the stick positions.)

Any idea on how to get 1.5gb total?

I really can't afford to keep buying more and more memory... and throwing
existing memory in the garbage.
I *ONLY* want/need 1.5gb total.

Do I need to change any BIOS settings? Move any jumper pins?

(I'm using Windows-XP with all the latest updates.)

Thanks.



Hardware is documented to different degrees, and it is harder to
give advice in some cases, than in others.

I got lucky this time, because I see a note on the Gigabyte specification page.
It says:

"Note2: Use of a Double-Sided memory module (the single chip is 512 Mbit)
is required if you wish to install a 1 GB memory module."

That sounds like they are recommending the following configuration.

(16) 64Mx8 chips (the chip density is 512 megabits each)

It is also possible for a DIMM maker, to use 1024 megabit chips. Gigabyte
is warning that this configuration would only be half-recognized.

( 8) 128Mx8 chips (the chip density is 1024 megabits each)

The addresses sent to chips, are multiplexed. There is a Row address and
a Column address, and they access the 128 meg address space. Row*Column must
have enough bits, to access 128 million locations. Gigabyte is hinting, that
there aren't enough Row*Column bits, to access into 128 meg addresses, but
there are enough for 64 meg addresses. If the hardware can only access 64
out of 128 million locations, the RAM will be half-detected.
512MB + (1/2 * 1024MB) = 1024MB total

My guess is, your 1GB module is single sided ? It should be double
sided, with 16 chips. I'm basing my answer, on that single sentence
on the Gigabyte site. (I do not expect to find documentation anywhere
from VIA, the chipset maker, on this matter. Intel, on the other hand,
has better documentation for their chipsets, which you can download.)

If you get your memory here (or buy this part number from some
retailer), you should get the correct formulation. In general, as
a consumer, you don't normally get a say as to how many chips
are used - only a few makers provide detailed info on their products.

http://www.crucial.com/store/listpar...M800PMD-775-RH

A single 1GB stick is $13 here. I don't know what shipping would be.
At this price, you could buy two.

http://www.newegg.com/Product/Produc...k=CT12864AA667

If I look on the Kingston site, this is as close as I can get to your model number.

http://www.ec.kingston.com/ecom/conf...sp?SysID=29258

"This system is limited to double-sided 16 chip DIMMs using 512Mbit (64Mx8) DRAM.
Kingston may ship single-sided 8 chip version DIMMs which are not compatible
with this board."

Note that they warn that they may accidentally ship the wrong product,
if you order a 1GB stick, since their supplier could provide an 8 chip or
a 16 chip module. Therefore, they chose not to sell a 1GB module in
this case! That is why that size is missing from their listing.
Kingston knows the board can take a 1GB stick, but their product
formulations are not precise enough in this case, to avoid a potential problem.

So that means, in this case, I'd "gamble" on the Crucial product, because
they seem to have their act together.

Even if you shop in person (i.e. went to the Best Buy and had a look at
their modules), products have enough packaging material around them,
you can't always tell what chip configuration is present. So no matter
where you buy this from, or how you do it, there will be some degree of
risk. Not all VIA chipsets are like this - I have a VIA chipset on my
motherboard, and it takes 1GB or 2GB sticks. So if I happened to pop in
an 8 chip, 1GB module, it would work fully.

HTH,
Paul

Thanks!

The 1gb stick I already have is this one:

http://www.newegg.com/Product/Produc...82E16820144151

So... do I buy another 1 of those... (and throw out BOTH of my 512mb
sticks).

Or do I buy your suggestion (and throw out BOTH of my 512mb sticks... and my
current 1gb stick too!)

Ugh.



"Paul" wrote in message
...
"A_Michigan_User" wrote:
GA-8VM800PMD-775-RH

I'm trying to mix a 1gb stick, with a 512mb stick. But I only get a
total of 1gb, instead of 1.5gb, as expected.

Can't find anything about "all sticks must be the same size" on the
Gigabyte site.

http://www.gigabyte.us/Products/Moth...ProductID=2333

Same problem with BIOS version F5... and BIOS F6.
(I also tried swapping the stick positions.)

