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New Memristor Makes Low-Cost, High-Density Memory

timothy posted more than 5 years ago | from the call-doctor-if-effects-last-more-than-four-hours dept.

Data Storage 86

KentuckyFC writes "A group of electronics engineers have discovered that a thin layer of vanadium oxide acts as a memristor, the fourth basic component of circuits after resistors, capacitors, and inductors that was discovered last year. At a critical temperature, a current passing through the layer causes it to change from an insulating state to a metal-like state, thereby changing its resistance (abstract). The effect lasts many hours — which is what makes the layer a memristor (a resistor with memory). The team says this could be scaled up to make resistive random access memory, or RRAM, at very low cost, from little more than layers of vanadium oxide."

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The flavour lasts forever. (0)

Ostracus (1354233) | more than 5 years ago | (#26406831)

"The effect lasts many hours â" which is what makes the layer a memristor (a resistor with memory). The team says this could be scaled up to make resistive random access memory, or RRAM, at very low cost, from little more than layers of vanadium oxide.""

If this was going into a SSD then it would have to last longer than that.

Re:The flavour lasts forever. (4, Insightful)

pines225 (1413303) | more than 5 years ago | (#26406845)

The article suggests use as resistive RAM rather than a solid state drive. As long as it doesn't need to be a permanent memory element it might be possible to refresh periodically on a schedule that's safely less than the lifetime of the state transition, i.e. boost the phenomenon every hour or two. Shouldn't cause much of a power hit.

The Memristor is NOT Fundamental (3, Insightful)

sup2100 (996095) | more than 5 years ago | (#26409073)

The memristor is is just a way to model nonlinear circuit elements and is one of many components in a nonlinear expansion for circuit modeling. See this paper [ieee.org] by Leon Chua, the memristor's inventor. Note that in this paper the fourth element of the four element torus is negative resistance and not the memristor. All of the publicity over the memristor has been (sucessfull) marketing by some researchers at HP. .

From the talk page for the memristor on wikipedia [wikipedia.org]

"Resistance, Capacitance and Inductance are regarded as fundamental because to each there corresponds a different picture of what is going on with the energy. Resistance refers to the loss of energy to Joule heating. Capacitance refers to storage of energy in the electric field. Inductance refers to storage of energy in the magnetic field.

If memristance is the "fourth fundamental" circuit element then memristors must do something with the energy they are imparted other than turn it into heat, or store it in electric or magnetic fields. So what do memristor supporters have to say about this? nothing. This is not surprising, since the concept of memristance stems from a purely mathematical argument bent on taming the current/voltage relationships of nonlinear circuit elements. The concept of memristance was invented out of convenience to avoid dealing with frequency-dependent (time-dependent) resistance, inductance, and capacitance. Thus the memeristor is not "fundamental", unless in your book fundamental is synonymous with convenient."

Re:The Memristor is NOT Fundamental (2, Informative)

Anonymous Coward | more than 5 years ago | (#26410035)

You sir are a troll.

Nobody is talking about resistance capacitance and inductance when talking about fundamental passive circuit elements.
There are four and only four passive circuit elements because they are defined in that paper as being the result of a relation between two of the four Fundamental Circuit Variables that are Voltage, Flow, Charge and Flux. So you can combine them in only 4 possible ways: The Resistor, The Capacitor, The Inductor and The Memristor(charge and flux).

So stop pretending thats all an HP Conspiracy.

Re:The Memristor is NOT Fundamental (1)

sup2100 (996095) | more than 5 years ago | (#26414487)

And you sir are a bigger troll. The paper I cited was not the original paper, but rather a follow up that puts the memristor in a better context. If you allow for various combinations of the derivatives and integrals of just current and voltage you can get a capacitor, resistor, inductor, and memristor. However you can also get the memductor and memacitor as well. This is all explained in that paper.

Why are combinations of voltage, current, charge and flux fundamental?

According to Websters fundamental is:
"a: serving as an original or generating source : primary (a discovery fundamental to modern computers) b: serving as a basis supporting existence or determining essential structure or function : basic"

Re:The Memristor is NOT Fundamental (0)

Anonymous Coward | more than 5 years ago | (#26416395)

And you sir are a bigger troll.

that's an implicit admission that you're a troll

Re:The Memristor is NOT Fundamental (1)

Teilo (91279) | more than 5 years ago | (#26420727)

Actually, I would like to understand this point as well, so I'm not sure about all this "HP Conspiracy" stuff. Sounds like an honest question to me.

If components were ever discovered that could couple Flow and Charge, or Flow and Flux, why would those not be "fundamental"?

Why was there only one missing circuit element? Why not three?

Re:The flavour lasts forever. (4, Informative)

drinkypoo (153816) | more than 5 years ago | (#26407015)

Not really; if you only have to refresh it once every four hours, you can do that for a LONG time on a battery. What I really see as the potentially big win is if they can get the speeds up to SRAM levels. Technically it would be a kind of DRAM, but if you only have to refresh it once every four hours, then just using a LRU scheme when allocating memory might mean that in practical usage you would never need a refresh anyway.

