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Intel Doubles Capacity of Likely Flash Successor

ScuttleMonkey posted more than 6 years ago | from the double-or-nothing dept.

Technology 91

Intel has announced a new technique that allows them to effectively double the storage capacity of a single phase-change memory cell without adding cost to the current fabrication process. "Phase-change memory differs from other solid-state memory technologies such as flash and random-access memory because it doesn't use electrons to store data. Instead, it relies on the material's own arrangement of atoms, known as its physical state. Previously, phase-change memory was designed to take advantage of only two states: one in which atoms are loosely organized (amorphous), and another where they are rigidly structured (crystalline). But in a paper presented at the International Solid State Circuits Conference in San Francisco, researchers illustrated that there are two more distinct states that fall between amorphous and crystalline, and that these states can be used to store data."

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Two Billion Transistors on Their Latest Chip (1, Redundant)

eldavojohn (898314) | more than 6 years ago | (#22294378)

I just read an article about them passing the two billion transistor mark [bbc.co.uk] for a single chip. The BBC announcement mentions many of these transistors are used for memory (the caches I assume). I am not a hardware expert although I wonder if this new phase-change memory is what they are using. Highly unlikely since this seems to be brand new research. If not, I certainly look forward to them integrating this into their chips and dies for use in caching--they could be blowing Moore's Conjecture out of the water! Exciting stuff for hardware nuts!

Re:Two Billion Transistors on Their Latest Chip (2, Informative)

networkBoy (774728) | more than 6 years ago | (#22294454)

PCM litho tech in not compatible with CPU litho tech.
So I doubt this will be happening any time in the near future.
-nB

Re:Two Billion Transistors on Their Latest Chip (-1, Troll)

Fat Wang (1230914) | more than 6 years ago | (#22295142)

Crap filled vagina! Right?

Re:Two Billion Transistors on Their Latest Chip (4, Informative)

SirLoadALot (991302) | more than 6 years ago | (#22294492)

No, processor caches are made out of SRAM, which is very fast but takes a lot of silicon space. Flash and phase-change RAM are totally different technologies. They aren't intended to be as fast as SRAM, but they are non-volatile, storing their contents without power. Phase-change or a similar technology may end up giving DRAM a run for its money if it gets fast enough, but I've never heard of an alternative to SRAM for internal processor caches.

Seems unlikely (1)

jd (1658) | more than 6 years ago | (#22295100)

The new memory is described as being as fast in reads as DRAM, which is an order or two of magnitude slower than register memory, which is what you need in a processor. Intel's adding of so much memory into the CPU interests me far more. Add enough, and the main memory is IN the processor (a-la the Transputer). From there, a little reorganizing reverses the arrangement - from memory in a CPU to a process in memory (PIM) architecture. Special-purpose PIM has been done before - Cray embedded a communications library in RAM before - but a full processor would be much more interesting.

Re:Seems unlikely (1)

dreamchaser (49529) | more than 6 years ago | (#22295246)

It's not register memory that's taking up half the die space; it's SRAM cache. There's nothing all that interesting about it. It's old technology that's just gotten tiny enough to fit a ton of it on a CPU die.

Re:Seems unlikely (1)

imgod2u (812837) | more than 6 years ago | (#22298306)

Depends on who you are. Intel is notorious in being able to pack tons of SRAM into a small die space and this isn't just due to process shrinks. There are all sorts of flavors of SRAM, the standard taking 6 transistors but some taking as little as 1 depending on density/speed trade-offs.

wtf mods? (0)

Anonymous Coward | more than 6 years ago | (#22295336)

Why was the grandparent post modded -1 Redundant? Seems like he/she was pretty stupid about the topic although you point out it is interesting to think about putting main memory sized dies on chips.

Is this what Slashdot has come to?

Captcha: neighbor

SLASHDOT SUX0RZ (-1, Troll)

Anonymous Coward | more than 6 years ago | (#22294382)

_0_
\''\
'=o='
.|!|
.| |
goatse man doubles capacity of own ass [goatse.ch]

Salt shaker please (2, Insightful)

techpawn (969834) | more than 6 years ago | (#22294472)

"Intel Doubles Capacity of Likely Flash Successor" this from a site that had a huge Intel logo on it for how many months?

It's neat tech, but as long as flash keeps getting bigger and cheaper we won't see it's 'Successor' for a while.

Long time in the lab (3, Insightful)

EmbeddedJanitor (597831) | more than 6 years ago | (#22294734)

Even when a technology becomes shippable it tends to take quite a while for it to catch on. It is easy to make small lab batches, but reliable low-cost high-volume production takes a lot longer. NAND flash was invented in 1988 but only really got going in around 2003 - 15 years later.

