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Low Voltage Is Key To Energy-Efficient Chip

kdawson posted more than 6 years ago | from the breaking-the-barrier dept.

Hardware 127

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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All well and good (5, Funny)

WiglyWorm (1139035) | more than 6 years ago | (#22314494)

But how well does it overclock?

Re:All well and good (1)

No2Gates (239823) | more than 6 years ago | (#22314520)


You can run one of these puppies for a week on a AA battery.

Don't fuck with the boys from MIT, they can kick (1, Funny)

Anonymous Coward | more than 6 years ago | (#22316318)



don't fuck with the boys from MIT

they can kick your ass using nothing but their brain waves from their slightly downturned head and funneled through their fingertips

laughing now, who is

Re:All well and good (2, Funny)

ccarson (562931) | more than 6 years ago | (#22314528)

Exactly! And can it run while submerged in beer?

Re:All well and good (1)

ruinevil (852677) | more than 6 years ago | (#22314670)

It's an interesting article. Even if the processing power of these radically redesigned processors is pathetic, you could maybe figure out how to distribute sufficiently threaded commands over an entire network of processors. It's not like they produce tons of heat. When I read the heading though... I just thought P=IV. It's sort of obvious...

Re:All well and good (4, Informative)

WiglyWorm (1139035) | more than 6 years ago | (#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 (4, Informative)

GigaplexNZ (1233886) | more than 6 years ago | (#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.

Re:All well and good (1)

Falstius (963333) | more than 6 years ago | (#22315002)

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 idle will be in the 10s of kHz range). That makes this roughly equivalent in clock rate to Eniac in 'low power mode', except it is running on microwatts instead of kilowatts and taking up the space of a pin head instead of a large room.

Nothing here is revolutionary, MIT is not the first group to do this. The interesting part is that they are teaming with TI to bring the design to market.

Re:All well and good (3, Informative)

Chris Burke (6130) | more than 6 years ago | (#22315406)

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

Re:All well and good (1)

Falstius (963333) | more than 6 years ago | (#22315624)

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:All well and good (2, Interesting)

Chris Burke (6130) | more than 6 years ago | (#22315800)

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 others did. It seemed to fall out of favor mostly for complexity and manufacturability reasons. And what is compared to that the minor problem that it makes silicon debug harder when you can't down-clock the chip too much because then the dynamic logic stops working. :P

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.

That's clearly where everything is headed. It is an interesting design problem for sure, but in my heart I like making chips that go fast. :)

More than exciting... electrifying! (0)

Anonymous Coward | more than 6 years ago | (#22316320)

Low power means great efficiency? I'm shocked to learn this!
[groan from crowd]

I'll be here all week, folks.

Re:More than exciting... electrifying! (1)

Mr2cents (323101) | more than 6 years ago | (#22318948)

Low power means great efficiency? I'm shocked to learn this!
Do you know what I'm shocked about? The fact that people, even on nerdish sites like this one, STILL don't know the difference between voltage and power. Man, you really made a fool of yourself!

Let me explain it once more: You can take running water as a model for electricity. In our model, the difference in height between two points correstponds to the voltage (U). The amount of water that is flowing between those two points corresponds to the current (I), and the resistance is.. well, the resistance (R) (determined by the size of the channel, its' smoothness etc).

Ohm's law states that U = I * R

The power, on the other hand, is the amount of energy per time-unit, and can be calculated as P = U x I. (Using Ohm's law we can substitute I by U/R, and we get P = U^2/R - the quadratic relation between power and voltage as stated in the article)

PS: That's why electricity companies charge you per KWh, i.e. (energy per time-unit) * time = energy consumed.

Re:More than exciting... electrifying! (1)

ShakaUVM (157947) | more than 6 years ago | (#22319326)

Hmm, in my college physics class, we always used V for Voltages.

Kinda weird, I know.

But yeah, even anyone who doesn't understand Ohm's law but tinkers with computers should know that lower voltage = less power, and higher voltage = 0vercl0ck. Or something like that.

Re:All well and good (1)

palegray.net (1195047) | more than 6 years ago | (#22318608)

I just use my servers to heat my home office. No, really, I'm not kidding; I live in Connecticut, have the office window half open half the time, and this room is still warm. Good thing power is included in my rent here...

Re:All well and good (1, Informative)

Anonymous Coward | more than 6 years ago | (#22315334)

The power is proportional to V^2/f, where V is the voltage and f is the frequency, so halving the voltage results in 1/4 of the power or in the case of 0.3^2/1^2 you get .09 or about 10% of the power usage. The amazing thing is that they were able to get the transistors to bias at that voltage.