Re:The flavour lasts forever. (1)

MathFox (686808) | more than 5 years ago | (#26407187)

For a memory device that operates near in a 335-340K (65 C) temperature range you need a power budget for heating...

On a second thought, if one taps into the excess CPUs and GPUs heat. Still a bad fit for a mobile device though

Re:The flavour lasts forever. (0)

Anonymous Coward | more than 5 years ago | (#26407399)

>On a second thought, if one taps into the excess CPUs
In before NSA steals your idea and contacts all major chip makers to get this implemented.

But seriously, this could be a rather sneaky way to spy on people.
You might think you turned your computer off and destroyed RAM contents, but woops, you forgot the super sneaky ninja memristor RAM.

But on a happier note, session backup anyone?
Power cuts are no longer a problem, just bootup the backup.
While session backups are possible now, this memristor is much better because it will barely need any power to keep it refreshed over time.
Hell, you wouldn't even need to hibernate anymore, just kill the power, it auto-hibernates.
I'd rather see this as a hardware function, rather than OS.
This would be a godsend for portable devices.
I love my PSP for being able to just pause everything and turn off, then continue at any time.

Re:The flavour lasts forever. (1)

TheLink (130905) | more than 5 years ago | (#26408027)

"You might think you turned your computer off and destroyed RAM contents"

Even for conventional RAM, most of the contents stay for quite a while after power loss. And the colder the chips are the longer the duration.

Try turning off your computer and then immediately booting something that lets you peek into the RAM.

See: http://www.freedom-to-tinker.com/blog/felten/new-research-result-cold-boot-attacks-disk-encryption [freedom-to-tinker.com]

Re:The flavour lasts forever. (1)

timnbron (1166139) | more than 5 years ago | (#26411075)

Not really. It's on a microscopic scale, and at that level, a tiny current can vaporise silicon quite easily. That's how EPROMs work, and also why you need to be careful with static when handling circuits.

SHORT-TERM MEMORY (0)

Anonymous Coward | more than 5 years ago | (#26407605)

Think of the short-term memory in your brain. Not everything needs to be stored for a long time. Many things only need to be stored transiently, and this could be fine for that purpose.

From the original announcment (4, Insightful)

Baron_Yam (643147) | more than 5 years ago | (#26406885)

I thought the more exciting announcement was that memristors could be tripled up to create transistors that were (despite being tripled up) still much smaller than a standard transistor.

Then, there were bits about them supporting more than just binary states, which would increase complexity and density yet again.

Denser memory may be the first pratical consumer product, but if the other possiblities work out, I'm pretty sure that memory will also be the least significant.

Re:From the original announcment (3, Interesting)

TheKidWho (705796) | more than 5 years ago | (#26407377)

Memristors sound like an interesting method for creating analog electronic neural networks that learn...

Re:From the original announcment (-1, Offtopic)

Anonymous Coward | more than 5 years ago | (#26416619)

But the truly exciting thing about memristors is that they will allow the construction of flying cars that consume no fuel and provide the occupants with delicious hot pizza.

Re:From the original announcment (4, Insightful)

BudAaron (1231468) | more than 5 years ago | (#26408535)

When I went through electronics school in the late 40s there were resistors, capacitors and inductors. The real news here is that we have a totally new circuit element and heaven only knows where that can take us.

How is this a new element? (1)

Roger W Moore (538166) | more than 5 years ago | (#26413359)

The real news here is that we have a totally new circuit element and heaven only knows where that can take us.

Exactly how is this a fundamentally new circuit element? Most ohmic resistors change their resistance when a current is passed through them because the current causes heating and the heating changes their resistance. Since they take time to cool down they will have a (rather limited) memory too.

Yes this is a very interesting device and vastly more practical in terms of applications for memory but, unless it does something the articles do not mention, calling it a new fundamental circuit component seems wrong because it appears that you can mimic its behaviour using a well insulated piece of wire. Of course the wire is not very useful in terms of applications (outside electric heaters and light bulbs) but nevertheless it is a resistance whose value depends on previous current history.

Re:From the original announcment (1)

Fjandr (66656) | more than 5 years ago | (#26409625)

It would be interesting to see greater-than-binary states, though it would require a serious paradigm shift in many of the basic concepts taken for granted in computing.

Re:From the original announcment (1)

Hucko (998827) | more than 5 years ago | (#26410575)

Imagine a base 4 code!

Re:From the original announcment (1)

Sparky McGruff (747313) | more than 5 years ago | (#26412863)

Imagine a base 4 code!

It's been done. It's called DNA.

Re:From the original announcment (1)

Hucko (998827) | more than 5 years ago | (#26463985)

Well I never! Who would have thought! Really?

