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Samsung to Produce Faster Graphics Memory
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.
Memo From Ki-Tae Lee
To: All Samsung Employees
CEO and President,
Samsung
December 3rd, 2007
Would someone tell me how this happened? We were the fucking vanguard of graphics memory in this country. Samsung's GDDR3 was on the card to own. Then the other guy came out with a GDDR3 graphics chip. Were we scared? Hell, no. Because we hit back with a little thing called XDR. That's GDDR3 on crack. For cokehead gamers. But you know what happened next? Shut up, I'm telling you what happened--the bastards went to GDDR4. Now we're standing around with our cocks in our hands, selling XDR & GDDR3. Cokehead gamers or no, suddenly we're the chumps. Well, fuck it. We're going to GDDR5.
Sure, we could go to GDDR4 next, like the competition. That seems like the logical thing to do. After all, three worked out pretty well, and four is the next number after three. So let's play it safe. Let's make a more crackhead gamer RAM and call it the XDR3SuperTurbo. Why innovate when we can follow? Oh, I know why: Because we're a business, that's why!
You think it's crazy? It is crazy. But I don't give a shit. From now on, we're the ones who have the speed in the memory game. Are they the best a man can get? Fuck, no. Samsung is the best a man can get.
What part of this don't you understand? If GDDR2 is good, and three is better, obviously five would make us the best fucking memory that ever existed. Comprende? We didn't claw our way to the top of the memory game by clinging to the GDDR2 industry standard. We got here by taking chances. Well, GDDR5 is the biggest chance of all.
Here's the report from Engineering. Someone put it in the bathroom: I want to wipe my ass with it. They don't tell me what to invent--I tell them. And I'm telling them to stick enough transistors on there to call it GDDR5. I don't care how. Make the chips so thin they're invisible. Put some on the handle. I don't care if they have to make the ram hang halfway off the motherboard, just do it!
You're taking the "safety" part of "safety electronics" too literally, grandma. Cut the strings and soar. Let's hit it. Let's roll. This is our chance to make memory history. Let's dream big. All you have to do is say that GDDR5 can happen, and it will happen. If you aren't on board, then fuck you. And if you're on the board, then fuck you and your father. Hey, if I'm the only one who'll take risks, I'm sure as hell happy to hog all the glory when the GDDR5 card becomes the gaming video card for the U.S. of "this is how we game now" A.
People said we couldn't go to three. It'll cost a fortune to manufacture, they said. Well, we did it. Now some egghead in a lab is screaming "Five's crazy?" Well, perhaps he'd be more comfortable in the labs at Sony, working on fucking electrics. Cell processing chips, my white ass!
Maybe I'm wrong. Maybe we should just ride in SanDisk's wake and make flash USB drives. Ha! Not on your fucking life! The day I shadow a penny-ante outfit like SanDisk is the day I leave the silicon game for good, and that won't happen until the day I die!
The market? Listen, we make the market. All we have to do is put her out there with a little jingle. It's as easy as, "Hey, shaving with anything less than GDDR5 is like playing Warcraft on a Commodore 64." Or "It'll be so smooth, I could snort lines off of your monitor." Try "Your frame rate is going to be so friggin' fluid, someone's gonna walk up and confuse it with a urinal."
I know what you're thinking now: What'll people say? Mew mew mew. Oh, no, what will people say?! Grow the fuck up. When you're on top, people talk. That's the price you pay for being on top. Which Samsung is, always has been, and forever shall be, Amen, GDDR5, sweet Jesus in heaven.
Stop. I just had a stroke of genius. Are you ready? Open your mouth, baby birds, cause Mama's about to drop you one sweet, fat nightcrawler. Here she comes: Put another microcontroller on that fucker, too. That's right. GDDR5, two microcontroll
Qimonda already released GDDR5 Article from November 2: http://www.pclaunches.com/other_stuff/qimonda_gddr5_memory_now_available.php
How does it consume volts?
it should be noted that it's 6Gbit per pin, not per chip.
Still waiting on Serviscope_minor to wake up to fucking reality and realize that Jessica Price isn't going to fuck him.
This lends a bit of credence to the rumored NVidia G9 series launch, although I still think February is unlikely.
Demented But Determined.
Will this make my porn look better--On second thought, I don't want to see Ron Jeremy any better. [shudders]. :P
Well, there's spam egg sausage and spam, that's not got much spam in it.
Samsung said the new chips consume 1.5 volts, making them about 20 percent more efficient than GDDR 3 chips.