Re:All well and good (2, Insightful)

unitron (5733) | more than 6 years ago | (#22317484)

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.

Re:All well and good (1)

aquila.solo (1231830) | more than 6 years ago | (#22318180)

But since I = V/R, you can also say P = V^2/R. Since current varies so much during a chips operation, it can be easier to have an average resistance value and the operating voltage to determine power consumption.

Re:All well and good (1)

Toast10101 (875553) | more than 6 years ago | (#22314900)

But how well does it blend?
There, fixed that for you.

Sigh, overclockers (1)

Artuir (1226648) | more than 6 years ago | (#22315482)

Why do overclockers have to ruin everything? Who gives a crap? Buy the chip and save some money from low power usage. It's obviously not engineered for overclocking so go out and rice out a Civic instead because it's the same damn thing. And yes. I just made a car analogy. There ya go.

Re:Sigh, overclockers (1)

kestasjk (933987) | more than 6 years ago | (#22315908)

Despite the interesting/insightful mod I think the GP's comment was a joke

IT'S OVER 9000! (1)

mnemonic_ (164550) | more than 6 years ago | (#22314508)

WHAT 9000

noshitposter (1, Insightful)

Anonymous Coward | more than 6 years ago | (#22314532)

I see someone tagged this "noshitsherlock". But this is a hard thing to do because the difference between "0" and "0.3" is smaller than "5" lowering the immunity to upsets like noise.

Re:noshitposter (1)

Traxxas (20074) | more than 6 years ago | (#22314550)

The tag is for the headline. Lower power from lower voltage?? No shit Sherlock!!

Re:noshitposter (1)

i_b_don (1049110) | more than 6 years ago | (#22318072)

This just in!! "Low Current key to Energy-Efficient Chip!"

Architecture is far more important (3, Insightful)

EmbeddedJanitor (597831) | more than 6 years ago | (#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.

Re:Architecture is far more important (0)

Anonymous Coward | more than 6 years ago | (#22314730)

Actually, the less potential difference there is in voltage between on state and off state for a transistor, the less time and energy it takes to switch it. Lowering transistor counts in CPUs just make them more efficient at the cost of functionality. Thats why your PC doesnt have an ARM CPU.

Unit of perception is far more important (0)

Anonymous Coward | more than 6 years ago | (#22314734)

Er no. The CISC chips are really RISC under the hood. You need to update your knowledge. TThis isn't the 8088 era.

You need to pervceive the right things... (2, Insightful)

EmbeddedJanitor (597831) | more than 6 years ago | (#22315024)

Sure some CISCs have a RISC under the hood, but that just means you need to have a "virtual machine" that emulates a CISC on top of the RISC. Those extra layers mean more internal operations which mean more switching.

Re:You need to pervceive the right things... (-1, Troll)

Anonymous Coward | more than 6 years ago | (#22316042)

clue + less = you

Re:You need to pervceive the right things... (0, Troll)

EmbeddedJanitor (597831) | more than 6 years ago | (#22316586)

ball + less = AC

No, I'm not the same guy as the AC. (1)

Xaositecte (897197) | more than 6 years ago | (#22317072)

Embedded + Janitor = Silly Nickname.

Re:Architecture is far more important (1)

Gat0r30y (957941) | more than 6 years ago | (#22314738)

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.

Re:Architecture is far more important (1)

unitron (5733) | more than 6 years ago | (#22317518)

I guess it has been too long since digital logic.

Or for some of us who suffered through it, not nearly long enough :-)

Re:Architecture is far more important (1)

marcansoft (727665) | more than 6 years ago | (#22318246)

Your typical SRAM memory cell uses a simple double inverter loop, plus transistors to switch the rows. To change the bit, you effectively short out the inverters using higher power transistors. This saves chip area but wastes power. I guess they moved to a design that doesn't depend on this "brute force" method and instead gracefully changes the state of the cell using logic.

Re:Architecture is far more important (4, Informative)

johnhennessy (94737) | more than 6 years ago | (#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 ...

Re:Architecture is far more important (0)

Anonymous Coward | more than 6 years ago | (#22318876)

Accidentally modded you redundant, wanted to mod you insightful, posting to revert my mods :D

in other news, high MPG key to better gas mileage (5, Funny)

pezpunk (205653) | more than 6 years ago | (#22314566)

aparently from the Bureau of Slowly and Painfully Working Out The Obivous.

