Basics of old school radio questions.

[jadaro2600]

In common / old school style crystal radios, a 'tank circuit' is used to create a resonant circuit.  A variable capacitor is used to tune the tank circuit to a desired frequency - and somewhere amongst the wiring a crystal earphone is tapped into the circuit across the diode so that energy in the system - when not passing through the diode, is passing through (out of ) the earphone - technically, half the signal is lost through this mechanism.

In such a circuit, is it solely the responsibility of the diode to create the mechanism of the tank circuit, or does the capacitor do this?

And if it's the diode responsible, couldn't two tank circuits be formed and for each diode-rectified half wave produced?  ...would an audible signal be heard for each half wave if two earpieces were used or would the tank circuit be lost due to the splitting.

The point would be to retrieve the entire sine wave frequency, possibly to rectify to DC outside that of the AC tank circuits or to hear the entire sine wave.

These question may have significance for tuning to resonance.   ..and getting that one diode to light from a remote source. 

My instincts tell me that i'm designing into the heterodyne / superheterodye circuits - yet i wish to simplify into the old school methodology.

[pese]

 deleted
I SEEN, I CANT HELP
TO UNDERSTAND THIS.
Pese

[gyulasun]

Hi jadaro2600,

...so that energy in the system - when not passing through the diode, is passing through (out of ) the earphone - technically, half the signal is lost through this mechanism...

I quoted your text I wish to address.  The energy in the system is enclosed in the LC tank circuit and only a certain amount of the energy is used up by the rectifier diode if the whole circuit is designed correctly.  This is because the earphone is the load and it shunts the tank circuit via the diode and if the loading effect is excessive, the tank may get attenuated too much and energy gets dissipated.  In correct designs the diode is connected to a tap on the coil i.e. impedance transformation is achieved (matching the load of your earphone to the resonant  tank if you like).  Matching is needed even at the input side of the tank where the antenna is connected, the point is to keep the resonant impedance of the tank as high as possible, ok?

So it is the resonant LC circuit which contains and enhances the input energy.  Energywise, an AM modulated signal consists of the carrier and two modulated sidebands, the LC tank is tuned to the carrier and the diode rectifies the carrier, you are left with a DC component and the modulation after the diode, this is analogous to a half wave rectifier case. You get a DC output after the diode and you get the "hum" (i.e the audio frequency in your case)  the difference is you use further filtering to get rid of the "hum" but in the AM demodulation case you wish to use filtering that removes only the carrier frequency "hum"  but does not attenuate the audio frequency.
See these links for further details:
http://en.wikipedia.org/wiki/Envelope_detector
http://en.wikipedia.org/wiki/Diode_detector

Pese's letter nicely includes these things.

rgds,  Gyula

PS  Now you may wish to rethink what your idea is in connection with all these?

[jadaro2600]

PS  Now you may wish to rethink what your idea is in connection with all these?

Some time today, i'm going to attempt a simple radio - but i'm going to have to use an audio amp or my sound card to listen to the output.  Having a vari-cap would be nice....

I have some simple diodes as well.  Also, I have a bunch of ceramic disk capacitors.

One of them says 561 1kv, the other says 221, ..I'm guessing that these are 561 and 221 pF capacitors?

[amigo]

One of them says 561 1kv, the other says 221, ..I'm guessing that these are 561 and 221 pF capacitors?

560 and 220pf, the 1 at the end refers to the multiplier: 1 = x10, 2 = x100, 3 = x1000, 4 = x10000

560 and 561 is basically the same value, just different convention.