Damn it! There is supposed to be a brain governing those eyes of yours!

Melts like a chocolate bar (2, Interesting)

trolltalk.com (1108067) | more than 5 years ago | (#26406889)

" At a critical temperature,"

"Gee, I had it stored on this memsistor chip - but I left it in my shirt pocket, and my data melted."

The article doesn't say what temperature, so there's probably an issue there. Until that issue is solved, it's about as useful as write-only memory.

Also, looking at the required voltage (50 volts @ 0.6 amp), this is NOT going to be either high-density, or portable,or particularly energy-efficient.

Re:Melts like a chocolate bar (1, Interesting)

Anonymous Coward | more than 5 years ago | (#26406935)

Yes, the combination of 240W/byte and "critical temperature" looks like a problem. I hope they find a way to scale it down a lot.

Re:Melts like a chocolate bar (5, Informative)

Anonymous Coward | more than 5 years ago | (#26406979)

correction:
it's actually 0.6 mA, so that would be 0.24W/byte, and only during the (very short) write pulse. Still some work to do, but it could possibly end up more efficient than flash memory.

Re:Melts like a chocolate bar (1)

trolltalk.com (1108067) | more than 5 years ago | (#26407007)

The problem is that you're going to need a significant amount of insulator material at 50 volts, which means that the physical spacing will be much further apart than on today's devices. So we will see much higher latency, much more bulk, and because of the physical separation, much higher internal current consumption on the interconnects.

You won't be seeing this particular implementation displacing flash.

Re:Melts like a chocolate bar (1)

cheater512 (783349) | more than 5 years ago | (#26410257)

Well I know that SD cards can draw a few amps for very small periods of time.
So 0.6mA is a definate improvement.

Re:Melts like a chocolate bar (2, Interesting)

who knows my name (1247824) | more than 5 years ago | (#26406947)

well, in its current state, it probably won't be very good for much; but small additions of other elements will probably give you a compound which does what you want. After all, the properties of most alloys change quite remarkably with small composition changes.

Re:Melts like a chocolate bar (2, Interesting)

hairyfeet (841228) | more than 5 years ago | (#26407209)

Not to mention according to Wikipedia its main component is HIGHLY [wikipedia.org] toxic to humans. Considering we've been phasing out solder for an inferior replacement(which I personally believe will lead to MORE E-waste due to whiskers) just to cut down the poisons in our electronics I can't seen the green folks jumping onto something that is highly toxic AND water soluble. So IMHO this will end up a non starter before it even gets off the ground.

Re:Melts like a chocolate bar (1)

David Gerard (12369) | more than 5 years ago | (#26407521)

This is why it was nice to see that TiO2 works as well.

Fuse (0)

Anonymous Coward | more than 5 years ago | (#26406933)

resistance changing based on past... fuses do that too!

calling it passive doesn't seem right.

Also, this doesn't seem that novel. That's why it may be great!

Re:Fuse (5, Funny)

trolltalk.com (1108067) | more than 5 years ago | (#26407021)

resistance changing based on past... fuses do that too!

Yes. Resistance is fusile.

fourth type? (1, Insightful)

Anonymous Coward | more than 5 years ago | (#26406959)

I've seen this statement repeated about memristors. Many devices cannot be duplicated with L's, C's and R's (diodes, fuses, etc.). Those parts all have one thing in common, they are nonlinear and/or time varying. Memristors are not LTI and therefore not a "fourth type" of circuit element.

Re:fourth type? (1)

who knows my name (1247824) | more than 5 years ago | (#26407131)

That's a bit of a weird conclusion. It would make sense for it to be the 4th type; so that all components can be made up of the passive circuit elements.

Re:fourth type? (1)

frieko (855745) | more than 5 years ago | (#26407677)

There are four fundamental circuit variables: current, voltage, charge and flux.

Resistance relates voltage and current.
Capacitance relates charge and voltage.
Inductance relates flux and current.
Memristance relates flux and charge.

Thus M is the 'missing relation'. I'm not so sure I buy it either, but that's where they're coming from. Perhaps LTI was just an incredibly useful coincidence.

Re:fourth type? (1)

brizzadizza (1195159) | more than 5 years ago | (#26407773)

I like thinking of a memristor as the REAL flux capacitor :)

Re:fourth type? (1)

Stormy Dragon (800799) | more than 5 years ago | (#26407895)

You're forgeting there's two physical laws covering the relationships between charge and current (dq = i dt) and flux and voltage (dphi = v dt). The reasons memristors are not LTI is because (unlike the other three) they don't involve either of the time invariant quanties of current or voltage. Of course, one wouldn't expect them to be entirely like the other three. If they were, you could just make a memristor from resistors, capacitor, inductors and it wouldn't be a fundamental component. To put it another way: the fact R's, C's, and L's are LTI and M's aren't is precisely why you need M's--not all electrical circuits are LTI either.

