0
\$\begingroup\$

I'm trying to understand these earings from ebay.

Here is my attempt to reverse engineer the schematic:enter image description here

If this is correct, I don't understand the function of the diode. If the diode is for demodulation how does going between ground and the filter work? This looks very different to a standard crystal radio: source https://en.wikipedia.org/wiki/Crystal_radio

I am also not sure why the resistor. Would it not reduce the efficiency?

\$\endgroup\$
  • 2
    \$\begingroup\$ Can you put the photos and schematics into your question so that we don't have to follow three links to understand it? It will also attract more help and ensure that the question still makes sense after the links die. Thanks. \$\endgroup\$ – Transistor Mar 21 at 14:56
  • 1
    \$\begingroup\$ Egad. I think I'm going to have to try and make something less hideous. \$\endgroup\$ – JRE Mar 21 at 15:02
  • 4
    \$\begingroup\$ Label the other side of speaker as ground. Flip your schematic over. Now what do you see? Pin 1 of speaker as ground. \$\endgroup\$ – Marla Mar 21 at 15:18
  • \$\begingroup\$ @Marla Right, so I just got ground on the wrong side, thank you. I'm still not quite sure what the resistor is for? \$\endgroup\$ – Edwin Shepherd Mar 21 at 15:22
  • 2
    \$\begingroup\$ Did you buy some to reverse engineer? Do they actually work? How inconvenient is it trailing a 10 metre antenna around with you when you go walkabout? \$\endgroup\$ – Andy aka Mar 21 at 15:46
1
\$\begingroup\$

Seems to be a standard circuit.
The placement of antenna and ground may be causing confusion. Both help convert radio signals to audio signals, but can be placed at various circuit positions. Without both, you'd have to be very close to a transmitting site to hear anything at all.

schematic

simulate this circuit – Schematic created using CircuitLab
Antenna & ground connect to either ?1 or to ?2. Doesn't matter which, since the headphone is floating into your ear.
There are two variations, depending on headphone:

  • A magnetically-driven headphone (left) is actually a coil of wire having significant resistance. The resistance of the coil ensures that audio currents can fall to zero when the radio signal amplitude dips to zero.

  • A piezo-active headphone (right) is open-circuit to DC so that it can charge up to a DC voltage and stay there instead of falling to zero. You don't want DC, you want an audio waveform that rises and falls.

    R2 is required to ensure that audio currents can discharge the piezo's capacitance to zero. Yours seems to be of this type. R2 and the piezo's capacitance form a low-pass filter that tends to attenuate audio above a few kilohertz. Yes, this resistor consumes some audio signal, but most goes into the piezo transducer XTAL_headphone.

Since the headphone is ungrounded, it doesn't matter whether ?1 connects to ground or to antenna. ?2 of course must be connected to whatever ?1 doesn't. Both ground and an antenna wire (with far-end unconnected) can receive radio signals.

There is another possible arrangement, where L1 is left out, and a big-area multi-turn loop of wire is connected between ?1 and ?2. This loop is an inductor which has enough turns to resonate with C1 at the desired frequency of the radio station. No ground or antenna connection is required. Awkward, because the loop should be large-diameter.

The very simplest version requires a diode + magnetic-type headphone of high-resistance. It is un-tuned, yielding a cacophony of radio stations, one of which is usually loudest. Note that diode direction in any of these circuits doesn't matter...cathode and anode can be swapped.

schematic

simulate this circuit
The whole idea of ear-ring broadcast-band radio receiver is quite ridiculous. An efficient method of receiving broadcast-band AM radio requires a long antenna wire, and a wire pegged to earth. Or two long antenna wires running in opposite or at least orthogonal directions.

This crystal radio could detect FM broadcast band signals, but the modulation method of FM doesn't produce audio with these diode detectors.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.