Perhaps John Madden Is Submitting Stories? (5, Funny)

eldavojohn (898314) | more than 6 years ago | (#22314708)

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!

Re:Perhaps John Madden Is Submitting Stories? (1)

Red Flayer (890720) | more than 6 years ago | (#22316062)

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's a trick question.

Re:in other news, high MPG key to better gas milea (1)

moosesocks (264553) | more than 6 years ago | (#22316094)

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 required for the chips to run. Although you've essentially got a 50/50 chance of being correct with this claim, the reasoning behind it is far from trivial.

Re:in other news, high MPG key to better gas milea (0)

Anonymous Coward | more than 6 years ago | (#22316244)

Power = Current^2 * Resistance. Reducing the voltage is good, reducing current is better since the resistance is determined by the processes and materials that make up the device.

Re:in other news, high MPG key to better gas milea (3, Informative)

Pulzar (81031) | more than 6 years ago | (#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.

Re:in other news, high MPG key to better gas milea (1)

v1 (525388) | more than 6 years ago | (#22316762)

This article seems counter intuitive. Power lines in the US are higher voltage, lower current, than local transmission lines, to reduce power loss on their primary feeds. Higher voltage means lower current, for the same power transmitted. Isn't it current passing through a resistance what causes power loss? So lowering current (and in turn raising voltage, so the power transmitted remains the same) the proper way to reduce power loss via transmission? Or am I missing something?

Re:in other news, high MPG key to better gas milea (2, Informative)

Malekin (1079147) | more than 6 years ago | (#22318026)

You are correct about power lines. The high voltage / low current reduces power lost due to the resistance of the wires. When you're dealing with long pieces of wire, the resistance adds up. Integrated circuits, however, are very small and though they are made of semiconductors (which are generally more resistive than metals) resistive losses aren't the big concern. In a semiconductor the important things are electric fields and charges moving about. Making a transistor work at low voltage means there are smaller potential barriers involved for charges to cross.

Basically, anyway.

What is voltage? (1, Informative)

Anonymous Coward | more than 6 years ago | (#22316686)

I think a lot of people have a minor misconception of what voltage is.... that is to say they have no fucking clue what it is. The "Well DUH.... less power used = more efficient" is less accurate than it is funny when applied to voltage. It is quite possible to use the same or even more power at a lower voltage given the required amperage.

Voltage in electronics is essentially the same thing as pressure in a pipe with water being pumped through. If you have a shitty-ass leaky pipe, higher pressure causes more water to leak from the shitty-ass pipe. Now imagine you have two of these shitty-ass pipes that you want to pump an equal amount of water through. One of them has a flow restricter on the end. If you pump the same rate of water through them, the one without the restricter will have lower pressure and less water will leak from the holes in the pipe. The pressure in the other pipe, however, is higher because of the restricter. This causes increased leakage from the shitty-ass pipe and overall more water will be required to get the same amount out of the end of the pipe (and yes I realize this uses resistance to get my point across, it's impossible to describe current, voltage, or resistance without referencing another of the three so preemptive STFU). Increased pressure also increases friction, blah, blah, blah.

This basically works the same way in electronics, except instead of leaking water (at least I hope not) electronics leak heat energy. Electronics are essentially shitty-ass leaky pipes, because it's hard to build real small things (and you can quote me on that). Sure, increased voltage generally means increased overall power usage. Voltage != Power, though.

This is meant to be a simple explanation for laymen. EE's STFU, I don't want to hear about how this or that is technically wrong. It coulda been worse, I could've used cars. And no, I didn't spellcheck, reread, etc. Go to hell spelling/grammar/regular type nazi. Done.

Bad Car Analogy strikes again (2, Informative)

Spy Hunter (317220) | more than 6 years ago | (#22317086)

Contrary to popular belief, voltage is *not* power. To use the analogy properly, what this article says is closer to "low horsepower key to better gas mileage". Which, while still obvious, is at least not a tautology.

It is possible for a low voltage system to transfer more energy than a high voltage one in the same amount of time if the low voltage one transfers more current (current is measured in amps, not volts). The exact relation is volts * amps = power (in watts). So if this chip ran at lower voltage but needed more amps, it could still use more power.

How can that work? (1)

Chemisor (97276) | more than 6 years ago | (#22314600)

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:How can that work? (2, Informative)

AuMatar (183847) | more than 6 years ago | (#22314656)

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.

