POST A COMMENT

24 Comments

Back to Article

  • Lonyo - Saturday, September 22, 2012 - link

    All the low power stuff coming "next year" keeps making me think Windows 8 is almost a year too early (while also being a year too late).

    I'm getting tempted to buy something relatively inexpensive this year (the lowest non-Atom tablet), and then save up for something better next year when everything lower power comes out.
    Reply
  • Penti - Saturday, September 22, 2012 - link

    Well Windows RT gets released before Office 2013 RT so good luck there with early devices. It's not delivering this year. Even techies haven't quite realized how limiting RT edition is, as people don't even believe you if you say that Outlook won't be available on Windows RT, or that it will have worse exchange support then stock Android, iOS, Blackberry etc. So they don't see why you would stress 8 Pro devices and full featured apps. They don't quite realize that there already is deployed MDM (management) tools out there for iOS, Android etc and that they integrate fine in an Windows Server/Microsoft Server-products environment and handles sharepoint, exchange, office-formats, Lync etc in most cases better then the mobile Windows products do even after the introduction of Windows 8, Windows RT and Windows Phone 8. Microsoft have to lift up other strengths than stuff people makes up themselves. Hardware isn't key here either. You can do portable stuff just fine.

    Their competitor Apple will also run Haswell etc next year. It's not like they will match up better just because new stuff is out. Here typically lower power hardware just means smaller batteries. Nothing radical here yet either software or hardware sadly.
    Reply
  • Impulses - Sunday, September 23, 2012 - link

    I've been having the same feeling, between new memory tech, Haswell/A15, new WiDi updates and Thunderbolt, next year seems like a much better time for some of these early Win 8 hybrid/tablet devices. I guess it's like most things the first gen is always sort of a rough draft (even the first IPhone was a very rough approximation of what a smartphone is today, people forget there was no app store etc). Reply
  • Impulses - Sunday, September 23, 2012 - link

    I'm really not sold on Win RT, specially with Clove Trail driving it... Just feels like a compromise on all fronts. Rather just upgrade my Android tablet with a cheap Nexus design than go that route. The higher end convertible/hybrid x86 designs intrigue me, but I don't think the first iteration of any of them is really gonna hit it out of the park. Reply
  • eanazag - Wednesday, September 26, 2012 - link

    Windows RT runs on ARM based hardware, so no Clover Trail or ATOM anything. Think Tegra 3. Reply
  • twotwotwo - Sunday, September 23, 2012 - link

    I mostly buy this theory for Pro (currently Ivy Bridge) tablets. In weight and battery life, they're more like touchscreen-only Ultrabooks than typical Android, etc. tabs. They're prolly going to be priced up there with some good ultraportables. But in a year, there'll be ones that are about as fast as today's, but significantly better at the whole mobile thing thanks to lower-power Intel ULV chips, etc. And you'd be plunking down a lot for a Surface Pro. So, all in all, there's more than usual reason to wait.

    (The Atom tabs may be 'mobile enough' already, but...$800, and you have Atom netbooks on one side and ARM tablets on the other. The Atom stuff looks really neat and unique in some ways, but that is some tough math to stomach.)
    Reply
  • eezip - Saturday, September 22, 2012 - link

    The article says "Power scales linearly with voltage", but that's untrue. P = V^2 / R, so power scales exponentially with voltage. Therefore, a small decrease in voltage has a much larger decrease in power consumption. Reply
  • eezip - Saturday, September 22, 2012 - link

    I could be more clear. Power does scale linearly when considering P = I * V, but it's unclear to me if the load impedance (and thus current consumption) of the memory changes with a change in voltage. I suspect current is relatively flat, and impacted more by the temperature increase from running at a higher voltage (greater leakage at higher temp) than anything else.

    http://www.powersystemsdesign.com/performance-vs-p... for example, states a 20% reduction in power when going from 1.35V to 1.5V. 1.35V is 90% of 1.5V, then showing that a 10% reduction in voltage leads to a 20% reduction in power.
    Reply
  • eezip - Saturday, September 22, 2012 - link

    And http://www.micron.com/~/media/Documents/Products/T... on page 13, corroborates my original post. By saying that reducing voltage reduces current by the same amount, that means the load resistance of the memory is virtually fixed. Therefore, P = V^2 / R is the correct equation and power does not scale linearly with voltage.

    I'm finding out there's no way to edit a post. Boo.
    Reply
  • Zink - Saturday, September 22, 2012 - link

    Wow, good job finding documentation on that right from Micron. It's well understood by enthusiasts that CPU power is related exponentially to voltage but I wasn't 100% sure about how memory behaves.