Re:fourth type? (1)

644bd346996 (1012333) | more than 5 years ago | (#26408007)

And what about the other two relations (voltage and flux, and charge and current)? Are they for some reason not missing?

Re:fourth type? (1)

frieko (855745) | more than 5 years ago | (#26408263)

As stormydragon pointed out, I included the passives but forgot the two supplies:

A current source relates charge and current.
A voltage source relates flux and voltage.

Re:fourth type? (4, Informative)

Stormy Dragon (800799) | more than 5 years ago | (#26407823)

There are four basic quantities that are of concern in an electrical circuit: charge(q), current(i), voltage(v), and magnetic flux(phi). Those four quantities can be matched into pairs 6 different ways. Two of those pairs are time constrained by basic physical law: dq = i dt and dphi = v dt. Three of the remaining are determined by the properties of resistance(R), capacitance(C), and inductance(I): dv = R di, dq = C dv, and dphi = L di. Resistors, capacitors, and inductors are ultimately just devices that have a lot of one of those three properites and nearly none of the others.

It was speculated in the early 70's that there must be a fourth property, called memristance(M), that describes the 'missing' relationship: dphi = M dq. The memristor, then, is the corresponding device that has a lot of memristance but none of the other three properties. While memristance has been previously measured in complex systems, no one figured out how to build an feasible isolated memristor until just recently.

The four are considered fundamental in that none of the four can be built from a combination of the other three (e.g. you can't make a resistor from some combination of capacitors, inductors, and memristors) but any device can be built from some combination of the four (e.g. you CAN make a diode from L's, C', R's, and M's).

4th element? (0)

Anonymous Coward | more than 5 years ago | (#26406995)

But how does memristor differ from NTC or PTC resistors that change resistance on temperature (i.e. caused by current flow) and the effect lasts for a while (ultil the resistor cools down)?

Re:4th element? (1)

who knows my name (1247824) | more than 5 years ago | (#26407117)

"Unlike those more familiar elements, the necessarily nonlinear memristors may be described by any of a variety of time-varying functions. As a result, memristors do not belong to linear time-invariant (LTI) circuit models. A linear time-invariant memristor is simply a conventional resistor." - Wikipedia

It's all about linearity.

Re:4th element? (0)

Anonymous Coward | more than 5 years ago | (#26416193)

It's all about linearity.

That's what she said.

Re:4th element? (2, Informative)

as400master (1110293) | more than 5 years ago | (#26407185)

It differs in the fact that NTC and PTC resistance value depends on a current condition as where memristors resistance depends on a past event. In this example's case once the vanadium oxide becomes an conductor due to trigger voltage and current that trigger can be removed and it will remain a conductor for several hours. So you can see this is vastly different from NTC / PTC devices.

This will make the spooks happy (3, Interesting)

jridley (9305) | more than 5 years ago | (#26406997)

No more need to supercool RAM on seized computers in order to extract passwords - the RAM will just naturally hold state for hours.

If they're going to use this, (some) people are going to want to have more secure operating systems that don't leak security data all over the place.

Re:This will make the spooks happy (1)

JSBiff (87824) | more than 5 years ago | (#26407079)

You know, this is the first thing that crossed my mind too. I don't *want* my memory to 'remember' what's stored in it for hours. How is that a good thing?

"If they're going to use this, (some) people are going to want to have more secure operating systems that don't leak security data all over the place."

How do you design an O/S that doesn't load security data into RAM? How do you design an O/S that doesn't load your encrypted files into memory when you decrypt them to view or work on them?

AFAIK, with memory which holds state for hours, there's no real possible way to secure your data, if you load it, until the computer's been turned off for hours ( Can you maybe wipe it sooner if you cool the memory with a gas or liquid, since it sounds like this tech might be based on thermal changes?)

Re:This will make the spooks happy (2, Insightful)

Anonymous Coward | more than 5 years ago | (#26407157)

The situation is still somewhat rare that someone can go in and get all the data they like off your computer. They still need physical access to the computer before they can have a prayer at touching the data.

And heck, since DRAM is also capable of storing data for a couple of minutes after power is cut, you have the exact same problem as someone who has the new memresistor based ram.

Re:This will make the spooks happy (1)

lysergic.acid (845423) | more than 5 years ago | (#26407939)

magnetic hard drives store data for years. so do you also abhor hard drives?

if you're that paranoid, it's fairly easy to have the OS wipe all user & program data from the memory at shutdown. just because it's non-volatile doesn't mean it's not erasable or re-writable (or it wouldn't be very useful for computer memory). but most people who actually have a need for data security would simply encrypt the data just like people do on their hard drives.

and this is a good thing because memristors can replace transistors which are several times larger. and non-volatile memory has always had lots of useful application. this would provide high speed non-volatile memory with much higher density than conventional hard drives--when this technology matures, it could be used to build hard drives with 100 Gb per square centimeter.

additionally, when memristors catch up to the speed of DRAM, they can be used to build instant-on devices, both due to the high data density as well as its non-volatile memory state. you could turn off the computer each night to save power and then flip it back on each morning like you currently do with your monitor.

