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Low Voltage Is Key To Energy-Efficient Chip

Posted by kdawson on Tue Feb 05, 2008 06:15 PM
from the breaking-the-barrier dept.
An anonymous reader writes in with news from the International Solid State Circuits Conference in San Francisco of a new energy-efficient chip designed by researchers at MIT. It's said to be able to run on 1/10 the power of current chips. Texas Instruments worked with MIT on the design, which is maybe five years from production. "The key to the chip's improved energy efficiency lies in making it work at a reduced voltage level, according to... a member of the chip design project team. Most of the mobile processors today operate at about 1 volt. The requirement for MIT's new design, however, drops to 0.3 volts."
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  • by WiglyWorm (1139035) on Tuesday February 05 2008, @06:18PM (#22314494)
    But how well does it overclock?
      • Re:All well and good (Score:4, Informative)

        by WiglyWorm (1139035) on Tuesday February 05 2008, @06:44PM (#22314810)
        I just finished reading the article, and it's actually got some exciting stuff. Having the processor scale its voltage when it's idle is a great idea. Current processors will change their FSB multiplier when idle so that they run at lower clocks and consume less energy, but a computer chip that could call on less voltage in a desktop machine, as well as lowering its number of clock cycles would be a huge energy saver. Though I do find the summary misleading. This processor will not run on 0.3v unless it is idle. Once you put a load on it, you have to increase the voltage.
        • Re:All well and good (Score:4, Informative)

          by GigaplexNZ (1233886) on Tuesday February 05 2008, @06:58PM (#22314978)
          Most current desktop chips do scale their voltage (such as the Core 2 Duo). The drop isn't all that dramatic, it drops from approximately 1.3V to 1.0V. But it does drop.
        • I believe my bios is currently capable of scaling the processor voltage based on demand.

          Your current multigigahertz processor relies on dynamic logic. Dynamic logic does not work at subthreshold (roughly below 1V). This chip almost certainly uses static logic and will not be as fast as a modern CPU no matter what the voltage. It is probably designed to be tied to low speed sensors where the chip never needs to run faster than the sensor can produce data, which may mean an upper limit of 1MHz (and the i

          • Re:All well and good (Score:4, Informative)

            by Chris Burke (6130) on Tuesday February 05 2008, @07:31PM (#22315406) Homepage
            Your current multigigahertz processor relies on dynamic logic. Dynamic logic does not work at subthreshold (roughly below 1V). This chip almost certainly uses static logic and will not be as fast as a modern CPU no matter what the voltage.

            Gigahertz speeds are not impossible for static logic, in fact most modern processors are in their vast majority (and perhaps entirety, though I couldn't prove it) static logic, and perform quite a bit of logic in a single clock using static circuits. 45nm transistors are really fast, they don't necessarily need the tricks (and design complexity, and manufacturing risk) of dynamic logic to get to high speeds. Maybe the double-clocked ALUs in the Intel P4 series used it for example, but otherwise static logic rules the day.

            Certainly you're right that it's unlikely that this chip would clock that high regardless of voltage. Static logic likes super-threshold voltages too. :P
            • Everything 'likes' super-threshold, the question is will they work. I admit I am an analog person and my digital design classes were a long time ago (in internet years). Sorry if my information is out of date.

              I also think these small super low power chips are far and away more interesting, and more important to our future lifestyles, than speed demon behemoths.

              • Re: (Score:3, Interesting)

                Everything 'likes' super-threshold, the question is will they work.

                True enough, there's certainly a different degree of "like" between dynamic and static in that respect.

                I admit I am an analog person and my digital design classes were a long time ago (in internet years). Sorry if my information is out of date.