Re:How can that work? (4, Informative)

austexmonkey (1108037) | more than 6 years ago | (#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:How can that work? (1)

unitron (5733) | more than 6 years ago | (#22317620)

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.

--unintentional pun alert--If you count silcon as a stone then actually it is. It takes between 0.6 and 0.7 volts to get a silicon PN junction to conduct, but that's Bipolar Junction Transistors, and Field-Effect Transistors are a different situation where you don't want the junctions to conduct.

Re:How can that work? (0)

Anonymous Coward | more than 6 years ago | (#22314680)

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

Re:How can that work? (2, Interesting)

Chemisor (97276) | more than 6 years ago | (#22315028)

> 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 "God of circuit design", I'll probably use lots of FETs too, just like the big guys...

Re:How can that work? (3, Funny)

Waffle Iron (339739) | more than 6 years ago | (#22316174)

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.

Not bipolar logic (1)

_merlin (160982) | more than 6 years ago | (#22314684)

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.

Re:How can that work? (3, Informative)

jhines (82154) | more than 6 years ago | (#22314704)

In Germanium the voltage is 0.3, if I remember correctly. So it depends on the materials used.

Re:How can that work? (4, Informative)

Durinia (72612) | more than 6 years ago | (#22314726)

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:How can that work? (0)

Anonymous Coward | more than 6 years ago | (#22314930)

And besides that, you lose one of the nice things of running at higher voltage: noise rejection. When you are seeing transients because of EMI and etc which are higher than your signal, you start not being able to detect the actual signal (low signal-to-noise ratio).

But then again, I'm sure they are working around this. I deal in high voltage/current power supplies and control environments where you are sparking off big flash lamps, and running things with 18V sure helps a lot towards not getting any wackiness.

Physics (1, Interesting)

Anonymous Coward | more than 6 years ago | (#22314646)

Hmm. P=V^2/R, so dropping the voltage from 1 to 0.3 drops the power by a factor of (1/0.3)^2 ~ 10. How many MIT researchers did that take?

Re:Physics (1)

compro01 (777531) | more than 6 years ago | (#22314840)

I'd say it took quite a few to figure out how to make it work at 0.3V.

Re:Physics (1)

jibjibjib (889679) | more than 6 years ago | (#22314972)

The electrical characteristics of a CPU are somewhat more complicated than those of a resistor.

Re:Physics (4, Insightful)

Chris Burke (6130) | more than 6 years ago | (#22315260)

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.

Re:Physics (1)

Falstius (963333) | more than 6 years ago | (#22316126)

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:Physics (2, Informative)

Falstius (963333) | more than 6 years ago | (#22316220)

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.

Re:Physics (1)

Chris Burke (6130) | more than 6 years ago | (#22316228)

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?

Re:Physics (1)

Falstius (963333) | more than 6 years ago | (#22316450)

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 and you start to use more energy per operation instead of less.

For a complete run down (if you have access), the paper 'Theoretical and practical limits of dynamic voltage scaling' Zhai, B.; Blaauw, D.; Sylvester, D.; Flautner in Proceedings 2004. Design Automation Conference goes over the details.

(I am not one of the authors nor affiliated with the research group, I did go to the first author's Thesis defense).

V Cube, Not Square (1)

Mateorabi (108522) | more than 6 years ago | (#22318256)

Power tends to be proportional to fCV^2, yes. But the achievable clock frequency f is actualy a function of V. Higher voltages = lower gate delay = higher freq, as many overclockers have discovered. So power tends to scale with V^3. (Assuming you are getting the optimal performance out of your process technology. I mean you could increase V and not take advantage of the bonus in f and only use ~V^2 more power, or decrease f wihtout decreasing V to use ~V less power, but that would mean you were wasting power for equivalent performance.) I've ignored where V gets small and static power starts to dominate dynamic power.

Will reduced voltage affect heat output? (1)

AbsoluteXyro (1048620) | more than 6 years ago | (#22314782)

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).

P = U*I (1)

sd.fhasldff (833645) | more than 6 years ago | (#22314888)

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.)

Re:Will reduced voltage affect heat output? (1)

amorsen (7485) | more than 6 years ago | (#22318868)

Will this reduction in voltage and increase in energy efficiency reduce the amount of heat generated by the chip?
In this house, we obey the laws of thermodynamics!

Always on (1)

owlstead (636356) | more than 6 years ago | (#22314812)

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 when an indexing service turns on. With voltage scaling like this, there is no need for a separate low power CPU within such a system. Screens are already powering up very quickly and OLED screens are on the horizon as well.

