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Super Talent's DDR3 module
Memory manufacturer Super Talent partners up with Samsung for some of the first DDR3

Super Talent sends us word that it has completed construction of its first DDR3 prototype modules and will be the first to market for commercial modules. As a member of JEDEC, Super Talent is devoting engineers and time to defining and finalizing the DDR3 specifications.

DDR3 is set to be the next PC memory standard and the successor to DDR2 memory. It holds a number of technological advantages over its predecessor. Power consumption has been decreased to 1.5 Volts, while clock speeds have gone up. Currently, DDR3 frequencies range from 800 MHz to 1600 MHz.

The 512MB module that appeared on Super Talent's website late last night uses a Samsung SKU that has not been publically announced yet.

Physically, DDR3 memory is very similar to DDR2 memory. Both technologies use Fine-pitch Ball Grid Array for their chip package. In addition, DDR3 modules will also use the same 240-pin form factor. They will not, however, be inter-compatible as DDR2 and DDR3 memory chips have the middle notch located at different positions.

Although JEDEC as of yet still needs to finalize the specifications for DDR3, which are expected to be publically available mid-2007, Super Talent has already been busy working on DDR3 prototypes. The recently developed Samsung-based DDR3 prototypes are currently being tested by Super Talent on various test platforms.   

The first motherboards supporting DDR3 are expected to begin appearing once Intel’s Bearlake-family of desktop chipsets launches. The new Intel desktop chipsets will support both DDR2 and DDR3 memory, though both can’t be used at the same time. Intel has already begun sampling its DDR3 memory motherboards, though confirmed ship dates for the boards are still flexible. AMD is expected to adopt DDR3 with the release of its quad-core architecture.  

Super Talent expects to begin shipping its DDR3 modules long before motherboards supporting them are actually released. Current ship time is estimated at mid-2007, but the date is subject to change with demand.


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.
By semo on 2/17/2007 8:31:24 AM , Rating: 2
quote:
Power consumption has been decreased to 1.5 Volts
this statement suggests that voltage is indicative of power consumption.

are these modules going to be available for purchase before mobos and chipsets are available or what.




RE: .
By A5 on 2/17/2007 9:29:37 AM , Rating: 1
V = IR
or
I = V/R

Power (very roughly) = I^2*R
or
(V/R)^2*R
which is the same as
P = V^2/R

Assuming the resistance is the same, lower voltage = less power.


RE: .
By Questar on 2/17/2007 10:14:24 AM , Rating: 3
Chips aren't resistive loads.

The OP is right, voltage is a measure of potential, not consumption.


RE: .
By masher2 (blog) on 2/17/2007 12:32:53 PM , Rating: 1
> "Chips aren't resistive loads....The OP is right, voltage is a measure of potential, not consumption..."

The OP was not correct. The simplest power model equation for a DRAM circuit is P = cfV^2, i.e. power is proportional to the square of the applied voltage, times the device frequency.


RE: .
By semo on 2/17/2007 1:01:43 PM , Rating: 3
oh common just read the statement.
quote:
Power consumption has been decreased to 1.5 Volts


power is not measured in volts and also if the eqation you quote is correct, v^2 is one of the terms and frequency too should be indicative (even more so by the looks of that equation) of power consumption. this of course can only be true for dram and not electronics in general.

i would also imagine power depends more on the memory capacity than voltage (0.5gb vs 2gb module for example). i doubt the ttl chips i used to build binary counters from were power hungry beasts only because i used 5v for the supply rail.


RE: .
By masher2 (blog) on 2/17/2007 1:24:45 PM , Rating: 1
> "power is not measured in volts ..."

True; I don't dispute that the sentence could have been phrased better. Still, voltage is the primary factor controlling power consumption.

> "and frequency too should be indicative (even more so by the looks of that equation) of power consumption..."

A squared term always predominates a linear one. Double the frequency and you double the power consumption....but double the voltage and you quadruple it.

> "i would also imagine power depends more on the memory capacity..."

Assuming same technology, its linear against memory capacity. But comparisons are done against equal capacity, so this point is moot.


RE: .
By Hare on 2/17/2007 5:48:33 PM , Rating: 3
quote:
Still, voltage is the primary factor controlling power consumption.
In your own words. It's a factor, not a power consumption unit like mW. So the OP was in fact correct. Voltage x current = power consumption. Voltage != power consumption.


RE: .
By masher2 (blog) on 2/17/2007 7:13:13 PM , Rating: 1
The OP's implication was that voltage was not "indicative" of power consumption. It is...it's the primary factor, in fact.