                Well a long time ago in Internet years might put that right around the time of the Alpha? It was one chip that I know made heavy use of dynamic logic in order to reach such high frequencies before ot
        • Re: (Score:3, Insightful)

          The amazing thing is that they were able to get the transistors to bias at that voltage.
          That was my first reflex thought (due to what I learned when), but I suspect that we're talking about Field-Effect Transistors where an electrostatic field affects the resistance of a unipolar channel and not Bipolar Junction Transistors where you need twice that much electrical pressure to get the base-emitter junction to conduct.
  • by EmbeddedJanitor (597831) on Tuesday February 05 2008, @06:22PM (#22314536)
    Less transistors switching per unit of work done means better power performance.

    That's why your cell phone has an ARM CPU and not an x86.

    • From what I gathered in the article, thats just what they did. Namely change the architecture of the memory cells so they could lower the voltage without loosing volatile memory to noise. I think it said they moved from 6 transistors per cell to 8 though I'm not clear on how the arrangement changed. Simple J/K latch or S/R latch changed how exactly? I guess it has been too long since digital logic.
    • by johnhennessy (94737) on Tuesday February 05 2008, @06:53PM (#22314896)
      Less transistors switching is only part of the story.

      Maybe a more signficant factor in determining the power consumption of a CPU is the technology process choice.

      Intel typically tune their process for performance, at the expense of leakage. This lets them squeeze out a couple of GHz in terms of clock speed, but it means that the power consumed when the chip is doing nothing at all (i.e. idling) is much larger. The CPUs that are put into cell phones (from companies like ST, TI, Broadcom, etc, etc) are normally fabbed with a "low power" or LP option. This reduces the maximum speed that you can get out of the processor, but reduces the leakage problem significantly. If the cell phone is only using the processor 1% of the time (think of how long it spends powered on in your pocket), then there is no point in having the best 3D games on your phone, if the stand-by time is 15 minutes.

      Switching between these standard (or GP) processes and LP processes is not quiet straight forward, as you need to design all your mixed-signal / analog blocks (think PLLs, bandgaps, regulators, etc) for both nodes. While I'm sure Intel could probably afford to do this, they would then have to turn around and support this process in their fabs, which would eat up their resources for their processor market.

      If you compare the numbers: Intel can sell their processors for hundreds of dollars. Phone manufacturers buy processors from the other Semicos at about 10-15 dollars each. Guess where the better margin is ...
  • aparently from the Bureau of Slowly and Painfully Working Out The Obivous.
    • aparently from the Bureau of Slowly and Painfully Working Out The Obivous.
      Either that or John Madden [thinkexist.com] is writing headlines for Slashdot. Can he really top this gem?

      "Hey, the offensive linemen are the biggest guys on the field, they're bigger than everybody else, and that's what makes them the biggest guys on the field." - John Madden
      And, as it turns out, yes you can. The key to being energy efficient is using less energy!
      • And, as it turns out, yes you can. The key to being energy efficient is using less energy!

        So if I leave Chicago heading eastbound at 9 PM and drive 5 miles at 7000 rpm in my convertible... and you leave NY at 10:30 PM and drive westound for 50 miles at 2000 rpms in your Canyonero... who has used less energy? But importantly, who was more energy efficient*?

        And most importantly, and what time does train A jump the tracks and decapitate me for bringing a car analogy to a football analogy fight?

        *[hint] it

    • Power = Current * Voltage

      To reduce power consumption, you either have to reduce the voltage or the current.

      If you shuffle your feet across the carpet, you'll generate static electricity at thousands of volts. The reason that this doesn't kill you is that the currents are absolutely tiny, making the power transmitted between your socks and the carpet also extremely small, and non-hazardous.

      These guys are claiming that we can most effectively reduce power consumption by focusing on reducing the voltage requi
      • by Pulzar (81031) on Tuesday February 05 2008, @09:09PM (#22316296)
        Power = Current * Voltage
        To reduce power consumption, you either have to reduce the voltage or the current.