Who needs all those problems with hibernate and suspend if your system is on all the time, while staying below a few volts?

Good, I'm switching off my main computer now, I cannot sleep with the fans and hard disk rattling away. I'll have to read the replies tomorrow after powering up for a few minutes (sleep & hibernate are broken due to some peripherals).

Re:Always on (2, Informative)

tknd (979052) | more than 6 years ago | (#22314948)

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]

Those MIT guys sure are advanced (1)

agraham (123576) | more than 6 years ago | (#22314824)

They've even managed to make 1/10=0.3

Re:Those MIT guys sure are advanced (1)

chillax137 (612431) | more than 6 years ago | (#22314884)

power scales with voltage squared. reducing the voltage by .3 reduces the power by .09, which is approximately .1 (1/10)

Re:Those MIT guys sure are advanced (1)

MLopat (848735) | more than 6 years ago | (#22314974)

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.

Duh (1)

InsaneProcessor (869563) | more than 6 years ago | (#22314830)

I guess they just figured out what the industry has known for years. Doesn't anyone notice that voltage requirements have been going down as power goes down?

Wow! (3, Funny)

StaticEngine (135635) | more than 6 years ago | (#22314860)

If they can just get this thing down to zero volts, this chip will run forever!

Re:Wow! (3, Funny)

ChrisMP1 (1130781) | more than 6 years ago | (#22315182)

0V processor [wikipedia.org]

Re:Wow! (2, Funny)

ichigo 2.0 (900288) | more than 6 years ago | (#22316688)

Nevermind that, they need to get it down to negative voltages, then all our energy problems would be solved!

Re:Wow! (1)

marcansoft (727665) | more than 6 years ago | (#22318302)

Power increases with the square of the voltage. (-x) is a positive number. You can already make CPUs that run on negative voltages: just swap all of the transistors (roughly). Or you could just take a normal CPU and rewrite the specs to put the 0 at the positive rail, call it ground, and mention that the chip uses inverted logic I/O. Voila, a negative voltage CPU.

0 Volts is easy. (1)

Kaenneth (82978) | more than 6 years ago | (#22317456)

Just make a dual core CPU, and have one core run on +5 volts, and the other on -5.

Re:Wow! (2, Interesting)

Spy Hunter (317220) | more than 6 years ago | (#22317604)

Actually, I seem to recall reading about a guy who had proven that there was no theoretical lower bound on the amount of energy it would take to do a given computation (assuming the computation was 'reversible' [wikipedia.org] ). Contrast this to an electric motor, where the desired result is mechanical power output, so obviously at least as much electrical energy must go in as mechanical energy comes out. When the desired output is merely 'computation', there may be no lower bound on the energy input required.

Process Counts (2, Interesting)

Colourspace (563895) | more than 6 years ago | (#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..

Leakage / Dynamic Scaling (1)

fpgaprogrammer (1086859) | more than 6 years ago | (#22314992)

subthreshold circuits with dynamic voltage scaling are fun and stuff, but the real issue at small geometries is leakage currents which constitute the majority of power consumption. Traditional dynamic voltage scaling doesn't help much because leakage is dominated by the threshold voltage not the power rail. multi-threshold CMOS uses enable signals to turn on and off high leakage paths. this does not help with active mode leakage when the system does not need to operate at full speed. to reduce active mode leakage we need to do dynamic threshold scaling requires body-bias modulation which is currently an expensive process in terms of IC production.

other sexy low power technologies: adiabatic logic (charge recovery), and asynchronous logic (the majority of power is consumed by the clock so let's get rid of it).

slashdot comment fetching (0)

Anonymous Coward | more than 6 years ago | (#22315030)

Can you please put a link in the sidebar that does the equivalent of showing every comment in full format? I am sick of having to click the "x more" link, wait a few seconds for it to load, click the link again and wait, click and wait, click and wait, click and wait. I like to browse through all the comments, and making me work for up to 2 minutes with ajax requests is annoying as hell.

Please, please give us a "all comments" link. If I still have to drag the bar to "Full" after doing so, that is fine. But honestly, I'm getting tired of clicking and waiting within every article just so I can read the comments.

P = V^2 / R (0)

Anonymous Coward | more than 6 years ago | (#22315246)

Funny how I learned that without spending $40K a year to go to a Cambridge school.

