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Intel's Haswell Moves Voltage Regulator On-Die

MojoKid writes "For the past decade, AMD and Intel have been racing each other to incorporate more components into the CPU die. Memory controllers, integrated GPUs, northbridges, and southbridges have all moved closer to a single package, known as SoCs (system-on-a-chip). Now, with Haswell, Intel is set to integrate another important piece of circuitry. When it launches next month, Haswell will be the first x86 CPU to include an on-die voltage regulator module, or VRM. Haswell incorporates a refined VRM on-die that allows for multiple voltage rails and controls voltage for the CPU, on-die GPU, system I/O, integrated memory controller, as well as several other functions. Intel refers to this as a FIVR (Fully Integrated Voltage Regulator), and it apparently eliminates voltage ripple and is significantly more efficient than your traditional motherboard VRM. Added bonus? It's 1/50th the size." Update: 05/14 01:22 GMT by U L : Reader AdamHaun comments: "They already have a test chip that they used to power a ~90W Xeon E7330 for four hours while it ran Linpack. ... Voltage ripple is less than 2mV. Peak efficiency per cell looks like ~76% at 8A. They claim hitting 82% would be easy..." and links to a presentation on the integrated VRM (PDF).

237 comments

  1. excited by Anonymous Coward · · Score: 5, Funny

    come guys, comment, so I know how excited I should be

    1. Re:excited by Anonymous Coward · · Score: 4, Funny

      The 1.2V regulator will actually produce 1.199988484939848 volts !

    2. Re:excited by girlintraining · · Score: 0, Offtopic

      come guys, comment, so I know how excited I should be

      Sorry, we were just reading about how the IT crowd will have one final episode, and really rather didn't care about Intel changing something about their product that likely nobody would have ever noticed if some marketdroid hadn't palmed the new Dice overlords a few scheckles to get it on the slashdot front page. So, umm, if you want to know how excited you should be, well... You know how happy a three year old is when they go pee-pee in the grownup toilet all on their own? Imagine you're mom, and the kid is thirty-five, retarded, and you take care of him.

      You should be as excited as mom is.

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    3. Re:excited by Anonymous Coward · · Score: 1

      Ahh Slashdot, the very finest in jokes that were funny in the mid 90's... And thus concludes my bi-monthly check to see if it has gotten any better.

    4. Re:excited by Lotana · · Score: 4, Funny

      All good guys, he is gone. We can go back to our regular insightful, interesting and funny posts again.

    5. Re:excited by Anonymous Coward · · Score: 1

      First post?

    6. Re:excited by Anonymous Coward · · Score: 0

      They measure it to the femtovolt?! You should be SUPER excited!

    7. Re:excited by Anonymous Coward · · Score: 0

      You have to admit that FIVR looks pretty similar to FDIV, though !

  2. sinking heat? by p51d007 · · Score: 4, Interesting

    with the on die regulator, won't that area of the chip be a tad warmer than the rest of the chip, or will the heat be a moot point?

    1. Re:sinking heat? by Anonymous Coward · · Score: 0

      with the on die regulator, won't that area of the chip be a tad warmer than the rest of the chip, or will the heat
      be a moot point?

      Heat hopefully won't be an issue. Let's hope heat output scales at least somewhat with component size that is 98% smaller.

      It'd be nice to see the VRM using the 22nm lithography process but when a processor is quoted as Xnm, this is the smallest feature size on the processor, not all features on the processor - unfortunately.

    2. Re:sinking heat? by dotgain · · Score: 2

      Heat hopefully won't be an issue. Let's hope heat output scales at least somewhat with component size that is 98% smaller.

      And why would it do that? A given voltage drop multiplied by the current through it equates to a certain wattage of heat dissipation, regardless of the size of the package.

    3. Re:sinking heat? by adolf · · Score: 2

      Your broad generalization is only if it is a linear regulator. Switch-mode regulators change the game. TFA doesn't seem to indicate which it is.

    4. Re:sinking heat? by Anonymous Coward · · Score: 0

      certain wattage of heat dissipation

      Watts is used to measure power, not heat.

    5. Re: sinking heat? by Anonymous Coward · · Score: 0

      that scaling is only true for linear regulators. if they managed to put a switching regulator on the die then the heat dissipation is based on the efficiency and the output power. that efficiency will scale with input voltage but definitely not by as much as a linear regulator. the efficiency is going to scale more with the topology of switching regulator they choose and their ability to limit the resistance of the regulator components and their leakage.

    6. Re:sinking heat? by __aaltlg1547 · · Score: 5, Informative

      It's a switch-mode (Buck) regulator. You can tell from the efficiency curve and the fact that it requires an inductor. It is more efficient than a linear regulator and less efficient than a good external Buck regulator. However, being on-chip it will regulate the voltage better because there won't be significant I*R drop between the regulator output and the load. And as they mention, the cooling fan will be right on top of it, so it is more effectively cooled than an external regulator typically is.

    7. Re:sinking heat? by AvitarX · · Score: 1

      Heat is energy, and it dissipates over time, energy / time = power.

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    8. Re:sinking heat? by viperidaenz · · Score: 3, Insightful

      The voltage regulation issue can easily be solved by having a feedback connection from the die to the external VRM.
      There are only two benefits I can see:
      1) Higher voltage in to the chip means lower current, which saves power. You I*R formula is slightly wrong, its actually I^2 * R, double the current means 4x the power loss.
      2) Lower system cost. the more crap that gets stuffed on the die/in the chip, the less is required on the board. That means fewer components, smaller board area and quicker assembly.
      There are of course other benefits that only benefit Intel
      a) Fewer external components means they can charge more for their chip without effecting system cost.
      b) smaller system = happier customer = will pay more
      c) If it does actually result in lower power, then you get more performance or more battery life = customer will pay more

    9. Re:sinking heat? by adolf · · Score: 1

      I figured that it must be that way, but with the power required by a CPU such a regulator must either very noisy and/or require substantial capacitance and/or use ridiculously high frequencies.

      So. Filtering? These might be the most expensive mass-produced caps in the world if they're also on-die.

    10. Re:sinking heat? by viperidaenz · · Score: 2

      Watts is used to measure heat.

      http://en.wikipedia.org/wiki/Thermal_resistance

      The unit is "degrees per watt". As in "5C/W" = "this heat sink will rise by 5 degrees dissipating 1 watt"

    11. Re:sinking heat? by cats-paw · · Score: 1

      seems like they could have beat the IR drop by simply bringing out sense lines.

      it's not obvious too me how this helps intel.

      unless of course the motherboard makers aren't doing a very good job with those external regulators because the intel chips now require such high performance regulation.

      that's a distinct possibility.

      --
      Absolute statements are never true
    12. Re:sinking heat? by SuricouRaven · · Score: 1

      I rather doubt Intel managed to cram the inductor on-chip too, so doesn't that defeat the point?

    13. Re:sinking heat? by kasperd · · Score: 1

      with the on die regulator, won't that area of the chip be a tad warmer than the rest of the chip, or will the heat be a moot point?

      The summary says 76% efficiency. That would mean 24% of the energy you put into the chip is turned into heat by the voltage regulator. Sounds like a valid concern to me.

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    14. Re:sinking heat? by wikdwarlock · · Score: 1

      Nope, they actually did get on-chip magnetics and inductors. They're the Intel folks, afterall.

      --

      "I must not fear. Fear is the mind killer." -Bene Gesserit Litany Against Fear
    15. Re:sinking heat? by Anonymous Coward · · Score: 0

      76% efficiency (24% waste) is for 1 cell i assume for 20 cells it would be 1.2% waste ( 98.8% efficiency)?

    16. Re:sinking heat? by ericloewe · · Score: 1

      Don't underestimate Intel's ability and will to put all kinds of things on-chip.

    17. Re:sinking heat? by kasperd · · Score: 1

      76% efficiency (24% waste) is for 1 cell i assume for 20 cells it would be 1.2% waste ( 98.8% efficiency)?

      What sort of construction reduce waste by adding cells?

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    18. Re:sinking heat? by Luckyo · · Score: 1

      Considering the significant heat output of the cores themselves, VRM's heat output is likely trivial in comparison and will be vented into the heat sink alongside the rest of heat.

    19. Re:sinking heat? by dingleberrie · · Score: 1

      You I*R formula is slightly wrong, its actually I^2 * R, double the current means 4x the power loss.

      You both have good points, but I think that Shavano didn't mean power.
      He meant the Voltage (I*R) drops from the external chip where R comes from the losses of extra traces and pins before it gets to the CPU. The reduction in R means that a variation in I will cause less variation in V as seen by the CPU.

    20. Re:sinking heat? by Meeni · · Score: 1

      Voltage regulators fail often. Wouldn't that decrease the operational life of the processor (I know, a concern only for people that keep operating old junk for a long time, the processor will be well past its prime when it would fail, but still a valid concern for long amortization systems).

    21. Re:sinking heat? by jcdr · · Score: 1

      Voltage regulators fail often.

      And you want to replace them with what exactly ?

      In my experience the capacitors and transistors are the two parts that can die early, mostly because of thermal stress, due to slow chemical change of some of there properties: the resistance tend to increase and the capacity tend to decrease. I really don't know anything about the chemical stability of the components integrated into the die, but I can see some interesting points: 1) The capacitors are composed of the metal layers of the die. There is no electrolyte involved, so there must be very reliable, like ceramic capacitors. 2) The transistors will probably be cooled more efficiently than in most actual motherboard.

      So there is possibility that this new design will increase the reliability. This argument is still speculation as there likely rely on a first stage standard switching power supply to lower the voltage at an acceptable level for the die...

    22. Re:sinking heat? by InvalidError · · Score: 1

      The voltage regulation issue can easily be solved by having a feedback connection from the die to the external VRM.

      Not quite. Part of the reason why Intel went for FIVR is due to response time.

      Haswell can change power states about 10X faster than SB/IB and needs a VRM that can adjust voltage outputs that much faster to minimize wasted power and time. The motherboard's VRM has a response time measured in tens if not hundreds of microseconds due to how far away it is from the load, how large its components are and how slow it is operating relative to its load - Haswell has time to wakeup/sleep a few times in the time it take most regulators to adjust to SB/IB's state changes. FIVR's much more tightly coupled components and load bring response time down to microsecond scale to better accommodate Haswell's highly dynamic power usage so the motherboard's VRM does not need to respond impossibly quickly.

      FIVR being only ~82% efficient (at least in Intel's experimental design) is somewhat of a disappointment but to be expected when converting low voltage to even lower voltage at high currents. Since there isn't much that can be done about reducing losses on the output side, the most obvious way to improve losses on the input side would be to increase input voltage. 3.3V and 5V seem like logical next steps up for desktop parts if they can be accommodated.

    23. Re:sinking heat? by Meeni · · Score: 1

      The only difference is that before you replaced the main board for a smaller price than getting a new processor. But yes, maybe the fab process will improve reliability overall. Just asking questions.

    24. Re:sinking heat? by InvalidError · · Score: 1

      Poorly designed regulators with poorly chosen components fail often. What kills most PSUs in my experience is failure to use capacitors with adequate AC current ripple rating which causes capacitors to heat up, dry up and inevitably fail. I have repaired a bunch of such PSUs over the years using capacitors rated for 3-4X higher AC ripple and none of them have failed afterward to my knowledge - two of these are in my PCs which have been on almost 24/7 for 5+ after the repair. The oldest of my refurbished PSUs is about 13 years old (11 since repair) and was still kicking when I retired the PC it was in late last year.

      It is pretty sad that less than $1 on the BOM separates a PSU/VRM that lasts 1-3 years from one that lasts 7+ years.

      Expertly designed regulators with quality components rarely fail unless there is a catastrophic failure upstream or downstream from them.

      My bet is that FIVR will lead to a net reliability improvement by taking the most delicate part of voltage regulation on-die where Intel has full control over quality using ceramic/glass capacitors, inductors and MOSFETs all tightly integrated in an equally tightly controlled process.

    25. Re:sinking heat? by InvalidError · · Score: 1

      The other efficiency quote says ~82% with planned design tweaks and even better efficiency hinted at for future iterations.

      In any case, if individual cells have 76% peak efficiency, it does not matter how many of them there are, the overall regulator cannot be more than 76% efficient. However, it can be less efficient if the load drifts too far off the cell's peak efficiency point which is 8A/cell in the prototype so FIVR would need to enable/disable cells to keep average load as close to that as possible or at least within the 5-12A range if FIVR behaves like the IVR prototype, which would keep efficiency above 72%.

    26. Re:sinking heat? by viperidaenz · · Score: 1

      True. The voltage drop can be compensated for by sensing at the point of load, not the regulator output though.

