How a Crystal Earpiece works

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  • เผยแพร่เมื่อ 5 ต.ค. 2024
  • Crystal- or Piezo-Earpieces are today only used for crystal-radios because they claim to have high sensitivity and present only a negligible load of over 10 Mohms.
    Well can this really be true?
    Roger tells the truth about the real impedance or load-resistance of crystal earpieces.
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ความคิดเห็น • 15

  • @alanmorris7155
    @alanmorris7155 4 ปีที่แล้ว +2

    This is an excellent explanation of the workings of a crystal earpiece. Just the right amount of information nicely supplemented with simple maths and actual measurements. Many thanks.

  • @KainkaLabs
    @KainkaLabs  8 ปีที่แล้ว +6

    What I forgot to mention is what the "Crystal" in "Crystal earpiece" means.
    Originally crystal earpieces were in fact made out of a specially shaped crystal of Rochelle salt (Potassium sodium tartrate tetrahydrate). Rochelle salt was the first known material to show piezoelectric properties (together with quartz and others).
    Modern piezo-transducers (piezo-discs) use a layer of Barium-titanate on a disc of metal. Barium-titanate is a "ferroelectric" material which is a sub-class of piezoelectric materials. Ferroelectricity is analogous to Ferromagnetism although there is no "Ferrum" (=iron) involved.

  • @AnthonyFrancisJones
    @AnthonyFrancisJones ปีที่แล้ว

    Excellent! I am reminded that similar setups are used in ultrasound sensors and there (as in this case possibly) the distance to the back of the case of the earpiece is important to so it deals with reflected sound waves effectively. Though, of course this device is much smaller than the wavelengths of sound it is required to produce.

  • @glennkrieger
    @glennkrieger 3 ปีที่แล้ว +1

    This was excellent sir. Thank you!

  • @stighenningjohansen
    @stighenningjohansen ปีที่แล้ว

    Thanks, for the precise info, received four of these today, part of a prepping effort, build the best crystal radio in the world, no less :) Very useful post, thanks.
    About prepping, I see dire straits ahead, things are very dark in these days and time, so, hold on to your butt :)

  • @putnik805
    @putnik805 2 หลายเดือนก่อน

    Very informative, thank you

  • @AndyPanda9
    @AndyPanda9 4 ปีที่แล้ว +1

    I'm not buying your exponential horn discussion pointing to the curve in the plastic housing - that curve is exactly opposite of a horn, maybe an "exponential funnel"? :) Or did you mean that the ear canal itself is your exponential horn?

    • @victor9501
      @victor9501 ปีที่แล้ว +1

      I think of it as an ear trumpet, those early types of hearing aids, where you held the narrow end near your ear.

  • @guliyevshahriyar
    @guliyevshahriyar ปีที่แล้ว

    Thank you very much

  • @t1d100
    @t1d100 ปีที่แล้ว

    Very interesting!

  • @michaelinzo
    @michaelinzo 2 ปีที่แล้ว +1

    Can this work in a regular smartphone?

    • @KainkaLabs
      @KainkaLabs  2 ปีที่แล้ว +1

      why not. Sound quality is of course very bad.

  • @ivanbiasutti4567
    @ivanbiasutti4567 7 ปีที่แล้ว

    if I connect two of these in series I get louder volume?

    • @KainkaLabs
      @KainkaLabs  7 ปีที่แล้ว +2

      Well, it depends on the output impedance of the signal-source:
      If the crystal-earpieces are connected to a low-impedance source (output of an amplifier-stage), then the voltage at each crystal-earpiece halfes and thus the power in each earpiece will be only one fourth compared to a single earpiece.
      If the crystal-earpieces are connected to a high-impedance source (output of a crystal-radio) then in fact the voltage (nearly) doubles over the total of the 2 crystal-earpieces.
      So each crystal-earpiece gets (nearly) 1,4 times the voltage and thus double the power.
      Both crystal-earpieces together then get four times the power.
      You can do the maths with Ohm´s law or simulate it in LT-Spice to understand it.
      It´s quite a good learning example :-)

    • @ivanbiasutti4567
      @ivanbiasutti4567 7 ปีที่แล้ว +1

      many thanks for explanation! btw great YTchannel :)