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New Chip Promises Longer Battery Life 188

Roland Piquepaille writes "It always happens when you need it the most: the battery of your cellphone just died. But now, researchers of the University of Rochester have developed a wireless chip that needs ten times less power than current designs. The new chip relies on a technology named injection locked frequency divider (ILFD) which dramatically reduces the time needed to check for transmission frequencies which are performed several billion times per second by your current phone. The new chip uses five transistors and can perform divisions by 3 instead of only 2 by previous circuits, allowing a perfect communication between two phones communicating at 2.0001 and 2.0002 gigahertz respectively."
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New Chip Promises Longer Battery Life

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  • Not A Big Deal (Score:5, Informative)

    by Bruce Perens ( 3872 ) * <bruce@perens.com> on Saturday April 22, 2006 @08:31PM (#15182557) Homepage Journal
    The PLL component this is supposed to replace is a small-signal component. It is not a major user of the power budget of a cell phone. The big power users are the transmitter and the microprocessor. The PLL is not heat-sinked and does not run warm. If it's not hot, it's not a power hog.

    Bruce

    • I don't use my cell phone much. Having several weeks of standby time would be convenient, even if talk time is not increased significantly.
      • Re:Not(?) A Big Deal (Score:5, Interesting)

        by green1 ( 322787 ) on Saturday April 22, 2006 @08:59PM (#15182658)
        the problem is, even in "standby" the phone does a lot of transmitting, and that transmitting is still a power hog.

        I'm not quite as negative as the grandparent poster, in that I'm happy if any component uses less power (every bit helps) but in reality, it's the transmitter that uses the lions share of the juice, not the reciever (even in standby).
    • Comment removed (Score:5, Informative)

      by account_deleted ( 4530225 ) on Saturday April 22, 2006 @09:12PM (#15182693)
      Comment removed based on user account deletion
      • by Dis*abstraction ( 967890 ) on Saturday April 22, 2006 @09:27PM (#15182740)
        Actually, I'm still a little confused. Could you try an analogy using cars instead? Thanks.
        • by alx5000 ( 896642 ) <alx5000&alx5000,net> on Saturday April 22, 2006 @09:41PM (#15182775) Homepage
          To explain in a slightly different way, we'll use the analogy of trying to accurately count a mountain of cars. The easiest way to do so, is to weigh the whole pile, and then divde by the average weight of a single car, and you get the total number of cars. The question is how you get the "average weight" of a single car. If you weigh just one car, and use that as the average, then you have some total inaccuracy X. If you instead weigh 10 cars and divde the weight by 10, the inaccuracy is much less: roughly X/10. This is how the old method of PLL circuit design worked. The greater the frequency, the more cars you used to find the average weight, and so the greater the accuracy you could get in finding out the total number of cars in the whole pile, or the exact frequency. The new method described in the Article is roughly analagous to modifying all of your cars to ensure that the variation in the weights of the cars is much lower, so you can rely on just one car to provide you with the precision needed to determine the total number in the pile.
          • what kind of car is it? are you going to average different kinds of cars or just stick with a single brand/model/make? that will introduce more errors. :P
        • How does this translate to libraries of congress?
        • ...try an analogy using cars instead?

          If an end-to-end mobile phone system was a used Portland taxi, the meter would never come off polling, the user would be the rear view mirror and you would be an old penny lost long ago under the right front seat.
      • Re:Not A Big Deal (Score:5, Informative)

        by chriso11 ( 254041 ) on Saturday April 22, 2006 @09:37PM (#15182767) Journal
        No, the digital circuitry does not run at the PLL frequency in a cell phone. The stable reference frequency from the crystal is upconverted to what is called the LO - this LO is mixed in with received signal from the antennea to downconvert the signal to a lower frequency. No digital processing occurs at 1.8GHz/1.9GHz on a cell phone - it is all much lower in frequency. That also goes for Bluetooth and WiFi.