    27. Re:sinking heat? by Anonymous Coward · · Score: 1

      Watts is used to measure heat.

      http://en.wikipedia.org/wiki/Thermal_resistance

      The unit is "degrees per watt". As in "5C/W" = "this heat sink will rise by 5 degrees dissipating 1 watt"

      No, watts are not used to measure heat. As the very wiki page you linked says, it's thermal resistance which is measured in units of degrees per watt. Thermal resistance is not the same thing as heat! Temperature isn't either.

      The SI unit used to measure heat (and all other forms of energy, including electrical and mechanical) is the joule. One joule is enough heat energy to raise the temperature of one gram of water by 0.24 degC. (A more precise relationship: 4.186J raises the temperature of 1 gram of water by 1C.)

      Watts are the unit of power, and they're a rate unit derived from joules. 1 W = 1 J/s. So if you run a 10W heater for 10 seconds and all that heat is captured by 100g of water, you've delivered 100J of heat to the water, which raises its temperature by 0.24 C since there's 1J per 1g water. (The general equation for this: Q = c * m * delta-T, where Q is the heat energy, c is a constant which varies by the material, m is the mass of the material, and delta-T is the change in the material's temperature.)

    28. Re:sinking heat? by __aaltlg1547 · · Score: 1

      Imperfectly. The additional inductance and delay result in poorer regulation.

    29. Re:sinking heat? by viperidaenz · · Score: 1

      delay not so bad when the regulator is running at ~1mhz.
      transient response for switching regulators comes mostly from the output caps.

    30. Re:sinking heat? by Anonymous Coward · · Score: 0

      You don't need inductors. Gyrators work just as well and they're easily fabricated in silicon for an integrated regulator. For external regulators a lump of wire is cheaper than the extra silicon (an analog inverter plus capacitor).

    31. Re:sinking heat? by Anonymous Coward · · Score: 0

      Not just the heatsink.

      All thermal junctions are rated this way (a heatsink is a thermal junction between metal and air). Techs look at all the intervening stages as well as the heatsink (there's a surprisingly high termal resistance between the chip and thermal spreader, f'instance), in order to enure that Tj(max) isn't exceeded under most circumstances - as well as looking at overall operational time, bearing in mind that a device's lifetime is halved for every 10C rise in Tj

      I know what kind of heatsink I'd like to see on a CPU (maximising surface area is the key), but I suspect it's impractical for mass production as well as being a dust trap.

  3. Heat by girlintraining · · Score: 4, Interesting

    Intel refers to this as a FIVR (Fully Integrated Voltage Regulator), and it apparently eliminates voltage ripple and is significantly more efficient than your traditional motherboard VRM. Added bonus? It's 1/50th the size."

    I have yet to come across a voltage regulator that doesn't run hot. Typically, it's one of the hottest components in an electrical circuit. And we're integrated this into a slab of silicon already well-known for getting so hot it can catch fire?

    Can someone please tell me why this is a good idea, because all of my experience in electrical engineering says that when things heat up, they become more unstable and prone to failure, and the one thing you do not want going critical is your voltage regulator. If that goes, the whole computer catches fire.

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    1. Re:Heat by Anonymous Coward · · Score: 3, Interesting

      Most likely, because it's integrated into the CPU itself, the voltage regulator can be made more efficiently and thus save power and heat etc. Discrete parts have their limitations, and doing it on-die might just mitigate that.

    2. Re:Heat by Anonymous Coward · · Score: 4, Informative

      You're going to run into that heat anyway, whether it's on the motherboard in general or on the CPU. You can't win. But at least it's better to have heat build-up near a heat-sink, so for high-power conditions it might actually be better to put it on the CPU. I'm also an electrical engineer, but thermals are really a mechanical engineer's realm, so I can't run numbers for you.

    3. Re:Heat by rrhal · · Score: 2

      Being 1/50th the size it will be welcome on mobile devices. Not sure that its a good thing for your gaming desktop.

      --
      All generalizations are false, including this one. Mark Twain
    4. Re:Heat by fuzzyfuzzyfungus · · Score: 5, Informative

      My suspicion(if only for die-space reasons, it isn't purely cosmetic that contemporary VRMs occupy a substantial amount of board space) is that is this a 'marketecture' summary of Intel moving some additional voltage adjustment and power gating functions on die, to support dynamic adjustment of power to the greater number of components(multiple CPU cores, possibly independently clocked, GPU, RAM controller, PCIe root, etc.); but we'll still see a bunch of chunky power silicon under serious heatsinks clustered around the CPU socket.

      Given that much of the contemporary power savings are achieved by superior idling, rather than absolute gains in maximum power draw, Intel is either going to have to keep moving power regulation on die, or start dedicating even more pins to tiny voltages at nontrivial currents, with the associated resistive losses; but that won't necessarily change the fact that the circuitry that brings the 12v rail down to what the CPU wants is a pretty big chunk of board.

    5. Re:Heat by Virtucon · · Score: 4, Interesting

      Well even at 10W I'm wondering how they'll address the heat.
      With the density of circuits in the adjacent silicon I would wonder how they're providing enough isolation to prevent it from becoming a very small brick.

      --
      Harrison's Postulate - "For every action there is an equal and opposite criticism"
    6. Re:Heat by RightwingNutjob · · Score: 1

      I can see the logic behind shortening the length of the wire carrying 'clean' power and getting it away from all the other components (read: noise sources) on the motherboard. It also takes the thinking burden away from the chip integrator and motherboard designer (which is a non-negligible bonus for both marketing and engineering).

    7. Re:Heat by Barlo_Mung_42 · · Score: 1

      I'm not an EE so I won't pretend to fully understand this particular case but I like it when tech companies reach a bit and try something hard. This may or may not be a good idea but I'm still excited about it.

    8. Re:Heat by Anonymous Coward · · Score: 0

      TFA says it has better efficiency, but it's not clear how much better.

    9. Re:Heat by citizenr · · Score: 2

      I'm not an EE so I won't pretend to fully understand this particular case but I like it when tech companies reach a bit and try something hard. This may or may not be a good idea but I'm still excited about it.

      Thats what she said.

      --
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    10. Re:Heat by Anonymous Coward · · Score: 0

      Perhaps it's vastly more efficient than a traditional voltage regulator? It is, you know, silicon fabbed by Intel. The undisputed leader in chip fabrication technology. (Seriously. They're conservatively at least 2 generations ahead of everyone else)

      Really, this is a pretty good thing. One of the weak points of many motherboards is the shit voltage regulators they come with. (And the biggest real selling point of premium motherboards) Pulling this in to the package will eliminate one more unknown and should allow Intel to better optimize their chips, not having to deal with whatever crappy noisy power being sent by the lowest-bid mobo the PC maker slapped in in to the case.

      Expect more things like this in the future.

      And less things like it in the further future. Systems integrate and de-intgergrate on a cycle as new technologies become prominent. We're currently in an integration cycle.

    11. Re:Heat by Anonymous Coward · · Score: 1

      Say you have the choice between using a different SoC (maybe an ARM based system) that needs only one power rail, and an Intel SoC that did not include these regulators. Which will cost less to engineer, and be a smaller, simpler solution?

      Remember the one of the many forms of Halswell is a low power solution, with TDP of 10 watts...

    12. Re:Heat by Anonymous Coward · · Score: 3, Insightful

      Considering that they've already started shipping an actual product, perhaps you should switch modes--from skeptic to sleuth. Start from the proposition that (a) it's possible or (b) they're leaving something out of the marketing jargon. There are a million ways they could do it wrong, and likely only a few ways to do it right. If you start from the proposition that Intel is shipping a working product, then it should be much easier to figure out.

    13. Re:Heat by girlintraining · · Score: 2

      My suspicion(if only for die-space reasons, it isn't purely cosmetic that contemporary VRMs occupy a substantial amount of board space) is that is this a 'marketecture' summary of Intel moving some additional voltage adjustment and power gating functions on die, to support dynamic adjustment of power to the greater number of components(multiple CPU cores, possibly independently clocked, GPU, RAM controller, PCIe root, etc.); but we'll still see a bunch of chunky power silicon under serious heatsinks clustered around the CPU socket.

      That's the only plausible thing I could come up with as well. The control logic could go into the CPU, but I don't see how pulling 12V down to fractions of a volt is going to happen on the die itself without it burning a hole through the board; heatsink or not, you can't escape Ohm's Law.

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    14. Re:Heat by sshir · · Score: 1

      May be that they still use discrete FETs, it's just control circuitry is on die now. (I'm speculating)

    15. Re:Heat by Anonymous Coward · · Score: 0

      you can't escape Ohm's Law.

      Actually you can. It's called a switching power supply.

    16. Re:Heat by Anonymous Coward · · Score: 0

      Because CPU nowadays runs much cooler then they use to. The regulator is more efficient (less heat) in the cpu and any extra heat should be easily managed. Most voltage regulators run hot because they often rely on limited cooling unlike CPU which contains quite a bit of cooling (thermal pad/paste, sizable heatsink, constantly running fan).

      There also this thing call temperature sensors which CPU already have which can basically dethrottled the cpu (less heat) or shut down in case of emergencies. Heat has always been an issue with CPU, one well researched and dealt extremely well with nowadays.

      Basically you have many pros for very little cons

      Pro:
      -More efficient (less heat and electricity)
      -Save space (no dedicated chip which takes up more room)
      -Save costs (no dedicated chip which would cost more)

      Con:
      -Less silicon space for CPU which can be used for like cache
      -More heat? (debatable since you could easily use that space for something else that generates as much heat)

    17. Re:Heat by Kjella · · Score: 5, Informative

      Can someone please tell me why this is a good idea

      The long story is here (PDF). Motherboard will still do the heavy lifting from 12V to 2.4V, but the integrated VRM will distribute it. Advantage is extremely clean, fine-grained, low-latency and flexible power supply to deliver exactly as much power to where it's needed and probably - this is just speculation on my part - allowing the CPU to work on a wider range of voltages since there's less noise and ripple so you don't need the same tolerance limits. It sounds perfect for smart phones, tablets and laptops that are primarily battery-limited, nice to have for average machines but potentially an issue for overclockers. All you need is cooling though, it shouldn't limit overclocking if you can keep the temp down.

      --
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    18. Re:Heat by girlintraining · · Score: 5, Funny

      you can't escape Ohm's Law.

      Actually you can. It's called a switching power supply.

      In other news, a Nobel Prize in Physics was awarded to Anonymous Coward of Slashdot today, after discovering that the laws of physics do not apply to switching power supplies... His next research proposal is on solving the energy crisis by designing keyboards to detect when someone is angry and then increasing the key resistance by piezoelectric effect to generate energy. While it would generate only marginal amounts of power when used by 99.975% of the population, it was recently discovered that the remainder are actually Linux and Apple fanboys who, if fed a regular diet of dismissives via their computer screen, will so furiously hit the keyboards that power for entire cities is easily achievable.

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    19. Re:Heat by __aaltlg1547 · · Score: 1

      Intel refers to this as a FIVR (Fully Integrated Voltage Regulator), and it apparently eliminates voltage ripple and is significantly more efficient than your traditional motherboard VRM. Added bonus? It's 1/50th the size."

      I have yet to come across a voltage regulator that doesn't run hot. Typically, it's one of the hottest components in an electrical circuit. And we're integrated this into a slab of silicon already well-known for getting so hot it can catch fire?

      Can someone please tell me why this is a good idea, because all of my experience in electrical engineering says that when things heat up, they become more unstable and prone to failure, and the one thing you do not want going critical is your voltage regulator. If that goes, the whole computer catches fire.

      It's cooled by your CPU fan.

    20. Re:Heat by __aaltlg1547 · · Score: 4, Informative

      Being 1/50th the size it will be welcome on mobile devices. Not sure that its a good thing for your gaming desktop.

      That 84 watts is going to rip through your mobile device's battery pretty damn fast.

    21. Re:Heat by __aaltlg1547 · · Score: 1

      May be that they still use discrete FETs, it's just control circuitry is on die now. (I'm speculating)

      That's an interesting question. They can make a FET with as big a voltage and current rating as they want by just making it many times the size of a logic FET. But they also need thicker gate oxide to prevent Vdg breakdown at multiples of the normal logic operating voltage. Not sure how they would do that. It could add a process step, which would make it expensive.

    22. Re:Heat by Anonymous Coward · · Score: 0

      Actually, not all devices obey Ohm's Law, so I wouldn't be surprised if switching power supplies don't. Inductors and capacitors, for example, are not ohmic.

    23. Re:Heat by ebno-10db · · Score: 2

      Being 1/50th the size it will be welcome on mobile devices.

      It's not clear how they measure "1/50th the size". I could be wrong but it sounds like marketing hype. With a switching regulator the inductors and capacitors generally take up much more real estate than the chip. If they have some magic way to reduce the inductor and capacitor sizes it isn't mentioned in the article (and that would be a much bigger deal than just putting the regulator on the die).