Any idea on how to get 1.5gb total?

I really can't afford to keep buying more and more memory... and throwing
existing memory in the garbage.
I *ONLY* want/need 1.5gb total.

Do I need to change any BIOS settings? Move any jumper pins?

(I'm using Windows-XP with all the latest updates.)

Thanks.



Hardware is documented to different degrees, and it is harder to
give advice in some cases, than in others.

I got lucky this time, because I see a note on the Gigabyte specification
page.
It says:

"Note2: Use of a Double-Sided memory module (the single chip is 512
Mbit)
is required if you wish to install a 1 GB memory module."

That sounds like they are recommending the following configuration.

(16) 64Mx8 chips (the chip density is 512 megabits each)

It is also possible for a DIMM maker, to use 1024 megabit chips. Gigabyte
is warning that this configuration would only be half-recognized.

( 8) 128Mx8 chips (the chip density is 1024 megabits
each)

The addresses sent to chips, are multiplexed. There is a Row address and
a Column address, and they access the 128 meg address space. Row*Column
must
have enough bits, to access 128 million locations. Gigabyte is hinting,
that
there aren't enough Row*Column bits, to access into 128 meg addresses, but
there are enough for 64 meg addresses. If the hardware can only access 64
out of 128 million locations, the RAM will be half-detected.
512MB + (1/2 * 1024MB) = 1024MB total

My guess is, your 1GB module is single sided ? It should be double
sided, with 16 chips. I'm basing my answer, on that single sentence
on the Gigabyte site. (I do not expect to find documentation anywhere
from VIA, the chipset maker, on this matter. Intel, on the other hand,
has better documentation for their chipsets, which you can download.)

If you get your memory here (or buy this part number from some
retailer), you should get the correct formulation. In general, as
a consumer, you don't normally get a say as to how many chips
are used - only a few makers provide detailed info on their products.

http://www.crucial.com/store/listpar...M800PMD-775-RH

A single 1GB stick is $13 here. I don't know what shipping would be.
At this price, you could buy two.

http://www.newegg.com/Product/Produc...k=CT12864AA667

If I look on the Kingston site, this is as close as I can get to your
model number.

http://www.ec.kingston.com/ecom/conf...sp?SysID=29258

"This system is limited to double-sided 16 chip DIMMs using 512Mbit
(64Mx8) DRAM.
Kingston may ship single-sided 8 chip version DIMMs which are not
compatible
with this board."

Note that they warn that they may accidentally ship the wrong product,
if you order a 1GB stick, since their supplier could provide an 8 chip or
a 16 chip module. Therefore, they chose not to sell a 1GB module in
this case! That is why that size is missing from their listing.
Kingston knows the board can take a 1GB stick, but their product
formulations are not precise enough in this case, to avoid a potential
problem.

So that means, in this case, I'd "gamble" on the Crucial product, because
they seem to have their act together.

Even if you shop in person (i.e. went to the Best Buy and had a look at
their modules), products have enough packaging material around them,
you can't always tell what chip configuration is present. So no matter
where you buy this from, or how you do it, there will be some degree of
risk. Not all VIA chipsets are like this - I have a VIA chipset on my
motherboard, and it takes 1GB or 2GB sticks. So if I happened to pop in
an 8 chip, 1GB module, it would work fully.

HTH,
Paul

I'm also finding that very few stores tell you the "actual number of chips".
(Why is that info missing? That's apparently extremely important.)

I apparently need "16 chips".

If a 1gb stick says "128m x 64" can I be certain that it will have 16 chips?


""A_Michigan_User"" wrote in message
...
Thanks!

The 1gb stick I already have is this one:

http://www.newegg.com/Product/Produc...82E16820144151

So... do I buy another 1 of those... (and throw out BOTH of my 512mb
sticks).

Or do I buy your suggestion (and throw out BOTH of my 512mb sticks... and
my current 1gb stick too!)

Ugh.



"Paul" wrote in message
...
"A_Michigan_User" wrote:
GA-8VM800PMD-775-RH

I'm trying to mix a 1gb stick, with a 512mb stick. But I only get a
total of 1gb, instead of 1.5gb, as expected.