        The article is really short on details. The real power hog in a cell phone is the transmitter - it will draw 3Amps of current - while the rest of the receiver and up-conversion components are maybe 10% of that. And transmitters are already quite efficient - generally, ~50% of the input DC power winds up going out as RF power.

        The lower power version of the PLL will be useful, since it needs to run constantly, even while not actively in a call.

        • I think part of the confusion is that this circuit is proposed to replace the PLL. The "digital circuitry" referred to that is running at the PLL frequency is the PLL itself, using a bunch of mixed-signal magic to take an input clock and spit out a very specific frequency. This new thigamabob proposes to take a very-high frequency quartz oscillator as an input signal and divide it down to a specific frequency using some analog magic. Although to say that this PLL replacement avoids digital logic entirely
      • You could probably hire mexicans to count them cheaper than you could buy a scale capable of weighing a mountain of pennies. Along with that massive scale you would also need front end loaders and cranes, and union workers. Ideally, you could get the mexicans to feed them into those coinstar machines, though capacity may be an issue.
      • no it doesn't... (Score:3, Informative)

        20W in use? Give me a break.

        Let's say I'm running at 1W (max for 1800/1900, half max for 850/900). I'm transmitting 1/8th of the time (due to TDMA slotting).

        Thus I would use 1/8Wh per hour just to transmit. My phone has a 3Wh battery (800mAh @ 3.8V). So I would have a talk time of 24h, if my phone didn't use power for anything else at all. It does, so the talk time on my phone is 8H.

        Now, let's try out your version. I'm using 22W when transmitting, 1/8th of the time. So I'm using 2.8Wh per hour. So, if my ph
      • Re:Not A Big Deal (Score:4, Informative)

        by Jeff DeMaagd ( 2015 ) on Saturday April 22, 2006 @11:26PM (#15183042) Homepage Journal
        Another issue with your claims is that the power needed to operate a CMOS digital circuit goes up not linearly but by the square. A circuit that operates at 20GHz would consume about 100x the power as the same circuit that operates at 2GHz. I'm not aware of any commercial digital IC that can operate at 20GHz anyway.
      • Re:Not A Big Deal (Score:3, Informative)

        by swiftstream ( 782211 )
        The question is how you get the "average weight" of a single penny. If you weigh just one penny, and use that as the average, then you have some total inaccuracy X. If you instead weigh 10 pennies and divde the weight by 10, the inaccuracy is much less: roughly X/10.

        Actually, you would expect it to be roughly X/sqrt(10). Standard error decreases with the inverse of the square root of the sample size.
      • The greater the frequency, the more pennies you used to find the average weight, and so the greater the accuracy you could get in finding out the total number of pennies in the whole pile, or the exact frequency.

        You were doing ok until that part. It stands as contradictory with the next content, however...

        The new method described in the Article is roughly analagous to modifying all of your pennies to ensure that the variation in the weights of the pennies is much lower, so you can rely on just one pe
      • Re:Not A Big Deal (Score:3, Informative)

        What a peculiar mish-mash of ideas. Where did you get them from? RF circuits don't work like CPUs. Just think about what you're saying: the CPU in your phone works at 2GHz? Yet the fastest CPUs in a PDA are about 500Mhz.
      • "roughly analagous to modifying all of your penis to ensure that the variation in the weights of the penis is much lower..."

        i'm very confused :-S

    • I would have said the LCD displays would also chew a fair bit of power - especially on phones like my Nokia N90 [nokia.com] with two of them, one a very high res one (352x416). Not to mention people who are high users of accessories, in particular MP3 players and cameras.
      • Re:Not A Big Deal (Score:3, Informative)

        by Bruce Perens ( 3872 ) *
        The backlight uses a lot of power, not the LCD. LCDs just modulate light, they don't emit it. oLED displays emit light. The backlit LCD or oLED display of a cell phone is an intermittent-use load with a duty cycle on the order of 1:1000, unless you use the phone as a PDA a lot.