RE: .
By semo on 2/17/2007 8:06:34 PM , Rating: 1
2 black boxes, 2 terminals each. all you have is a dvm.
you take measurements of both and each time the reading is 12v to 2 s.f.

which one is the car battery?
next scenario...
my teacher once made an observation that a lot of my fellow students wore watches and other kinds of jewelry. he told me that back in his days, tvs had supply rails in the 10s of kvs. all that metal junk dangling would almost surely get you a nasty shock on occasions when you had to work on live electrical equipment (old crts in this case). iirc rail tracks on the london underground rail are _only_ 600v dc.

bearing in mind that 10000v can give you a "nasty" shock, would you piss on underground rail tracks?


RE: .
By RyanHirst on 2/18/2007 4:45:56 AM , Rating: 2
It doesn't seem anyone aknowledged your correct presentation of the math.
So I, too, weigh in on the side of middle school algebra.

_____________
For the die-hard skeptics, here is your 1AM quick review:
y = a*x^2
is a second order equation. Both a and x may be variable (adj), but x is the independent variable (n).
A is a scalar multiplier.
People can be as willing as they like to bet that other factors are just as, or more important to, the value of y; y is no less a function of x. Today's lesson: if your hunch is at odds with the fundamental theorem of algebra, you are wrong.
_________________
Back to your regularly scheduled programming.
Tonight's headlines:
SONY declares its farts smell like roses. Playstation3 powered by miniaturized man surrounded by a wall of slowly exhaling synthetic vulvas. Product team states idea came to them when high on mescalin.
"I talked to God. He said, 'put a vulva in the ps3'"


RE: .
By bkiserx7 on 2/19/2007 2:42:56 PM , Rating: 3
who cares about all the power talk


RE: .
By Icepick on 2/17/2007 11:00:54 AM , Rating: 3
To put it more simply, Power = Voltage * Current

or

P=VI


RE: .
By semo on 2/17/2007 11:48:21 AM , Rating: 2
please read my comment again and i will not thank you for the elementary ohm's law (and arithmetic) lesson.

i know what the assumptions and you're only supposed to use ambiguity in formal discussions with colleagues or like minded peers that you know they understand you and even then this statement sounds wrong.


RE: .
By VIAN on 2/17/2007 12:04:52 PM , Rating: 2
Power is a term that thrown around quite a bit. There's the real definition of power, P=IV, which is directly porportional to Voltage.

Of course, then if you're buying power drills, power is talked about only as Voltage. Higher Voltage coming from the battery gives the drill more power.

Then there's Vacuum cleaners and lawnmowers where power is talked about only as Amps.

It's all related, but it's all part of marketing. More people would know the voltage of a battery compared to how much current it provides. Same thing with the memory modules. People are only allowed to control the voltage. And higher voltage is synonymous with heat. Heat is synonymous with more power.


RE: .
By ddopson on 2/17/2007 5:02:16 PM , Rating: 2
Except that the battery in my smoke detector is 9V and the battery that starts my car is only 12V.

Oh, I believe someone already said this, but the simplistic formulation for the dynamic power in a chip is P=C*f*V^2, where C is capacitance, f is frequency, and V is voltage. There is also a leakage current which is more resistive in nature: P=V^2/R. Things aren't quite as simple when things like clock gating and dynamic voltage and frequency scaling come into play. However, DRAMs are still pretty simple circuits, basically big transistor + capacitor grids, and the simple formula is pretty accurate. All things being equal, dropping voltage has a huge impact on power consumption due to the second order term in the power equations.


RE: .
By JeffDM on 2/18/2007 10:32:01 AM , Rating: 2
That's better than my bicycle, that can only do three square yards to the mile. The units are related but they aren't the same and don't necessarily make sense.

Yes, units matter. The sentence was just wrong at least one way. It could have been better stated by saying:

quote:
Power consumption has been decreased through a voltage reduction.


The concept is what was intended without looking ignorant. Power and voltage are related, but the original sentence basically tried to say they were the same thing. As such, it's either a grade school error in science, a grade school error in English, or both.


RE: .
By MrTeal on 2/19/2007 9:46:53 AM , Rating: 2
Voltage is the main determinant of the chip's power consumption, but the phrase still should be reworded to something like "Power consumption has been decreased by the move to 1.5 volts"
Also, because volts are named for a person, the word is spelled with a lowercase v. The abbreviation is uppercase (1.5 V).


Upgrade bug
By Icepick on 2/17/2007 11:09:10 AM , Rating: 2
I'm going to be bit by the upgrade bug again and want to get some of these new babies into my rig. That means I'll need a new mobo too and maybe even a CPU. And the cycle continues...

Guys - Can someone tell me why I'm not able to vote on the quaility of these posts anymore? I was able to do this when Dailytech first started and lately the option is not available. What happened?




RE: Upgrade bug
By Warren21 on 2/17/2007 3:49:48 PM , Rating: 2
As soon as you post