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I'll be visiting home soon after school lets out, and I want to build an AM crystal radio with my little brother. This is the design we'll use:

http://crystalradio.net/beginners/

I'll be doing my best to explain to him how this works. I understand that when a radio ray passes over an antenna, it generates an alternating electrical field in the antenna wire. If the antenna wire is the correct length for the given radio ray, the electrons in the wire will bounce back and forth in sync with the alternating field, moving faster and faster with each oscillation (resonance). I also understand how audio can be embedded in an AM signal using amplitude modulation.

However, I do not understand the diagram in the diy linked above.

  • Where do the electrons flow?
  • Doesn't splitting the circuit in the middle of the coil create two "antennas", each with a different resonant frequency?
  • What is the purpose of the resistor? I have looked up the internal design of piezoelectric earpieces, and it does not appear that current flows all the way through them.

The circuit I've chosen is similar to, but different from, the circuit shown here:

wikipedia https://en.wikipedia.org/wiki/Crystal_radio#Design

I was able to find some helpful material on the howstuffworks pages for AM radios, but it is not a one-to-one mapping of the design proposed in the first diagram.

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  • $\begingroup$ Looks like you have the basics. The electrons come from the air, via the antenna. btw, half the fun of building one is to actually use a galena detector. Using a diode works but doesn't help explain why. Now if you REALLY want to have fun with a youngster, try the Fox Hole radio, where all you need is an earphone, some wire, a razor blade and a match. It is in Wikipedia but I like this link: n6cc.com/crystal-radios-it-started-here $\endgroup$
    – SDsolar
    Apr 17, 2017 at 15:41
  • $\begingroup$ There are superior crystal radio designs than this one, that are more sensitive and selective. Depending on your proximity to AM stations, their ERP, and how close they are in frequency to each other, you may or may not be happy with this design. AE7KU is an expert on crystal radios, and I uploaded some of his stuff to w0btu.com/files/Crystal_Radio. But there is far more info on the web. Google is your friend. :-) $\endgroup$
    – Mike Waters ♦
    Apr 19, 2017 at 18:14
  • $\begingroup$ @SDsolar I agree on using the catwhisker detector. However, if he's never built a crystal radio before, it might be better if he started with a 1N34 or other diode. That's especially the case with such a basic circuit which may not work well regardless of how good the diode is. If it works, then trade the diode for a more challenging galena crystal, razor blade, etc. :-) $\endgroup$
    – Mike Waters ♦
    Apr 19, 2017 at 18:19
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    $\begingroup$ I totally agree, @Mike. He'll see that it works without batteries, then you can up the game at your leisure. I think it is great you are doing this with him. Pretty cool. $\endgroup$
    – SDsolar
    Apr 19, 2017 at 22:12

1 Answer 1

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I'll try to address your concerns a bit! I'll be simplifying here and there, but I think you'll notice where I'm doing that.

I'll be doing my best to explain to him how this works. I understand that when a radio ray passes over an antenna, it generates an alternating electrical field in the antenna wire.

Close, but not quite :) I'll correct you here, because it makes things easier to explain below.

We talk of radio waves, not rays. The point is that the wave actually is the alternating electrical field in the air. It induces a current in the antenna. So, any piece of metal that you have lying around is an antenna, and there's electricity flowing through it, because there's radio waves of all kinds of sources (some natural, some man-made, some intentional, some accidental) everywhere! But: these antennas don't work overly well, and also, there's not much power at most wavelengths (see below).

If the antenna wire is the correct length for the given radio ray,

Here's why I mentioned the wave naming: If the wave's wavelength happens to be such that the antenna connects a "deep" and a "high" point in the electrical field, you get an especially strong current. So, the length of your antenna has to be in a specific relation to the wavelength of the radio signal you want to capture. If it's a bit "off", that's not too bad. But you can't properly receive, for example, a 3 MHz (wavelength: 100 m) signal with an antenna that's meant to receive a 2.4 GHz (wavelength: 1/8 m).

the electrons in the wire will bounce back and forth in sync with the alternating field,

Exactly! The electrical field that is the wave "pulls" and "pushes" the electrons in the antenna. If we make sure the antenna is sized and shaped just right to push and pull the electrons with the most force possible, we get the strongest reception!

moving faster and faster with each oscillation (resonance).

The speed is fixed (it's given by the frequency of the radio wave, which is just the speed of light divided by the wave length), but with resonance, we can make much stronger currents.

I think this calls for an analogy: get a (maybe just in your head) bucket of water. Hold it between your hands.

Slightly push it back and forth. For some amounts of times pushing the bucket back and forth per second, the ripples on the water aren't really big. But for others, the water gets really troubled and starts making extremely high peaks! You've hit the resonant frequency of your water/bucket system.

