Low Voltage Is Key To Energy-Efficient Chip
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."
But how well does it overclock?
That's why your cell phone has an ARM CPU and not an x86.
Engineering is the art of compromise.
aparently from the Bureau of Slowly and Painfully Working Out The Obivous.
i could live a little longer in this prison
In Germanium the voltage is 0.3, if I remember correctly. So it depends on the materials used.
My work here is dung.
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.
If they can just get this thing down to zero volts, this chip will run forever!
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.
The enemies of Democracy are
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.
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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.
My server
You need to buy them in bulk. For example, Intel will sell you about 500 million FETs for only $200.
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.
Never underestimate the bandwidth of a 747 filled with CD-ROMs.
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