Samsung to Produce Faster Graphics Memory 138
Samsung has announced a new line of GDDR5 chips that will supposedly be able to deliver data at speeds of up to 6 Gbps. In addition to faster data delivery the new chips also claim to consume less energy than previous versions. "Samsung said the new chips consume 1.5 volts, making them about 20 percent more efficient than GDDR 3 chips. Samples of the GDDR 5 chips began shipping to graphics-processor makers last month, and Samsung plans to begin mass production of the chips during the first half of next year. GDDR 5 memory should first appear in high-end gaming systems where users are willing to pay a premium for better graphics. Samsung did not disclose pricing for the chips.
Rumors (Score:5, Interesting)
Re:Consuming Volts? How about actual Wattage pleas (Score:2, Interesting)
The reason that you see a trend to lower and lower voltages is not because lower voltage = lower power. It's because in order to make transistors faster, you generally want to make them smaller (less area = lower capacitance) and you want to run them at a lower voltage (= less total charge to move to switch between a 1 to a 0). The problem with smaller, lower-voltage transistors is that they are more fragile (they have thinner gates, which higher voltages would fry), and higher leakage current (which makes them more inefficient - sometimes much more so). We've seen a lot of companies trying to reduce the power usage of the current generation of devices. That is not because they're running at lower power. It's because transistor companies have been spending tons of money doing materials and device physics research trying to find new ways to make transistors in order to overcome the problems caused by trying to make them smaller and faster. There have been some major recent advances [iht.com].
Re:Informative? Where are the EE's in slashdot? (Score:3, Interesting)
First of all context is everything. If I say that this line is at 5V, then someone in the power field of EE would think there was really 7.07 V (peak) on it since they alway deal with RMS. Different fields can make different assumptions: in the digital field I can roughly assume that the voltage in my circuits is in 1 of two states (well mostly [wikipedia.org]).
The leakage losses occur because silicon is an imperfect insulator. Even when transistors are 'off' (or switching) some current leaks through to the drain and bulk. This doesn't depend on the amount of current in the transistor doing useful work or even on the switching frequency, but only on the voltage. The actual power loss depends on the layout and operation of each transistor (with really complex interactions among them). There isn't really a simple resistor (in fact most models include 3 or more), but I was trying to give the layman's version. I may have goofed with the formula and I should have written V^2/R (though this is still far from accurate).
The power lost through switching is not reactive power. Reactive power is useful if you are planning a power distribution network, but not so much when you are calculating heat generation (since reactive power doesn't create heat). The switching losses occur because of the way CMOS logic works [wikipedia.org]. When the pair of transistors changes from a 1 to a 0 the charge built up on the source of the NMOS transistor is dumped to ground (and from 0 to 1 with the PMOS dumped to Vcc). This charge is due to (among other things) the capacitance across the transistor, and when it is dumped through the NMOS transistor, all that energy is lost. The formula for energy in a capacitor is 0.5 C V^2, and since this amount of energy is lost every switching cycle, the power lost is 0.5 f C V^2. This is not the complete picture as there are other losses (and some devices shutoff portions of the chip not in use [clock gating, etc]).
Yes, the current will vary with the voltage, but there's really not need to over-specify. As long as you are using silicon processes the parameters are going to be roughly the same (though it would have been nice if they mentioned the fab process or the scale). If you can calculate the power with P=V^2/R and everybody knows R, why bother to provide I? Also since the second power of voltage is in all those equations, halving the voltage means you can more than quadruple the frequency with the same power consumption (okay not really, due to the transistor switching times), so small changes in frequency are insignificant compared to changes in voltage.