Can't find anything about "all sticks must be the same size" on the
Gigabyte site.

http://www.gigabyte.us/Products/Moth...ProductID=2333

Same problem with BIOS version F5... and BIOS F6.
(I also tried swapping the stick positions.)

Any idea on how to get 1.5gb total?

I really can't afford to keep buying more and more memory... and
throwing existing memory in the garbage.
I *ONLY* want/need 1.5gb total.

Do I need to change any BIOS settings? Move any jumper pins?

(I'm using Windows-XP with all the latest updates.)

Thanks.



Hardware is documented to different degrees, and it is harder to
give advice in some cases, than in others.

I got lucky this time, because I see a note on the Gigabyte specification
page.
It says:

"Note2: Use of a Double-Sided memory module (the single chip is 512
Mbit)
is required if you wish to install a 1 GB memory module."

That sounds like they are recommending the following configuration.

(16) 64Mx8 chips (the chip density is 512 megabits
each)

It is also possible for a DIMM maker, to use 1024 megabit chips. Gigabyte
is warning that this configuration would only be half-recognized.

( 8) 128Mx8 chips (the chip density is 1024 megabits
each)

The addresses sent to chips, are multiplexed. There is a Row address and
a Column address, and they access the 128 meg address space. Row*Column
must
have enough bits, to access 128 million locations. Gigabyte is hinting,
that
there aren't enough Row*Column bits, to access into 128 meg addresses,
but
there are enough for 64 meg addresses. If the hardware can only access 64
out of 128 million locations, the RAM will be half-detected.
512MB + (1/2 * 1024MB) = 1024MB total

My guess is, your 1GB module is single sided ? It should be double
sided, with 16 chips. I'm basing my answer, on that single sentence
on the Gigabyte site. (I do not expect to find documentation anywhere
from VIA, the chipset maker, on this matter. Intel, on the other hand,
has better documentation for their chipsets, which you can download.)

If you get your memory here (or buy this part number from some
retailer), you should get the correct formulation. In general, as
a consumer, you don't normally get a say as to how many chips
are used - only a few makers provide detailed info on their products.

http://www.crucial.com/store/listpar...M800PMD-775-RH

A single 1GB stick is $13 here. I don't know what shipping would be.
At this price, you could buy two.

http://www.newegg.com/Product/Produc...k=CT12864AA667

If I look on the Kingston site, this is as close as I can get to your
model number.

http://www.ec.kingston.com/ecom/conf...sp?SysID=29258

"This system is limited to double-sided 16 chip DIMMs using 512Mbit
(64Mx8) DRAM.
Kingston may ship single-sided 8 chip version DIMMs which are not
compatible
with this board."

Note that they warn that they may accidentally ship the wrong product,
if you order a 1GB stick, since their supplier could provide an 8 chip or
a 16 chip module. Therefore, they chose not to sell a 1GB module in
this case! That is why that size is missing from their listing.
Kingston knows the board can take a 1GB stick, but their product
formulations are not precise enough in this case, to avoid a potential
problem.

So that means, in this case, I'd "gamble" on the Crucial product, because
they seem to have their act together.

Even if you shop in person (i.e. went to the Best Buy and had a look at
their modules), products have enough packaging material around them,
you can't always tell what chip configuration is present. So no matter
where you buy this from, or how you do it, there will be some degree of
risk. Not all VIA chipsets are like this - I have a VIA chipset on my
motherboard, and it takes 1GB or 2GB sticks. So if I happened to pop in
an 8 chip, 1GB module, it would work fully.

HTH,
Paul

"A_Michigan_User" wrote:
I'm also finding that very few stores tell you the "actual number of chips".
(Why is that info missing? That's apparently extremely important.)

I apparently need "16 chips".

If a 1gb stick says "128m x 64" can I be certain that it will have 16 chips?


That is the problem. The 128m x 64 is actually meaningless.
It is a "window dressing" that does not help customers.
It is intended to make the advert look technical.