With electrons, antennas and a tuned receiver, the same happens: a specific wave frequency (that is, a specific wavelength) is able to rock the electrons just at the right rate so that the peaks get really high.

Now, in your bucket experiment, the bucket is fixed, and you find the frequency that resonates well.

Unlike you with your bucket, the frequency is fixed (because you want to receive a specific station!), so you have to vary your bucket: Essentially, tuning your receiver is the same as shrinking or growing your bucket. The bigger the bucket, the lower the resonant frequency becomes.

I also understand how audio can be embedded in an AM signal using amplitude modulation.

Great! In essence, staying in the bucket picture: The audio signal would just be how high the peaks got, which is just a result of how strongly you pushed (maintaining the same frequency, of course, that's where it gets difficult with the bucket).

However, I do not understand the diagram in the diy linked above. Where do the electrons flow?

diagram from DIY article

So, the electrical field pushes and pulls the electrons in the antenna back and forth – that push/pull extends down the antenna wire up into the coil – which basically acts like a reservoir of water. You push in a little current, the next moment you pull it back out, and so on.

Now, the size of that reservoir determines at which frequencies we are in resonance. Which means, the amount of windings that are effective in the coil determines how much current can flow into the coil, get its energy stored in the magnetic field of that coil, and then flow back.

By selecting less or more windings with the alligator clip, you get less or more reservoir – thus, you change the resonant frequency of the antenna/coil system!

Doesn't splitting the circuit in the middle of the coil create two "antennas", each with a different resonant frequency?

Kind of, yes! But the other half (not attached to the antenna) is attached to ground, so nothing is received there. Keen observation!

What is the purpose of the resistor? I have looked up the internal design of piezoelectric earpieces, and it does not appear that current flows all the way through them.

Exactly. Piezoelectrics react to changes in voltage, and other than that, basically isolate. To allow net current flow from the diode to ground, you'll need something that always passes current.

Now, using this resistor means that you won't get all energy that you've received to drive your ear piece, but that's not all that bad.

I hope that gave you a start :) Would love to hear your feedback, and see you back around here :)

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  • $\begingroup$ This is great. Thanks very much for the detailed reply. I thought that the purpose of the coil was only to change the length of the antenna, but if I understand you correctly the purpose of the coil is actually to store the "momentum" of the electrons with inductance, do I have that right? Also, I read about the diode. I see that electrons only flow from the side with the black ring to the other side. So electrons can only enter the antenna? Do they leave through the "dead" half of the antenna? $\endgroup$
    – BitPusher16
    Apr 17, 2017 at 14:27
  • $\begingroup$ A few suggestions: (1) Neither circuit has a capacitor in parallel with the phones. The resistor should be replaced with a small disc ceramic capacitor, IIRC about .0005 to .001 uf as an RF bypass. (2) The headphones/earphones must be high-impedance (in the 1000s of ohms); common low-Z earbuds for a stereo will produce no sound. (3) The antenna in this circuit has capacitance to ground, and so that and the coil form a resonant (tuned) circuit. That means if the antenna is too long or short, it may have too little or too much capacitance to resonate with the coil in the AM band. $\endgroup$
    – Mike Waters ♦
    Apr 20, 2017 at 17:48
  • $\begingroup$ @MikeWaters: The ceramic earbuds I've seen behaved more like capacitors than resistors, and would thus need parallel resistance. Some other earphones use coils and thus behave like a resistor or (to a slight extent) inductor, and would thus benefit from parallel capacitance. That having been said, I can't think when I last saw ceramic earbuds for sale. $\endgroup$
    – supercat
    Aug 20, 2017 at 18:28
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    $\begingroup$ @MikeWaters: My 1970s 75-in-one and 150-in-one kits both came with a ceramic earbud, and I've used those in crystal radios without a bypass cap. What's necessary is to build something that can act as a peak detector, which in turn essentially requires a diode and something that acts like a parallel RC. BTW, there can also be advantages to adding some parallel capacitance on the other side of the diode, but knowing what cap to use would require knowing the inductance of the coil. $\endgroup$
    – supercat
    Aug 20, 2017 at 18:47
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    $\begingroup$ @MikeWaters: I also commented on the question of AM vs FM. The styles of AM receivers I'm familiar with are based on glorified peak detectors, which are sensitive to noise at all parts of the incoming wave even though they can only be sensitive to signal during parts of the wave where signal is actually present. I didn't feel confident enough to answer because I don't know how receiver technology has advanced since my 150-in-one-kit days. $\endgroup$
    – supercat
    Aug 20, 2017 at 19:00

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