        Thanks

        Bruce

        • Well, yes... the backlight. I would query the ratio of the duty cycle, though. My latest phone, I /do/ use as a PDA - but even before then, my cell phone would see a lot more than a minute of backlit activity a day. As a moderate user, I'd estimate 20-30 times that, at least, 1:50.
    • That heuristic only works for devices that don't emit anything else, and even then it ignores a lot of important factors.

      Furthermore, it doesn't tell you whether a component is responsible for high overall power consumption; in order to be responsible for high overall power consumption, a component doesn't need to use a lot of power itself.
      • You mean looking for heat is a bad hueristic for parts that radiate some other energy? Well, if you have really efficient components. Even if RF amplifier transistors are run as switching rather than linear devices, they are not 100% efficient and make some heat. If you run them in their linear region, they are going to spend a lot of time acting like resistors and will make a lot more heat. LEDs warm up a bit, too.

        Consider that microprocessors are CMOS digital devices, we're not unused to getting some heat

  • Would "ten times less power" be anything like "one tenth as much power"?
  • by LiquidCoooled ( 634315 ) on Saturday April 22, 2006 @08:32PM (#15182562) Homepage Journal
    Dude: Hui Wu invented this new chip that saves loads of power.

    Bloke: Who?

    Dude: Yes

    Bloke: so who invented this chip.

    Dude: Hui did.

    Bloke: Thats what I'm asking you.

    Dude: Yer I know, Hui did.

    Bloke: Quit it and tell me who invented the chip.

    Dude: Im not joking, Hui did.
  • Battery power (Score:1, Flamebait)

    by cr@ckwhore ( 165454 )
    "It always happens when you need it the most: the battery of you cellphone just died. But now, researchers of the University of Rochester have developed a wireless chip that needs ten times less power..."

    Ok, but that still doesn't solve the "I need my phone now but I was too lazy to charge it last night" problem. So what, this chip can run from a dead battery? No.

    It really doesn't matter how much power the phone uses... the fact is that it still uses power. Consuming power from a limited source mea
  • But now, researchers of the University of Rochester have developed a wireless chip that needs ten times less power than current designs. The new chip relies on a technology named injection locked frequency divider (ILFD) and permits to dramatically reduce the time needed to check for transmission frequencies which are performed several billion times per second by your current phone.

    Out of curiousity, why have we not yet figured out how to wirelessly power devices? I mean, we can send lots of RF energy t

    • I mean, we can send lots of RF energy through the air. Why can we not use that same energy to power the device as well as send it information? I can see where it would be a problem for something that requires lots of power, but for small devices this should be possible, no?
      Dude, you just re-invented RFID tags! You'll make me smile next time I unlock the doors at work.
    • If you have a 2-way radio (I don't know if a cellphone would work), go to Radioshack and get a low-current light bulb (not LED) and connect solid leads about 8" long to each lead of the bulb. Make them both into a coil, slip one over the 2-way radio antenna, and transmit. The lightbulb should glow bright. Unfortnately RF energy decreases as inverse square.
    • by Com2Kid ( 142006 ) <com2kidSPAMLESS@gmail.com> on Saturday April 22, 2006 @09:31PM (#15182751) Homepage Journal

      Out of curiousity, why have we not yet figured out how to wirelessly power devices?

      Short answer: We already have, it is just so inefficient that nobody uses it. (in fact it was invented over 100 years ago!)

      Long answer: Electromagnetic waves radiate outwards. Either you have a simple non-directional antenna that radiates in all directions at the same time (in a sphere basically) and you lose power REALLY fast, or you have a directional antenna that radiates power in a cone at a target destination.

      The omni-directional radiators suck so much that they are absolutely useless. Inverse square means 1/(x^2). Basically (and this is crappy math but gets the point across) if you have 10 watts at 1 feet, you would have 10*(1/(2^2)) = 2.5 watts at 2 feet. At 3 feet you would have 10*(1/9) = 1.11 watts. Please ignore that you would use meters instead of feet and that all my units are all messed up in various other ways as well. The point is that your power drops off REALLY fast.

      So what about those directional antennas?

