Memory power supply

From what PSU rail does this motherboard take the power
for the memory? +3.3V or +5V?
http://www.asus.com/products/mb/sock...d/overview.htm

As ram uses around 2.5-8 then I would say the 3.3 volt one.

On Sat, 12 Jun 2004 08:57:14 +0100, "Nero" wrote:

As ram uses around 2.5-8 then I would say the 3.3 volt one.

The problem is that both options are told at different sites.

The CPU today use the +12V rail and the CPU gets only 1.4 - 1.7V

If the 3.3V are used, then PSUs are not able to manage over
1GB of RAM. 1GB RAM takes 64W, that's about 20A from +3.3V
The graphic card takes a lot of power from 3.3V too.
Most PSUs are only able to put out max 30A from +3.3V
That's only 99W.
From +5V rail the PSU are able to mange 40A and it's 200W.

------------------------------------------------------
http://www.pcpowercooling.com/maxpc/index_cases.htm

http://www.firingsquad.com/guides/po...pply/page2.asp

http://www.ocworkbench.com/2002/ecs/...de/psuFAQ2.htm
From the site:
Most mobo’s are designed to take most power out of the 5.0V
rail and used it for the 3.3V and the 2.5V for the dimms.
But other mobo’s uses the 3.3V rail for the dimms and also
for the rest of the mobo circuits. A mobo who takes 3.3V
and 2.5V power out of 5.0V will be more stable and easier
to power with compare to those who uses the 3.3V rails.
------------------------------------------------------

In my new computer I start with 1GB of RAM, and it's capable
to have 4GB RAM, but the problem is the PSU capacity.
I am sitting an doing some calculation on that.
I hope this motherboard use the +5V rail for the memory (RAM).
http://www.asus.com/products/mb/sock...d/overview.htm

In article , Ken
wrote:

From what PSU rail does this motherboard take the power
for the memory? +3.3V or +5V?
http://www.asus.com/products/mb/sock...d/overview.htm

As the board only has a switching converter for Vcore, I would have
to guess it is 3.3V. To make 2.5V, without a switcher, and provide
four amps of current or so, is a stretch even for a linear regulator,
as four amps times (3.3-2.5) give 2.8 watts, and that would make
a MOSFET get pretty hot, if it is only cooled by the copper plane
it is soldered to.

I tried to trace it. Start with this doc (pg.3 bottom)

http://download.micron.com/pdf/datas...128x64ag_a.pdf

Notice how the pins are colored, and the red pins are connected
to +2.5V. Near the key for the DIMM, you can see pin 143 is red.
Looking at the PDF version of the Asus manual, the back of the DIMM
is on the left hand side of the DIMM slot, and pin 143 lines up with
an electrolytic cap. The electrolytic cap is to the left of
DIMM_A1 and located roughly in the center of the DIMM. My guess is
that cap filters +2.5V, and I used that to verify I might have
the correct pin (143).

Plugging my meter into pin 143, I probed MOSFETs on the board, looking
for a match. The lower leg of the MOSFET below the DIMM_A1 slot is
shorted to pin 143, so I suspected that MOSFET might be feeding both
the I/O voltage on the Northbridge, and the VDD and VDDQs on the
DIMMs. But, the trouble is, when I probed the other leads of that
MOSFET and touched +3.3 or +5, I didn't find a direct short. So,
I cannot be sure whether that is the MOSFET doing the regulating
for +2.5 or not. The board has at least two LM324 quad op amps, so
the board has plenty of op amps to build regulator circuits with.

For the board to make a 2.5V source at say four amps, from +5V,
means wasting 10 watts if done with a linear regulator, and without
a large heatsink, that would cook something. So, I cannot see
how +5V can be used. Since some of the P4C800/P4P800 series of
boards have a second switching regulator, those could use
+5.0V to make DIMM power, but on this board in front of me, it is
likely to be a linear powered from a lower voltage like +3.3V.
I presume that would make it hard to volt mod the DIMM voltage,
if that is what you are planning on hacking. (The Intel reference
schematic for 875, uses a switching regulator for +2.5V, so is of
no help in deducing what Asus is doing.)

HTH,
Paul

In article , Ken
wrote:

On Sat, 12 Jun 2004 08:57:14 +0100, "Nero" wrote:

As ram uses around 2.5-8 then I would say the 3.3 volt one.

The problem is that both options are told at different sites.

The CPU today use the +12V rail and the CPU gets only 1.4 - 1.7V

If the 3.3V are used, then PSUs are not able to manage over
1GB of RAM. 1GB RAM takes 64W, that's about 20A from +3.3V
The graphic card takes a lot of power from 3.3V too.
Most PSUs are only able to put out max 30A from +3.3V
That's only 99W.
From +5V rail the PSU are able to mange 40A and it's 200W.

