Ken said:
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.
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http://www.pcpowercooling.com/maxpc/index_cases.htm
http://www.firingsquad.com/guides/power_supply/page2.asp
http://www.ocworkbench.com/2002/ecs/k7s5aguide/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.
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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/socket478/p4c800-e_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/datasheets/modules/ddr/DDA16C32_64_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/misc/DDR_Power_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 (
[email protected])
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 (
[email protected])
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 (
[email protected])
1GB DIMM at idle:
100% idle(VDD x IDD2F) = 1.00*2.6V*0.880A = 2.29W (
[email protected])
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