      Well, you have to find some way to really accurately track someone's cell phone position, and have a world-wide array of directional antennas so that you can beam power to them no matter where they are at.

      Oh and remember to keep those power levels low, else you will fry anything that gets in the way.

      People worry about cell phones causing cancer as it is, directional power beamed at your head WOULD cause some serious issues!

      Wireless power is possible, just not feasible!

      • Is it that all that 7.5 Watts is dissipated by the transmission distance increasing from 1ft to 2ft? Or is it that the EMF at the receiver reduces by distance^2?
      • Just sayin'.

        But I suppose we'd need to violate plant patents to be able to interface with its system...
    • by darthwader ( 130012 ) on Saturday April 22, 2006 @10:34PM (#15182918) Homepage
      A while ago, Mythbusters did a "free energy" show. They collected a bunch of plans from "the Internet", built the devices, and tested them.

      One of the devices that surprised me was a 50' long aerial, attached to some simple circuitry. The aerial absorbed RF energy, and the electronics converted it into a somewhat useful DC power supply. I think it was producing somewhere around 1 volt, no idea how much current, indoors. IIRC, they said it was "almost as good as a AA battery".

      So, not only is is possible in theory, it's possible in practice. But it's still wildly impractical.

      I think it's episode 24 (http://dsc.discovery.com/fansites/mythbusters/epi sode/episode_06.html [discovery.com]).

      • they said it was "almost as good as a AA battery"

        eghads Brain... imagine a beowulf cluster of these things. You could use it to decode the very rf signal that powers them thus creating a sustainable way to spy on everyone and thus please both the liberals and the conservatives at the same time... and then TAKE OVER THE WORLD.
    • the trouble is especailly with omnidirectional antennas its fantasically inefficiant and that efficiancy drops off with the SQUARE of the distance.

      and you need enough power to get the signal back. so your equation would be something like

      new transmit power at base station= old transmit power at source station divided by old receive power at phone multiplied by old transmit power at phone = massive.

      and thats not accounting for the losses inherent in going from radio to electricity and back.

      its feasible for ve
  • Billion? (Score:1, Offtopic)

    by sapgau ( 413511 ) *
    Like in gazillion?
  • Two/Three (Score:5, Funny)

    by samkass ( 174571 ) on Saturday April 22, 2006 @09:15PM (#15182697) Homepage Journal
    The new chip uses five transistors and can perform divisions by 3 instead of only 2 by previous circuits


    Bender: "Ahhh, what an awful dream. Ones and zeroes everywhere... and I thought I saw a two!"
    Fry: "It was just a dream, Bender. There's no such thing as two."
  • One idea (Score:2, Interesting)

    by thePig ( 964303 )
    The transmitter would be the one which would be using the max power in any cellphone.
    In that case, make the antenna directional.
    But then, we do *not* know the direction to which I have to sent the signal.
    That can be done by maybe -
    1. Changes needed for Towers
    Sent downstream a small pilot signal of the same freq as the upstream signal which the phone emits for that call.
    2. Changes needed in the Cell
    Have a direction sensor in your mobile for this pilot signal. Once
    • Re:One idea (Score:3, Interesting)

      by planetmn ( 724378 )
      Is a antenna which can change direction depending on a signal already there ? If not the idea wont work at all.

      A combination of multiple "antennas" with a 120degree coverage (for three) rather than a single antenna with 360 coverage, and phased array (look at phased array radars) could make this possible. Power savings though, might not happen because of the processing required.

      -dave
    • Is a antenna which can change direction depending on a signal already there ? If not the idea wont work at all.

      Yes, it is possible to have a directional antenna without it physically having to move in order to change directions. I think they have been around for a long, long time. Mutliple antennas/elements are required. (phased array?) . But the consumer wants a cell phone, not a porcupine.

