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AM cyrstal radio and high imp earphones

Discussion in 'Electronic Design' started by Paul Burridge, Feb 13, 2006.

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  1. Nope! You want all your (tiny) amount of power to be dissipated in the
    headphones, not wasted in some half-arsed resistance network.
     
  2. Guest

    Hi all!

    taken from here :
    http://www.electronics-tutorials.com/receivers/crystal-radio-set.htm
    I quote :

    "Headphones: This is by far the hardest part to obtain. The type used
    for hi-fi WILL NOT work here. Ideally you need high impedance 2,000 ohm
    types but these are nearly impossible to find. You can sometimes buy
    1,000 ohm crystal earpieces (no not the usual transistor radio type).
    The headphone is a high impedance load for the crystal set and as we
    are working on free power from the air we can't load it down. If you
    used the 8 ohm hi-fi type it would be like trying to run your electric
    toaster off a very small battery. The power is not available. Remember
    we're using free power from the sky."

    could one not just use normal earphones,say 8ohm but put them in
    parralel with a 40k ohm resistor in the circuit ?
     
  3. GregS

    GregS Guest

    There are plenty of Piezo speakers at Digi-Key or other sources.
    You could easily replace the dynamic element of most
    headsets with these, and then you can have that awfull
    old fashioned sound.

    greg
     
  4. Leon

    Leon Guest

    Won't work. Try an audio transformer like the ones that used to be used
    in the output stages of transistor radios. They were something like
    1k:4R, and should work OK with ordinary low-impdeance headphones.

    Leon
     
  5. Guest

    cheers
     
  6. Guest

    just a detail,with a transformer you dont mean directly in the circuit
    do you?
    because total power in and out is the same , right?
     
  7. Pooh Bear

    Pooh Bear Guest

    The attenuation wil be 20*log10(40,000/8) = 74 dB.

    It will make the sound inaudible.

    Graham
     
  8. Pooh Bear

    Pooh Bear Guest

    Were the turns ratios as high as that ?

    I susepct the low-Z headphones still won't be efficient enough though.

    Graham
     
  9. Tim Wescott

    Tim Wescott Guest

    Yes, he does:


    <------. ,----->
    To radio )|( To earphones
    )|(
    <------' '----->
    (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)


    Total power in = total power out, which is good. Depending on your
    8-ohm earphones this may not be good enough: the old 2000 ohm earpieces
    were carefully made to be very power efficient; good sound was
    sacrificed for high efficiency.

    --

    Tim Wescott
    Wescott Design Services
    http://www.wescottdesign.com

    Posting from Google? See http://cfaj.freeshell.org/google/
     
  10. Tim Wescott

    Tim Wescott Guest

    http://www.tubesandmore.com has carried them in the past, I'd be at
    least mildly surprised if they didn't still have them.

    --

    Tim Wescott
    Wescott Design Services
    http://www.wescottdesign.com

    Posting from Google? See http://cfaj.freeshell.org/google/
     
  11. No, but using an impedance matching transformer with a 16 ohm
    rare-earth headphone will probably work better than the usual ceramic
    type. Google around, I'm sure people have solid data on this.

    However, your easiest path is to just order the antique thing from
    Mouser for the price of a coffee, but they're backordered on them atm.


    Best regards,
    Spehro Pefhany
     
  12. Guest

    thanks for the info guys!! :D
     
  13. Pig Bladder

    Pig Bladder Guest

  14. Chris

    Chris Guest

    If I might add to the climenole chorus:

    1,000 ohm dynamic earphones
    Mouser 25DE139 -- $1.63 USD ea.

    20 Meg ceramic earphones
    Mouser 25CR060 -- $1.85 USD ea.

    Both in stock, and ready for budding DXers.

    73
    Chris
     
  15. Tim Wescott

    Tim Wescott Guest

  16. Rich Grise

    Rich Grise Guest

  17. Donald

    Donald Guest

    Yes, but do you understand DC circuits ??

    What would be the voltage drop on the 8 ohm device with a 40K ohm
    resistor in series, if the voltage across them both is 40 microvolts ??

    donald
     
  18. Rich Grise

    Rich Grise Guest

    Well, yianvn did say, "parallel"[1], but ohm's law will still apply - the
    effect of a 40K in parallel with 8 ohm phones will be negligible.

    He still won't get any sound, however. :)

    Cheers!
    Rich
    [1] OK, he said, 'parralel', but I'm not going to rag on him about it.

    Yet. >:->
     
  19. Donald

    Donald Guest

    Opps :-~
     
  20. Robert Baer

    Robert Baer Guest

    Parallel??? That is supposed to increase the impedance? No way!
    And in series does decrease the loading to the crystal set, but
    decreases the drive to the headphone.
     
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