What poor science reporting. Nothing "consumes volts." Volts measure voltage -- difference in potential. Devices consume Joules -- units of energy. Also acceptable would be Watts -- energy per unit time. It would have been really nice to be given the Watts per Bandwidth per Size (W/Gbps/bits), but I realize that's asking way too much of the Times.
everything in moderation
I'm thinking that was not written by an EE (or anybody else that commonly speaks about electronics and efficiency.) I would suppose they operate at 1.5 volts and consume some number of amps. Volts alone doesn't say much about their power consumption... (P=IE)
JW
Watts that?
I take my children to see Madonna(..), but I never for once ever thought I was in the same business.Chris Rea.
But volts is a good unit of efficiency- it says nothing about power, but if a chip can run with less difference between vdd and ground, that's less of a potential that has to be available for it.
but what exactly about this story is novel. The article was pretty bland. With Moore's law anybody can pretty much predict that chipsets and RAM are going to get faster. Now what would have been interesting is to see price comparisons, predictions on what this will mean for the consumer, testing of the product, discussion of the architecture, insight into the manufacturing process, etc. I guess it's just the fact that it was from a mainstream source, but this article was about as useful as a press release, ie not very.
I got a catholic block.
No, volts are not a good metric of efficiency. Voltage and current would be, but that's because Volts * Amps = Watts. If you need a real example to understand why, consider a 1.5V chip that draws 3 Amps and a 5V chip that draws 100mA, then tell me which is more "efficient" and how you'd know that from the voltages alone.
everything in moderation
*Results may vary for AAA, C, and D batteries. B batteries are only a myth, anyone who believes in them probably also believes that P=IE is some EE's recipe for a tasty tart.
"Trolls they were, but filled with the evil will of their master: a fell race..." -- J.R.R. Tolkien on Olog-hai
How many people here keep up with the latest and greatest graphics card? I know that the newest may get you more fps in some game, but are there really that many people who regularly go spend $400+ for the latest and greatest in graphic cards?
Ninjas don't carry tic tacs
What is the CAS Latency of said chips? The article didn't say.
You overestimate us, sir.
Well if Current = Voltage / Resistance, how much does resistance vary in practice in electronics?
I dunno, you tell me -- slashdot user who took 6 units at a junior college and works 20 hours a week at Best Buy -- what *aren't* you qualified to talk about?
;)
As for me, I'm an EE so I'm qualified to talk about anything that matters
everything in moderation
In addition to faster data delivery the new chimps also claim to consume less energy than previous versions. Could be. Chips Outscore College Students on Memory Test. Most likely. I'm getting tired.
You mean across different devices? As much as current and power consumption do, which is to say, a lot. Many orders of magnitude. That's kind of a strange question that belies a weak understanding of electronics. We don't really speak of the "resistance" of a chip, because that's pretty much meaningless. You could calculate an "effective resistance" by R = V / I, but it's not very meaningful or useful to do so.
everything in moderation
The only slight mitigation here is that P = V^2/R, or I = V/R so reducing voltage reduces current too. Of course that does make TFS accurate.
Engineering is the art of compromise.
Samsung has taken a page from God's book, and intelligently designed these chips with a special digestive system just for consuming volts. Its multi-chambered, kinda like ruminents, but they don't have to barf the volts back up and eat them again. Samsung: 1, God: 0.
Actually, they did make B batteries.
The difference is we are talking about semiconductor devices. Losses from these semiconductor devices are primarily due to leakage and switching. As long as we're still using silicon, leakage will be roughly 0.5 V^2/R, no matter how much current you pump through the transistors. Switching losses occur in when logic changes from 1's to 0's due to the capacitance of the transistors. The power lost here is roughly 0.5 f C V^2, where f is the switching frequency and C is the capacitance (material dependent). The V^2's means that reducing the voltage has a significant impact on losses. If we note that R and C are completely determined by the material (silicon) and the fabrication process, we can see that as long as the frequency is held constant, the voltage is a reasonable metric for comparing power consumption in silicon devices.
Of course this analysis is purely approximate since there are a lot of there things going in the devices. And I'm assuming complete capacitive discharge (independent of switching frequency), and didn't consider the changes in refresh rate to this DRAM device. But suffice it to say voltage is still a pretty good metric for comparison (until you actually build the thing and test it).
Ohm my God.
What a terrible pun.
------
beware he who would deny you access to information, for in his mind he dreams himself your master
B Batteries were real.
Difference between Vdd and ground means nothing in terms of efficiency. Modern switch-mode DC voltage converters work with efficiencies exceeding 98%. In fact, in modern power supplies, virtually all of the various voltage levels your system needs are generated by stepping up or down a different DC voltage.
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.
... never use Google's Korean->English translation for company memos/press releases/important documents...
Seems like execs keep getting burned this way.
--Coming up with something clever... please wait...
Missing Floppies?