      I *think* something might be done like this in current MIMO research. I believe the problem of finding the direc
    • Is a antenna which can change direction depending on a signal already there?
      The technical term for what you want is retrodirective array [google.com].
  • by Anonymous Coward
    What the hell does "ten times less" mean? If it uses 1 watt now, does that mean it now uses 1 - (10 * 1) = -9 watts? So using htis actually generates energy?
  • What a crock (Score:4, Informative)

    by amjohns ( 29330 ) on Saturday April 22, 2006 @09:55PM (#15182815)
    This is mostly BS. First off, the PLL is a small fraction of the power consumed by a modern phone, even though it is running all the time. Far more power is consumed in the rest of the receiver chain, from the LNA (low nose amplifier) and the digital demodulator. And no, this does not do a thing to minimize the demod, as it is running all the time too, to detect an incoming call notification.

    Second, the statement that a "phase-locked loop multiplies the pulse from a highly-stable reference clock, such as a quartz crystal oscillator, up to the desired frequency" is 100% false. The function of a PLL is to lock (in phase...) a divided down version of a totaly independent RF oscillator, called a VCO, to a divided down version of the reference clock. The distinction may appear subtle, but it's enormous. Multipliers are large, power consuming IC's, while dividers are fairly small and efficient. There are NO multipliers in a PLL, period. Also, PLL's can already do split division, it's called a fractional-N PLL.

    Mobile, battery powered electronics will never achieve decent battery life beyond a few GHz. There are several effects coming into play, from cosmic noise to H2O and O2 molecular resonances to increased multipath effects, and most importantly path loss. RF power spreads in a spherical wavefront, so there is a 1/R^2 power falloff. BUT, you need to recognize that this is in terms of wavelength (lambda), which is mathematically equal to C/f (speed of light / frequency). The net result is that doubling the frequency on a radio link incurs a 4-fold power fallof for a fixed distance.

    So if I want to go from say just under 2GHz w/ a current GSM system to say 8GHz, then I need an effective 16 times the power output from my transmitter. I say effective, because you can use antenna gain, but not in the mobile handset (it needs to be omnnidirectional), and base stations directionality is very limited, since they need to support many users on the same antenna, and can't steer the beam to all of them simultaneously. You wouldn't be allowed ot put out that much powr form a safety perspective, never mind the power consumption and heat requirements in the power-amplifier. Handsets are at 600 milli-watts now, we're not going to put out >10 watts!
    • So, if I may ask, why do you say that "battery powered electronics will
      never achieve decent battery life beyond a few GHz"? It would seem that
      as base stations grow in density of coverage we will be able to drop
      power requirements. Imagine a base station every 10 m (like e.g. in
      every lamppost). Already today cell phone coverage is only good in
      civilized places, i.e. where roads go, so this would not drop
      quality of service compared to what we have now.
      • i've never been to america so i dunno how poor thier coverage is, but i've only been to a few places here in the uk where mobiles didn't work

        whereas with a 10M range like you propose even if you put one in every lamp post you'd lose coverage as soon as you walked to the middle of a small park or down an unlit sideroad.

    • Every once in awhile, someone comes along with a post that restores your faith in /.
      • Re:What a crock (Score:3, Informative)

        by randyest ( 589159 )
        I hope you're not referring to the parent, since he's totally wrong. PLLs certainly can, and do, include multiplers (and/or dividers.) They're called . . . wait for it . . . "multiplying PLLs" (as opposed to "clock-insertion-delay removing PLLs." He also botched his bit about the inverse square law (on multiple levels.)
    • Re:What a crock (Score:3, Informative)

      by zippthorne ( 748122 )
      inverse square law is proportional to wavelength? Where did you ever get this wacky idea?