*******

When I looked at your Kingston item on Newegg, then looked
at the picture, I see the back of the module has no chips.
So that one is an 8 chip module. An 8 chip 1GB module,
on that motherboard, will only be half-detected.

(No chips on back of module)
http://images17.newegg.com/is/image/newegg/20-144-151-S02?$S640W$

Buy the Crucial one instead. You can buy it straight from
Crucial, as they're more likely to help you if there is a
problem with the product. The Newegg picture, did not remove
the module from the packaging, so we can't look at the back
of it. But if Crucial says it works, then it works.

Paul

Paul,

Maybe it would help if I understood the math instead.

128m x 64... in order to get 1gb... wouldn't it *HAVE TO BE* 16 chips?
No?

I thought 128m x 64 was the memory available in each chip.
No?

I've already bought my "wrong" memory from NewEgg... so (if they allow me to
exchange it)...
I'd like to buy the "correct" stuff also from NewEgg.

I really hate to shop by "what the picture might look like"... but I guess
I'll have to in this case.

I was *SO* careful to try and buy the right stuff:
1gb (all on 1 stick, not 2)
DDR2
DIMM
Avoid laptop memory
533/400
Unbuffered
Not dual-channel
Not ECC
240-pin
PC2-4200
Big name manufacturer
Big name store
Low price
Free shipping
I even made sure it was 1.8 volt

Results? The wrong memory anyway.

Ugh.






"Paul" wrote in message
...
"A_Michigan_User" wrote:
I'm also finding that very few stores tell you the "actual number of
chips".
(Why is that info missing? That's apparently extremely important.)

I apparently need "16 chips".

If a 1gb stick says "128m x 64" can I be certain that it will have 16
chips?


That is the problem. The 128m x 64 is actually meaningless.
It is a "window dressing" that does not help customers.
It is intended to make the advert look technical.

*******

When I looked at your Kingston item on Newegg, then looked
at the picture, I see the back of the module has no chips.
So that one is an 8 chip module. An 8 chip 1GB module,
on that motherboard, will only be half-detected.

(No chips on back of module)
http://images17.newegg.com/is/image/newegg/20-144-151-S02?$S640W$

Buy the Crucial one instead. You can buy it straight from
Crucial, as they're more likely to help you if there is a
problem with the product. The Newegg picture, did not remove
the module from the packaging, so we can't look at the back
of it. But if Crucial says it works, then it works.

Paul

"A_Michigan_User" wrote:
Paul,

Maybe it would help if I understood the math instead.

128m x 64... in order to get 1gb... wouldn't it *HAVE TO BE* 16 chips?
No?

I thought 128m x 64 was the memory available in each chip.
No?

I've already bought my "wrong" memory from NewEgg... so (if they allow me to
exchange it)...
I'd like to buy the "correct" stuff also from NewEgg.

I really hate to shop by "what the picture might look like"... but I guess
I'll have to in this case.

I was *SO* careful to try and buy the right stuff:
1gb (all on 1 stick, not 2)
DDR2
DIMM
Avoid laptop memory
533/400
Unbuffered
Not dual-channel
Not ECC
240-pin
PC2-4200
Big name manufacturer
Big name store
Low price
Free shipping
I even made sure it was 1.8 volt

Results? The wrong memory anyway.

Ugh.


The thing is, *you* were careful, but the manufacturers were not. They're the
ones getting the "epic fail" here.

I guess I didn't explain the size very well. The 128m x 64 is a "total array"
size for the entire DIMM. It means the module is 128 million locations deep
by 64 bits wide. 64 bits is equal to 8 bytes. 128M * 8 bytes = 1GB. In effect
it says the module has a total of 1GB of storage space.

Several generations of memory modules have been 64 bits wide. There isn't a
choice in the matter. They're all going to be 64 bits wide. *Every* advert
will have 64 for the second number.

Since 1GB = 128m x 64bits, we can work out that number without ever
looking at the module. Say, for example, you had a 2GB module. Without
seeing any chips, I can tell you the advert will say "256m x 64". You
haven't learned a damn thing, except how to divide 2GB by 64. So the
"256m x 64" is as meaningful as saying "2GB". In effect, the advertisement
has stated the module capacity *twice* in the same ad, by saying both
"256m x 64" and "2GB".