Re:Long time in the lab (2, Insightful)

networkBoy (774728) | more than 6 years ago | (#22295410)

It is easy to make small lab batches
I see you don't work in an R&D lab doing PCM...

Before I left my former job we were working on PCM.
It was anything but easy to make in small batches in the lab. Our average yield of 100% good die was under 1 die/wafer.
We had plenty of 50% dice, but very little fully functional ones.
-nB

Re:Long time in the lab (1)

Mattsson (105422) | more than 6 years ago | (#22304520)

He choose a poor phrasing.
He should have said something like:
"Relative to each other, it's easier to make one functional unit in a lab than to make 1000 functional units a day"

Except... (2, Interesting)

Khyber (864651) | more than 6 years ago | (#22298990)

...the reliable low-cost high-production facilities already exist, as the process deviates very little from the CMOS manufacturing process. It's the same material that is used in rewritable optical media, and on top of that, it's basically just glass. Where you once needed stable unchanging silicon for memory/data storage, now we're just using different states of glass. Most of your concerns are addressed in this technology, and this is why I'm watching it very closely. Go read up a bit here. (PDF WARNING) [ovonic.com]

Oh, it also does have the theoretical capability to replace SRAM and DRAM. But in order for it to do that, it would need to be a little faster and we would have to be able to fully exploit all four states that it can be in for data. Also, read/write cycles would need a few more orders of growth to be used as a processor cache or extended RAM replacement, but as it is they're great for hard disk usage.

Re:Salt shaker please (4, Informative)

Microlith (54737) | more than 6 years ago | (#22294938)

as long as flash keeps getting bigger and cheaper we won't see it's 'Successor' for a while.

As I understand it, flash (nand) capacity grows with the shrinking of the trace size. It's also cheap because it's produced in mass quantities.

Everything that has made flash high capacity and cheap can be applied to PCM, only PCM has a number of advantages:
- more durable, since it doesn't force high voltages over blocks to erase them
- smaller cells, allowing more to be packed in the same space
- rewriteability. You don't have to erase a block to change a single byte. It's more like RAM or hard disks in that respect.

So what will likely happen is a slow change from FLASH to PCM as the major flash manufacturers transition their products to this technology. It'll still have the same form factor, and most people won't notice aside from an increase in capacity.

IANAPCMEBIWNS (I am not a pcm expert but I work near some...)

Re:Salt shaker please (1)

imgod2u (812837) | more than 6 years ago | (#22297966)

I'm not sure about the density of PCM vs flash but one of the biggest advantages of flash right now is that it uses the same fabrication process as other semiconductors. It's still CMOS.

Consequently, you have economies of scale that translate from the other microelectronics markets. More importantly, you have one monolithic chip with control, interface, encryption logic, etc. all on one chip with one fab run. Many of our chip designs use small pockets of flash memory here and there (specially available from our fab vendor) for non-volatile storage. This not only makes flash more convenient, it also means that as process technology is pushed forward, so will flash.

PCM seems limited to non-volatile storage.

Re:Salt shaker please (0)

Anonymous Coward | more than 6 years ago | (#22298442)

PCM has a number of advantages

You didn't list speed, which the article claims is the most important one (replacing SDRAM and flash with one technology).

Re:Salt shaker please (1)

adrianmonk (890071) | more than 6 years ago | (#22302466)

rewriteability. You don't have to erase a block to change a single byte. It's more like RAM or hard disks in that respect.

While I don't disagree with your point overall, isn't that exactly unlike a hard disk? On a hard disk, you must rewrite an entire sector to change a byte, and not only that, you must wait until the platter spins around to the right spot again in order to do it.

Double the bandwidth for Duplicates? (1)

mosel-saar-ruwer (732341) | more than 6 years ago | (#22295476)


Does this mean twice as many stories at Slashdot [slashdot.org] ?

Re:Double the bandwidth for Duplicates? (1)

Microlith (54737) | more than 6 years ago | (#22295958)

Too bad they're not dupes.

SLC/MLC Flash is a different issue than PCM (1)

billstewart (78916) | more than 6 years ago | (#22298674)

The story you're referencing is about advances in Single-Level Cell NAND NAND flash memory, which is the more expensive longer-wear-cycle cousin of the Multi-Level Cell flash memory that's been taking over the USB Flash market. MLC is more dense, and the price/gig has dropped by about 75% since last summer. Unfortunately, MLC technology only supports about 10,000 write cycles per cell (vs. about 100,000 for SLC), so you need to use wear-leveling drivers to keep it from wearing out, but that's still good enough for most current applications.