      The inverse square law is so because it describes the effect of the expanding wave front as it propogates through space. The energy of any particular shell is constant, but as the shell expands the energy becomes more spread out. The square law is a consequence of our three dimensional space. The area of a sphere (the pattern of a so-called isentropic radiator) is pi*r^2, so the unit density will be {something}/pi*r
      • See the Friis Transmission Equation [wikipedia.org]. Doubling the frequency results in a 6 dB loss at the receiver. So the signal strength is proportional to the square of the wavelength.
        • Your application of the Friis equation is incorrect. Antenna gain in the directional antenna depends on wavelength with a 1/lambda^2 factor.
          • I don't see how. A given antenna, say dipole or isotropic radiator, becomes less effective as the frequency is increased. I'm assuming that is the source of the frequency dependent part of the equation. So other things being equal, increasing the frequency also increases the path loss.
            • Slashdot does not have convenient math notation, which would be required for the lengthy explanation of what's going on. But your assumption is incorrect. The key words to look for are "aperature size" and "effective aperature size" and note what happens in the case of isotropic radiator. It is also important to keep in mind that only one of the antennas in the cell phone/tower loop is isotropic.
    • Re:What a crock (Score:3, Informative)

      by thestuckmud ( 955767 )

      RF power spreads in a spherical wavefront, so there is a 1/R^2 power falloff. BUT, you need to recognize that this is in terms of wavelength (lambda), which is mathematically equal to C/f (speed of light / frequency). The net result is that doubling the frequency on a radio link incurs a 4-fold power fallof for a fixed distance.

      Sorry, but this last point is wrong. The inverse square law for power is, indeed, in terms of power, not wavelength. Actual radiated power depends on the power input to the fin

    • by dtmos ( 447842 ) on Sunday April 23, 2006 @06:32AM (#15184037)
      I'm considering the devotion of the rest of my professional career to the eradication of the "propagation loss increases with frequency" myth.

      Repeat after me:

      Propagation loss does not increase with frequency!
      Propagation loss does not increase with frequency!
      Propagation loss does not increase with frequency!

      Think about it: If the propagation loss of an electromagnetic wave increased in proportion to its frequency, there would be so much so much attenuation at the THz frequency of light that we'd never see sunlight--or stars. Propagation loss is independent of frequency, except for scattering due to molecular and atomic resonances that are insignificant at the frequencies we're discussing. (There are also changes in scattering behavior that become relevant in indoor applications, like propagation around corners.)

      What is dependent on frequency, however, is the performance of the antennas we use to transmit and receive electromagnetic waves. Antennas can be characterized by a parameter called effective area. Returning to the sunlight example, recognize that the output power of a solar panel is proportional to its physical area; the larger this area, the greater the fraction of the incident power transmitted by the sun is received by the solar panel and converted to available output power. Receiving antennas, and antennas in general (even wire antennas), have an effective area; it's the area required to produce the measured output power, based on the density of transmitted power (watts/unit area) at the location of the receiving antenna.

      Antennas can also be characterized by their gain, a function of their directivity and efficiency.

      Interestingly, based on these two parameters any given antenna can be placed into one of two categories: There are constant-area antennas, the effective area of which is constant with frequency, and constant-gain antennas, the gain of which is constant with frequency. Constant-area antennas have gain that increases with frequency; constant-gain antennas have effective area that decreases with frequency.

      The source of the myth is that most portable consumer wireless products use constant-gain antennas, usually some variant of a dipole. While the gain of a resonant dipole is constant with frequency, as the frequency goes up its physical length, and therefore its effective area, goes down. 2.4 GHz dipoles are physically smaller than 900 MHz dipoles. They therefore have less effective area, and recover less power from the incident wave. It seems like the path loss at 2.4 GHz is greater, but it's really just a result of the antenna choice in the product design. If consumer products used constant-area antennas, like a parabolic dish of fixed physical dimensions, exactly the opposite result would be found: Since constant-area antennas have gain that increases with frequency, the recovered power at 2.4 GHz would be greater than that at 900 MHz, and we could start a myth that propagation loss decreases with frequency.

