Nope! You want all your (tiny) amount of power to be dissipated in the headphones, not wasted in some half-arsed resistance network.
Nope! You want all your (tiny) amount of power to be dissipated in the headphones, not wasted in some half-arsed resistance network.
Hi all!
taken from here :
"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 ?
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
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
The attenuation wil be 20*log10(40,000/8) = 74 dB.
It will make the sound inaudible.
Graham
Were the turns ratios as high as that ?
I susepct the low-Z headphones still won't be efficient enough though.
Graham
-- Tim Wescott Wescott Design Services http://www.wescottdesign.com Posting from Google? See http://cfaj.freeshell.org/google/
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
-- "it\'s the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com
thanks for the info guys!! :D
They're all over the place:
Have Fun! Rich
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
If they still have them the ones from Antique Radio Supply are real honest-to-god (or perhaps honest-to-Taiwan) _head_phones, not just earplugs.
That doesn't mean they work well, but they sure look like the ones in my old books.
-- Tim Wescott Wescott Design Services http://www.wescottdesign.com Posting from Google? See http://cfaj.freeshell.org/google/
OOpps! Guess I had one too many booldy maries - this was suppposed to have been juts plain me! %-}
Thanks! Rich
-- bloody marys, just plain - my drunken proofreader homunculus told me to leave them in, because he thinks they\'re cute. Go Figure! ;-D
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
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. >:->
Opps :-~
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.
Use a trandformer dugh.
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