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I am building a crystal radio for my science fair project, and I am having some issues. For my earpiece, I am using a telephone receiver which may be the issue. For my ground, I am using a water spigot in my backyard which is completely metal. My diode is a germanium diode, which works. My coil is a bottle wrapped in enamel coated magnet wire that is between 18 and 22 gauge. Instead of sanding a section of the wire and using a wiper blade, I have taps with the enamel sanded off. I have tested all of my parts with a voltmeter, so the circuit isn't the issue. Any advice? What might I be doing wrong?

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    \$\begingroup\$ Add a photo and a schematic. There are buttons for both on the editor toolbar. Also show what design calculations you used for the coil (diameter, lenght and number of turns). \$\endgroup\$
    – Transistor
    May 10 '16 at 22:29
  • \$\begingroup\$ IME using a variable capacitor is much eaier than a variable inductor as your instructions seem to specify. Otherise, your problem is headset impedance, as wbeaty describes. \$\endgroup\$ May 11 '16 at 8:05
  • \$\begingroup\$ Recently I had to setup 30 crystal radio sets for a STEM electronics class I was teaching and we used these high impedance crystal ear pieces that seem to be designed for crystal radio projects and they do the job for a low price protechtrader.com/… Seems like the OP could've used the same earphone unless he had to build the earphone himself for the science fair. \$\endgroup\$
    – Ajk Tek
    Jun 13 '18 at 17:02
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The problem involves the "impedance" concept, so it's usually not discussed in beginners' crystal radio projects.

Crystal radios need a high-resistance earphone. This is the type that produces tiny sounds when driven by several volts, while only drawing a few hundred microamps. For AC signals it appears to be a large resistor, 5K or higher.

A standard 8-ohm earphone appears as a much smaller resistor for AC; roughly eight ohms, and it expects a signal of hundreds of millivolts, while drawing a few tens of milliamps.

Yet the RF signal from a crystal radio must be high voltage at low current, so it can well exceed the 0.3V detector voltage. It's designed to produce few-volts DC output, not few-tenths. A standard earphone will just short out your receiver, and won't convert very much of the DC output into sound.

There of course is a simple cure. Connect a small audio transformer to your earphone. You want to step down the few-volts output by a factor of 20 to 50, converting 8ohms into many Kohms. Small transformers like this are available, called audio matching transformers, or audio output transformers for old-style transistor radios. Connect one of these to your tiny earphone, so the high-volt, high-ohms side connects to your crystal radio output.

The above will work, but unfortunately this transformer uses up some energy (as wire heating,) as does the coil in your earphone. A piezo-crystal earphone is far more efficient than a coil/magnet earphone.

So, for the same milliwatts of EM energy being received by your radio, a crystal earphone will sound distinctly louder than a coil earphone, even if exactly the right audio transformer is being used. The missing audio ends up as coil-heating.

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  • \$\begingroup\$ IME a magnetic headphone can be as efficient as a crystal earpiec. The problem is impedance (as you also mention), not so much efficinecy. I my younger days I built dozens of 'crystal' radio's using high-impedance magnetic headsets. Alas, those son't seem to be common any more. \$\endgroup\$ May 11 '16 at 8:04
  • \$\begingroup\$ Lacking an audio transformer, as small mains transformer could work, but is has to have a very low secondary voltage (even more so in 120VAC versions than in 220VAC versions). \$\endgroup\$ May 11 '16 at 8:06
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I doubt that a telephone receiver is sufficiently sensitive to work with a crystal radio. It needs a significant amount of power to operate, which you cannot get from a crystal radio. You will have to use a crystal earpiece, such as this.

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the problem is the lack of power, a crystal radio draws its power wirelessly from the signal that it's receiving. Regular speakers require quite a bit of current to operate (and a reasonable voltage - hence high power). So unless you have an antenna the size of a house, it's probably not going to collect enough energy to drive a regular speaker. Crystal earpieces need almost no power to operate (they also need almost no current - which is the harder to generate anyway), that's why they're used, because a regular crystal radio only generates enough power to drive a crystal earpiece. If you had a large enough antenna you could power a full hifi system, but the antenna would be impractically large. If you're interested in working out roughly how much power might be available, measure the voltage at the speaker (with the speaker disconnected) using the AC volts range, then measure the AC uA, multiply the two and that should give you a rough idea of how many microwatts you've got to play with

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The heart of the crystal set (as in any other radio receiver) is a parallel tuned circuit. This parallel tuned circuit is attached to the antenna and the antenna behaves like a very short dipole. This means it looks like a capacitor to earth from the perspective of the tuned circuit.

Why does it behave like a short dipole - at 1 MHz the wavelength is 300 metres and half of this is 150 metres and, does your xtal set antenna have an antenna this long? Almost certainly not because nobody has that amount of space.

Hence the parallel tuned circuit formed by the coil and parasitic capacitance (or tuning capacitance) has added to it a bit more parallel capacitance due to the antenna.

At resonance (the tuning point) the tuned circuit behaves like an open circuit and any current taken from it causes both selectivity and sensitivity to be reduced so, for a reasonable load (~20 kohm), the sensitivity (size of signal) might half and the selectivity (ability to exclude close by unwanted signal) may also half. For 10 kohm load this gets nearly twice as bad and as load resistance drops it becomes worse and worse.

In other words, the tuned circuit wants zero energy taken from it by any load in order to be sensitive and selective.

Xtal earpieces that I've seen have an impedance quoted at about 20 kohm so there is some loading but, it's not that bad if the tuned circuit is designed/built correctly. Some loading is actually required to avoid selectivity value that excludes the full spectral width of the station being received.

Anything below 1 kohm is going to be fairly rubbish and anything below 100 ohms just won't work.

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  • \$\begingroup\$ Just read your answer and trying to be sure if I get it right. Is the reason for using high impedance earphone like 20 kohm to obtain a better Q-factor and better selectivity? What I understand from plots in LTspice that in an LCR circuit LC part sets the tuning freq. and the series resistor defines the sharpness of the band-pass and Q-factor. But how about for a crystal radio using a usual 8 ohm headphone in series with a 20k resistor instead of a piezzo headphone? Why wouldn't it work since no one uses this way? \$\endgroup\$
    – user16307
    Oct 4 '16 at 16:55
  • \$\begingroup\$ Because an 8 ohm in series will receive 34 dB less power than the 20 kohm and as the power will be low and fairly quiet on a 20kohm earpiece, the 8 ohm will be 34 dB quieter. Each ten dB reduction is a halving of loudness so 30 dB reduces perceived loudness by a factor of 8 to 1 so 34 dB will be about a 10:1 reduction in loudness and you might not hear anything because it's so quiet in the first place. \$\endgroup\$
    – Andy aka
    Oct 4 '16 at 17:35
  • \$\begingroup\$ Oh i see and putting 20k in parallel wouldnt work either since the R_equivalent will be less than 8ohm \$\endgroup\$
    – user16307
    Oct 4 '16 at 18:03
  • \$\begingroup\$ Correct @user16307 \$\endgroup\$
    – Andy aka
    Oct 4 '16 at 19:00

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