    24. Re:Heat by Charliemopps · · Score: 1

      Who said they're going to keep it at 12 volts? A VRM can also be the transformer, but it doesn't have to be. They could ramp down the voltage on the board just like they always did and just have a VRM on the chip that is maintaining a steady voltage. If you read the article they are barging about how little fluctuation they are getting. So it seems like what they are doing here is adding basically an extra regulator on chip so they can have extremely stable voltage. I'm guessing as small as things are getting now, just the trip across the motherboard can have noticeable fluctuations on supply voltages due to EM interference and temp fluctuations, so having a regulator on the chip lets them get more precise. Maybe that precision will let them do some magic on the chip to increase performance or something?

    25. Re:Heat by ebno-10db · · Score: 1

      Perhaps it's vastly more efficient than a traditional voltage regulator? It is, you know, silicon fabbed by Intel. The undisputed leader in chip fabrication technology. (Seriously. They're conservatively at least 2 generations ahead of everyone else)

      Maybe one generation ahead, but that's in digital chips. While things like switching regulators can be built on digital processes (it's been done before) it's generally not the optimal process. Maybe they've come up with some clever ways to build better switchers on a digital process than previously, but it's hard to believe it's better than a process designed for stuff like this.

    26. Re:Heat by petermgreen · · Score: 4, Interesting

      You can reduce the inductor and capacitor sizes a lot by increasing the switching frequency. Of course doing so will likely increase your switching losses but it may still be worth it if it lets you put the regulator closer to the load. Especially given the ever lower voltages that modern chips are running at.

      --
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    27. Re:Heat by Anonymous Coward · · Score: 0

      The control logic could go into the CPU, but I don't see how pulling 12V down to fractions of a volt is going to happen on the die itself without it burning a hole through the board; heatsink or not, you can't escape Ohm's Law.

      You should look up how switching power supplies work, which gets around this problem quite well. Hobbyists may still use their 7805 and 7812 because of how simple they are, but even for years now a diy electronics person can make an equivalent switching circuit for fractions of a dollar. Efficiencies are typically 70+%, and can go upwards of 95% And like many semiconductor devices, it is nonlinear and doesn't follow Ohm's law (although you can sometimes still define a local impedance for the voltage to current ratio...). The efficiency in many designs gets high with larger voltage drop. Where a linear 12 V to 1.2 V regulator would only be 10% efficient and burning up say 108 W if delivering 10 A on the output, a well built switching supply delivering the same current and same voltage drop can dissipate only 1 W or less in the regulator itself.

    28. Re:Heat by ebno-10db · · Score: 4, Interesting

      You can reduce the inductor and capacitor sizes a lot by increasing the switching frequency.

      But you can do that w/ an external regulator too. Apparently the secret is on-chip inductors. Now that's impressive. I'm surprised that some of the "analog" companies making switchers didn't come up with that first. I know Intel has good fab tech, but this seems more like the sort of funky thing analog guys would come up with first.

      http://www.psma.com/sites/default/files/uploads/tech-forums-nanotechnology/resources/400a-fully-integrated-silicon-voltage-regulator.pdf

    29. Re:Heat by Anonymous Coward · · Score: 0

      The size of external components on any switcher is made by the frequency of the switcher! Multi-megahertz switchers have tiny discreets.
      If I were to guess, I would say that they are incorporating a very high frequency switching core that would require small external discreets. Really, there is no other way that I know of to do this. IOO, there will be a switching controller and a big ole FET or two. The discreets (cap and inductor) will be external.

    30. Re:Heat by Anonymous Coward · · Score: 0

      The CPU is already dissipating tens of watts. A switching mode power supply can easily achieve their claimed ~80% efficiency. This means overall there is only a roughly 20% increase in total heat dissipation and most of this is likely offset by improvements that can be made to the chip because of the integrated Vreg.

    31. Re:Heat by Anonymous Coward · · Score: 0

      I'm fairly certain that they were refering to the space occupied by the regulator... stripping all of the casings, leads, and the length of interconnects would shrink the circuit considerably.

    32. Re:Heat by Anonymous Coward · · Score: 2, Informative

      Actually, no real device is ohmic at all. Even a resistor will heat up with increasing current causing an increased resistance that is non-linear. For us EEs, ohmic devices are our massless pullies and frictionless inclines.

    33. Re:Heat by marcosdumay · · Score: 1

      Thanks, that gets the overall picture.

      So, the idea is that they'll get some very nice inductors on die, capable of replacing some much more expensive external ones. Also, they can distribute the load to a lot of paralel circuits, creating the right tension for each part of the chip, and reducing the loss of each circuit.

      But really, at 90W, a embebing a 76% efficient (not really an exceptional result) conversor means that you'll need to dissipabe other 28W at peak power. Well, I can't say if this is worth it. Certainly, Intel engineers can say, but won't, and the marketeers will lie anyway.

    34. Re:Heat by Anonymous Coward · · Score: 0

      Can someone please tell me why this is a good idea

      1. Precise voltage control as implemented by CPU manufacturer.
      2. Less net system heat.
      5. Less total power consumption.
      3. Fewer discrete board components.
      4. Smaller board size.
      6. Lower total cost.

      Not necessarily in that order. Don't be thick headed. FPUs were once discrete components. Are you still butt-hurt about FPUs being integrated?

      Never bet against integration because eventually you're wrong. Integration is why computers succeed.

    35. Re:Heat by Omega+Hacker · · Score: 5, Informative

      Ohm's law is completely irrelevant to this situation *in the form you describe*. "Burning a hole through the board" would be possible and a simple function of Ohm's law only if they were using a linear regulator to generate the Vcore. But VRM's have been switching DC/DC converters since the 486 days. They achieve a voltage conversion by switching the incoming voltage on and off *very fast*, which results in an output voltage that's a function of the input voltage and the duty cycle of the on/off switching. An inductor (current-smoothing) and capacitor (voltage smoothing) give a nice clean DC voltage.

      The differences between on-motherboard VRMs and this new in-package (it's technically a separate chip...) are significant. First off, physically moving it closer means that you're not sending 100+ Amps of current over the 3-4 centimeters of generally very thin copper traces on the PCB, they're sent millimeters through die-bond wires, or even through a solid substrate (no idea what Intel does at that level). There's your Ohm's law coming into play at that level, but the power losses there are relatively minimal since you're talking maybe a few tenths of an ohm. Die-bond wires are going to drop that to 10's of milli-ohms probably, so nothing major but still a positive effect.

      The main reason this will generate a lot less heat is because of the *frequency* of the switching. Because this on-board VRM is so much smaller, it can switch the input faster (shorter wires, less parasitic capacitance, less ringing, etc.). This in turn means smaller value components required, e.g. the switch from the monster inductors seen on the motherboard (at maybe 1-2MHz switching) in the slide to the tiny chip-scale inductors on the FIVR (at 10's or 100's of MHz). The end result of all of this is that switching losses get significantly smaller. It's those losses that create heat local to the regulator. If they can for example go from an 80% efficient VRM to an 90% efficient FIVR for a 100W CPU load, they reduce the switching losses from 25W to 11.1W.

      --
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    36. Re:Heat by Anonymous Coward · · Score: 0

      Well yeah, but most EE's don't deal with circuits at that level unless they're doing special stuff. For most non-sensitive circuits you don't have to worry about the frequency response of a resistor.

    37. Re:Heat by girlintraining · · Score: 4, Insightful

      Actually, no real device is ohmic at all. Even a resistor will heat up with increasing current causing an increased resistance that is non-linear. For us EEs, ohmic devices are our massless pullies and frictionless inclines.

      I'm not a certified EE, but I have built electronic circuits. I know there's a lot of ways to 'cheat' on paper; switching power supplies don't get rid of ohm's law though, they're simply more efficient. Ohm's law is about the relationship between resistance, voltage, and current. Those relationships are derived from the physics about electron exchange between different materials. Now yes, capacitors and inductors both run 90 degrees out of phase between voltage and current so it can appear to be violating ohm's law, but if you apply a correction factor you'll see it's pretty close to parity. When you get down to really small discrete components, like a transitor for example, measurement inaccuracy and time domains will really start to screw with you, but ohm's law still holds even down to that scale.

      Ohm's law is the reason for these changes Intel is making: An attempt to remove parasitics from the circuits, which all boil down to resistance; Whether it's phase-shifted forward because of capacitors, or backwards because of inductors, or because of components that create those effects, doesn't really matter.

      Now you're right, a purely ohmic device doesn't exist. Even resistors can generate small amounts of phase shift. But that doesn't make them "massless pullies" or "frictionless inclines". Ohm's law is still useful for the same reason the OSI 7 layer model is still useful, despite no network yet having been designed that perfectly adheres to it...

      --
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    38. Re:Heat by l3iggs · · Score: 1

      But at least it's better to have heat build-up near a heat-sink

      True, that it's a good thing that the CPU heatsink is nearby to cool the regulator, but having it on dye also means that the CPU is nearby which seems to be a bad thing from a system stability standpoint.

      Something tells me the folks at Intel have carefully considered all this and that the extra capacity the CPU heatsink now needs to keep the CPU stable is preferable to having an on motherboard Vreg, which is relatively far away from the CPU and that they can't trust to be cooled properly.

    39. Re:Heat by swalve · · Score: 1

      Sure you can. Voltage regulators aren't just resistors any more. You divide the 12v @ 1 amp into 1v @ 1amp at 12 different spots. More or less. Think of it like TDMA, if that helps. Switching voltage regulators are super efficient. And even if there isn't an efficiency gain in the VRM, they will likely be one since the processor will be operating at tighter voltage tolerances. The VRM will be closer to the load and be able to react to load shifts quicker, meaning the processor spends less time slightly over-volted. (If the processor needs 1.2 volts or it crashes, an external regulator might have to have a setpoint of 1.3 volts so it never goes under. But that means it wastes a lot of wattage on heat. If it can instead have a 1.225 setpoint, you are saving power that would normally be shed as heat.

    40. Re:Heat by Anonymous Coward · · Score: 0

      Ohm's law is the reason for these changes Intel is making:

      Ohm's law as applied to the connection between the regulator and the load, not because of how Ohm's law applies (or doesn't apply...) to the regulator.

      I know there's a lot of ways to 'cheat' on paper; switching power supplies don't get rid of ohm's law though, they're simply more efficient.Ohm's law is about the relationship between resistance, voltage, and current.

      And there are vast categories of devices that are not ohmic, not even in the more generalized AC circuit sense using complex impedance. Basic semiconductor devices can be very non-linear relation between current and voltage, or not even have a pure functional relationship between the two due to hysteresis or other dependences. You have the first sentence backwards, as the on paper cheat is the idea that everything is broken up into simple, linear ohmic (complex or not) components, while the real world has a lot of disregard for that.

    41. Re:Heat by swalve · · Score: 1

      Some of the pentium 4 chips dissipated over 100 watts. I think they know how to move heat off of silicon.

    42. Re:Heat by viperidaenz · · Score: 5, Interesting

      If you core requires 1V and 90 watts you need to transfer 90A through your PCB traces, up in to the chip, across the bond wires (if there are any) and on to the die.
      If your die has a regulator on board and accepts 12V instead, and is 80% efficient you only need to transfer 9.4A. You've just lowered your resistive losses by about 100x. If the connection between the external VRM and die is 0.001ohms, at 90A you waste 8.1W. at 9.4A you waste 0.088W.

    43. Re:Heat by Anonymous Coward · · Score: 0

      I've got a Corsair H80 for my i7-3770. That VRM could straight up burn FIFTY watts, right where it is, and it'd be a direct match for the TDP of an AMD FX-8350, which ships with an *air cooler*. I can pull so much heat off of that one spot that it's not even funny, and there IS NO better place in a computer system for that component. This is a step forward-- period.

    44. Re:Heat by InvalidError · · Score: 1

      The big problem with bringing 12V on-chip is not Ohm's law. It is silicon's breakdown voltage at 22nm.

      Trace-to-trace and channel/gate breakdown voltages can be addressed by patterning high-voltage areas with wider insulation gaps but that does not address insulation between metallization layers. It would be feasible but it would also make the fabrication process a fair bit more complex to accommodate the different insulation thickness in the high-voltage area vs the digital stuff.

      I'm guessing the ~2.5V happens to be the safe maximum working voltage for Intel's insulation layers. For mobile applications, it would be nice if they brought this up to the 3-4V range so the CPU could be powered right off a single-cell lithium battery instead of requiring a pre-regulator.

      If Intel wanted to bring straight 12V into 22nm CPUs, it could most likely be done at the expense of more wasted space around high-voltage vias between MOSFETs and high-voltage layers. I see no obvious reasons why they couldn't do this in future iterations - the extra fraction of a mm^2 it may cost in surface area would likely be easily recovered from eliminating several Vcc/GND pads.