Phase Change Memory (PCM)'s a different animal entirely - much faster, with much different physical and chemical design, and it's also had technology advances recently.

Only Double? (0)

CastrTroy (595695) | more than 6 years ago | (#22294478)

Is it really only doubling the capacity. Let's say you have a piece of memory that can hold 2 bits. Now, if you are using binary storage, and each bit stores 1 of 2 values, then you have the possibilities of: 00,01,10,11. Which is 4 different values. Now if you have 4 states for each bit (which I guess wouldn't, by definition, be a bit anymore), then you have 00,01,02,03,10,11,12,13,20,21,22,23,30,31,32,33. So, you have have squared the amount of information you can store. Unless I'm completely misunderstanding what they are doing here.

Re:Only Double? (4, Informative)

nonsequitor (893813) | more than 6 years ago | (#22294556)

It's double the number of bits. If you look at the largest value a 16 bit number or a 32 bit number can store, its not "double" in size. When it comes down to it, they're just bits, how you use them is up to you.

Re:Only Double? (2, Insightful)

deanlandolt (1004507) | more than 6 years ago | (#22294822)

It's double the number of bits. If you look at the largest value a 16 bit number or a 32 bit number can store, its not "double" in size. When it comes down to it, they're just bits, how you use them is up to you.
I suppose that's a lot like saying IPv6 "quadruples" the number of addresses of IPv4 -- you know, 128 bits versus 32 bits. I mean, they're just bits, right?

Re:Only Double? (1)

treeves (963993) | more than 6 years ago | (#22298806)

That's not what he said: he specifically said that it's "not 'double' in size". 32 bits *is* double the *number of bits* of 16 bits.

Re:Only Double? (0)

Anonymous Coward | more than 6 years ago | (#22299402)

Well, 128 bits can be used to express 4x 32-bit numbers which can then be added together to give you the resultant number, quadrupling the range of numbers which can be expressed...

This isn't a particularly good way of using these bits, but it does depend on how they are being used.

Re:Only Double? (2, Informative)

Dan Posluns (794424) | more than 6 years ago | (#22295422)

Another way of looking at it:

Say we use a bit to store the result of a coin toss. True for heads, false for tails.

With two bits, we can store the results of two coin tosses. There are four possible outcomes when two coins are tossed, ranging from neither of them being heads, to only the first or the second being heads, or both of them being heads.

If we double the number of bits, we can store the result of four coin tosses. There are now sixteen possible outcomes, but we're still only storing the result of four tosses.

(Note that this example assumes we're interested in storing off the result of each coin toss. If we're only interested in counting the total number of heads or tails and don't care which coin was tossed in what order, then we can use our bits to store the total number of successful tosses rather the result of each toss, which is a much more efficient use of our bits but carries less information.)

Dan.

Re:Only Double? (0)

Anonymous Coward | more than 6 years ago | (#22295434)

Rephrased slightly - the grandparent offered...

  • 4 possible states for 1 element in memory. It takes 2 bits to represent a value from 1..4.
  • 16 possible states for 2 elements in memory. It takes 4 bits to represent a value from 1..16.


They've only doubled the bits...

Re:Only Double? (0)

Anonymous Coward | more than 6 years ago | (#22294568)

Sheesh. The number of states is distinct from the number of bits used to represent the states. The bits per cell are doubled; that's it.

Re:Only Double? (1)

SleptThroughClass (1127287) | more than 6 years ago | (#22294576)

Maybe the four states are not independent, so we're moving from "two" states to "four" storage states. 2 to 4 is a doubling. That is implied by the phrasing that these two new states lie between the other states.

Re:Only Double? (4, Informative)

SirLoadALot (991302) | more than 6 years ago | (#22294594)

Currently the phase change RAM can only store 1 bit per element, in two possible states -- 0 or 1. They are changing this four possible states, which corresponds to two bits -- 00, 01, 10, 11. Hence, the amount of data that can be stored is doubled. The number of bits held per element increases by one every time the number of possible states is doubled.

Re:Only Double? (1)

sxeraverx (962068) | more than 6 years ago | (#22296836)

No! They are in fact doubling the number of bits. A bit is a BInary digIT. '0'=1 bit '00'=2 bits.

To be more precise, what they're doing is changing bits to qits (possibly pronounced 'kits'?), or Quaternary DigITs. There are the same number or qits as there used to be bits.