      Interestingly enough, if the transmitter has a constant-gain antenna and the receiver has a constant-area antenna (or vice-versa), the recovered power at the receiving antenna terminals would be independent of frequency (i.e., constant), and we could avoid the generation of propagation loss myths entirely.
  • by 3flp ( 172152 ) on Sunday April 23, 2006 @12:45AM (#15183266)

    I don't post here very often, but this time I couldn't handle this. (Maybe I should drink less coffee). There was probably some paper at that uni, talking about an incremental improvement in frequency divider design. Ok, cool ... we may or may not see in in a PLL chip in a few years. But the news release (TFA) and RP's writeup are rubbish. Actually, after a bit of Googling, it's all over the net. Next thing I expect, my PHB will ask me to change my totaly unrelated design to use ILFD. My signature notwithstanding, I'll try to pick out some of the c***p, and put some actual information in. BTW, I design 3G mobile terminal circuitry full time. And yes, I am an arrogant SOB. That doesn't make me wrong.

    "...But now, researchers of the University of Rochester have developed a wireless chip that needs ten times less power [GC] than current designs."

    So far so good.

    The new chip relies on a technology named injection locked frequency divider (ILFD) which dramatically reduces the time needed to check for transmission frequencies which are performed several billion times per second by your current phone.

    This statement is wrong 2 times. First of all, the time needed to check for transmission frequencies depends on PLL settling time. Nothing to do with divider technology. Even broader scope, it is a rare occurence in 3G that the phone needs to change RF frequency. It's WCDMA, so all cells from a given operator transmit on the same channel. Secondly, tthe checking for transmission does NOT occur "several billion times per second". The RF carrier frequency is several billion cycles per second (ie several GHz). But the carrier frequency is changed on every 10ms roughly, even when it needs to happen. That's 100 times per second. GSM is different, as it does frequency hopping normally, but that doesn't change the point: nothing to do with divider technology.

    The new chip uses five transistors and can perform divisions by 3 instead of only 2 by previous circuits

    OK, agreed. Anyway, who gives a f**k. A modern PLL chip has a programmable divider, settable from 3 to several thousand. Yes, 3, because it is different technology.

    ..., allowing a perfect communication between two phones communicating at 2.0001 and 2.0002 gigahertz respectively.

    That's not how mobile phones work. Mobiles establish connection with the cell (base station), then remain frequency locked to it, to compensate for temperature dependant frequency variation of their reference reference crystal oscillators - and Doppler shift, if they are moving. A "perfect" communication hardly ever depends on this. And frequency locking does not happen via changing PLL settings in this case anyway - too coarse steps, so other techniques are used.

    Anyway, as other people posted already, the frequency synthesizer is not significant contributor to mobile terminal power consumption. Even old PLL chips only use a few milliamps [national.com]

    The ILFD technology seems to be good for building efficient frequency dividers at higher microwave frequencies. That will probably not affect current mobile phones anyway, because all the current systems work around 1-2GHz. Higher up, it's difficult to achieve coverage. Again, other people already pointed this out.

    If you want real news in this area, go to sites like this [rfdesign.com], or this [mwjournal.com]. Slashdot's editorial quality has degraded in the last few years so much that I am thinking about deleting it from my bookmarks.

    [/rant]
    • Slashdot's editorial quality has degraded in the last few years so much that I am thinking about deleting it from my bookmarks.

      Calm down... Most people come to slashdot for the comments. The articles themselves only serve as a starting point for a discussion, which is often valuable since there are always people like you who really know what they're talking about.

    • Slashdot's editorial quality has degraded in the last few years so much that I am thinking about deleting it from my bookmarks.

      Take it from an old-timer. Slashdot's editorial quality has remained pretty much consistent for about its entire existence.
  • Does this mean it actually supplies enough power to run 9 regular chips?

    Can I buy a thousand of these new chips and use them to power my electric car?

  • Phones do not communicate with phones! Phones communicate with base-stations. If this adjustment was really a power issue, then it could be done in tha base-station. However it is not. The power issue is the sending power. If you put out 2 Watt of RF, then you have to drain at least 2 Watt from your battery. There is no way around that in this universe.

    Personal guess: Sloppy journalism and a marketing depatment working hand in hand. This is non-news and none of the stated benefits is even possible.

    Bad slash
  • There was an excellent thread here [slashdot.org] some time ago.

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