        While your formula is right, it's not too applicable for chip power usage because current is not a constant. The formula you will normally see is

        P = P-switching + P-leakage

        Now, P-switching = fCV^2, so you can reduce it by reducing the clock frequency, voltage, or the number of transistors. But, P-leakage actually increases exponentially as the gate threshold voltage is reduced -- so, reducing the voltage too much will not help, either. There's only so far you can go before leakage power becomes the dominant one and reducing voltage further doesn't help.
  • I don't get it. As far as I know, transistor Vbe is still around 0.7V. How do they build circuits when the supply voltage is less than that? I mean, how can you fit in resistors and stuff when you have no room to drop anything?
    • Re: (Score:3, Informative)

      You don't use resistors in CMOS logic. You take a transistor and wire source to gate. This turns it into a constant load, more or less the equivalent of a resistor of 10-100K ohms.

      The activation voltage of a transistor is variable- it's a property of the materials its made of. .7 is a common one and thus used in a lot of texts, but it isn't set in stone.
      • by austexmonkey (1108037) on Tuesday February 05 2008, @11:23PM (#22317216)

        Dear God, how did this get modded Informative? The parent is confusing CMOS logic with NMOS logic (you do NOT use static loads with CMOS logic), and FETs do not have a parameter called "activation voltage".

        For a description of CMOS logic that's actually accurate, check out the wikipedia article here:

        http://en.wikipedia.org/wiki/Cmos [wikipedia.org]
      • Re: (Score:3, Interesting)

        The activation voltage of a transistor is variable- it's a property of the materials its made of. .7 is a common one and thus used in a lot of texts, but it isn't set in stone.

        It's also a property of the doping levels of the silicon. Basically, you need to bring a certain amount of charge under the channel to turn the device on. This depends on the substrate material, but also depends on how much charge is available (i.e. doping).

        In a given process, you can different flavors of transistors, each with it

    • Well you only need to exceed Vbe (and the concept of Vbe only exists) if you have bipolar switching transistors. They're using IGFETs of some kind. I'm guessing that the way they do this is by making the channel and the gate insulation really thin, so you only need a tiny electrical field to switch it. I bet the noise immunity and rejection of external electrical and/or magnetic fields is really poor.
    • by jhines (82154) <john@jhines.org> on Tuesday February 05 2008, @06:35PM (#22314704) Homepage
      In Germanium the voltage is 0.3, if I remember correctly. So it depends on the materials used.
    • by Durinia (72612) on Tuesday February 05 2008, @06:37PM (#22314726) Homepage
      In this case, they're operating the transistors in a sub-threshold voltage environment. A full channel never opens for the transistor, but energy will trickle through at different rates.

      Instead of the typical "open/closed water pipe valve" model of the transistor, imagine having a leaky bucket, and then determining 1 vs 0 on how many drops get through.

      It's a tough area to design circuits in because of the very delicate balance. It doesn't take many electrons (or much process variation) to bust up your circuit.
      • Re: (Score:3, Interesting)

        > who the hell still uses BJT's?!?!?!?!?

        Pretty much everyone who uses them for fun :) You can get 2N3904s for 3c each, so it doesn't bother me if I accidentally let the smoke out of one. FETs are much more expensive, are easy to fry if you aren't extra careful to ground before touching, and are present in far fewer circuits you can find online. Then there's the fact that my old Horowitz and Hill only has one chapter on them and so I am just not as familiar with their properties. Eventually, when I'm a "G
        • FETs are much more expensive

          You need to buy them in bulk. For example, Intel will sell you about 500 million FETs for only $200.

  • Will this reduction in voltage and increase in energy efficiency reduce the amount of heat generated by the chip? It would be nice to have a powerful laptop that I could actually use in my lap (without fear of roasting my dangly bits).
    • Power is equal to the voltage multiplied by the current, so if the current stays the same and the voltage drops to 1/3, well, so does the power.

      (Yes, I'm well aware that's only ohmic power, so shoot me.)
  • Great!