This has already been done before (2, Interesting)

trajan96 (901853) | more than 6 years ago | (#22315348)

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 (4, Informative)

AdamHaun (43173) | more than 6 years ago | (#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.

P = V^2/R (1)

usul294 (1163169) | more than 6 years ago | (#22315570)

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.

And in breaking news.... (1)

BuhDuh (1102769) | more than 6 years ago | (#22315672)

Researchers have proved that the secret to longevity is to continue to live without dying.

This is more interesting than TFA makes it sound (4, Informative)

CTho9305 (264265) | more than 6 years ago | (#22315808)

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.

Re:This is more interesting than TFA makes it soun (2, Informative)

fpgaprogrammer (1086859) | more than 6 years ago | (#22316316)

"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.

almost like silverthorne... (1)

mczak (575986) | more than 6 years ago | (#22316594)

(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 at even lower voltages than does silverthorne (which seems to have an operating range of 0.7V-1.0V) but if they need 5 years to get it to market it might be too late...
(I've taken the silverthorne info from http://www.heise.de/newsticker/meldung/103038 [heise.de] , in german, but it can now or soon likely be found elsewhere)

Leakage Power! (2, Interesting)

borowcm (1233996) | more than 6 years ago | (#22317182)

From what I can remember from my Low Power VLSI class...

1. Dropping Vdd to a CMOS transistors requires you to drop the threshold voltage to maintain performance.
2. As the two voltages approach each other, theres an increase in the current in the substrate (the current which flows between n-wells in a typical CMOS transistor).
3. This substrate current ends up contributing to massive amounts of leakage current.

I couldnt resist - the handy eq. from my VLSI Design for Deep submicron book says something along the lines of
Isubstrate =u0*cox*(w/l)*Vt^2 *e^((Vgs-Vth )/n*Vt)
u0 : carrier mobility
Cox: gate oxide cap
w&l: transistor dimensions
Vt : thermal voltage
n : some tech parameter
Vgs: Voltage between Gate and Source
Vth: Threshold Voltage

Re:Leakage Power! (1)

jmv (93421) | more than 6 years ago | (#22319334)


1. Dropping Vdd to a CMOS transistors requires you to drop the threshold voltage to maintain performance.
2. As the two voltages approach each other, theres an increase in the current in the substrate (the current which flows between n-wells in a typical CMOS transistor).
3. This substrate current ends up contributing to massive amounts of leakage current.


I suspect they work around that by using a High-k dielectric [wikipedia.org] . That means they can use a move a large number of charge carriers in/out of the channel without having to apply a large voltage.

How efficient... (1)

slap20 (168152) | more than 6 years ago | (#22317324)

How efficient is something like the Asus eeePC? Are they using a low voltage chip, or just making use of CPU scaling? I know the Flash Disk in it saves a fair amount of power, is the CPU saving much as well? Just curious... might be looking into one.

-Eric-

A few thoughts (1, Interesting)

Anonymous Coward | more than 6 years ago | (#22317994)

I'd have to read the paper to know for sure, but I'd think they'd have to do some fundamental device process work to make a chip work at that low of a supply voltage. The threshold voltage (or "turn-on voltage" as some people call it) of a transistor doesn't scale with reducing device size, although there are steps that can be taken to adjust the threshold voltage in process, like ion implantation of various types. It's either that or they're operating it in subthreshold. The only problem with subthreshold operation is that it's incredibly slow. I guess it's conceivable though, for low-power, low-intensity applications where speed isn't a factor.

Let's assume they when with adjusting the threshold voltage. In order to make a low threshold voltage possible and practical, you'd have to be able to set the threshold voltage very precisely over process and temperature, etc, or else you'd get a zero or negative threshold voltage, resulting in an depletion-mode transistor (google it) that was always on, which wouldn't exactly be good for power consumption.

On a different note, it seems they use an integrated DC-DC converter to adjust the transistor supply voltage seen by the transistors on the fly. DC-DC conversion reduces the efficiency of a circuit, but 90%+ efficient switching converters are available, and for an ninefold increase in efficiency, that would be worth it. Unfortunately, switching regulators are also noisy. The little spikes caused by a buck or buck-boost converter could conceivably cause some of the transistors to unpredictably flip, especially if they're operating at such a low supply voltage.

Apparently they've solved all these issues, if it made it to ISSCC. ISSCC is the big leagues of circuit - they don't let snake oil or unproven claims in.

New math (0)

Anonymous Coward | more than 6 years ago | (#22318032)

1/10 = 0.3
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