Re:This will make the spooks happy (1)

TheLink (130905) | more than 5 years ago | (#26408117)

"it's fairly easy to have the OS wipe all user & program data from the memory at shutdown"

If you're paranoid that doesn't help. People can just yank the power cord, or even pull the chips from their sockets.

Re:This will make the spooks happy (1)

ion.simon.c (1183967) | more than 5 years ago | (#26409363)

If you're *that* paranoid, you have your hardware locked in a good safe.

Re:This will make the spooks happy (1)

TheLink (130905) | more than 5 years ago | (#26414549)

If you are paranoid about keeping secrets, thermite is your friend.

It'll be a while before they develop tech for recovering stuff erased with thermite.

How to use it or rig it up is beyond the scope of this discussion :).

"Security RAM" (1)

Burz (138833) | more than 5 years ago | (#26408569)

if you're that paranoid, it's fairly easy to have the OS wipe all user & program data from the memory at shutdown.

IMO that is no solution. A system can be easily reset/halted before an OS has a chance to neatly shut down.

Creating "Security RAM" modules would be more effective: Equipped with a capacitor for power, they could self-wipe at the hardware level when they detect a reset signal or power interruption. Given the precarious nature of info esp. on laptops, one would think this category of RAM would have already been developed.

This can't be used for hard drives. . . (1)

JSBiff (87824) | more than 5 years ago | (#26440877)

I don't think this 'memristor' is suitable, at least in its current incarnation, to create Solid-State hard drives. Why not? It only retains your info for several hours. So, that essentially limits it to use as RAM. But RAM shouldn't store data for several hours.

The distinction here is that, on the hard drive, the data is always encrypted, but in RAM, it must be decrypted. I don't mind the hard drive storing the *encrypted* data for years, but I don't want my memory storing decrypted data for hours after I shut off my computer.

Now, that said, I could see, possibly, one application for this technology - used as a very large, very high-speed 'cache' in an otherwise conventional hard drive. That is, instead of the computer transferring data directly to/from the HD platters, it would read the data and write changes to the cache, then the hard drive would synchronize any changes in the cache back to the platters. Would, no doubt, speed up games and databases tremendously having a 100 GB high-speed cache. The key here is, if you have encrypted data, it is always encrypted in this cache (since the cache is just part of the hard drive), while your RAM is of a more 'conventional' nature that loses its state a few seconds or minutes after shutdown. Heck, with this tech, if your computer loses power, if the power comes back on within an hour or two, the hard drive could finish writing changes from the semi-persistent cache to the disk before bootup, so you minimize data loss.

Re:This will make the spooks happy (0)

Anonymous Coward | more than 5 years ago | (#26410529)

How do you design an O/S that doesn't load security data into RAM? How do you design an O/S that doesn't load your encrypted files into memory when you decrypt them to view or work on them?

The xbox 360 does encryption & checksumming of data in protected memory pages; it's fairly smart (and also fast enough that they can run fairly performant code on it). So it's not impossible, it might just be a bit slower on existing x86 chips

Re:This will make the spooks happy (0)

Anonymous Coward | more than 5 years ago | (#26407089)

You only need to write over the same memory again to erase the previous one.

Re:This will make the spooks happy (1)

Delwin (599872) | more than 5 years ago | (#26407585)

you'd think that would work for hard drives too wouldn't you? non-volatile memory can be read dozens of writes back. No reason to believe memristors would be different.

Re:This will make the spooks happy (1)

ion.simon.c (1183967) | more than 5 years ago | (#26409375)

No reason to believe memristors would be different.

Why?

Re:This will make the spooks happy (1)

andy_t_roo (912592) | more than 5 years ago | (#26412465)

non-volatile memory can be read dozens of writes back.

[citation needed]

seriously, i would interested in reading a paper where people have managed to do this - the closest thing to "dozens of writes" back that i know of is noting that different OS's had been installed, due to different formatting patterns, but i don't remember seeing anyone reporting the ability to extract that level of info from any hard drive.

Re:This will make the spooks happy (1)

Bigjeff5 (1143585) | more than 5 years ago | (#26419863)

Er, that's kinda the reason the DOD wipe spec requires three wipes at the bare minimum. One all 0's, one all 1's, and one a random sequence of 1's and 0's.

I don't remember where the paper was for this, but the reason is that when the magnetic arms write a 1 to the disc, it doesn't actually write a full 1, it's more like a .9, and when it gets written back to 0 it drops it to 0.1 (from a full 1). A significantly more sensitive magnetic reader can see for exampe a 0.8 as a 0.8, instead of picking it up as a 1 like your standard harddrive reader arm will. So by analyzing the pattern of charges on the disc, the data can be successfully re-built.