------------------------------------------------------
http://www.pcpowercooling.com/maxpc/index_cases.htm

http://www.firingsquad.com/guides/po...pply/page2.asp

http://www.ocworkbench.com/2002/ecs/...de/psuFAQ2.htm
From the site:
Most mobo’s are designed to take most power out of the 5.0V
rail and used it for the 3.3V and the 2.5V for the dimms.
But other mobo’s uses the 3.3V rail for the dimms and also
for the rest of the mobo circuits. A mobo who takes 3.3V
and 2.5V power out of 5.0V will be more stable and easier
to power with compare to those who uses the 3.3V rails.
------------------------------------------------------

In my new computer I start with 1GB of RAM, and it's capable
to have 4GB RAM, but the problem is the PSU capacity.
I am sitting an doing some calculation on that.
I hope this motherboard use the +5V rail for the memory (RAM).
http://www.asus.com/products/mb/sock...d/overview.htm


Your second post makes it clear what you are after.

Some resources for DRAM power calculation:

Table of IDD values (pg. 16 IDD values for 512MB DIMM)
http://download.micron.com/pdf/datas...4_128x64AG.pdf

Micron application note on doing power calcs.
http://download.micron.com/pdf/technotes/TN4603.pdf

Micron spread sheet for automating the calculation.
http://download.micron.com/downloads...er_Calc_10.xls

Rather than bother with trying to figure out the spreadsheet,
try page 3 of this document. They like the following estimate:
http://www.flomotion.com/technical_papers/t322.pdf

63% read(VDD x IDD4R + IO(w/50% data toggle)) and
27% write(VDD x IDD4W) and 10% idle(VDD x IDD2F).

The I/O power calculationm in the Micron TN4603 pg. 7 is a device
power dissipation calculation. I.e Micron is working out how warm
the memory chip gets, whereas we are interested in the total power
consumption, from the power supply. I'm going to assume the 16.8ma
flows from +2.5 power supply, whether making a logic one or a logic
zero, as the termination power supply is probably coming from the
same +2.5V master source. This makes the power for the I/O
2.6V*0.0168A * 72 signals (64 data, 8 DQS signals) during reads.

63% read(VDD x IDD4R) = 0.63*2.6V*1.632A = 2.67W
63% read(I/O Power) = 0.63*(2.6V*0.0168A*72) = 3.14W
27% write(VDD x IDD4W) = 0.27*2.6V*1.512A = 1.06W
10% idle(VDD x IDD2F) = 0.10*2.6V*0.960A = 0.25W
Total estimated power = 7.12W )

For an idle DIMM, a refinement would be to take idle power
plus a refresh component, but I'll just take the idle power
for now. Refresh happens once in a blue moon, and occupies a
low percentage of the total time.

100% idle(VDD x IDD2F) = 1.00*2.6V*0.960A = 2.5W )

On any given memory channel, only one DIMM can be drawing the
active power number, while the other DIMMs remain idle. As a
result, adding DIMMs to a channel doesn't make the power scale
directly. Only additional idle power is drawn.

With two DIMMs in dual channel, current is 2*2.74A = 5.48A
With four DIMMs in dual channel, current is 2*2.74+2*0.96 = 7.4A

Note that for the active power, I consider the t322 paper is
still a bit pessimistic.

Now, 7.4A is not going to scare a 20A power source :-)
It does make me wonder though, why the linear regulation
on my motherboard is cool as can be. It does suggest the
calc is still on the high side.

Note - that calc is for 4x512MB memory. Using the 1GB data
from Micron gives:

63% read(VDD x IDD4R) = 0.63*2.6V*1.56A = 2.56W
63% read(I/O Power) = 0.63*(2.6V*0.0168A*72) = 3.14W
27% write(VDD x IDD4W) = 0.27*2.6V*1.56A = 1.10W
10% idle(VDD x IDD2F) = 0.10*2.6V*0.880A = 0.23W
Total estimated power = 7.03W )

1GB DIMM at idle:

100% idle(VDD x IDD2F) = 1.00*2.6V*0.880A = 2.29W )

Yes, the 1GB is actually drawing less power than the 512MB :-)
Perhaps the 1GB uses a smaller geometry transistor.

And, as far as the calculations go, your finger is the best
judge of all. Imagine a 60W light bulb for a moment - if
you put your finger tip on it, the skin would burn off and
coat the glass on the bulb. Now, open your current computer
and stick your finger on a DIMM. You have to sum the thermal
contributions of all the chips, for a fair comparison, but
I think you'll agree that the DIMMs cannot possibly be consuming
64W, they would have a much higher surface temperature.