Either way you look at it, however, you're doubling the data density, and therefore doubling the maximum storage capacity in a theoretical standard-sized hard drive without changing the price point.

Re:Only Double? (5, Informative)

wizardforce (1005805) | more than 6 years ago | (#22294638)

Now, if you are using binary storage, and each bit stores 1 of 2 values, then you have the possibilities of: 00,01,10,11. Which is 4 different values. Now if you have 4 states for each bit (which I guess wouldn't, by definition, be a bit anymore), then you have 00,01,02,03,10,11,12,13,20,21,22,23,30,31,32,33. So, you have have squared the amount of information you can store.
simply put, no. 1 bit can either be 0 or 1 *not* 00, 01, 10, 11, that would be TWO bits. doubling the number of combos for one "bit" would likely be stored as 00,01,10,11 although what you're actually doing here is storing 0001 and 1011 together hence two "bits" of info can be stored per unit in a 4 state system.

Re:Only Double? (0)

Anonymous Coward | more than 6 years ago | (#22294906)

In fact, while each cell could previously only store two values (0 and 1), while they can now store 4 (1, 2, 3, 4, no zero value due no equivalent of 'lack of charge' in a solid state system). This corresponds by the Riemann equation to a bit-per-byte value of 4^8 or 65535 "characters" per cell. That still needs 5 for the integral divisor register.

Happy to explain further.

Mod points please

Re:Only Double? (1)

Mr2cents (323101) | more than 6 years ago | (#22300034)

But a bit is a contracion of binary digit, and binary means two. In communications, you have the same distinction between bits per second and baudrate. Bits per seconds give you the amount of ones and zeros transmitted, while the baudrate gives you the amount of symbols per digit. If a symbol can be one of four possible tones (or phases), you get 2 bits per baud, and that's what happening here again.

Re:Only Double? (1)

Mr2cents (323101) | more than 6 years ago | (#22300060)

the baudrate gives you the amount of symbols per digit.
Sorry for the mistake: that should be symbols per second, not per digit.

Re:Only Double? (0)

Anonymous Coward | more than 6 years ago | (#22296536)

>Now, if you are using binary storage, and each bit stores 1 of 2 values, then you have the possibilities of: 00,01,10,11. Which is 4 >different values. Now if you have 4 states for each bit (which I guess wouldn't, by definition, be a bit anymore), then you have >00,01,02,03,10,11,12,13,20,21,22,23,30,31,32,33. So, you have have squared the amount of information you can store.

I was thinking the same thing although this falls more under the under the heading of "hardware compression options with a 4 state memory cell". Or can we move to base 4 math?

Re:Only Double? (1)

ThaNooch (1186931) | more than 6 years ago | (#22297506)

I agree, so we are in base 4. So there is double the exponential growth rate, so using this type of memory we can store not twice the amount of data, but rather the square of the amount of data.

So while it's not an exponential growth, it's not linear either (twice the amount). Linear would look like 2*2^n. What we have is 4^n, or (2*2)^n, which is the same as (2^(n*2)), which is the same as (n^2)^2.

So it's actually a polynomial increase in overall storage, not just a linear doubling. This squaring is the effect of 4 / 2 (base 4 to base 2). In order to square the amount of data again, we would need 8 new states (base 8 from base 4), which is a byproduct of this being only a polynomial increase of exponential space.

Re:Only Double? (1)

imgod2u (812837) | more than 6 years ago | (#22298210)

No, it's twice the amount of data. If you can store 1 bit, you have:

0, 1

Two bits:

00, 01, 10, 11

In order to double it again, we simply add another bit:

000, 001, 010, 011, 100, 101, 110, 111

Each binary digit doubles the data capacity. Just like each decimal digit results in 10x the capacity.

For any given number of bits per cell, n, we have 2^n combinations. This technology added one more bit, so it increased the amount of data storage by:

2^(n+1) / 2^n = 2

It only "doubled" because we started with n = 1. There's nothing inherent about how many more physical states are available that would suggest we have to be able to find them in orders of 2. The next breakthrough will most likely just add another state, i.e. another bit.

Re:Only Double? (1)

ThaNooch (1186931) | more than 6 years ago | (#22300704)

Unless the blurb is incorrect i read that there were 2 extra states that were discovered. Once we get more than 2 states per "unit" we stop counting in binary, ones and zeros are purely a representation of "on" and "off" (and typically cited in theory since it's the smallest positive integer exponentiation).