    45. Re:Heat by fluffy99 · · Score: 1

      Who said they're going to keep it at 12 volts? A VRM can also be the transformer, but it doesn't have to be. They could ramp down the voltage on the board just like they always did and just have a VRM on the chip that is maintaining a steady voltage. If you read the article they are barging about how little fluctuation they are getting. So it seems like what they are doing here is adding basically an extra regulator on chip so they can have extremely stable voltage. I'm guessing as small as things are getting now, just the trip across the motherboard can have noticeable fluctuations on supply voltages due to EM interference and temp fluctuations, so having a regulator on the chip lets them get more precise. Maybe that precision will let them do some magic on the chip to increase performance or something?

      Or they are finding those Chinese made MB motherboards have such poor regulators, that they have to do some final regulation on the die to keep things stable?

      This is probably also driven by a desire to start putting higher horsepower CPUs into smaller things like tablets.

    46. Re:Heat by fluffy99 · · Score: 1

      The big problem with bringing 12V on-chip is not Ohm's law. It is silicon's breakdown voltage at 22nm.

      From the linked PDF - "90 nm technology for test devices". It looks like it's not on the same silicon as the actual processor, but rather stacked on top..

    47. Re:Heat by fluffy99 · · Score: 2

      . This in turn means smaller value components required, e.g. the switch from the monster inductors seen on the motherboard (at maybe 1-2MHz switching) in the slide to the tiny chip-scale inductors on the FIVR (at 10's or 100's of MHz).

      From the linked pdf - Programmable switching frequency 30MHz to 140MHz

    48. Re:Heat by Anonymous Coward · · Score: 0

      Now yes, capacitors and inductors both run 90 degrees out of phase between voltage and current so it can appear to be violating ohm's law, but if you apply a correction factor you'll see it's pretty close to parity.

      I assume that you're talking about complex impedances, which yes, are a generalization of Ohm's Law. But I can make even more generalizations than capacitors and inductors. What about thermistors? There really is no easy way to describe a thermistor in terms of a complex impedance.

      Actually, I'm glad you mention transistors. Think about a MOSFET. For a given gate-source voltage, you're going to have a maximum current through the drain before the device becomes saturated. No complex impedance is going to give you the behavior of a MOSFET. Ohm's Law is certainly fundamental in that you use the basic principle a lot as an EE, but by no means is it a fundamental limitation on all electronic circuits.

    49. Re:Heat by Anonymous Coward · · Score: 0

      I would hope most EEs have done circuits at a level where you need to worry about non-linear IV curves. Pretty much anyone who has gone into electronics beyond an intro physics course should have come across a diode or transistor. And an EE should have seen much more complicated circuits than that. At the very least, I would hope an EE would know why a light bulb makes for a very bad linear resistor (although bonus points for those that know when a light bulb is useful as a "bad resistor").

    50. Re:Heat by Anonymous Coward · · Score: 0

      How does that escape Ohm's Law? What the hell are you babbling about and where did you get your ideas from? Stop spreading them.

    51. Re:Heat by allanw · · Score: 1

      As an EE, I like to think of Ohm's law as a definition for resistance. The resistance can be non-linear for active devices in which case using V = IR is quite useless.

    52. Re:Heat by allanw · · Score: 2

      So what stops someone from taking the switching frequency really high, like into the hundreds of megahertz? In switching regulators, there is both conduction loss and switching loss. Conduction loss occurs from resistance in the power supply path, including switch resistance. It can be reduced by increasing the switching frequency. However, this increases the switching loss -- you have to switch the power FET gate capacitance more often. The most efficient system is achieved when conduction loss is balanced with switching loss. It is a complex engineering problem. By making a tiny package integrated solution, the inefficiencies of switching can be reduced so the frequency can go up. This cannot be easily done with a traditional discrete-based system like on current motherboards.

    53. Re:Heat by Virtucon · · Score: 1

      And I have a couple of Bulldozer AMDs that run up to 125W that still doesn't mean that you can't cook something or shorten its life. Since these will be 10W processors it will be interesting to see how well they stack up against their predicessors because if they are favorable in comparison then it will be a lot less heat and a lot less noise in desktops.

      --
      Harrison's Postulate - "For every action there is an equal and opposite criticism"
    54. Re:Heat by Anonymous Coward · · Score: 0

      I'm the AC above you, and I'm also an EE. I guess I communicated poorly, I meant that in most cases, a *resistor* behaves ohmically, not a diode or transistor or anything like that obviously, since the person I responded to was talking about non-ideal effects of resistors as well. And in most cases they will behave ohmically, unless you're really pushing the energy that's going through them or you're intentionally using thermistors.

      And yes, temperature swings introduce different resistances to the same device. For your bonus points, one use of thermistors is current-limiting since high current will raise the temperature/resistance which lowers the current etc. You can also use them to sense temperature if you have a very precise model of the resistance as a function of temperature. There are a lot more, but I don't want to write a dictionary entry ;).

    55. Re:Heat by dgatwood · · Score: 5, Informative

      Intel might be the first to do it on a CPU die, but they're not the first to do on-silicon inductors by any stretch. Switching regulators with inductors on silicon have been commercially available for several years now. The R-78 and MIC33030, for example, are drop-in replacements for linear regulators, with all components on die.

      The real question in my mind is why anyone still uses linear regulators for anything, but I digress.

      --

      Check out my sci-fi/humor trilogy at PatriotsBooks.

    56. Re:Heat by Anonymous Coward · · Score: 0

      "An inductor (current-smoothing) and capacitor (voltage smoothing) give a nice clean DC voltage."

      You have obviously way better switching regulators than I have ever seen. Might be nice, but i've never seen a clean one. There are always some harmonics that may or may not be troublesome.

    57. Re:Heat by SuricouRaven · · Score: 2

      That R-78E3.3-0.5 looks useful... the raspberry pi uses a rather inefficient linear regulator to take the 5v input down to 3.3v. An NCP1117. Can't just replace it with an R-78, as the minimum input voltage is 6V, but if you're running it off of something higher... yes. This component may be of use to me.

    58. Re:Heat by SuricouRaven · · Score: 1

      Ohm's law goes out the window as soon as inductance becomes a nontrivial factor.

    59. Re:Heat by Joce640k · · Score: 1

      I have yet to come across a voltage regulator that doesn't run hot. Typically, it's one of the hottest components in an electrical circuit. And we're integrated this into a slab of silicon already well-known for getting so hot it can catch fire?

      Gee. If only Intel had some proper engineers like you who could think of clever things like that...

      --
      No sig today...
    60. Re:Heat by AmiMoJo · · Score: 1

      Actually the analogue guys like Texas and Linear do make switching regulators with on-board inductors that are pretty small. Small enough that they could be integrated into a typical desktop CPU form factor.

      Unfortunately TFA is light on details. 1/50th the size of what? A mains transformer? A SOT-23 IC?

      --
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      SJW, n: "Someone I don't like, and by the way I'm a fuckwit" - AC
    61. Re:Heat by Anonymous Coward · · Score: 2, Informative

      Sorry, but I design quite a lot of switch mode regulators for my own hardware design, and there are several concerns here:
      - the efficiency of 76% they claim is abysmally low, my regulators are never below 80% in their operating range, and often above 90%, with peaks in the 96-97% range.
      - switch mode regulators need an inductor, and inductors need ferrite or iron cores, which is not going to happen on a silicon die. External inductors are much better
      but low loss inductors for large currents are large, for fundamental physical reasons.
      - the fastest acheivable switching frequency is in the low tens of MHz, and this is with considerable losses. Lowering the frequency decreases the losses which happen while the swictch(es) change state, which are dominating at higher swictching frequency. At lower switching frequencies, losses are dominated by I2R
      in chokes and traces.
      - increasing switching frequency allows to increase loop bandwidth and transient response, but another way to improve transient response is to increase output capacitance, which is relatively cheap at low voltage and is best attained by a parallel combination of ceramic capacitors (class II dielectrics like X5R/X7R, never use class III like Z5U or Y5V, they have horrible temperature and voltage dependent characteristics) for high frequency filtering and low ESR tantalum (like
      AVX TPS, designed for this task) for absorbing transients while the regulator adapts within the limits of the loop bandwitdh.
      - loop bandwidth can never be much above one tenth of the switching frequency, to avoid excessive phase shifting due to the sampling that makes the loop unstable
      - an efficient way of minimizing ripples is to have several regulators in parallel with the same clock phase shifted. I've got exceptionnally low ripple in this case
      (could not measure it with a good scope, not with a voltmeter whose bandwidth included the switching frequency).
      - for the large currents, the regulator will be dierctly connected to a ground plane and to an output voltage plane, voltage drops on plane should be low enough
      if you put enough vias to the component (BGA/LGA), and you can always sense the voltage back from the largest consumer.

    62. Re:Heat by wikdwarlock · · Score: 1
      --

      "I must not fear. Fear is the mind killer." -Bene Gesserit Litany Against Fear
    63. Re:Heat by wikdwarlock · · Score: 1

      Check the next-to-last slide on the linked presentation. 50X smaller than typical VR components usually placed next to the CPU. http://www.psma.com/sites/default/files/uploads/tech-forums-nanotechnology/resources/400a-fully-integrated-silicon-voltage-regulator.pdf

      --

      "I must not fear. Fear is the mind killer." -Bene Gesserit Litany Against Fear
    64. Re:Heat by Anonymous Coward · · Score: 1

      I have seen you post on slashdot a lot. You are not a nice person. Perhaps you should work on that.

    65. Re:Heat by Anonymous Coward · · Score: 0

      An inductor (current-smoothing) and capacitor (voltage smoothing) give a nice clean DC voltage.

      That is highly application specific. For some RF stuff, switching supplies still suck.

    66. Re:Heat by Anonymous Coward · · Score: 0

      quote: "switch mode regulators need an inductor, and inductors need ferrite or iron cores, which is not going to happen on a silicon die"

      Yes I agree, except they have put an inductor on top of the die see http://www.psma.com/sites/default/files/uploads/tech-forums-nanotechnology/resources/400a-fully-integrated-silicon-voltage-regulator.pdf (pages 7-9)

      Looks like using "thin film magnetics" (see their publication http://www.psma.com/sites/default/files/uploads/tech-forums-nanotechnology/resources/400a-fully-integrated-silicon-voltage-regulator.pdf ) they have achieved the magnetic equivalent of a supercapacitor.

      I couldn't work out the geometry of the inductor from their publication. ..

      Agree the efficiency is low, full stop. And my feeling is they are setting up a heap of problems down the line - eg power electronics and magnetic fields on a sensitive electronic device. Also how does this fit with stacked RAM, and speculated future technologies such as magnetic RAM..

      Suprised this is going into production, from ignorance I would have assumed it was a research project.

    67. Re:Heat by serviscope_minor · · Score: 2

      Most likely, because it's integrated into the CPU itself, the voltage regulator can be made more efficiently and thus save power and heat etc. Discrete parts have their limitations, and doing it on-die might just mitigate that.

      I'm not sure that follows. For transistors doing computation, the efficiency saving is by reducing the amount of current that they have to switch. In this case, all they're really doing is building a very large MOSFET onto the die itself along with a bit of other gubbins.

      Also, the efficiency figures they claim aren't very high compared to other buck regulators. For example, a quality PSU exceeds the efficiency figures they claim.

      In fact, using a really tiny process size for power components is no advantage since you start to suffer on things like the short channel effect. As such the process won't provide any advantage since they'd have to fab the power device to be much larger than the process size anyway.

      All in all though, they're moving 10W or so from an external power device to an internal one. It's not going to be a huge change either way.

      --
      SJW n. One who posts facts.
    68. Re:Heat by Anonymous Coward · · Score: 0

      "why anyone still uses linear regulators for anything"

      Because they create no switching noise .. Or because they are cheap ..

    69. Re:Heat by serviscope_minor · · Score: 2

      But you can do that w/ an external regulator too. Apparently the secret is on-chip inductors. Now that's impressive. I'm surprised that some of the "analog" companies making switchers didn't come up with that first. I know Intel has good fab tech, but this seems more like the sort of funky thing analog guys would come up with first.

      They did.

      I did a semiconductor course on analog CMOS as an undergrad over 10 years ago. It's interesting because the CMOS processes are typically adapted for digital electronics, and can't give things like BJTs without modification. The radio (phone, wifi, bluetooth) companies want to make analog circuits as cheaply as possible, which means using older gen CMOS to do analog.

      Firstly you put down a bunch of metal in a spiral shape to make the inductor. That bit is easy.

      Getting the Q up is interesting, because especially on a bulk silicon process there is a lot of stray capacitance.

      Due to some weirdness of surface chemistry which is beyond me you can etch preferentially along certain cleavage planes, which means with care you can actually dig under the spiral using careful placement of SiO2 and bare silicon patches.

      I doubt intel do that though since there's no need to get such high Q outside of radio receivers.