With that accuracy, they can go back many, many writes to re-build the data. It is expensive, and dificult, but it is doable. That is how data-recovery services work, and it's also why they cost upwards of $5,000 to retrieve your data.

They tell you not to write over your harddrive when you want data back because it is significantly easier to re-build data that has only been written over once or twice, not because it is impossible to rebuild data that has been written several times.

In fact, regular old data recovery software can get back stuff that has been written over once or twice. The only way I know of to completely eliminate the possibility of retrieval is to de-magnetize the drive. Though there is a program that will wipe it 30 times randomly, and that's pretty darn close to impossible to recover.

Re:This will make the spooks happy (1)

Daimanta (1140543) | more than 5 years ago | (#26407119)

Then we probably get software/OS that "wipes" the RAM when it shuts down.

Not vaporware... (4, Informative)

Menkhaf (627996) | more than 5 years ago | (#26406999)

As the memristor was developed in HP Labs while working on fabrication techniques for "normal" memory, the fabrication technology is already here. It'll only be a short while before we'll see memristors in consumer products.

"HP prototyped a crossbar latch memory using the devices that can fit 100 gigabits in a square centimeter.[10] HP has reported that its version of the memristor is about one-tenth the speed of DRAM.[27]"

http://en.wikipedia.org/wiki/Memristor#Potential_applications [wikipedia.org]
[27] http://www.nytimes.com/2008/05/01/technology/01chip.html [nytimes.com]

Re:Not vaporware... (5, Funny)

trolltalk.com (1108067) | more than 5 years ago | (#26407053)

As the memristor was developed in HP Labs while working on fabrication techniques for "normal" memory, the fabrication technology is already here. It'll only be a short while before we'll see memristors in consumer products.

"HP prototyped a crossbar latch memory using the devices that can fit 100 gigabits in a square centimeter.[10] HP has reported that its version of the memristor is about one-tenth the speed of DRAM.[27]"

So, knowing HP, we can expect memristors that need a new cartridge to "refill the memory" every few weeks.
And your initial memsistor will have just a "starter cart" that only accesses 1/4 the data.
And for best performance, you should only use genuine HP Brand electricity.
And random blocks of memory in the memristors won't be accessible under linux. Especially when you try to send data via a wireless connection.

Re:Not vaporware... (2, Funny)

daveime (1253762) | more than 5 years ago | (#26408977)

You forgot to mention the HP Memristor (tm) driver software, which despite being about 335 bytes in size, will come bundled in an installed package that is 37MB, just so HP software can show pointless splash screens and randomly create services and daemons that appear to serve no purpose whatsoever, while STILL not being able to cancel the printing of a document without cycling the power.

Re:Not vaporware... (2, Interesting)

ScrewMaster (602015) | more than 5 years ago | (#26409517)

You forgot to mention the HP Memristor (tm) driver software, which despite being about 335 bytes in size, will come bundled in an installed package that is 37MB, just so HP software can show pointless splash screens and randomly create services and daemons that appear to serve no purpose whatsoever, while STILL not being able to cancel the printing of a document without cycling the power.

And which will periodically send copies of your memory to a remote HP server in order to "improve the customer experience."

HP is not what it once was. Thanks for that, Carla.

Re:Not vaporware... (1)

Glonoinha (587375) | more than 5 years ago | (#26407249)

Although the speed is 1/10th that of DRAM, it appears that they can crank the density up and reduce the refresh rate. It still has effectively zero latency, which puts it squarely in the space between core memory and hard drive storage.

Personally I see this as a future replacement for the iRAM and other such 'ramdrives' - with the bandwidth limits of the SATA interface, a device of these interlaced across four or eight banks would make a very sweet solid state working storage area. It would need to be transient data, of course, such as a swapfile, temporary file storage for your browser, tempdb like space for your database, working storage for your graphics apps, etc - but still, that's where I see it coming into play.

Re:Not vaporware... (1)

dimethylxanthine (946092) | more than 5 years ago | (#26409441)

with the bandwidth limits of the SATA interface"

Right, cos who would need transfer speeds faster than 300MB/s...

--- "Determine never to be idle...It is wonderful how much may be done if we are always doing." -Thomas Jefferson

Wow, you learn something every day! (0, Redundant)

EmagGeek (574360) | more than 5 years ago | (#26407017)

I didn't know that resistors, capacitors, and inductors were only discovered last year. Wow, that's really cool! Now we can start making really cool electrical gizmos as opposed to what we've been using these last hundred years or so.

Practical relevance of this: Zero (4, Informative)

Bender_ (179208) | more than 5 years ago | (#26407205)

Sorry to be so harsh, but the specific experiment reported here is of little to none value outside of science. Why?