If they can represent 4 different states in the same physical component then each bit becomes a 0, 1, 2, or 3. So we will count up as 0, 1, 2, 3, 10=4, 11=5, 12=6, 13=7, 20=8, 21=9, 23=10, 23=11, 30=12, 31=13, 32=14, 33:15, 100, 101, 102, 103, 110, 111, 112, 113, 120, 121, 122, 123, etc.. as we see 2 bits can

Similar to how we represent roughly a 5v current as high, and a .5v current as low, if one could reliable generate and observe signals of .5, 2, 3.5, and 5, then we would be able to compute in base 4.

If it requires the technology more space (or time) to switch between the space then the gains will be diminished or even counterproductive but from a purely discrete theory perspective with no performance analysis and extra state changes the alphabet, and the resulting language gets more done with shorter strings (finite length members of the source alphabet).

Re:Only Double? (1)

Hao Wu (652581) | more than 6 years ago | (#22298048)

I think I recall reading that a three-state system is best because it's the closest possible number to e (2.718), which was mathematically proven to be the most efficient state to process.

Since there is no known way to represent e physically due to its fraction, "3" is as good as it gets.

Re:Only Double? (1)

corsec67 (627446) | more than 6 years ago | (#22299594)

I think what the GP meant was that you have more bits per symbol (or memory cell), then you would double the capacity.
A similar thing is why the Baud rate [wikipedia.org] isn't equal to the bit rate on 56.7K modems.

Re:Only Double? (0)

Anonymous Coward | more than 6 years ago | (#22294682)

No, the number of states is not proportional to the storage (the relationship is exponential). That's why a long can store the square of any int, despite the fact that it's only twice as long. 4 states are represented by 2 bits, so in your case the storage would be doubled.

Re:Only Double? (1)

nleaf (953206) | more than 6 years ago | (#22295000)

No, its still only double. Instead of thinking of bits, just think of memory cells. Each cell can store two values under the original method. The new method allows for four values. So, with two values you get 0 or 1, => 1 bit. With four values you get 0, 1, 2, 3, => 00, 01, 10, 11 => 2 bits. So, two bits for every one bit previously is just a doubling.

Re:Only Double? (0)

Anonymous Coward | more than 6 years ago | (#22295058)

They doubled the amount of bits. That is all. So instead of a chip holding 8 bits per unit of area its holding 16. That is twice as much. Not 256 times as much. A 32 bit integer holds twice the information as a 16 bit integer, not 65536 time as much.
 
The number value held by a 32 bit integer is much greater than the number held by a 16 bit integer is much greater, but if I have a PCM stream I can still only hold twice as many samples n 32 bits as I could in 16.

Re:Only Double? (1)

TheLostSamurai (1051736) | more than 6 years ago | (#22295572)

This has nothing to do with bits of memory, but rather cells of memory. Each cell of memory has four states and the four states are mutually exclusive. It can only be in one state at a time. Therefore rather than holding only a 0 or 1 as it previously did, it can now hold 0,1,2 or 3. Hence, double the capacity of before.

Your logic is flawed (1)

Skapare (16644) | more than 6 years ago | (#22296032)

You are counting the number of VALUES that could be stored, not the equivalent number of bits needed to store those values. With your logic, starting with 8 slots that store 2 states, going to 8 slots that store 4 states, we'd be going from 256 to 65536. But that's not 256 times the capacity of bit storage; it's still just a doubling of the capacity. For every slot that can store 4 states, that's just the equivalent of 2 bits. That's certainly twice as much as before, and it means this technology of memory can come in twice the capacity for about the same fabrication costs (and you know a corporation will take as much of that as profit as the market can handle). Still, it's a better thing to have if these states are reliable (there is a risk that the new states may not be as reliable under adverse conditions like voltage error, thermal stress, and radiation, as the original two states).

Yes (1)

jgoemat (565882) | more than 6 years ago | (#22302714)

Each item that could store two bits can now store twice as many bits of data. Before you had one item that had two states: (0, 1) Now you have one item that has four states (0, 1, 2, 3). With two states, you would need two items to make four possibilities: (00, 01, 10, 11). So you get the same amount of information with 1/2 the number of storage items, hence double the capacity. Using your example, you would need four bits to store the same possiblities as two items with four states: (0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111).

No longer binary? (1, Interesting)

sm62704 (957197) | more than 6 years ago | (#22294498)

Will we now have computers that do base 4 arithemetic rather than base 2? At leat the memory of them? Or is this exactly what the INtell engineers are saying?

Could this new technology be used for CPUs as well, or only memory?

Re:No longer binary? (0)

Anonymous Coward | more than 6 years ago | (#22294848)

luckily 4 has the rather unique property of being a power of 2, so binary is still safe. for now...