Paul

The 3.3V rail as if there is a problem with it (too low or too high)then the
ram starts acting up.
My old psu was putting out a lot less than the 3.3v it should(3.3v rail) and
my memory played up.
When a better psu with better 3.3v rail was used the problem went away.

On Sat, 12 Jun 2004 10:12:45 -0400, (Paul) wrote:

I think you'll agree that the DIMMs cannot possibly be consuming
64W, they would have a much higher surface temperature.

64W/GB is what most sites told me.
Thank you for helping me.
I was very concerned about the power consumption.

In article , Ken
wrote:

On Sat, 12 Jun 2004 10:12:45 -0400, (Paul) wrote:

I think you'll agree that the DIMMs cannot possibly be consuming
64W, they would have a much higher surface temperature.

64W/GB is what most sites told me.
Thank you for helping me.
I was very concerned about the power consumption.

Here is another bound on the calculation.

ftp://ftp.asus.com.tw/pub/ASUS/Bareb...4_pundit-r.pdf
ftp://ftp.asus.com.tw/pub/ASUS/Bareb...0_pundit-r.pdf

This is a barebones system. It uses the ATI RS300, and based
on the line drawing of the motherboard, has a two slot dual channel
layout.

The power supply in this product has:

+3.3V 8.0amp
+5V 4.0amp
+12V 9.5amp
-12V 0.2amp
+5VSB 1.5amp

If the DIMMs were to be operated from +3.3V, then they would
be sharing with other PCI chips on the motherboard. The 5.48A
I calculated for two active DIMMs (because the board might be
dual channel - the only picture of the motherboard in the manual
is not of the actual board, but of its predecessor) just fits
into the 8amp limit of the supply. If this board can do it with
8 amps, then 20 amps should handle four slots of memory with ease.

Paul

The primary power supply for the CPU in Intel based systems comes from
the +12 volt power supply. That is the whole, entire purpose of the
square 4-pin "ATX 12V" connector. By drawing the CPU Vcore power from
the 12 volt rail, for any given POWER requirement, the current
requirement is minimized (since power = current x voltage (12 volts, in
this case)).

What actually happens is that the 12 volts is the input to a second
switch-mode power supply that is on the motherboard itself. This power
supply takes the 12 volts (in an Intel system) and creates the Vcore
supply for the CPU, which is in the 1.5 volt range (but at tremendous
current, up to 70 amps).

I'm not sure what AMD systems are doing, or if they use the 12 volt
supply or the ATX 12V connector, since I normally don't build AMD based
systems. But I do know that they still get their Vcore from a dedicated
switching power supply on the motherboard.

In theory, it's possible to run the onboard Vcore switching power supply
from 3.3 volts or 5 volts, but the power levels required make this a
poor choice. If you were to run such a supply from 3.3 volts instead of
12 volts, you would need approximately 4 times more current, which most
[main] PC Power supplies could not provide.


Ken wrote:

On Sat, 12 Jun 2004 08:57:14 +0100, "Nero" wrote:


As ram uses around 2.5-8 then I would say the 3.3 volt one.


The problem is that both options are told at different sites.

The CPU today use the +12V rail and the CPU gets only 1.4 - 1.7V

If the 3.3V are used, then PSUs are not able to manage over
1GB of RAM. 1GB RAM takes 64W, that's about 20A from +3.3V
The graphic card takes a lot of power from 3.3V too.
Most PSUs are only able to put out max 30A from +3.3V
That's only 99W.
From +5V rail the PSU are able to mange 40A and it's 200W.

------------------------------------------------------
http://www.pcpowercooling.com/maxpc/index_cases.htm

http://www.firingsquad.com/guides/po...pply/page2.asp

http://www.ocworkbench.com/2002/ecs/...de/psuFAQ2.htm
From the site:
Most mobo’s are designed to take most power out of the 5.0V
rail and used it for the 3.3V and the 2.5V for the dimms.
But other mobo’s uses the 3.3V rail for the dimms and also
for the rest of the mobo circuits. A mobo who takes 3.3V
and 2.5V power out of 5.0V will be more stable and easier
to power with compare to those who uses the 3.3V rails.
------------------------------------------------------

In my new computer I start with 1GB of RAM, and it's capable
to have 4GB RAM, but the problem is the PSU capacity.
I am sitting an doing some calculation on that.
I hope this motherboard use the +5V rail for the memory (RAM).
http://www.asus.com/products/mb/sock...d/overview.htm