    1.5V to 1.35V then is a 19% reduction in power draw.
    Reply
  • CyberAngel - Sunday, September 23, 2012 - link

    quadratically Reply
  • JarredWalton - Sunday, September 23, 2012 - link

    I was going off P = V * I, but didn't consider the reduction in current as part of the package. Generally speaking, reducing voltage also reduces current (I believe), so yes it's quadradically reduced power in most cases. However, if current remains constant, power scales linearly with voltage. I have updated the text to mention current just to keep everyone happy. :-) Reply
  • Penti - Sunday, September 23, 2012 - link

    I = V / R R = V / I V = I*R P = V^2 / R

    If a chip uses 5 W at 1.5V it draws 3.33A. If it follows Ohm's law it would be 4.05W at 1.35V. I.e. 19% reduction. That's the same number memory manufacturers use. If we don't have any laws of physics it would be just at 4.5W. I.e it's not linearly. Double to voltage quadruple the power. Current is a function of voltage and resistance. Resistance is virtually constant, but dependable on temperature. Of course if you like to design a chip drawing 200W at 1.15V it will draw very high currents at up to 174A from the power source so it's not like we lower voltage to lower the current draw. A 1.75V 200W chip would be at 114A. That would be a lot easier to power. It stems from manufacturing, materials, leakage etc. If we can lower it below the normal threshold you do save power of course. You do save even more power by constructing it in another way so you utilize less space (silicon, also why you process shrink stuff), uses other properties like reducing leakage etc. Like transistor types. That could be from materials or physical differences etc. If you shrink stuff then of course you won't be able to feed it with the same voltage as you lower the threshold voltage of the transistors due to oxide thickness. Thus it depend largely on design. There is no good reason to feed 30nm DRAM at 1.5V. It's not because it's low power however. One of the first DDR3 DRAM's was produced at 80 and 90 nm. About 90 nm was the target back then. If you like to save power you also have to have other power saving features like being able to shut off part of the logic. LPDDR2 can use up to about 70% less power then a DDR3-module alone. LPDDR2 still is powered by a 1.2V operating voltage. Most of the reduction comes from other stuff then the lower voltage there.
    Reply
  • MrSpadge - Sunday, September 23, 2012 - link

    That's quadratically, not exponential. But thanks for catching the "linear" error in the article and your further posts, saves me some time :) Reply
  • jjj - Saturday, September 22, 2012 - link

    "The result is that while you can get the best power characteristics out of LPDDR, the volume is much lower as it’s generally not used in high volume markets like laptops and PCs with 8GB or more RAM. "

    Right so most decent smartphones right now are using LPDDR2.
    In Q2 some 150-155 mil smartphone units shipped ,25 mil tablets and 87mil traditional PCs.
    About a week ago IHS said that in Q2 PC RAM was 49% of the market while smartphones and tablets were 14.1%. and they expect Q4 2013 to be 42.8 % for PC and 26.7% for mobile devices.
    So yeah not really used in " high volume markets".
    Reply
  • SunLord - Sunday, September 23, 2012 - link

    Well if you consider that just about all those smartphones and tablets usually only have 1 chip maybe 2 for some of the new high end SoC compared to 8 chips per so-dimm in laptops and 8-16 chips per dimm in desktop those pcs have a lot more ram chips compared to the smart phones.

    So with global smartphone and tablet sales were just under 300Million last year so that probably 310M LPDDR2 chips compared to global desktop/laptop sales of around 370M which works out to about 4.4+Billion DDR3 chips which is a far more massive amount and its not even taking into consideration servers
    Reply
  • Beenthere - Saturday, September 22, 2012 - link

    There won't be a rush to DDR4 because it's primarily designed for servers though it may be useful in some portable devices also. The anticipated change in topology to point-to-point for DDR4 means you install all the memory when you build and you replace all the memory if you want to upgrade to higher speed/density DRAM.