Re:No longer binary? (1)

jimicus (737525) | more than 6 years ago | (#22295208)

unique

You keep using that word. I do not think it means what you think it means.

Re:No longer binary? (1)

anotherone (132088) | more than 6 years ago | (#22295702)

He's telling a joke- I realize that it's probably hard for you to understand since your concept of humor is limited to Princess Bride quotes, but just try to accept it.

Re:No longer binary? (0)

Anonymous Coward | more than 6 years ago | (#22295868)

Surely he means no harm...

He's just a little short on charm.

Re:No longer binary? (3, Informative)

disassembled (977342) | more than 6 years ago | (#22294854)

Will we now have computers that do base 4 arithemetic rather than base 2?
A base-4 digit is the same as two base-2 digits, information-wise, so it doesn't really matter how the information is stored. If you want to store a byte in some piece of hardware, you can store it as 4 base-4 values or 8 base-2 values, and it'll all look the same to the CPU.

Re:No longer binary? (1)

MaskedSlacker (911878) | more than 6 years ago | (#22295804)

Not really. Logic circuits use binary voltage levels so the single voltage output from the pcm representing two bits has to be converted into two separate signals sent to the CPU. So it does matter how its stored, because you have to convert the base 4 to base 2 before sending it to the CPU. This isn't a challenge, but it does need to be done.

Re:No longer binary? (0)

Anonymous Coward | more than 6 years ago | (#22298130)

Congratulations, you win the "DUH!"-award of the week.

Now please proceed to meet Mr. Bush - you'll be his new technical advisor.

Re:No longer binary? (1)

earlymon (1116185) | more than 6 years ago | (#22295004)

Those are good questions.

I'm terribly out of date on this, but - in the old days with multiple chips for a memory bank, there was address decoding circuitry that would point you to the right chips/pins. If - and I'm clueless here - the same thing is still being done but it's all been reduced to fewer chips, then this _might_ imply that you still do the address decoding as before, but you have fewer address wires to route. In other words, we used to need two cells for four - now we have four in one cell - so the complexity in reading the states may be offset by the reduced address decoding.

I'm guessing.

Re:No longer binary? (1)

imgod2u (812837) | more than 6 years ago | (#22298540)

Each cell, being just a variable resistor, will output a certain amount of voltage. Let's assume a 1.2V supply:

00 = 0.0V
01 = 0.4V
10 = 0.8V
11 = 1.2V

You'd then do a crude A-D conversion (it's crude and small/lower power because it only needs to handle 4 distinct voltage levels). Off the top of my head, the smallest would be two pairs of N-FET and P-FET. One will have the threshold voltage biased at 0.2V and the other biased at 1.0V using a voltage divider from VDD to GND.

The output of the first pair of FETs would be bit 1 and the output of the second pair bit 0.

You can then either send these two bits in parallel or serialize them depending on the memory controller interface.

You'd have 1 less address decode line, yes but you'd have to store data 2 bits at a time and you'd have the extra latency of the "A-D conversion" circuit above.

Re:No longer binary? (1)

imgod2u (812837) | more than 6 years ago | (#22298944)

Actually, scratch that. You'd probably want to bias the voltages coming out of the cell. Two paths, one to bias it by Vt-0.2V and the other Vt-0.4V and have them go to separate FET pairs.

Binary is easier to work with (1)

m50d (797211) | more than 6 years ago | (#22295658)

On the very low hardware level, yes, the memory will be base 4. However, on anything but that very low level there's no reason for anything to care - it's trivial to convert between base 2 and 4, and it's a lot easier and more sensible to program a computer that works in base 2 than to make your opcodes do something sensible on 4-state things.

Re:No longer binary? (1)

gaelfx (1111115) | more than 6 years ago | (#22295692)

probably not since I doubt the elements would interact in a mathematically coherent way that transistors do. Even if you figured out a way to make sense of it, it would likely involve the use of something simpler than the element itself, thereby making it essentially the same as using transistors, at least from my understanding. But with the way that computers actually "evolve, it's unlikely that such technology would take hold in a marketable fashion in our lifetime, I mean, how do you think Microsoft is still around? Too much is invested in current infrastructure, and most people are too scared to be weened off of something that currently works for something that MIGHT work better. How many applications do you use that actually utilize a 64-bit processor in any meaningful way?

Re:No longer binary? (0)

Anonymous Coward | more than 6 years ago | (#22296054)

Oh, that'd be fantastic! Instead of our typical logic gates that respond in binary fashion, you'd have two possible values between "yes" and "no": 0=NO, 1=MAYBE, 2=GUESS_SO, 3=YES

Sounds like Intel during the Pentium fiasco to me.