Hysteretic resistive switching in metal oxide systems is a well known phenomenon (RRAM) and occurs in all transition metal oxides with noble eletrodes. This is what has been recristened as "Memrestor" by HP. It is widely agreed upon that this switching mechanism is due to a redox reaction where oxygen is added or removed from the insulator. The specifics (filament, interfacial barrier lowering etc.) are still subject of current research though.

The experiment in the paper takes a slightly different approach: vanadium oxide has a very interesting property where its resistance switches apruptly by orders of magnitude at a certain temperature due to a reorganisation of its electronic structure. This phenomenon is known as metal to insulator (MTI) transition and has been research for at least 50 years.

The MTI has a hysteretic behavior which means that it retains its state if you vary the temperature only a little above or below the critical MTI temperature Tc. The researchers have now shown that if you keep the temperature of the system close to Tc, you can use an additional electric current to switch the resistivity of the system. A possible explanation could be self heating.

Why is it useless for practical application?

1) The phenomenon instrinsically only works at a certain temperature. Deviations by fractions of degrees K will destroy all information.

2) As far as I can see they only demonstrated electrical switching into one direction. To erase the memory both would be required.

All in all a nice experiment, but again with typical university style hype, piggybacking on the Memristor craze.

I am also relatively certain that current driven MTI switching has been reported before. I am aware of a couple of experiments where a field switched MTI transition was proposed for transistors. Those devices should exhibit exactly the same hysteresis and "memory" properties.

Re:Practical relevance of this: Zero (1)

Bender_ (179208) | more than 5 years ago | (#26407221)

s/MTI/MIT

It is NOT a fourth basic component (0)

Kupfernigk (1190345) | more than 5 years ago | (#26407225)

This is pure marketing hype. The three basic components - resistance, capacitance and inductance - are time invariant and their values in a theoretical circuit are constant. The behaviour of an invariant circuit topology to an electrical stimulus can be completely predicted and will always be the same for that stimulus.

These things are not invariants. They are in the same class as magnetic core memories/hard drives or DRAM, i.e. they retain for a time a change in their electrical state. They may be a better future short term memory technology than either, though this is yet to be shown. But there is nothing fundamentally theretically new about them.

Re:It is NOT a fourth basic component (5, Informative)

Hodapp (1175021) | more than 5 years ago | (#26407343)

i = current
q = charge
V = voltage
phi = magnetic flux

dq = i dt (current)
dphi = V dt (voltage)
dV = r di (resistance)
dq = C dv (capacitance)
dphi = L di (inductance)
(see http://www.spectrum.ieee.org/may08/6207 [ieee.org] )
It was hypothesized that some device should exist that connects charge and flux, and follows the relationship: dphi = M dq. This is "memristance." It was predicted in 1971 as the "fourth basic circuit element"; see: http://ieeexplore.ieee.org/xpl/freeabs_all.jsp?arnumber=1083337 [ieee.org]
They were fundamentally theoretically new then. They just had not been physically realized and connected with that theory until recently.
Please don't dismiss them as "pure marketing hype" without some research.

Re:It is NOT a fourth basic component (2, Insightful)

brendank310 (915634) | more than 5 years ago | (#26407441)

The GP was commenting on the fact that a memristor is not linear time invariant element. So it isn't as basic as the other 3. The other three can be analyzed using fairly simple analysis, while the memristor seems to introduce state to a circuit.
Circuits that are time linear, and time variant are more difficult to analyze on a broad scope than those that are LTI.

Re:It is NOT a fourth basic component (1)

Hurricane78 (562437) | more than 5 years ago | (#26407543)

So according to you,

V dt = dphi = L di?

Or is this wrong? And if yes, why?

P.S.: Slashdot, please fix your UTF-8 support! I have phi on my keyboard, but your system ignores it. And I also tried φ, which did not work either.

Re:It is NOT a fourth basic component (2, Insightful)

Hodapp (1175021) | more than 5 years ago | (#26407661)

So according to you,

V dt = dphi = L di?

Or is this wrong? And if yes, why?

Well...
V dt = L di
(V dt)/dt = (L di)/dt
V = L di/dt ...which is the standard formula they give for inductors, isn't it?
Or did I totally miss your point?

Re:It is NOT a fourth basic component (1)

neomunk (913773) | more than 5 years ago | (#26408145)

I think the GP was doing his or her math homework, and you just gave him/her the answer to the extra-credit question. :-)

Re:It is NOT a fourth basic component (1)

Hurricane78 (562437) | more than 5 years ago | (#26412251)

No. I actually have no idea of what he's talking, because I have nearly no knowledge of electrical engineering. (I'm a software guy.)

I just recognized that you could combine those two equivalents that he wrote, and wanted to find out if I were right about EE at least once in life. ;)

Re:It is NOT a fourth basic component (1)

Bender_ (179208) | more than 5 years ago | (#26407627)

not been physically realized and connected with that theory until recently.