      Can some practicing semiconductor engineer chip in?

      --
      SJW n. One who posts facts.
    70. Re:Heat by Anonymous Coward · · Score: 0

      - an efficient way of minimizing ripples is to have several regulators in parallel with the same clock phase shifted. I've got exceptionnally low ripple in this case
      (could not measure it with a good scope, not with a voltmeter whose bandwidth included the switching frequency).

      You meant "Could measure it with a good scope, not with a voltmeter whose bandwidth included the switching frequency", correct?

      A good scope can measure ripple off a DC battery.

    71. Re:Heat by serviscope_minor · · Score: 1

      The control logic could go into the CPU, but I don't see how pulling 12V down to fractions of a volt is going to happen on the die itself without it burning a hole through the board; heatsink or not, you can't escape Ohm's Law.

      I think that may be the point: at some point you have to pull 100W down to fractions of a volt (i.e. hundreds of amps). By doing it on die, it is closer to where you want it to be so you don't have the problem of shipping hundreds of amps quite such long distances.

      --
      SJW n. One who posts facts.
    72. Re:Heat by thegarbz · · Score: 2

      What an absurd statement. Just because a device has non-linear characteristics doesn't mean that V=IR is any less useful.

      LEDs are non-linear yet we use V=IR as way to determine the required compliance of a current source, or what resistor you need if you go down the easy voltage source route. Or an example closer to home, the MOSFET may have a non-linear and time changing characteristic while turning on, but the regulator's power efficiency is none the less determined by the I^2R losses. That doesn't change just because R changes with time and voltage.

      The only time it becomes useless is when power is radiated out of the circuit as in antenna theory.

    73. Re:Heat by domatic · · Score: 1

      I've heard of light bulbs used as current limiters in speaker stacks. Put something like a motorcycle tailight in series with a bass driver.

    74. Re:Heat by fnj · · Score: 2

      Well even at 10W I'm wondering how they'll address the heat.

      Are you serious? Essentially all of the power going into the CPU is coming out of it as heat already. That's 35-100+ watts of heat already being dissipated. And you're worried about another 10 watts?

    75. Re:Heat by dlcarrol · · Score: 1

      I suspect he meant, "Could not measure it with a good scope, nor with a voltmeter ..."

    76. Re:Heat by fnj · · Score: 1

      Perhaps you can explain where you are getting that "10W" figure from. Haswell spans 10W (ULV) to 37W (mobile) to 100+W (desktop). The referenced article has a slide talking about 120A and 400A, which is a far cry from the current that would support only 10W.

    77. Re:Heat by Anonymous Coward · · Score: 0

      An EE usually won't use Ohm's Law to determine the power through a MOSFET. Ohm's Law applies for constant resistances. If you allow R to be anything at all in all time, of course V = IR is true, but that doesn't tell you *anything*. No, to determine the power through the MOSFET EE's will use P = IV (which is decidedly not Ohm's Law) and the IV characteristic of a MOSFET.

      To reinforce my point above, if R(t) = V(t) / I(t), then you're telling me that R(t) is defined for every device ever and that every device follows Ohm's Law. Now *that* is an absurd point. Oh, by the way, you use V = IR when using LEDs because you need a current-limiting resistor. LEDs have a decidedly non-linear IV characteristic (in fact, exponential).

    78. Re:Heat by Anonymous Coward · · Score: 0

      But that 5 minutes (8wh battery) of gaming will look fabulous!

    79. Re:Heat by ebno-10db · · Score: 1

      The Micrel data sheet says "internal inductor", not on-chip, and the Micrel web site puts it in the switching module category, so I doubt this is on-chip. Nice small pkg (planar magnetics?) but not the same as on-die. Similarly for the R-78, which is listed as a module.

    80. Re:Heat by Anonymous Coward · · Score: 0

      That 84W figure was for a 75W desktop processor with a FIVR. Which means, assuming it scales linearly with the power it needs to supply, a 25W high-end mobile processor now draws 28W, and a 10W ultramobile uses 11.2W. And that's taking some of the power load from the motherboard, so total power consumption at the very least stays still, supposedly goes down through increased efficiency.

    81. Re:Heat by ebno-10db · · Score: 1

      People have been building inductors on-die for many years. The trick is getting a high enough inductance (and max current) for a switcher. According to the Intel presentation they have magnetic material on die (Ni80 Fe20 WTF?) which AFAIK is new.

    82. Re:Heat by ebno-10db · · Score: 1

      The discreets (cap and inductor) will be external.

      The inductors are on-chip, which is what makes this so interesting.

    83. Re:Heat by Anonymous Coward · · Score: 0

      Both of the regulators you link to there just contain conventional wire wound inductors epoxied into the same package as the silicon voltage regulator. What is really impressive is creating commercially viable ferritic core inductors entirely via the lithographic process. One implication of this is the number of different inductors one can reasonably fit on a chip -- Intel actually has dozens of independent regulators on a chip, and each regulator uses dozens of inductors in order to reduce voltage ripple. Another important reason to do this is package height -- the regulator can fit under the same heatsink as the CPU.

    84. Re:Heat by Anonymous Coward · · Score: 0

      LEDs are non-linear yet we use V=IR as way to determine the required compliance of a current source, or what resistor you need if you go down the easy voltage source route.

      But you don't use it for the LED itselt, you instead pick a current, then figure out what voltage is needed from the spec sheet.

      but the regulator's power efficiency is none the less determined by the I^2R losses.

      Well, Ohmic losses ... and switching losses... and core losses depending on the type of transformer used and its size.

    85. Re:Heat by ebno-10db · · Score: 1

      You make some good points, but the input voltage is 2.4V, not 12V. No way a semi process like this could take 12V. Also, the caps are external, so there will be some resistive losses with the ripple currents.

    86. Re:Heat by ebno-10db · · Score: 1

      you can't escape Ohm's Law

      In all the hoopla about on-die inductors, they forgot to mention the detail about room temperature superconductors.

    87. Re:Heat by ebno-10db · · Score: 1

      If only Intel had some proper engineers like you who could think of clever things like that

      If only other companies had engineers like Intel, who never make mistakes.

    88. Re:Heat by Anonymous Coward · · Score: 0

      If power consumption is not a major concern, linear regulators are used to limit EMI issues. Super efficient switching regulators run at high frequencies which can create all kinds of havoc with FCC B limits. If you don't have a chamber on site, this can rack up a lot of expenses and destroy a schedule going to an outside test house. Been there done that....

    89. Re:Heat by Anonymous Coward · · Score: 0

      you don't need the same tolerance limits.

      and this is the crucial part as intel pushes towards 10nm process.

      the smaller things get the trickier it becomes to produce. they need higher tolerances at smaller feature sizes to avoid rebinning half their product as pieces of shit^H^H^H^H^H^Hcelerons selling for a small fraction of what they could have.

    90. Re:Heat by Anonymous Coward · · Score: 0

      So, it talks?

    91. Re:Heat by geekoid · · Score: 1

      " my experience in electrical engineering says"

      and the sentences says you don't have a lot of experience.
      It's going to be smaller and use less energy.

      --
      The Kruger Dunning explains most post on /. http://en.wikipedia.org/wiki/Dunning%E2%80%93Kruger_effect
    92. Re:Heat by geekoid · · Score: 1

      Gosh, if it's was 2.4V that would be fine, to bad it's not..oh wait, yeah, 2.4 volts, it's right in the pdf, not 12 volts.

      "but I don't see how... "
      that's because you didn't look.

      --
      The Kruger Dunning explains most post on /. http://en.wikipedia.org/wiki/Dunning%E2%80%93Kruger_effect
    93. Re:Heat by geekoid · · Score: 1

      If it could get enough power to power a cordless keyboard, you could make millions.

      --
      The Kruger Dunning explains most post on /. http://en.wikipedia.org/wiki/Dunning%E2%80%93Kruger_effect
    94. Re:Heat by geekoid · · Score: 1

      I wouldn't call it very bad. In fact, in some circuits they are useful.

      --
      The Kruger Dunning explains most post on /. http://en.wikipedia.org/wiki/Dunning%E2%80%93Kruger_effect
    95. Re:Heat by geekoid · · Score: 1

      right now some EE at intel has read you post and is slapping his forehead. Why didn't we address the heat? of my, oh my.

      --
      The Kruger Dunning explains most post on /. http://en.wikipedia.org/wiki/Dunning%E2%80%93Kruger_effect
    96. Re:Heat by Virtucon · · Score: 1
      --
      Harrison's Postulate - "For every action there is an equal and opposite criticism"
    97. Re:Heat by tlhIngan · · Score: 1

      I have yet to come across a voltage regulator that doesn't run hot. Typically, it's one of the hottest components in an electrical circuit. And we're integrated this into a slab of silicon already well-known for getting so hot it can catch fire?

      Can someone please tell me why this is a good idea, because all of my experience in electrical engineering says that when things heat up, they become more unstable and prone to failure, and the one thing you do not want going critical is your voltage regulator. If that goes, the whole computer catches fire.

      Voltage regulators typically run hot because of the pass transistor. In a linear supply, the pass transistor is what dissipates the difference in voltage from input to output. In a switching supply, the pass transistor is the chopper (the one that converts the DC to AC for the transformer or magnetics). Most VRMs in computers are actually cooled just passively - if you look, the tab is soldered to the motherboard and a huge copper plane on the PCB (usually via-stitched to carry heat to the ground or power plane as appropriate).

      HOWEVER, there is one thing the on-die VRM has over the on-motherboard one - a heatsink!

      The main CPU core has to be cooled anyhow, so the VRM has a very easy access to the heatsink and can thus dissipate the heat much easier.

      I would guess one of the reasons for doing it on board is because of the currents and voltages involved - power supply is becoming a big issue when you have multiple voltage rails (Vcore, Vgpu, Vio (x a dozen or so depending on the IO), etc). Each voltage rail requires at least one pin or multiple pins, and even though we have 1100+ pins, it still means the CPU is pin-limited (in ASICs, you can be pin-limited, where what you can stuff onto an IC is limited by the number of pins you can have, or silicon-limited, where what you can stuff in is limited by silicon area. SoCs and such are typically pin-limited, while memories are silicon limited. Similar concept to IO bound and CPU bound programs, really.)

      If instead of dedicating easily 100 pins to power (if your core is 1.8V and you're drawing 90W, you're wanting 50A, and each pin may take .75-1A max meaning you're looking at 50-75 pins, plus power to other onboard stuff), you can dedicate 25 to higher voltage single rail, it's a lot less current per pin, so lower IIR losses, and a lot more pins freed up. If you need 90W, and you feed in say, 15V, you only need 6A, so instead of the 50-75 pins to carry 50A, you only need say, 10 pins. Plus, if you want other voltages, instead of dedicating more pins, the onboard VRM gives you more flexibility - say you want another 60W, that's 150W, at 15V, that's 10A, or 10-15 pins. Still a lot less and giving you 80-odd more pins for further I/O.

      80 pins is a lot - you can add a number of additional PCIe channels and lanes with that, for example. Or additional functions.

      And you've simplified the motherboard design a lot and gotten rid of a potential source of issues - poor power supply design by motherboard manufacturers (applies mostly to cheap budget boards).

    98. Re:Heat by InvalidError · · Score: 1

      While the paper is about a "test device" produced several months ago, Haswell is a shipping 22nm product coming out early next month that already integrates this on-die and calls it FIVR - Fully Integrated Voltage Regulator. If the IVR was stacked die, it would not be quite "[b]Fully[/b] Integrated", wouldn't it?

      Stacked die would also have a problem with dissimilar die areas: Haswell itself will be over 140mm^2 while the IVR powering it would only be around 15mm^2. That does not stack very well unless you want to waste over 125mm^2 per 90nm stacked chip. Not to mention all the thermal resistance stacking the VRM on the CPU would add.

      If Intel's experimental IVR was intended to remain discrete, they would likely have chosen some higher voltage than 2.4V for the feed voltage since they would not have had to worry about keeping it compatible with their CPU process. I'm betting that they will increase the input voltage it as they refine the design, techniques, process and materials... maybe Broadwell will step it up to 3.3-5V to be more PC-friendly and efficient through reduced input I2R losses.

      The main reason Intel used 90nm for their experiment most likely is that 90nm is much cheaper to experiment with than 22nm and high-voltage/high-power electronics does not really care about features below 100nm anyway: at 2.4V, you would need about 100nm insulation between input and ground/low(er)-voltage structures so 90nm has just about the right resolution to provide representative results.

    99. Re:Heat by jcdr · · Score: 1

      Or they are finding those Chinese made MB motherboards have such poor regulators, that they have to do some final regulation on the die to keep things stable?

      Mod parent up.

      This is a very interesting argument, especially now that chip require multiple voltages with dynamic setting and precise tolerances.