The numbers I'm interested in, are "internal" numbers.

(16) 64Mx8 chips (the chip density is 512 megabits each)
( 8) 128Mx8 chips (the chip density is 1024 megabits each)

In the case of the first module, they're arranged like this. This is a two rank
or "2R" module. I think that is one of the terms JEDEC might use for it. Only
one rank is "active" or drives the bus, at a time. There is a chip select
for each rank, and that is how the motherboard gets their attention. Eight
chips would be on one side of the module, and eight more on the back.

+--------+ +--------+ +--------+ +--------+ +--------+ +--------+ +--------+ +--------+
| 64mx8 | | 64mx8 | | 64mx8 | | 64mx8 | | 64mx8 | | 64mx8 | | 64mx8 | | 64mx8 |
+--------+ +--------+ +--------+ +--------+ +--------+ +--------+ +--------+ +--------+
| | | | | | | |
/ 8 / 8 / 8 / 8 / 8 / 8 / 8 / 8
| | | | | | | |
+----------+--- data-----bus-------on----------edge------connector------------
| | | | | | | |
/ 8 / 8 / 8 / 8 / 8 / 8 / 8 / 8
| | | | | | | |
+--------+ +--------+ +--------+ +--------+ +--------+ +--------+ +--------+ +--------+
| 64mx8 | | 64mx8 | | 64mx8 | | 64mx8 | | 64mx8 | | 64mx8 | | 64mx8 | | 64mx8 |
+--------+ +--------+ +--------+ +--------+ +--------+ +--------+ +--------+ +--------+

The other module design (the one Kingston said they might mix up accidentally, if they
had chosen to sell you a module), looks like this. This module is only half detected
on your motherboard.

+--------+ +--------+ +--------+ +--------+ +--------+ +--------+ +--------+ +--------+
| 128mx8 | | 128mx8 | | 128mx8 | | 128mx8 | | 128mx8 | | 128mx8 | | 128mx8 | | 128mx8 |
+--------+ +--------+ +--------+ +--------+ +--------+ +--------+ +--------+ +--------+
| | | | | | | |
/ 8 / 8 / 8 / 8 / 8 / 8 / 8 / 8
| | | | | | | |
+----------+--- data-----bus-------on----------edge------connector------------

Both modules have an "external dimension" of 128m x 64, but that number doesn't tell
you how they're arranged internally, or the size of each chip. There are other
ways I can make modules if I want. Here is another module. (Chips are available
in 4, 8, 16, and even some 32 bit wide ones. Plenty of choices.)

+--------+ +--------+ +--------+ +--------+
| 64mx16 | | 64mx16 | | 64mx16 | | 64mx16 |
+--------+ +--------+ +--------+ +--------+
| | | |
/ 16 / 16 / 16 / 16
| | | |
+----------+--- data-----bus-----+
| | | |
/ 16 / 16 / 16 / 16
| | | |
+--------+ +--------+ +--------+ +--------+
| 64mx16 | | 64mx16 | | 64mx16 | | 64mx16 |
+--------+ +--------+ +--------+ +--------+

Again, the advert for that module says "128m x 64". It doesn't even hint that 16 bit
wide chips are being used, or any other internal detail you might want to know.

So there is plenty missing from the description.

The industry has had plenty of time to fix this, but they choose not to.
This is even a problem for people with ten year old 440BX motherboards,
buying PC133 modules. There are two ways to make 256MB SDRAM modules,
and one is "half-detected". So the problem has existed for ten years,
and customers are still getting the wrong memory.

Paul

Paul,

Thanks again.

Looks like I should just switch to two sticks of 1gb each. (2gb total)

Crucial says this is what I need: CT2KIT12864AA667

Crucial has it for $38:
http://www.crucial.com/store/partspe...2KIT12864AA667


NewEgg has it for $25:
http://www.newegg.com/Product/Produc...82E16820146526


So I guess I'll get the NewEgg.