Actually that is not true. Resistive switching had been demonstrated even before 1971 (there are some examples on Wikipedia).

Let's see, what happened here?

1) Someone found resistive switching
2) Someone developed a trivial algebraic model of a resistor with a memory effect
3) Someone with good marketing skills and connections combines 1) and 2) and manages to ram a paper into Nature.
4) Lots of press hype ensues
5) Profit? nope. Novel? nope. Publicity? Way too much, given for the nonevent. Application? Was already there at 1) in form of RRAM.

Did anybody notice that a new class of high temperature superconductors was found recently? Or multiferroics? Or commercial availability of phase change memory (PCRAM), which may render RRAM obselete before it went anywhere?

Assuming you are correct, please explain (2, Interesting)

Kupfernigk (1190345) | more than 5 years ago | (#26409365)

OK, postulating the component you describe, I understand that in simple terms it would store or release charge (like a capacitor)as the magnetic flux through it changed, or it would modify a magnetic field based on the charge added to or removed from it. As field strength and charge are potentials, under static conditions it would have a particular magnetic field around it based on charge stored. Is this correct? you state that the equation is d(phi)=M*dq.

How does this relate to a resistor which undergoes a discontinuous resistance change under critical conditions? Can you explain how it relates to the advertised device? Where is the charge being stored? Please continue to assume that I'm stupid, and explain the reasoning. My electromagnetic theory is thirty five years in the past now.

Re:Assuming you are correct, please explain (2, Interesting)

dkf (304284) | more than 5 years ago | (#26410109)

How does this relate to a resistor which undergoes a discontinuous resistance change under critical conditions? Can you explain how it relates to the advertised device? Where is the charge being stored? Please continue to assume that I'm stupid, and explain the reasoning. My electromagnetic theory is thirty five years in the past now.

No electronic component has truly discontinuous behavior under critical conditions. Even physical switches have rather complex transients (which is why they need a debounce circuit), and transistors are interesting analog devices. It's just that they're non-linear devices and (in computers) they're mostly used biased so that the circuits have (to a good approximation) binary behavior; the prerequisite for that is non-linearity.

Now, if there was an effect that was previously a theoretical one, or at best a lab curiosity, and HP have managed to figure out how to turn it into something practical, then more power to their elbows. It'll be very interesting to see what the electronic engineers of the world do with it.

I'm sorry, I meant step function (1)

Kupfernigk (1190345) | more than 5 years ago | (#26414527)

You are of course correct. However, you seem to have missed what this is all about. My original comment was that the effect described (a switch between two resistance states) is not a "fourth electronic component". The GP says that the "memristor" in fact follows the rule d(phi) = M*dq. This is a linear function relating charge stored to magnetic field, and is nothing at all to do with a two state resistive device.

It looks to me as though there is a confusion. The original vanadium oxide film when nonconducting can store supplied charge in a magnetic field. But the switching behaviour doesn't seem to be anything to do with this, unless there is some explanation not given in the article.

I guess someone from man.ac.uk ought to be interested in electronic storage, seeing as how you started off using cathode ray tubes:-)

Acessing the iee journal article (1)

shadeshope (1341571) | more than 5 years ago | (#26575449)

Thanks for citing the ieee paper http://ieeexplore.ieee.org/xpl/freeabs_all.jsp?arnumber=1083337 [ieee.org] . Unfortunately I am not a current member. Can the entire paper be found anywhere else online? Thanks this is a very interesting discussion.

Re:It is NOT a fourth basic component (0)

Anonymous Coward | more than 5 years ago | (#26407407)

Please downmod the misinformative parent post.

Grammar (0, Offtopic)

matt_martin (159394) | more than 5 years ago | (#26407811)

> resistors, capacitors and inductors that were discovered last year

Wow, I don't know how we made ANYTHING until last year !

Patentability (2, Insightful)

kimvette (919543) | more than 5 years ago | (#26408191)

. . . just posting this in the very unlikely chance that a competent USPTO employee (I know you're out there) is not only reading this thread but also is assigned related patents.

IMHO this kind of development is worthy of a patent; it includes a brand-new type of component, with no prior art in a single component appearing to exist, and a method by which it is manufactured.

Now, I expect patent trolls will start the patenting insanity with "it's a PDA, but with memristors" and "it's a phone, but with memristors" and "it's an instant-on PC, but with memristors" and in all of those cases I would say that the patent should not be allowed, because those are "innovations" which are obvious to those skilled in the art.

Also, the software to store to memristors should not be patentable. "method by which data is semi-permanently stored in a memristor-based storage device" should not be patentable, because that skill (putting data in memory or storage) is obvious to every literate computer user, let alone software engineers.

Mr cpcty... (0)

Anonymous Coward | more than 5 years ago | (#26409355)

at the cost of dropped vowels?

----
"A strong positive mental attitude will create more miracles than any wonder drug."
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