FCC approves first 'power-at-a-distance' wireless charging device

Discussion in 'Amateur Radio News' started by W4RAV, Dec 28, 2017.

ad: L-HROutlet
ad: l-rl
ad: Subscribe
ad: l-assoc
ad: L-MFJ
ad: Left-2
ad: Left-3
ad: MessiPaoloni-1
  1. KI7HSB

    KI7HSB XML Subscriber QRZ Page

    Sure it is... and funny too.

    The point is, just because something is patentable and functional, doesn't mean that it's a good idea, nor does it guarantee that it will catch on and be popular or profitable... If it were so, we would all be wearing fart proof underwear.
     
    W8OKX likes this.
  2. W1YW

    W1YW Ham Member QRZ Page

    Nope. I couldn't care less about that. The objective of a patent is to claim a novel invention and teach the art of making it. I read other folks' patents to have them teach me that art. Whether a minuscule fraction of issued patents is 'foolish' or not is irrelevant to me. I don't read foolish patents. In fact, I have never run across one in 35 years of patent searching. Guess my interests don't coincide with ones that lend themselves to foolishness.

    I am not interested in learning that. I'm not interested in every QSO on the ham bands either. Surely we must concede that a non-trivial fraction of those are foolish, too.

    Obviously you must be interested in mitigating the aroma of flatulence or you wouldn't have brought it up.

    Frankly, that concept doesn't rock my world. As I approach 63 I am only interested in solving real problems, and my digestive tract is in unusually good shape, thanks to psyllium, carrots, and proper bacterial flora.

    The patent library is the most important source and documentation of progress in the human condition.
     
    Last edited: Jan 2, 2018
    W4RAV likes this.
  3. AA7EJ

    AA7EJ Ham Member QRZ Page

    Patent no patent , I am from Missouri , show me state.
    No more voodoo guess work.

    Assume that the wireless charger spec is to TRICKLE charge with 1 mA current and we still have contains of 1W RF output and load located 3 feet ( app 1 meter ) from the source.
    Show me the intensity of RF - measured by microvolts per square meter at the distance of 1 meter.
    Show me , with measure units of you choice , "level " of RF at the receiving antenna terminal.
    Show me the rough outline of the "power converter" using measuring units of your choice at the OUTPUT of the "converter".
    Does the output meets the spec required 1 mA of trickle charge?

    As a bonus question - show estimated efficiency of each component of this setup.

    I am too old , OK lazy, to do this high school math.

    73 Shirley

    PS As far as I know obtaining a patent DOES NOT require actual demonstration of the "idea".
     
  4. W1YW

    W1YW Ham Member QRZ Page

    Ideas are not patentable.

    Patents are all subject to the requirement of enablement.
     
    KQ6XA likes this.
  5. KI7HSB

    KI7HSB XML Subscriber QRZ Page


    Wow... You may be a brilliant physicist, but you apparently know nothing about the human condition... and don't even realize it.


    Fail-Yoda.jpg
     
  6. W1YW

    W1YW Ham Member QRZ Page

    Sometimes, I wish I knew a lot less.

    But nice try anyway.
     
  7. AA7EJ

    AA7EJ Ham Member QRZ Page

    OK, I used wrong word.
    Let's use different word since English is such rich language - what is the subject of this "patent" in general?
    Or - do you have a real patent number so I can laugh at my own leisure?

    As far as "enablement" - if Henry Ford had a patent on Model T it enabled to convert can of gas to few miles of road passage.

    In this case we are trying to "enable" 1 mA of current into a battery.
    Compare to Henry's patent it is child's play.

    73 Shirley
     
  8. W1YW

    W1YW Ham Member QRZ Page

    Shirley,

    I am just informing you a bit on how patents are set up. Enablement is a basic patent term.

    I dont define the ground rules.

    On the OP topic, I guided us to look at the patents, to see how the charging system works. Thats all.

    73
    Chip W1YW
     
    KQ6XA likes this.
  9. G4KHU

    G4KHU Ham Member QRZ Page

    However convenient wireless charging of devices "up to three feet away" the laws of physics clearly show an inverse square signal loss so that even at 900 MHz a device being charged at 3 feet is approx. three wavelengths away and will only receive a fraction of the transmitted power. This means the power the consumer takes from the supply will be vastly more than the power transferred to the wireless device. What's the point of that when most of us are trying to reduce energy consumption and be a little more green in our approach to energy use. Basically don't be so lazy plug just phone in the wall to charge it and save energy!
     
    KQ6XA likes this.
  10. W1YW

    W1YW Ham Member QRZ Page

    I do want to correct a naive rule of thumb that many people invoke on the Friis Equation. So this is a generic response to the post, and not targeted at any person.

    First, the Friis equation only applies to the far field. That means that moderate or high gain antennas require a minimum distance before it applies, and that distance is not 1/16 or 1/2 wave--it depends on the apertures used. Next, the medium is not invoked in the Friis equation. So you've got that attenuation to deal with.

    The naive rule of thumb on the Friis Equation is that the received power goes down dramatically as the frequency goes up. Well, sure, if there is no wavelength dependence on the antennas. IOW if you choose antennas of a given gain, and don't change the gain(s) with increasing frequency, the rule of thumb applies.

    But it is a bad rule of thumb. Here why: the antenna gains INCREASE, for fixed apertures, as the frequency increases, even though the NUMERATOR lambda-squared term applies.

    Want to see that dependence? It may surprise many (unless you took a signals and systems course in college):

    Gain = 4 pi Ae/ lambda-squared

    Ae= pi D squared/4 (roughly, for an assumed circular aperture, diameter D)

    If D is fixed then

    Gain=k/ lambda-squared where k is a constant

    And, as a good example, if you have the same antenna apertures at RX and TX, then

    Gt =Gr

    So in the Friis equation, you end up with a factor of lambda -squared IN THE DENOMINATOR, with NO lambda factor IN THE NUMERATOR.

    IOW, the received power GOES UP as you increase frequency.

    It always disappoints me that we tend to take even simple equations linke the Friis equation and relegate them to boxes on an excel sheet, without thinking about what the equations MEAN anymore:-(

    Pity.
     
    Last edited: Jan 3, 2018
    W8YIH likes this.

Share This Page