    100. Re:Heat by nerdbert · · Score: 3, Informative

      I do SoCs with integrated regulators now.

      Their inductors are on-chip using extra thick metal levels. But "extra thick" levels on sub-20nm chips are still pretty damn thin, meaning high r/square, so the Q they can get out of the inductors is pretty low, especially since the configuration they use (linear coupling rather spiral) will also limit their available Q. That's what's driving their efficiencies down from what you're used to in discrete buck regulators.

      The big advantage of integrating this is that you don't get the all the nasties of the pads. You usually get the power routed to the corner pins on an SoC and a big chip can easily generate 20+ nH and 1 pF on the pins of a wirebond SoC, and even a flip-chip will still see more than 10 nH typically. That's a problem when trying to deal with power transients, so the on-chip regulator really helps get the ripple down since it can sense/adapt to the voltage at the pin.

      Personally, the big eye-opener to me was doing 400A of DC power on chip. Even at sub-1V the electromigration issues they have must be killer.

    101. Re:Heat by nerdbert · · Score: 2

      Discrete buck regulators lose efficiency above roughly 1 MHz due to the parasitics associated with the pass devices and pads. You'll see fully integrated solutions that run at 5 MHz or so if they're meant to supply other chips simply because the parasitics will generally limit the performance to around that frequency. Specialized systems (like the regulators to previous Intel CPUs) can run higher than 5 MHz, but in general the return isn't that good for increasing the frequency.

      Putting the regulator on the same chip changes the ball game since the pad parasitics in particular are avoided. There you can run much, much higher in frequency and the Intel guys go up to 167 MHz/phase in their presentation.

    102. Re:Heat by cellocgw · · Score: 1

      I've heard of light bulbs used as current limiters in speaker stacks. Put something like a motorcycle tailight in series with a bass driver.

      Cool. Reminds me of the trick some pinball designers used (back in the EM days) to generate a one-shot long pulse, typically to fire off a buzzer. Run the solenoid enable line thru a blinker bulb, so the enable remains active until the bulb warms up. First "blink" and the circuit opens.

      --
      https://app.box.com/WitthoftResume Code: https://github.com/cellocgw
    103. Re:Heat by girlintraining · · Score: 1

      What an absurd statement. Just because a device has non-linear characteristics doesn't mean that V=IR is any less useful.

      I assume you meant to reply to the GP, not my post. My position is the same as yours...

      --
      #fuckbeta #iamslashdot #dicemustdie
    104. Re:Heat by viperidaenz · · Score: 1

      Ok, so the factor is only 4x then.

    105. Re:Heat by thegarbz · · Score: 1

      But you don't use it for the LED itselt, you instead pick a current, then figure out what voltage is needed from the spec sheet.

      You mean the spec sheet which shows graphs for I against V?

      Don't be daft. You do use it directly. It may be time varying for high current pulses but the fundamentals are still the same.

    106. Re:Heat by Anonymous Coward · · Score: 0

      all of my experience in electrical engineering

      Here's your problem: you don't have any. This post is a case in point. You're just waving your hands around and saying "I, girlintraining, am Officially Worried! Lookatme!! FIRE!!!!1!" You don't have enough EE experience to actually calculate and attempt to figure things out on your own. You don't even have enough to take the hints in what you quoted (such as: higher efficiency implies that less power gets converted into heat in the voltage regulator).

      In fact, this phenomenon is not limited to your commentary on EE topics. Your posts are reliably terrible. But because Slashdot, they get modded up a lot anyways. +4 Interesting for this almost entirely content free post, whee!

    107. Re:Heat by Anonymous Coward · · Score: 0

      The big problem with bringing 12V on-chip is not Ohm's law. It is silicon's breakdown voltage at 22nm.

      Trace-to-trace and channel/gate breakdown voltages can be addressed by patterning high-voltage areas with wider insulation gaps but that does not address insulation between metallization layers. It would be feasible but it would also make the fabrication process a fair bit more complex to accommodate the different insulation thickness in the high-voltage area vs the digital stuff.

      You're making a mountain out of a molehill. They (and the rest of the industry) have been varying insulation and metal thickness for a long time. Lower metal layers are finer pitch, thinner, and are paired with thinner insulation layers, upper layers are much thicker (for power distribution). Here's a cross sectional image of Intel's 22nm high perf and SoC processes to show what I'm talking about, taken from chips which don't have these integrated regulators:

      http://1.bp.blogspot.com/-WmRa2AgocN0/UMd0t4HPx5I/AAAAAAAAAT0/0_kiHmAbqWw/s1600/Fig+2.png

    108. Re:Heat by Anonymous Coward · · Score: 0

      (Ni80 Fe20 WTF?)

      called http://en.wikipedia.org/wiki/Permalloy

  4. But did they fix the TIM? by Anonymous Coward · · Score: 0

    I don't want to delid my Haswell CPU.

  5. On-Die die risk? by Anonymous Coward · · Score: 0

    As long as incorporating logical features to a chip is nice, is it prudent to incorporate such risky function to it? Wouldn't it become more vulnerable to highly out of control voltage variations? Will it become more easy to burn?

    1. Re:On-Die die risk? by eclectro · · Score: 1

      I suggest the tag for this story be 'whatcouldpossiblygowrong"

      --
      Take the cheese to sickbay, the doctor should see it as soon as possible - B'Elanna Torres, "Learning Curve"
  6. Interesting by Anonymous Coward · · Score: 0

    Interesting possibilities but as TFA says, it could cause a lot of pain for overclockers.

  7. Full presentation by AdamHaun · · Score: 5, Informative

    You can find the full slide set in PDF format here.

    If I read this right, it really is a fully on-chip switching regulator, inductors and all. They already have a test chip that they used to power a ~90W Xeon E7330 for four hours while it ran Linpack. (Or a virus -- it says Linpack in the summary page.) Voltage ripple is less than 2mV. Peak efficiency per cell looks like ~76% at 8A. They claim hitting 82% would be easy, and there are "additional advancements that cannot be reported at this time" planned for the future.

    The slides have bunch of other technical details about testability features, too, which is always neat to see.

    --
    Visit the
    1. Re:Full presentation by girlintraining · · Score: 2

      hey already have a test chip that they used to power a ~90W Xeon E7330 for four hours while it ran Linpack. (Or a virus -- it says Linpack in the summary page.)

      Respectable viruses take issue with your comment that Linpack is anything like them. Viruses do useful work.

      --
      #fuckbeta #iamslashdot #dicemustdie
    2. Re:Full presentation by ebno-10db · · Score: 1

      inductors and all

      Now that's impressive, and I suspect the real secret to this. Not to say the semi design is trivial, but without the on-chip inductors you wouldn't have much. They weren't clear about it, but perhaps it means getting rid of the big filter caps, and relying on smaller caps for each regulator. It also explains the fast response time with a bunch of smaller regulators.

    3. Re:Full presentation by Anonymous Coward · · Score: 0

      From a switchmode supply point of view, they are shitty. 75% at 10A Yes, they can do a much bigger version, but it would eat up lots of die real estate.
      So from a cost point of view, Intel is better off using their silicon area for logic and hang off POL around the board.

      What is interesting is that, they can make them on die, low ripple etc and it uses a lot of digital technology.

      See here for _MUCH_ better parts from their analog competitors http://www.linear.com/product/LTM4620
      LTM4620 - Dual 13A or Single 26A DC/DC Module Regulator with integrated magnetics
      At 10A output, you get close to 90% efficiency and they take 5V or 12V input directly. The two outputs can be wired in parallel to get the 26A output.

      P.S. I use a lot of switch mode power supplies in my designs in module and discrete forms..

    4. Re:Full presentation by allanw · · Score: 2

      An Intel CPU has a TDP of 90W+ running under 1V. That's 100A+ from the switching power supply. With resistive loss, and inefficiencies from multi-phasing the regulators, efficiency are worse than you say. The cost is also high -- having all of this integrated into the package saves on the platform cost.

    5. Re:Full presentation by allanw · · Score: 4, Informative

      The term "virus" in this context means a power virus -- which is an artificial workload designed to draw as much power as possible from the chip. For example, normal CPU burn stress tests might only activate 90% of the chip's power consumption, but a specially designed power virus would be able to activate all of it. In some cases designing the thermal and power integrity solution to support the chip's full power consumption under a power virus needlessly adds extra costs to a product, because it will never see that workload in real life. It's a virus because a malicious person might be able to activate this mode and melt down your CPU, so typically they _do_ have to design the system to support it.

    6. Re:Full presentation by jcdr · · Score: 1

      Keep in mind that the Intel solution use a 30MHz to 140MHz switching frequency, compared to the 0.5MHz of the LTM4620.

    7. Re:Full presentation by ebno-10db · · Score: 1

      From a switchmode supply point of view, they are shitty. 75% at 10A

      See here for _MUCH_ better parts from their analog competitors http://www.linear.com/product/LTM4620 LTM4620 - Dual 13A or Single 26A DC/DC Module Regulator with integrated magnetics At 10A output, you get close to 90% efficiency and they take 5V or 12V input directly.

      Yeah, and the Intel thing will need a double conversion (12V to 2.4V external, then 2.4V to 0.9V? on-chip) so the total efficiency will be even lower. Not an efficiency improvement for the overall system.

    8. Re:Full presentation by Anonymous Coward · · Score: 0

      There was an old speed test program (I think it was Norton SI) that used to be pretty good at stressing chips. In the early 90's I had gotten in some PC's with 386 -25Mhz chips and we installed the optional 387 math co-proc's in them. In those days, we didn't bother with heat sinks, so when I fired-up the speed test to see how fast my "shiny new PCs" were, they ran fine for a minute or so:
      beep-beep-beep... until they heated-up then went beep-beep-bloop-blehhh and died. I wonder what happened and noticed the CPU's were HOT! So leaned-over and blew on it and then it started back up (beep-beep-beep...)
      The PC vendor had an engineer from the MB manufacturer fly-in to see (because they didn't believe me) and of course I could not replicate the issue on the day he arrived! He did see how hot those things were, so I think he had enough info to work off.
      Funniest thing was when I told the vendor about the issue and they said "We installed a bunch of those in a local hospital and they had no issues." to which I replied "Did they have math co-pro's in them?" They replied "No" and I told them "Well you may have an issue in the future if they do!"

    9. Re:Full presentation by Anonymous Coward · · Score: 0

      This saves on platform cost, yes, but does this not also mean in the event the regulator dies, you're now out the chip instead of a board? I think I would have preferred the other outcome, but perhaps I'm too partial to my chips...

  8. Not a bad idea by EmagGeek · · Score: 1

    An on-die power controller still needs to have external capacitors, especially at the power levels we're talking about.

    But, the problem areas in a switching supply are EMI and stray inductances on the board slowing down the turnon of the mosfet. Because the turnon is slowed down, the mosfet spends time in the ohmic region, which creates excess heat.

    An on-die gate driver routed directly to the gate with no trace length has no stray inductance, and an on-die gate probably also has less capacitance than a discrete component. So, switching times are much much faster, and there is far less loss per cycle in the ohmic region. That's a Good Thing(TM).

    EMI will also be reduced just because there aren't a bunch of high speed traces running around, and the thing can run in the multi-MHz range, so less energy is switched per cycle, and maybe even the switch inductor can even be small enough to simply be drawn in silicon.

    I would not be surprised to see efficiency in the 95+% range even coming from 12V down to 0.9V or whatever voltage this thing runs at, so you'll be throwing an extra 6W or so into a 120W package. Not bad.

    1. Re:Not a bad idea by ebno-10db · · Score: 1

      even coming from 12V

      Way too high of an voltage for these sorts of semi processes. I think it starts at 2.4V.

    2. Re:Not a bad idea by serbanp · · Score: 1

      I would not be surprised to see efficiency in the 95+% range even coming from 12V down to 0.9V or whatever voltage this thing runs at, so you'll be throwing an extra 6W or so into a 120W package. Not bad.

      Bollocks! Since the internal VR uses the same process as the CPU itself, it can't sustain high input voltages, therefore a one-stage 12V to 0.9V conversion is just a pipe dream.

      The longer pdf presentation actually shows the motherboard-level 12V to 2.2V VR, which would be still rated for the full power (85W plus margin). OTOH, it's quite impressive that the 22nm process has support for 2.2V CMOS.

      As others mentioned already, Intel is just trying to solve the power distribution issue, not eliminating the main down-conversion stage, which will *always* be external.

      As a CPU VR designer myself, I'm very interested in seeing how this concept will play out. Beside the issues of more peak power dissipation on the CPU die and increased EMI, there is a long-term reliability issue involved, especially on the power stage; switching inductive loads creates ringing, which will degrade in time (through HCI) the switching transistors.

    3. Re:Not a bad idea by ebno-10db · · Score: 1

      switching inductive loads creates ringing, which will degrade in time (through HCI) the switching transistors

      What's "HCI"?