    Since current typical desktops are not bottlenecked with DDR3 running at ~1333 MHz, there really is no need for the added bandwidth of DDR4 for years to come. With DDR3 being converted to low voltage similar to DDR4 and higher frequency - again similar to DDR4, the updated DDR3 DRAM is an advantage and cheaper than DDR4 will be.

    http://en.wikipedia.org/wiki/DDR4_SDRAM
    Reply
  • UpSpin - Sunday, September 23, 2012 - link

    If I read the Wiki article correctly it makes no sense to use DDR3 at all, because DDR4 is much better suited for smartphones and tablets:
    - lower power consumption from the start on
    - higher bandwidth, which is important, if you consider that both CPU and GPU use the same memory and at the moment, memory bandwith is the biggest limiting factor in smartphones (see iPhone 5 boost thanks to improved memory bandwidth)
    - The to point-to-point topology is the current way SoCs connect to DIMM, or am I wrong? So no change here.
    - Support for 3D stacking, the only way to increase memory in smartphones/tablets is stacking. So only DDR4 allows to increase the amount of memory, without increasing required space and power draw (long routes if you don't stack it)

    So DDR4 will consume less power (see above chart), be cheaper and allows higher densities (3D stacking), and will remove the current bottleneck (low bandwidth). And because manufactures had to redesign their SoC completely for ARM A15 and new GPU and DDR4 is available already, I hope they just skipped DDR3 completely.
    Ultrabooks will also benefit, because they also integrate both CPU and GPU in a single die and thus share the same memory, which is a limiting factor for IGPs right now.
    Reply
  • Penti - Sunday, September 23, 2012 - link

    Mobile DDR already uses 3D-stacking. 3D stacking has also been used on Deskop and Server DDR3. None 3D stacking is used in any DRAM before the 3D-technology for that matter. DRAM usually comes in DDP and QDP-packages i.e. several dies. Reply
  • ssj3gohan - Sunday, September 23, 2012 - link

    Regular DDR3, 1.5V stuff, according to Micron runs at 7mA per package in self-refresh mode. For my 5.9W ivy bridge desktop computer (http://ssj3gohan.tweakblogs.net/blog/8217/fluffy2-... with 16GB RAM (32 packages in two SODIMMs) that works out to be, theoretically, 0.336W. i actually went and measured this, and it turned out to be roughly 0.4W so that is pretty much spot on by Micron. The 'extra' power use is probably down to interface termination current and frankly measurement error.

    With DDR3L-RS, it's running at 1.2V and the current requirement is roughly 2 times lower. That means that for the same configuration, we're looking at 0.135W. Theoretically about 0.2W lower, in practice maybe even more because of the termination and interface drivers also requiring less current.
    Reply
  • macuser2134 - Sunday, September 23, 2012 - link

    You are saying DDR4 will be half the power? 0.4W -> 0.2W ? Hmm it doesn't seem like much of an active power draw when running in a typical notebook to begin with. Let's say we are conservatively drawing 10W total on a brand new Haswell ultrabook. And that includes all the partf of the computer. That saving = 2% of the overall system savings. So on a regular notebook the extension to battery life would be even less. No wonder many people just upgrade with the cheapest 1.5v RAM modules in their MacBook Pros if that is the case.

    Of course, having said that about the active power draw, it is a very different case for the standby/suspend power mode savings where we can imagine a huge improvement.

    Another question: (its not quite mentioned in the article): Do Intel discuss their new procesor architectures with these memory makers? (Micron, Corsair, etc). I mean, ahead of time so that the memory manufacturers can also develop their side of the technology early enough to be ready for the new Intel Memory Controllers? It would be kindda silly if they manufactured millions of brand new Haswell processors, and there was no memory to go with them. So don't they have earlier discussions that we are not privvy to?
    Reply
  • MrSpadge - Sunday, September 23, 2012 - link

    > You are saying DDR4 will be half the power? 0.4W -> 0.2W ?

    No, he measured the power needed for self refresh of his DDR3 DRAMs when idle or standby. And Micron says this power loss will be halved with DDR3L-RS. Which sounds plausible based on his data and some fundamental equations.
    Reply
  • ssj3gohan - Sunday, September 23, 2012 - link

    This difference in power draw is *only* under the action of self-refresh, i.e. the computer is on but completely idle (or the computer being in S3 suspend-to-RAM). Power draw in active (C1) mode is higher, depending on the OS usage of the memory (i.e. ASLR uses more power than linear memory allocation), I/O speed, etc. Here the -RS memory has no impact at all, the only thing that helps power draw under active circumstances is better OS memory management and lower voltage memory. Reply
  • raok7 - Thursday, September 05, 2013 - link

    well this processor plays a very important role in computer technologies....
    http://www.jupiterelectronics.com/
    http://www.steelrange.com/heavy-duty-racks.html
    http://www.bajeria.com/
    http://www.genesis-gifts.com/
    http://www.opportune.in
    http://www.fivebrosforgings.com/
    http://www.aimaxprovider.com/index.php/magento-web...
    Reply

Log in

Don't have an account? Sign up now