AM cyrstal radio and high imp earphones

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

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

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

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

thanks for the info guys!! :D

They're all over the place:

Have Fun! Rich

If I might add to the climenole chorus:

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

Both in stock, and ready for budding DXers.

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.

OOpps! Guess I had one too many booldy maries - this was suppposed to have been juts plain me! %-}

Thanks! Rich

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.