    4. Re:Not a bad idea by ebno-10db · · Score: 1

      a one-stage 12V to 0.9V conversion is just a pipe dream. The longer pdf presentation actually shows the motherboard-level 12V to 2.2V VR

      Do current CPU VR's do a one-stage jump from 12V to 0.9V? If so it seems they'd have an efficiency advantage by avoiding a double conversion. 12V to 0.9V seems like a big jump for a buck converter, but perhaps there's another way.

    5. Re: Not a bad idea by Anonymous Coward · · Score: 0

      Hot carrier injection

    6. Re:Not a bad idea by serbanp · · Score: 1

      Yes they do. Peak efficiency is usually above 90% and these days it depends mostly on the quality of the power switches and inductors, and the switch drivers strength.

      There have been computer systems (especially notebooks) that did a double conversion (Vbatt to 5V, then 5V to whatever voltage was needed, CPU, memory etc), but it did not caught on.

      There is also the "hidden" double conversion scheme in which the final VRs are powered from the battery, while the charger acts as a first VR,converting the adapter voltage to the battery voltage. Systems with this type of power tree are becoming more popular.

    7. Re:Not a bad idea by jcdr · · Score: 1

      Bollocks! Since the internal VR uses the same process as the CPU itself, it can't sustain high input voltages, therefore a one-stage 12V to 0.9V conversion is just a pipe dream.

      The longer pdf presentation actually shows the motherboard-level 12V to 2.2V VR, which would be still rated for the full power (85W plus margin). OTOH, it's quite impressive that the 22nm process has support for 2.2V CMOS.

      The actual FIVR don't use the same process as the CPU. It's a separete die using a 90nm process. Read the page 7 of http://www.psma.com/sites/default/files/uploads/tech-forums-nanotechnology/resources/400a-fully-integrated-silicon-voltage-regulator.pdf for the details.

    8. Re:Not a bad idea by ebno-10db · · Score: 1

      It says 90nm for test devices. It's not clear how the production device works.

    9. Re:Not a bad idea by jcdr · · Score: 1

      An on-die power controller still needs to have external capacitors, especially at the power levels we're talking about.

      From the presentation PDF, the capacitors are builds from the the metal layers of the die.

    10. Re:Not a bad idea by jcdr · · Score: 1

      Intel is just trying to solve the power distribution issue, not eliminating the main down-conversion stage, which will *always* be external.

      Maybe the first stage can still be integrated into the package if not in the die. Only putting the transistors into the package will allow them to dissipate in a far better way than is most existing motherboard designs. Some high end server motherboards and some high end GPU cards uses state of the art switching regulators with high frequency, small magnetic and small ceramic capacitors. There small size can possibly make realistic an integration into the package.

      Now just stack the SDRAM as planned and enjoy a package with almost only I/O signals...

  9. Yawn by Dachannien · · Score: 0

    Wake me up when they move the keyboard on-die.

    1. Re:Yawn by wbr1 · · Score: 2, Funny

      Apparently your e-peen is already small enough to fit on die.

      --
      Silence is a state of mime.
    2. Re:Yawn by Dachannien · · Score: 2

      Seriously? And here I thought I was being clever. I bow to the master.

    3. Re:Yawn by wbr1 · · Score: 1

      Couldn't resist. The door to the barn was wide open. No harm intended though.

      --
      Silence is a state of mime.
    4. Re:Yawn by Anonymous Coward · · Score: 0

      "Yer dick is small" is a clever witticism that trumps irony any day, and easier to understand. A clever retort is to kick him in the balls, which is funniest of all.

  10. From a former power supply designer - Neat! by jimmyswimmy · · Score: 5, Interesting

    That's some amazing work. The current state of the art in CPU power supply designs hasn't changed in 15 years. 12V in, low voltage out, and the output voltage has been moving lower and lower for years, with designs below 1 V. If you figure you had a few percent of tolerance in the early years when everything ran off 2.5V and that few percent remains constant, then at 1 V you have almost no room for slop. So there are a lot of output capacitors there, both those electrolytics (you always hear people complaining about them but they're CHEAP) and ceramics. The ceramics cost a fortune and you need a lot of them to get your tolerance down - the first half microsecond of a load step is entirely the ceramic capacitor's response, not the controller or anything else. Moving part of the VR onboard allows them to reduce the parasitics significantly and they can probably tolerate a little higher tolerance as a result, but moreover they can get rid of some of those ceramics in the whole system - ultimately many of those on the motherboard.

    So this is taking a lot of cake out of company mouths. Analog, Intersil, IRF, ON, who else - manufacturers of controllers, MOSFETs. Inductors, ceramic and 'lytic vendors are all going to lose out a bit here. Potentially Intel can reduce the platform cost vs. AMD as well, which is interesting. There is still an onboard VR but it will be 12 - 2.4 V, wherever they think the sweet spot is for efficiency and size. And the first real change in this industry for a long time. Cool work.

    --

    Just my $0.55 (US inflation, 1774-2008, for $0.02)
    1. Re:From a former power supply designer - Neat! by gr8_phk · · Score: 1

      Where did you see the input voltage for this thing being 2.4V? I was going to ask what it is.

    2. Re:From a former power supply designer - Neat! by Anonymous Coward · · Score: 0

      The current state of the art in CPU power supply designs hasn't changed in 15 years

      This is only true if you ignore the entire ARM ecosystem.

    3. Re:From a former power supply designer - Neat! by petermgreen · · Score: 1

      Afaict it's not explicitly stated but it's implied by a caption on page 24 of the slides pdf that someone linked above.

      --
      note: i'm known as plugwash most places but i screwd up registering that here somehow in the past and now can't register
    4. Re:From a former power supply designer - Neat! by allanw · · Score: 1

      ARM chips are more like 10W, not 100W like high-performance CPU's. These power supplies are much more complicated to design. This is actually at the leading edge of R&D -- no other chip maker has made a commercial product with integrated voltage regulator in the 100W range. It's only been done in academia recently.

  11. I am totally impressed by overshoot · · Score: 1

    ... that they can incorporate the inductor and capacitors for a 90 W switchmode regulator onto silicon. This is a breakthrough in physics, not just in semiconductor processing.

    --
    Lacking <sarcasm> tags, /. substitutes moderation as "Troll."
    1. Re:I am totally impressed by Anonymous Coward · · Score: 0

      This! I'm excited too. Great for the industry.

    2. Re:I am totally impressed by ebno-10db · · Score: 1

      This is a breakthrough in physics, not just in semiconductor processing.

      What sort of a breakthrough in physics? Have they found a way around Maxwell's equations or something?

    3. Re:I am totally impressed by Anonymous Coward · · Score: 0

      Realize that the necessary inductor for this kind of supply will be much smaller than you would need from normal switched mode power supply (smps) chips. The reason for this is that a fairly normal/average smps will run at a frequency between.. 100kHz on the low end to may a couple of MHz on the high end. This VR runs anywhere from 30MHz to 140MHz (programmable range). As frequency increases the needed inductance drops. So you can see, even compared to the high end chips at a slow speed the needed inductance will be 15 times smaller. Comparing high end and high speed it is 70-100 times less inductance. Higher switching frequencies do come with the drawback of harder to control EMI and more significantly lower efficiency.

      Also, you need less capacitance since the regulator will be so close to the transistors on the chip due to having less parasitic inductance due to the closeness.

      I'm not surprised they went this way. They are called /integrated/ circuits after all.

    4. Re:I am totally impressed by jimmyswimmy · · Score: 2

      Not such a big breakthrough as you'd think. As you increase the switching frequency you can decrease the value of inductor and capacitors required. Last CPU supply I built - 10 years ago! - used 100 nH inductors at 300 kHz per phase. I skimmed the PSMA article but there was mention of MHz operating speeds, not at all unheard of these days, so the components ought to be much smaller. A 10 nH inductor and some hundreds of pF of capacitance seems very feasible without stretching the bounds of silicon technology at all.

      --

      Just my $0.55 (US inflation, 1774-2008, for $0.02)
    5. Re:I am totally impressed by InvalidError · · Score: 2

      Nothing new about physics there. It has been known for decades that a piece of wire starts behaving like an inductor at high frequencies and parallel wires or planes behave like capacitors. Both notions have been in use for high-speed analog and RF ICs and PCBs for a long time. For capacitors, there is even a whole class called "Multi-Layer Chip Capacitor" which is basically an IC with several metallization layers connected at alternating ends.

      This is simply the somewhat unexpected but logical application of well-known principles to an old problem: making PSUs smaller.

      The real breakthrough here is a VRM solution capable of operating reasonably efficiently at 30+MHz with a multi-phase architecture that brings ripple frequency over 500MHz; within the realm of what can be filtered on-chip.

    6. Re:I am totally impressed by Anonymous Coward · · Score: 0

      you are about a factor 1000 off, 100uH is more in the range for 300kHz

    7. Re:I am totally impressed by jkflying · · Score: 1

      Frequencies of 30 - 140 MHz. And to think that it has only been 20 years since our CPUs weren't even that fast.

      --
      Help I am stuck in a signature factory!
    8. Re:I am totally impressed by serbanp · · Score: 1

      *You* are either completely off or are talking about something totally different; the subject is VR for CPUs.

      In a buck converter, the inductor's value is computed from the desired current ripple, which usually is a fraction of the TDC value. Although a 100nH inductor is a little extreme for 300kHz switching when vout is about 1V, its certainly doable; you'll get high inductor ripple, crappy efficiency but also fast transient response and such inductor is cheap.

      The 100uH inductor at 300kHz is appropriate for a high-voltage converter, with an output voltage of e.g. 400V and requiring very small ripple.

  12. Time to call my broker, by Fengpost · · Score: 1

    and dump my Maxim stock!

    --
    The purpose of writing is to inflate weak ideas, obscure poor reasoning, and inhibit clarity....Calvin
    1. Re:Time to call my broker, by Anonymous Coward · · Score: 0

      I dump a stock of manjuice while reading Maxim!

    2. Re:Time to call my broker, by Anonymous Coward · · Score: 0

      Good god why? Have you never heard of Internet Porn?

  13. 2010. Not news, just unattributed slides. by Anonymous Coward · · Score: 0

    So we have some slides for a February 2010 presentation, about a widely known technology thats going to be included in some upcoming parts.

    Now here is something more news like: Today http://hothardware.com/ posted an article calming to be news, but was actually just a portion of an 2010 Intel slide deck with the date and copyright removed. Tech fans seem excited by their content, but are ignoring the illegal complete lack of attribution regarding the stolen slides, as well as the fact that there is no actual news there.

  14. Compared to a traditional regulator? by perpenso · · Score: 1

    Being 1/50th the size it will be welcome on mobile devices. Not sure that its a good thing for your gaming desktop.

    That 84 watts is going to rip through your mobile device's battery pretty damn fast.

    Don't we need to compare it to a traditional regulator implementation before we come to that conclusion? Assuming pretty damn fast means faster than current Atom based devices.

  15. Rotten idea for performance by ChrisMaple · · Score: 0

    CPU chips are performance-limited by heat. Adding the regulator on-chip, dumping heat into the chip without adding processing capability, is a net loss.By making the chip bigger, it decreases yield and makes it more expensive to produce. Lose-lose.
    For moderate-performance applications where CPU yield is already high, the cost reduction achieved by simplifying the motherboard might make this a winner.

    --
    Contribute to civilization: ari.aynrand.org/donate
    1. Re:Rotten idea for performance by Anonymous Coward · · Score: 2, Funny

      You should contact Intel - I bet they didn't even consider this.

    2. Re:Rotten idea for performance by 0123456 · · Score: 1

      CPU chips are performance-limited by heat.

      My old Pentium-4: 130W
      My new i7: 75W

      There's plenty of headroom to output a few more watts without having to underclock the chip.

    3. Re:Rotten idea for performance by allanw · · Score: 1

      If they can cut down platform costs by a few tens of dollars then it is a huge win. This solution removes a lot of discrete chips sourced from a lot of outside companies. Current voltage regulators for these 100W chips are multi-phase regulators, which means something like 4 - 12 parallel voltage regulators with their own inductors, etc. Also, the required area is 50x smaller according to their presentation, which directly affects form factor and cost of these systems.

    4. Re:Rotten idea for performance by fnj · · Score: 1

      No, unfortunately it doesn't remove any exterior components. The on-CPU regulator is only 2.4-to-1V. The off-CPU regulator just changes from 12-1V to 12-2.4V, but it is still there.

    5. Re:Rotten idea for performance by InvalidError · · Score: 1

      Delivering power at 2.4V is still far more efficient than delivering it at ~1V thanks to significantly reduced I2R and contact losses. While it does not eliminate off-chip pre-regulation, it does eliminate the need for having several regulators for various chip circuits and it reduces the remaining regulator's output requirement from ~100A to ~40A.