    I'm waiting for several years now for a system that is completely silent, uses very low power and does not heat my room. And can be used and accessed all the time. And of course, one that does not make the performance penalties that VIA makes in their current EPIA offerengs (otherwise I would be there).

    Fortunately this seem to be going to happen in the very near future. Chipsets and CPU's are partially powering down where ever possible, and with a flash SSD's there is no spin-up or (loud) rattling whe
    • Re: (Score:3, Informative)

      With the latest hardware and fully integrated chipsets, you can already build an incredibly power efficient system for as low as 20watts idle, and yes, it will perform better than the VIA platforms. Here's one example. [silentpcreview.com]
  • Wow! (Score:4, Funny)

    by StaticEngine (135635) on Tuesday February 05 2008, @06:50PM (#22314860) Homepage
    If they can just get this thing down to zero volts, this chip will run forever!
  • Process Counts (Score:3, Interesting)

    by Colourspace (563895) on Tuesday February 05 2008, @06:59PM (#22314986)
    It's very simplistic to say that with voltage drops comes power efficiency - process geometry and materials play a part here too (and I'm not even going to mention the issues with noise tolerance and problems with SSO - Simultaneous Switching Outputs at the 0.3v level). So called 'current' (90nm) geoms are a nightmare for power leakage due to the the relatively small atom thickness that goes to make the gate of the switching transistors. You need to look at such tricks as gate oxides and other power mitigating technologies... BTW - When I say 90nm is current, I know people are doing 65nm, 45nm, 32nm and beyond (which are, given process geometry/power efficiency/newer techniques slightly better in some ways) but the lower geoms are slightly ahead of the curve somewhat..
    • Re: (Score:3, Insightful)

      The core voltage and the I/O voltage (which is where SSO is a concern) need not be the same, and rarely are in advanced processes. I'm sure the I/O's are not 0.3V. The rest of your comment was similarly confusing: using gate oxides aren't a "trick" (they're pretty much a requirement,) 65nm and under are more than "slightly" better then 90nm "in some ways," and I don't know what curve you're talking about.
  • There have been 150-200mV microcontrollers (pdf) at the University of Michigan for some time now: http://wimserc.org/research_highlights/Submiminal_Processor_Research_Highlight.pdf [wimserc.org] Conference paper 3: http://vlsida.eecs.umich.edu/resource.php?grp=1 [umich.edu] what is new is TI and MIT are involved in a commercial low voltage product. But thats still 5 years out. MIT is good at getting press.
  • Power consumption (Score:5, Informative)

    by AdamHaun (43173) on Tuesday February 05 2008, @07:30PM (#22315400)
    Power consumption in a digital circuit can be approximated by the formula:

    Pavg = N*f*C*Vdd^2 + Pleak

    where N is the probability of a gate switching during one clock cycle, f is the clock frequency, C is the average gate capacitance, Vdd is the supply voltage, and Pleak is the power loss due to current leakage. Since power is proportional to the square of the voltage but directly proportional to everything else, reducing the voltage has a much greater impact on total power consumption. Going from 1V to 0.3V implies a >10x dynamic power reduction.

  • The more voltage the more power, old computers and logic used 5V, 0.3V will be very hard to use because noise in the computer and in chips may reach 0.15 Volts, the absolute minimum resolution for the circuitry to distinguish between high and low voltage at 0.3 volts.
  • by CTho9305 (264265) on Tuesday February 05 2008, @08:05PM (#22315808) Homepage
    TFA isn't very techincal, and makes it sound like the MIT team isn't doing anything very interesting (they mention 8-transistor SRAM cells, but even regular CPUs sometimes have to use them). The interesting story here is that the chip is being operated at a voltage below the voltage where the transistors are normally viewed as being "on". In this region, transistors operate more like amplifiers than digital switches.

    One cool thing about this is that the leakage power will be negligible. Leakage currents are generally exponential with respect to voltage.