(But neither company says "number of chips".... so I'm again "guessing at
all this".)



"Paul" wrote in message
...
"A_Michigan_User" wrote:
Paul,

Maybe it would help if I understood the math instead.

128m x 64... in order to get 1gb... wouldn't it *HAVE TO BE* 16 chips?
No?

I thought 128m x 64 was the memory available in each chip.
No?

I've already bought my "wrong" memory from NewEgg... so (if they allow me
to exchange it)...
I'd like to buy the "correct" stuff also from NewEgg.

I really hate to shop by "what the picture might look like"... but I
guess I'll have to in this case.

I was *SO* careful to try and buy the right stuff:
1gb (all on 1 stick, not 2)
DDR2
DIMM
Avoid laptop memory
533/400
Unbuffered
Not dual-channel
Not ECC
240-pin
PC2-4200
Big name manufacturer
Big name store
Low price
Free shipping
I even made sure it was 1.8 volt

Results? The wrong memory anyway.

Ugh.


The thing is, *you* were careful, but the manufacturers were not. They're
the
ones getting the "epic fail" here.

I guess I didn't explain the size very well. The 128m x 64 is a "total
array"
size for the entire DIMM. It means the module is 128 million locations
deep
by 64 bits wide. 64 bits is equal to 8 bytes. 128M * 8 bytes = 1GB. In
effect
it says the module has a total of 1GB of storage space.

Several generations of memory modules have been 64 bits wide. There isn't
a
choice in the matter. They're all going to be 64 bits wide. *Every* advert
will have 64 for the second number.

Since 1GB = 128m x 64bits, we can work out that number without ever
looking at the module. Say, for example, you had a 2GB module. Without
seeing any chips, I can tell you the advert will say "256m x 64". You
haven't learned a damn thing, except how to divide 2GB by 64. So the
"256m x 64" is as meaningful as saying "2GB". In effect, the advertisement
has stated the module capacity *twice* in the same ad, by saying both
"256m x 64" and "2GB".

The numbers I'm interested in, are "internal" numbers.

(16) 64Mx8 chips (the chip density is 512 megabits each)
( 8) 128Mx8 chips (the chip density is 1024 megabits
each)

In the case of the first module, they're arranged like this. This is a two
rank
or "2R" module. I think that is one of the terms JEDEC might use for it.
Only
one rank is "active" or drives the bus, at a time. There is a chip select
for each rank, and that is how the motherboard gets their attention. Eight
chips would be on one side of the module, and eight more on the back.

+--------+ +--------+ +--------+ +--------+ +--------+ +--------+
+--------+ +--------+
| 64mx8 | | 64mx8 | | 64mx8 | | 64mx8 | | 64mx8 | | 64mx8 | | 64mx8
| | 64mx8 |
+--------+ +--------+ +--------+ +--------+ +--------+ +--------+
+--------+ +--------+
| | | | | | |
|
/ 8 / 8 / 8 / 8 / 8 / 8 / 8
/ 8
| | | | | | |
|
+----------+---
data-----bus-------on----------edge------connector------------
| | | | | | |
|
/ 8 / 8 / 8 / 8 / 8 / 8 / 8
/ 8
| | | | | | |
|
+--------+ +--------+ +--------+ +--------+ +--------+ +--------+
+--------+ +--------+
| 64mx8 | | 64mx8 | | 64mx8 | | 64mx8 | | 64mx8 | | 64mx8 | | 64mx8
| | 64mx8 |
+--------+ +--------+ +--------+ +--------+ +--------+ +--------+
+--------+ +--------+

The other module design (the one Kingston said they might mix up
accidentally, if they
had chosen to sell you a module), looks like this. This module is only
half detected
on your motherboard.