      So Haswell's FIVR does allow cutting off-chip regulators to less than half or using components with half the current rating, which should save up to maybe $3 on parts. It definitely allows for some savings, just nowhere near Rotten's "tens of dollars."

    6. Re:Rotten idea for performance by csumpi · · Score: 1

      Good catch. I'm pretty sure Intel missed the heat issue. You should call them.

    7. Re:Rotten idea for performance by jcdr · · Score: 1

      Mod parent up.

      This will also make the whole system more reliable to cheap off-chip regulators.

  16. So much for overvolting.... by Anonymous Coward · · Score: 0

    'Nuff said.

  17. This switcher responds in nanosecond to load step by Anonymous Coward · · Score: 0

    With the 100 MHz cycle and sixteen phases, the output capacitors have to hold up the output for less than a nanosecond, as there
    is a new pulse every 10 nS / 16 = 0.625 nS.

    Very impressive indeed.

    Many thanks for the link to the PDF of the presentation.

    This technology could have application in AM broadcast band radio transmitters, and in audio power amplifiers.

    Peter Traneus Anderson

  18. yay house fires by Anonymous Coward · · Score: 0

    nothing says progress like a good ol house fire

  19. Big whoop by Anonymous Coward · · Score: 0

    Now, if only the performance increase over Ivy Bridge was more than a handful of percent....

    It's a shame AMD and Intel are both coasting along at the moment. TBH there's very little incentive for desktop users to upgrade from Sandy Bridge, since then performance increases have been miniscule.

    1. Re:Big whoop by EmagGeek · · Score: 1

      The need for more performance is also minuscule. Right now there is way too much computer and not enough problem to solve with it. Right now at least it's all about the GPU, which is where almost all computational work is being done. CPUs have become glorified bureaucrats in the shuffling of information around.

  20. No it's about ripple by dutchwhizzman · · Score: 4, Insightful

    If you'd read at least the summary, the benefit would be less ripple. Because it takes time to get the feedback voltage to the external VRM, there would always be ripple if power demands would fluctuate fast enough. In a typical CPU on a typical load, you get a lot of power load changes, so you'd get a lot of ripple. Ripple means that ultra low power circuitry will be harder to implement and hit limits earlier, since it is more dependent on precise voltages.

    Power saving wouldn't be relevant, if you are looking at the power loss in the circuit board traces to the CPU. The efficiency of the internal regulator is lower than that of external voltage regulators so it would probably consume even more power.

    System cost would be higher. Other components on the main board still require regulated voltages, so no components would be saved there.

    --
    I was promised a flying car. Where is my flying car?
    1. Re:No it's about ripple by thegarbz · · Score: 2

      System cost would be higher. Other components on the main board still require regulated voltages, so no components would be saved there.

      This is actually something you're wrong about. A modern CPU require 3 distinct voltages separate from all other devices on the motherboard. The bus, northbridge, memory, and every other non-cpu component will run at different voltages. About half of the regulators that take up real-estate directly around the CPU serve only the CPU. These components could be saved.

    2. Re:No it's about ripple by viperidaenz · · Score: 1

      Regulator efficiency isn't the whole picture. When you're converting to low voltages and high current there can be significant losses in the PCB traces and chip packaging between the regulator and the die. Moving the regulator closer can minimise these losses.
      and like another reply to your post stated, the CPU core voltage isn't shared with any other component.

    3. Re:No it's about ripple by Anonymous Coward · · Score: 0

      If you'd read at least the summary, the benefit would be less ripple.

      So Slashdot article summaries are the final word? Sure you don't want to think that one over a bit?

      Power saving wouldn't be relevant, if you are looking at the power loss in the circuit board traces to the CPU.

      As others have pointed out, the power lost in a resistor is I^2 * R. And power is I * V. So consider a thought experiment. Say you must deliver 100W to a load through a 1-ohm conductor. In one design, you're targeting 100V across the load, 1A current (a bit like a 100W incandescent light bulb, if it was supplied by 100V DC instead of 120/240V AC). In an alternate design, you're going to put 1V across the load, at 100A current (this is something more like a modern CPU). How much power do you lose in the 1-ohm conductor in each scenario?

      Well, that's easy to calculate: I^2 * R. 1*1*1 for the lightbulb, or 1W. Livable. But it's 100*100*1 for the CPU, or 10kW. Holy shit.

      Needless to say, the PCB power plane and CPU pins and CPU package power planes which deliver low-volt high-amp power to a modern CPU core don't even approach 1 ohm in practice. That has to be a very low resistance path.

      Which brings us to why modern CPUs have over 1000 pins. Most of those are power and ground connections, not signal pins. The resistance of a conductor is proportional to its length divided by cross-sectional area. Adding more power pins is equivalent to increasing area while keeping length roughly the same.

      Another way to attack the problem is to minimize conductor length by moving the regulator physically closer to the load. And that's a big part of what this new scheme is about. It also means the input to the integrated regulators can be significantly higher voltage and lower current. If you keep the pincount roughly the same (and as far as I can tell Haswell is doing that), the losses distributing power to the CPU socket are going to drop a lot.

      The efficiency of the internal regulator is lower than that of external voltage regulators so it would probably consume even more power.

      I'm not sure that's a safe assumption at all. Even ignoring the I^2 * R issue.

    4. Re:No it's about ripple by Anonymous Coward · · Score: 0

      Less Ripple???

      There goes the market in the seedier parts of town.

  21. No significant cost savings by dutchwhizzman · · Score: 1

    There are more components on the main board that need voltages regulated. You may be able to skip some parts for specifically the CPU, but the rest of the main board needs clean voltages too. For all the peripheral chips and the PCI bus, you still need all the rest of the voltage regulators and discrete components to make those work.

    --
    I was promised a flying car. Where is my flying car?
    1. Re:No significant cost savings by Anonymous Coward · · Score: 0

      Hence the shift to SoC, so the number of those IC's and their associated passives external to the CPU die decreases. Ultimately, this is about reducing potential manufacturing costs.

  22. No it's not. by dutchwhizzman · · Score: 1

    The heat is one factor in a list of many that limits performance. Others are how precise the voltages are, clock speed, die size and probably more. If you can increase voltage precision and gain more with that than the hit you take with adding the heat, you're still winning overall. Intel clearly got to the point where they could take the heat penalty and still come out winning because of the better voltage regulation.

    --
    I was promised a flying car. Where is my flying car?
  23. virus == power virus by girlinatrainingbra · · Score: 1

    re:

    The term "virus" in this context means a power virus -- which is an artificial workload designed to draw as much power as possible from the chip. ... ... It's a virus because a malicious person might be able to activate this mode and melt down your CPU, so typically they _do_ have to design the system to support it.

    Wow, thanks for the very informative post. It makes sense that being able to deal with full thermal stress would be useful. I've had my quad-core shut down on me once at 20-30 seconds into the boot-up sequence, and then I realized that the heat-sink was not fully applied to or in contact with the CPU surface. I reseated everything and the boot-up sequence went just fine. So I'm really glad that there was an on-die thermal danger detector that shut down my quad-core as the thermal level went to dangerous failure levels. The fact that it's alsodesigned and tested to be actually able to withstand a full power-load draw for a fixed interval is even more impressive.

    Here's a link to the wikipedia page about a "Power Virus". And there's an Intel link about "Thermal performance challenges from Silicon to systems" there on the wikipedia page which is a dead link: http://download.intel.com/technology/itj/q32000/pdf/thermal_perf.pdf .

  24. Can they move the CPU away from the die ?!? by ctrl-alt-canc · · Score: 1

    Given the performance of the on-die voltage regulator, the chip could very useful for designing miniature power supplies. Unfortunately the CPU and the associated digital crap ruin what seems to be a very succesful design of an innovative power supply regulator.

    1. Re:Can they move the CPU away from the die ?!? by MachineShedFred · · Score: 1

      It depends.

      The initial launch of Haswell will be a 2-chip setup. At this time, it's unclear what that second chip is - perhaps it's the fancy-schmantzy VRM. The one-chip Haswell is due in Q3.

      --
      Slashdot still doesnâ(TM)t support Unicode after it was added to the HTML standard in 1997.
    2. Re:Can they move the CPU away from the die ?!? by InvalidError · · Score: 1

      The second chip in the IVR demonstration is the regulator and is integrated in the Haswell die as FIVR - Fully-Integrated Voltage Regulator.

      The second chip seen in some Haswell articles is the 64MB eDRAM for Intel's GT3e IGP.

  25. wtf? first post with no mention of hosts!? by crutchy · · Score: 0

    (Score:-1, Unslashdotian)

  26. It's not about resistive losses. by Anonymous Coward · · Score: 0

    Guys, the real reason why it's a good idea has to do with (parasitic) trace inductance and not so much with resistive losses!
    CPUs don't draw steady current. The average may be 30A or so, but the peaks can be much much higher. And that's the real problem: sudden changes in current (delta I) create time varying magnetic fields around the power traces/pins/wires according to Faraday's law of electromagnetic induction which, by Lenz's law, oppose the changes in currect that created them. The effect is that the supply voltage seen by the processor sags whever its current demand quickly rises (and, conversely, the voltage will jump whenever the current deman falls abruptly!).
    You can mitigate this to some extend by adding decoupling capacitors, but this gets you only so far before it becomes more efficient to just move the regulator on chip.

    1. Re:It's not about resistive losses. by Anonymous Coward · · Score: 0

      Wow, someone is taking his first college electromagnetism class!

  27. Why is this moded up? by Anonymous Coward · · Score: 0

    Why is this moded up when it is mostly incorrect or irrelevant?

    Those relationships are derived from the physics about electron exchange between different materials.

    Ohm's law in most cases is an empirical relationship, not derived from physics of electron properties. And as such, it is an approximation, great in some contexts, inaccurate in others, and flat out useless in many other situations.

    The talk of complex impedance is pretty much irrelevant. A switching power supply is not going to be Ohmic, complex impedance or not. E.g. increasing the voltage supplied to a switching supply will reduce the current drawn from the source.

  28. What are you all talking about? by ckatko · · Score: 2

    I looked through everyone's comments, hoping to see this important issue and everyone's too busy debating silly shit like heat. When is the end-effect of paradigm shifting ever the same as the issue a company portrays to the public? Did everyone think "Microsoft Open Technologies" was a true attempt at embracing open source software?

    Let's look at what's actually going on:

    • 1. Overclocking allows for people to buy cheaper processors that do the same thing as more expensive ones.
    • 2. Increased voltage is required for overclocking.
    • 3. Placing the voltage regulator on the CPU instead of the motherboard removes the control from the motherboard manufacturer and end user.

    The end result?

    • 4. Overclocking can only be achieved by "official overclocking CPUs" that cost much more.

    Intel is working to take away the control people have over their processors. Whether this is the final step, or just a means to an even bigger end, we should be asking more questions.

    1. Re:What are you all talking about? by Anonymous Coward · · Score: 0

      Allow me to supply you with a free clue: you're an idiot.

      That's not enough for you? Okay. Let me put it to you this way: Intel has been deliberately selling overclocker chips which explicitly permit overclocking for years now. They'll be doing the same with these.

      Also: 0v3rC10x0R kiddiez are not as important a part of the market as they believe themselves to be. The vast majority of computer buyers don't overclock. Easy proof: virtually nobody overclocks notebooks. Notebooks are more than 50% of the market. If only half of desktop buyers overclock (the actual percentage is much lower than this) that means at least 75% of computer buyers don't OC. So Intel doesn't much care about eliminating OC -- it's already a tiny fraction of the market. They're not losing real money to overclockers either, except in warranty return costs since the privileged little shits love to break things and then return them. But they've figured out how to cover that anyways, by channeling OC'ers into "K" model unlocked CPUs. They sell for more but cost the same to make as lower cost CPUs.

      Also: in this context, handwringing about "means to an even bigger end" is really dumb. Speaking as an actual engineer, if Intel wanted to shut down overclocking cold they could've done it a very, very long time ago. No on-die voltage regulators required.

      You are indulging in dumb -- and, dare I say, silly -- conspiracy thinking. You're inventing stupid-movie plots to over-explain things which have already been adequately explained.

    2. Re:What are you all talking about? by ckatko · · Score: 0

      Speaking as an actual engineer,

      As an "actual engineer" myself, I must ask what your degree and field of expertise is considering that only two processor generations ago, Athlon XP's were sold that could be "unlocked" by soldering/desoldering bridges on the top of the CPU package.

  29. Faster Smaller More by Anonymous Coward · · Score: 0

    Faster and smaller and more ... the shrinking industry ... soon they'll get down to the atom ... and then the subatomic ... it all based on photography, the principles thereof ... the universe is vast, the understanding of it ... "For the heavens are higher than the earth, so are my ways higher than your ways, and my thoughts than your thoughts." Isa 55:9