How come we never see the insides of graphics cores the way that CPU manufacturers release pics of the internals of their CPUs? I looked and couldn't find any. Now that AMD makes both CPUs and GPUs will it start do so? What gives?
This has been understood in the industry for decades: in a given silicon process, power consumption fits roughly within the envelop V^2 * F, where F represents frequency. Given a process shrink, this relation might or might not hold true. For a long time it was a good rule of thumb, but then in the era of excessively high leakage current it did not hold true, more recently with better control over leakage, the relationship is again a good rule of thumb. The upshot is that, over two decades, almost every reduction in voltage for a given class of part corresponds to a significant increase in power efficiency.
What the article failed to explain is this long history of voltage serving as a proxy for power efficiency.
The other relationship is that a given part will usually demonstrate a relationship where lower frequencies are stable at lower voltages. If increasing the voltage by 20% allows you to overclock a processor from 2GHz to 3GHz, you can estimate your increased power draw as 1.2^2 * 1.5, about double where you started.
It's almost pointless to convert this measure into watts, as so many other variables change in tandem. The new part has different bandwidth, different latency, different leakage, different dynamic consumption. There's no simple number that gets you apples vs apples. Most of the time, however, voltage is fair proxy. Peak consumption figures are mostly worthless from an efficiency perspective, except for sizing your power and cooling requirements.
On a side note, I'm wondering when we hit the floor on practical CMOS voltage levels. Surely the band-gap will come into play in the near future, and then what? Does the efficiency graph suddenly develop a crimp and stagger forward on a much reduced slope? This happened with hard drives, where there was a period of accelerated capacity increase (PRML/GMR/pixie-dust era) only to return to the more sedate curve once again later on. It wasn't long ago that F hit thin air (due to thermal issues) and now F is increasing at half the rate it sustained for a least a decade prior.
Long ago apparently respectable sources used to proclaim "silicon will hit the brick wall at 0.1um". In turns out S-curves hardly ever play out that way. The curve begins to taper downward when the easy gains are exhausted. The phrase "peak oil" is another one of those conceptual nightmares, much like the chimeric brick-wall on photo lithography. It's not going to be a peak, is it? It's going to be a wavy plateau. On any particular graph, you can point to a "peak" (though none of the graphs will agree), it's just that there won't be a momentous Alderan-disturbance that ripples though planet earth as the precocious metaphor suggests. Much like the silicon people had to finally confess, driving F higher and higher as your primary performance metric (at the cost of absolute efficiency) makes about as much sense in the long run as a single-occupancy air-conditioned Hummer in rush hour traffic.
Speaking of which, engine displacement is roughly as fair as a measure in the automotive sector as voltage in silicon. It's the nature of the internal combustion engine that these engines are far from their peak efficiency at low to medium throttle, which is why having a lot of power you rarely use is no free lunch. If you accept that a typical 2 liter engine is more efficient than a typical 3 liter engine, why would voltage as a proxy for power be any different?
I don't know--I thought it was a Joule.
There's no place I can be, since I found Serenity.
Get a voltmeter, place the black lead on ground and the read lead on the power lead of the chip. 1.5 Volts going in. Check.
...um... ground)
Now move the red lead to the other power lead on the chip - Zero volts coming out! Where are these volts going? But it not accurate to say they are being *consumed* per se. The volts are stored in the form of magic smoke. When the chip is full of volts (as magic smoke), it stops working. Be careful, too many volts going in and you can burst the smoke tank, then you're really screwed!
(While performing this test, don't let the amps leak out of the power supply and onto the
"B" batteries in the context of vacuum tube circuits refer to their function, not their form, so they are not contemporaries of AAA-, C- and D-sized cells
If you read some of the other articles on the subject, it sounds like A and B at least where almost always of a certain size as well.
The sizes today don't just dictate form, they also tell you other characteristics such as voltages.
Hey, if people couldn't abuse units of measurement then Han Solo would never have been able to do the Kessel Run in less than 12 parsecs.
It is by the juice of the coffee bean that thoughts acquire speed, the teeth acquire stains. The stains become a warning
EE was ten years ago, so I'll trust you on the formulas, but all you're saying is that *for the same chip*, power consumption will go down as the square of voltage, and up linearly with frequency (that's assuming that the characteristics of the material id not affected by those, which we'll just assume is correct).
However, if it's a different chip, R and C are different, so it's meaningless to use voltage to compare power consumption with existing designs (or processes).
How is this a troll, exactly?
everything in moderation
The summary reminds me of a commercial from several years back that purportedly allowed you to give your own car a jump start through the cigarette lighter. One of the selling points for the device was how it was supposedly better than the actual battery in your car. To prove it, they hooked a multimeter up to each one, while the voiceover said, "Your normal car battery has only 12 volts of energy. But {our product} has 48 volts of energy!"