+--------+ +--------+ +--------+ +--------+ +--------+ +--------+
+--------+ +--------+
| 128mx8 | | 128mx8 | | 128mx8 | | 128mx8 | | 128mx8 | | 128mx8 | | 128mx8
| | 128mx8 |
+--------+ +--------+ +--------+ +--------+ +--------+ +--------+
+--------+ +--------+
| | | | | | |
|
/ 8 / 8 / 8 / 8 / 8 / 8 / 8
/ 8
| | | | | | |
|
+----------+---
data-----bus-------on----------edge------connector------------

Both modules have an "external dimension" of 128m x 64, but that number
doesn't tell
you how they're arranged internally, or the size of each chip. There are
other
ways I can make modules if I want. Here is another module. (Chips are
available
in 4, 8, 16, and even some 32 bit wide ones. Plenty of choices.)

+--------+ +--------+ +--------+ +--------+
| 64mx16 | | 64mx16 | | 64mx16 | | 64mx16 |
+--------+ +--------+ +--------+ +--------+
| | | |
/ 16 / 16 / 16 / 16
| | | |
+----------+--- data-----bus-----+
| | | |
/ 16 / 16 / 16 / 16
| | | |
+--------+ +--------+ +--------+ +--------+
| 64mx16 | | 64mx16 | | 64mx16 | | 64mx16 |
+--------+ +--------+ +--------+ +--------+

Again, the advert for that module says "128m x 64". It doesn't even hint
that 16 bit
wide chips are being used, or any other internal detail you might want to
know.

So there is plenty missing from the description.

The industry has had plenty of time to fix this, but they choose not to.
This is even a problem for people with ten year old 440BX motherboards,
buying PC133 modules. There are two ways to make 256MB SDRAM modules,
and one is "half-detected". So the problem has existed for ten years,
and customers are still getting the wrong memory.

Paul

"A_Michigan_User" wrote:
Paul,

Thanks again.

Looks like I should just switch to two sticks of 1gb each. (2gb total)

Crucial says this is what I need: CT2KIT12864AA667

Crucial has it for $38:
http://www.crucial.com/store/partspe...2KIT12864AA667


NewEgg has it for $25:
http://www.newegg.com/Product/Produc...82E16820146526


So I guess I'll get the NewEgg.

(But neither company says "number of chips".... so I'm again "guessing at
all this".)


As long as either company treats you right, if there is a problem
with the product, then go with the cheaper one. Make sure you
know what the Newegg return policy is - how many days you have to
test it and so on.

After installing the RAM, your first boot should be a copy of memtest86+.
Run two passes, to prove the memory isn't total crap.

http://www.memtest.org/

Keep the old memory around, just in case. If you ever have a problem,
and need to verify the computer still works, you can plug in that
512MB for a test. You don't have to immediately sell it on Ebay (if
you could - the demand for it might not be there).

Paul

Wow, this just gets stranger and stranger.

Now even with the "correct" memory (as suggested by crucial.com)... my 2x1gb
sticks
are only seen as a total of 1gb.... instead of 2gb.

(My original memory worked fine. 2x512mb sticks.)

Memory: CT2KIT12864AA667 (matched set of 2 sticks @ 1gb each)
Motherboard: Gigabyte GA-8VM800PMD-775-RH
BIOS: F6 (the very latest)
OS: Windows XP Pro (with all the latest patches)

What else could be causing this odd problem?

Thanks.

"Paul" wrote in message
...
"A_Michigan_User" wrote:
Paul,

Thanks again.

Looks like I should just switch to two sticks of 1gb each. (2gb total)

Crucial says this is what I need: CT2KIT12864AA667

Crucial has it for $38:
http://www.crucial.com/store/partspe...2KIT12864AA667


NewEgg has it for $25:
http://www.newegg.com/Product/Produc...82E16820146526


So I guess I'll get the NewEgg.

(But neither company says "number of chips".... so I'm again "guessing at
all this".)


As long as either company treats you right, if there is a problem
with the product, then go with the cheaper one. Make sure you
know what the Newegg return policy is - how many days you have to
test it and so on.

After installing the RAM, your first boot should be a copy of memtest86+.
Run two passes, to prove the memory isn't total crap.

http://www.memtest.org/

Keep the old memory around, just in case. If you ever have a problem,
and need to verify the computer still works, you can plug in that
512MB for a test. You don't have to immediately sell it on Ebay (if
you could - the demand for it might not be there).

Paul