    Another cool thing is that the chip can actually operate at the low voltage. It's not too hard to make a chip retain state at very low voltages, but as soon as you want to do anything you usually have to raise the voltage back up before execution resumes. Any task that requires a small amount of work frequently will benefit from something like this. A contrived example of where this make a big difference is in a poorly-architected MP3 player in which the CPU has to shuffle a few thousand bytes per second to a sound chip, but in very small chunks (this poorly-architected sound chip has a very tiny buffer), hundreds of times per second. A normal chip would be constantly jumping to a high voltage and going back to sleep; depending on how long the voltage transition takes, it might have to stay in a higher voltage state constantly. This chip, on the other hand, could operate continuously at the "sleeping" voltage.

    The catch is that transistors operating in the subthreshold regime are going to be pretty slow, so for any tasks that require high performance you'll have to bump the voltage back to a more normal range.
    • "One cool thing about this is that the leakage power will be negligible. Leakage currents are generally exponential with respect to voltage." leakage is more dependent on threshold voltage than Vds. running a chip subthreshold means you are relying on leakage to charge up capacitance. we've had this research going on for years at MIT.
  • (if you don't know, Silverthorne is intel's next-gen low-power chip for ultra-mobile applications)
    The article states it goes down to 0.3V at idle - so it doesn't actually _run_ at that voltage (just preserve register contents). Compare this to Silverthorne which has a C6 Deep Power Down State - coincidentally at 0.3V... The article also states that this cpu uses 8-bit sram cells instead of the usual 6-bit sram cells - Silverthorne also uses 8-bit sram cells for its caches.
    Granted maybe this design works a
    • I'd say it took quite a few to figure out how to make it work at 0.3V.
    • The electrical characteristics of a CPU are somewhat more complicated than those of a resistor.
      • Re:Physics (Score:5, Insightful)

        by Chris Burke (6130) on Tuesday February 05 2008, @07:19PM (#22315260) Homepage

        The electrical characteristics of a CPU are somewhat more complicated than those of a resistor.
        True, but in fact a chip's power does scale with the square of the voltage. At a gross level you can approximate the chip as a certain constant resistance for static power, aka leakage, and as an RC circuit with a given constant for dynamic power, which scales linearly with frequency as well. Nobody actually does that, they just measure the power consumption and know that they the number is proportional to voltage squared and frequency.

        Of course I just knew some jackass was going to use this fact to try to downplay the achievement. Okay, yeah, every computer engineer knows that to reduce power by four you drop the voltage by half, but the trick is actually making this work. That's why not every chip runs on 1E-20 Volts, Mr. Anonymous Idiot.
        • That's why not every chip runs on 1E-20 Volts, Mr. Anonymous Idiot.

          At subthreshold, power draw from leakage current begins to become more important than transient switching power and the V^2 factor no longer dominates. Then further dropping the voltage increases the energy used to accomplish tasks.

          • Re: (Score:3, Informative)

            self correction/clarification: in subthreshold leakage current beings to become more important, eventually you stop gaining from dropping the voltage. That can be well into subthreshold, I've seen chips which run at 0.2V (a 45nm process has a threshold on the order of 0.5V). I didn't mean to imply that any drop into subthreshold was self defeating.
          • Oh I thought the leakage power component still could be approximated as a resistive circuit, but like I know anything about sub-threshold circuits. What's the scaling factor?
            • Its complicated .. in short, power is always equal to voltage times current. In deep subthreshold, the transistors don't turn off very well and so there is more leakage current. This relative current goes up exponentially as the voltage drops, where as the voltage is dropping linearly, so the energy lost to leakage does not drop as fast. The active power (power used to switch transistors) does continue to drop, but the gate delay increases. At some point, dependent on the process, the two curves cross a
    • Then again, if you could have gone to MIT, you'd know that the equation is actually P=V^2/R and therefore we're looking at (1/0.3)^2 which is approximately a factor of 10.