This was almost as good as the one where one of those ionic air filters was "electrostatically charged... like a magnet!"
Eventually, one of these advertising agencies got tired of being ridiculed. When they made a commercial for one of those "shake vigorously to charge" flashlights that has a magnet that moves through a coil when you, er, shake it, they actually put Ampere's Law in integral form on the screen to prove how smart they are.
Gosh its nice to see some non fuzzy maths that actually mean something.
Same reason there was no such thing as Arctic Silver 4 thermal paste. "Arctic Silver replied to me regarding my question of why they decided to skip the name of Arctic Silver 4 and decided to go straight from AS3 to AS5. The number 4 is not very comfortable with Asians since it sounds very similar to the word d**th and therefore to keep the interest of their Asian customers, they decided to not name the next AS compound AS4 but to AS5. Being Chinese myself, 4 is very badly connotated and it was a wise marketing strategy in both the customers and Arctic Silver's interest to skip naming their Arctic Silver compound Arctic Silver 4." http://www.slcentral.com/thermal-compound/page2.php
Oh right, I'm an EE.
The difference is we are talking about semiconductor devices. Losses from these semiconductor devices are primarily due to leakage and switching.Ok. That really doesn't change anything.
As long as we're still using silicon, leakage will be roughly 0.5 V^2/R, no matter how much current you pump through the transistors.What the hell do you mean, 'no matter how much current you pump through the transistors?' You gave me R for a reason. Here's where your formula comes from: P = V*I, I=V/R, transistor is only on 50% of the time (50% duty cycle). 0.5 * V * (V/R). I don't know R for that memory chip, nor do I know the current, so I don't know the power loss.
The power lost here is roughly 0.5 f C V^2, where f is the switching frequency and C is the capacitance (material dependent).That comes from the reactive power, and it's actually 0.5 * 2*pi * f * C * V^2. The impedance of a capacitor is 1/(2*pi*f*C). Thus the power equation is the same equation as above, but the impedance of a purely resistive circuit was just R.
The V^2's means that reducing the voltage has a significant impact on losses.Yes, but the voltage applied influences the current drawn, so it's not that current is irrelevant.
If we note that R and C are completely determined by the material (silicon) and the fabrication processWhich are pretty important variables. Do you know anything about the fabrication process of this new memory? Or the number of transistors? More transistors = more devices drawing power.
we can see that as long as the frequency is held constant, the voltage is a reasonable metric for comparing power consumption in silicon devices.Also a very, very bad assumption, considering that the whole point of the thing seems to be that this memory can be clocked faster.
Warning: Opinions known to be heavily biased.
I'm also an EE, and while this is not my area (it's not yours either!), I think you are simply trying to be over-pedantic.
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).
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. 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.
"Of course this analysis is purely approximate since there are a lot of there things going in the devices."
Absolutely. For one, dynamic memories are quite different from microprocessors. We can figure that there will be GDDR interface logic active the whole time, but that power consumption probably pales in comparison to the rest of the chip, with data coming in and out at gigabit speeds. The capacitors used for storage will be drained and filled with every read cycle -- but only one row/column will be active per cycle. (Well, no, I guess there's prefetching and other such things.)
Also, the bus interface also uses power. GDDR5 features bus inversion: since GDDR5 lines are terminated to Vdd, reducing the number of zeros transmitted saves power.
And is it more or less of a heat-distance-volume-power-mass measurement if the LOC is in English vs. Esperanto?
The problem with quotes on the internet, is that nobody bothers to check their veracity. -- Abraham Lincoln
1 LOC == Negative Square Root of knots traveled times the volts of energy in one cubic ounce of free hydrogen.
Further, LOCs decrease in both mass and energy as square of the distance from any funding source.
The problem with quotes on the internet, is that nobody bothers to check their veracity. -- Abraham Lincoln
this nation, under God, shall have a new birth of freedom. -- Lincoln, Gettysburg Address
I may have managed to completely miss it, but what happened to GDDR4? All I've seen is GDDR2 and GDDR3, at least on the higher-end GPUs. So when did GDDR4 come out? Did they decide that only prime numbers are good enough to be used as GDDR version numbers?
Everything is subjective.
I find your lack of Star Wars knowledge disturbing.
Fear it.
If you can calculate the power with P=V^2/R and everybody knows R, why bother to provide I?
If "everybody" knows the effective R for every different device then why don't you go ahead and tell it to us so we can calculate the power for this device?
(I'm also an EE, this is my area, and you're oversimplifying to the point of error, as has been explained many times to you and elsewhere in this thread.)
everything in moderation