Re:No longer binary? (1)

sm62704 (957197) | more than 6 years ago | (#22306786)

I think you were going for "funny" but in case you weren't, the first column is base 10, the second base 4, the third base 2 (binary):
10 04 02
---------
00 00 000
01 01 001
02 02 010
03 03 011
04 10 100
05 11 101
06 12 110
07 13 111

It will be converted to binary (2, Insightful)

Skapare (16644) | more than 6 years ago | (#22296136)

You can label the now total of 4 states however you like, such as 00/01/10/11 or 0/1/2/3 or A/B/C/D or T/A/C/O. But whatever they are, Intel would need to, at some point, convert this all back to 2 bits with states 0/1 when interfacing with external binary circuits. If they don't know how to do that they are welcome to "Ask Slashdot".

Re:No longer binary? (1)

gardyloo (512791) | more than 6 years ago | (#22296948)

An interesting piece I ran across many years ago about ternary (and other bases -- try base-e!) systems, and how they _can_ be better at some things than binary.
http://www.americanscientist.org/template/AssetDetail/assetid/14405?&print=yes [americanscientist.org]

Re:No longer binary? (1)

BungaDunga (801391) | more than 6 years ago | (#22298328)

I'm partial to base i myself.

Re:No longer binary? (1)

imgod2u (812837) | more than 6 years ago | (#22298816)

Mathematically speaking, e is the most efficient base. Imagine a decode being a search tree and imagine the base being the number of children per node. A decode function is simply traversing the tree. The first level of the tree being the first bit, second level being the second bit, etc.

To find any number N, one would need a tree of L levels and B children per node at each level. B^L must then be >= N in order to guarantee that the number N is represented in the tree. We can then take this to mean that L = ln(N)/ln(B).

To do a full search of the tree in the worst case time T, would take T = (B^L)*S where S is the amount of time it takes to search and recognize each node. Substituting L:

T = S*B^(ln(N)/ln(B))

To find the base that would result in the lowest search time, take the double derivative of T with respect to B and find the roots (peak and valley where search time either maximizes or minimizes).

Turns out that there is only one positive root (negative base makes no sense) and that's at B = e and it's a minimum.

Re:No longer binary? (1)

gardyloo (512791) | more than 6 years ago | (#22299298)

T = S*B^(ln(N)/ln(B))

To find the base that would result in the lowest search time, take the double derivative of T with respect to B and find the roots (peak and valley where search time either maximizes or minimizes).
First, I believe you mean roots of the _first_ derivative (not second) to find extrema. Second, I don't believe the heuristic model, because S*B^(ln(N)/ln(B)) has no dependence on B!

Re:No longer binary? (2, Interesting)

imgod2u (812837) | more than 6 years ago | (#22298012)

You'd be surprised how much of your computer isn't "binary" per se. If you have a modem, I think the standard is a base-16 transmission code. Flash memory currently contains 2-bits-per-cell cells. Hell, the quad-pumped signal going from memory to processor (if you have a Core 2 or P4) isn't "binary" per se.

Re:No longer binary? (0)

Anonymous Coward | more than 6 years ago | (#22307144)

This isn't a new idea.

ISDN used 2B1Q (Two bits, one quaternary) years ago.

Binary is just a lot easier to work with. We could make a base-10 computer, for example, but the complexity and tolerances involved isn't worth it.

Re:No longer binary? (1)

sm62704 (957197) | more than 6 years ago | (#22321798)

At one time they had analog computers, where the values were represented by voltages. I built a very, very primitive one in the 7th grade (actually it was more like an electric slide rule), but grownups made big complex ones.

The advantage of analog computers was that they had no rounding errors. The disadvantage was noise.

Where's the beef ? (0)

Anonymous Coward | more than 6 years ago | (#22295042)

More PR about potential of this technology, but no product released.
They promised a chip last year - where is it ?
Something smells bad...

Scientists double the speed of starships!! (1)

joebob2000 (840395) | more than 6 years ago | (#22295154)

The article says that Intel has just doubled the size of PRAM, which is nice, but PRAM will not be commercially viable for some time to come, so I the article, or at least the headline is somewhat sensational. I guess science journalists are still journalists.

When I am working on a design, I guess i could say that I increased the capability by an infinite amount at the moment when the first prototype is verified functional.

Just like the MLC flash. (1, Informative)

Anonymous Coward | more than 6 years ago | (#22295158)

The principle is similar to MLC flash (multi-level cell NAND flash), that also stores 2 bits per data element by using 4 different voltages. The tech behind this memory is completely different though.

Three states (1)

Yetihehe (971185) | more than 6 years ago | (#22295180)

If they found only one middle state, they could implement { true, false, file_not_found [thedailywtf.com] } enum

I didn't know Silverlight HAD storage capacity! (4, Funny)

ZombieRoboNinja (905329) | more than 6 years ago | (#22295392)

N/T

Bits and States Explained (1)

KiwiCanuck (1075767) | more than 6 years ago | (#22295498)

One bit of material can be switched b/w four states. Thus Base 4 logic is possible. However, the article indicated that each cell will be used for 2 bits. In this arrangement capacity is only doubled.

Re:Bits and States Explained (1)

AdrocK (107367) | more than 6 years ago | (#22296048)

If you take each cell individually, there are 4 states it can be in. (2 bits worth of states). But when you look at that in the whole array you have a base4 number.

I can see using this to jam more storage onto the device, then making a simple ciruit to convert the base4 to base2, but I don't ever see this being usable outside of storage; Unless there are some sort of quarternary logic gates that I dont know about.

Re:Bits and States Explained (1)

SuiteSisterMary (123932) | more than 6 years ago | (#22305952)

Unless there are some sort of quarternary logic gates that I dont know about.

Binary logic - yes, no.

Quarternary logic - Yes, probably, possibly, no.

Hmmm. 50 percent of the choices there are indeterminate; better stick with the trinary 'yes, maybe, no' model.

step backwards (1)

RiotingPacifist (1228016) | more than 6 years ago | (#22295870)

Storing the information in physical states of atoms, while it may lead to a practice increase in storage, means the theoretical limit of storage/volume is alot smaller. So unless we want to go one step forward and two steps backwards this looks like a major mistake as an area to do research in!

Re:step backwards (1)

Bender_ (179208) | more than 6 years ago | (#22296464)


And the alternative would be...?

Re:step backwards (2, Informative)

imgod2u (812837) | more than 6 years ago | (#22298386)

Not really. If you think about it, fundamentally there just as many states a bunch of atoms can arrange itself in as counting electrons. You're really only bound by Plank's constant in how resistive or conductive a collection of atoms are. Assuming you had a device sensitive enough to detect the variation.

The trick is, of course, in how fast you can change those states. I would imagine electrons are much easier to move than whole atoms. I understand how read speed for PCM is faster than a transistor but writing....I don't know.

Re:step backwards (1)

RiotingPacifist (1228016) | more than 6 years ago | (#22304254)

With electrons (im not sure how small the groups of atoms that can hold the electrons in a circuit can be but theoretically they could get down to 10s of atoms ( hell they could get down to storing it in 2 atoms (not 1 due to quantum effects)). But when you store the data in the arrangement of atoms you instantly loose the edges because they go crazy, then you need enough to store the pattern which i suppose wont be that many as its the just the energy levels of the electron see changing, but for it to be an electron see not just a multi-atom bond you'll need atleast 4 (probably 10s).

but thinking about it these limits are unlikely these limits can be reached for any sort of production.

the more I think of chemistry i wonder if anybody has tried spectroscopic storage, it would at low resolution it would give about 5/6 states for an electron to be stored in, at higher resolution people can reliable detect about 30 states. i suppose the problem is that its too hot at room temperature to keep the data stored for long ( although it is possible )

Another innovation? (1)

Mr.Fork (633378) | more than 6 years ago | (#22296266)

Oh look, another innovation that we will probably never see. I love new ideas - but instead of reporting on possible uses of technology science, how about actual science- stuff you will see this in 2 months reports. The "LOOK - SUPERBATTERIES WITH CARBON NANOTUBES - CHARGE IN SECONDS - 5 BAZILLION HOURS USE" kinds of articles are interesting, but not are they slashdotworthy. In Seinfeld's world, it definitely would not be sponge worthy.

Re:Another innovation? (1)

networkBoy (774728) | more than 6 years ago | (#22303180)

I've seen it.
In fact I have a USB key with PCM on it... A whopping 32K.
-nB

I for one... (1)

mac1235 (962716) | more than 6 years ago | (#22297234)

Welcome our new bulging fore-headed overlords! Oh, you meant the other kind of memory...

two more states, eh? (1)

porky_pig_jr (129948) | more than 6 years ago | (#22297464)

Shall we call them 'amorlline' and 'crystphous'?

Star Trek (1)

Meneth (872868) | more than 6 years ago | (#22304310)

So now we're actually gonna see storage capacity measured in "gigaquads"?
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