Antenna ferrite loopsticks verses air core?

Why do you care what I do?

Well, maybe not in the case of a crystal radio. The "noise" portion of the SNR in a crystal radio also practically includes ambient audio noise and the noise between your ears (for those of you with tinnitus). Cranking up the RF signal and the RF noise in equal proportions can still help.

Please follow your own advice. You and Joerg are the only posters here who are saying the high Q of the coil will impact the bandwidth. We have patiently explained to you that the Q of the receiver will *NOT* be as high as the Q of the antenna and will not adversely impact the bandwidth of the received signal.

You are doing nothing but making a fool of yourself.

Possible answer:

Because, if you actually need Q=500 for selectivity, an LC tank with an unloaded Q of 500 leaves you no power to drive an earphone? And any improvement in unloaded tank 'Q' translates directly into more power available for audio output?

Cheers, James Arthur (who has *no* dog in this fight!)

Or looking at it another way, you want the loaded Q to be high enough to pass the signal but not the out-of-band noise. Since you're loading the tank with a head set, you need the unloaded Q much higher.

Only cats here.

James, thanks.

That's the sanest answer yet and goes some way to explaining what the xtal guys are after.

However, the single-diode detector is a very assymetric load. That's going to affect things a bit, don't you think? By which I mean that the coil will "see" the load during rising amplitude, but not during falling amplitude. The coupling to a crystal earpiece (generally capacitive, but highly reactive load) is gonna make quite things weird, by pulling the resonant frequency around a lot, dependent on the audio signal...

I should pull out my old xtal earpieces and measure the capacitance. I suspect they'd detune a high-Q front end quite a bit, and that really calls into question the use of Q to limit bandwidth.

Clifford Heath.

I'm unclear what you are describing. When you say "detune" you are not describing an effect on the Q, but rather an effect on the tuned frequency. If the capacitance of the earpiece impacts the tuning of the receiver, the adjustment of the tuning capacitor would make up for that when operated.

Or do you mean detune as a way of describing the reduced Q from the loading of the circuit by the earpiece? How exactly would the capacitance of a crystal earpiece affect the loading?

When you measure the capacitance of the earpiece, also measure the resistance at audio frequencies. The capacitance is only significant in that context.

I believe a serious crystal set operator will use high end headphones however. I believe various military headsets are valued for this use due to their high sensitivity. So the functioning of an inexpensive peizo earpiece may not be so relevant, but interesting anyway. When I read about the detail explored to characterize the detecting diode I was very impressed. So I can see every part of such a simple radio being optimized to the nth degree.

Both. When the diode conducts (as it must to transmit power) it connects the filter capacitance and the capacitance of the earpiece to the tuned circuit. This will happen whenever the diode is forward biassed, i.e. when the audio signal is rising.

The extra capacitance on rising audio signals will pull the tuned circuit to a much lower frequency. As soon as the diode turns off (when the audio signal starts to fall), the tuned circuit will jump to a higher frequency, perhaps receiving a different station, and hence affecting the audio signal once again.

I would expect a double-hump response in the overall receiver, and the gap between the humps will depend on the filter and earpiece capacitance.

Fair enough. I think I've given some answers above.

Perhaps. The earpiece still has to present some load, or it will receive no energy. For optimum power transfer, you want its audio impedance to roughly approximate the equivalent parallel resistance across the tank - which is why Q gets halved. Or not... perhaps that only happens on audio half-cycles.

Anyhow, those are the reasons for my skepticism about the whole project. If the detector and earpiece presented a purely resistive load, or even a fixed capacitive+R load, regardless of audio polarity, then you wouldn't get a double hump. But I think that's unlikely.

Clifford Heath.

Maybe you could build one and test that?

"Presenting a load" is a description of the power transfer. Of course that has to happen or you don't hear a signal. But how much of that power gets converted to sound depends on the earpiece used.

Theory is nice, but it is easy to miss important aspects. That's why people build crystal radios, to learn what is important and what is not.

It would need a fairly ornate test setup compared to what I have available; specifically two different signals, one fixed, one swept, to see at what frequencies the swept one gets in. Possible, but I can't easily do it at present.

It seems easier to just ask: with your high-Q receiver, do you ever hear two stations at the same time that are on different frequencies? What do the community report?

The crystal set I built as a kid seemed often to get multiple stations, but it probably didn't have enough pixie dust.

Clifford Heath

If your going to build a high quality crystal set, then your most likely going to use high end earphones. Some of the most sensitive are the, "WWII RCA Big Cans MI-2045E US Navy Deck Talkers Headset' They are magnetic headphones and as best I can find are fairly low impedance at 300 ohms. That means you need a transformer to match you high Q LC to the low impedance headphones. Ben Tongue did work on that, here;

Page down past Table 7 and see, "A UTC O-15 'Ouncer' transformer"... In that section he tests a two transformer arrangement that transforms a 1.3Meg impedance down to 8 ohms, 300 ohms, 1.2k ohms and 10k ohms. The transformer loss at 8 ohms is 1.2 db at 1kHz, and I suspect a little less at 300 ohms. And then there is the sensitivity of different headphones,

Mouser single old style xtal ear plug - 103

RCA big cans - 128 (before alignment) Mikek

This is a good example that even a "simple" Crystal radio is not so simple if you dig deep enough into the details. M

I'm not a crystal radio guy (RFI on my powerline is so bad that I get *no* AM stations)(*), but these pages seem to offer a decent overview:

This one-paragraph introduction lays out the big picture:

A short discussion of detector diodes:

Detailed empirical measurements of diode detectors:

At low signal diodes are square-law detectors, quadrupling output for every doubling in signal voltage.

These guys' coupling transformers ought to transform (lower) the effective capacitance of any earphone. And any residual capacitance is simply compensated by re-tuning, anyhow.

The old crystal earphones were amazing. The crystal radio guys seem to prefer impedance matching with transformers to magnetic earphones.

(*) Hey, I just scanned the AM band and the years-standing RFI is gone!! Who knew! Cheers, James Arthur

That's good to know, but not relevant to the dual impedance caused by the detector's polarization.

I mentioned the peaky resonances in crystal earpieces, but they're not the main problem.

Interesting articles - I just read all three - but none of them mention the effect I'm suggesting, and none of the experimental setups would see it. Ben Tongue gets close in his paragraph "Tuned circuit loss and bandwidth considerations", but doesn't acknowledge the assymetric load characteristic.

I bet I could construct an LTSpice simulation that would show it clearly.

Yes, but that capacitance is only coupled to the antenna during half-cycles of the audio, while the diode is conducting. So you have two frequencies being tuned. While the diode is off, the alternate frequency is the one which will turn it back on again and slew the tuned frequency to the one you hear.

That would explain a lot of what I recall hearing in my crystal earpiece, anyhow; effectively audio intermodulation from adjacent stations.

I guess that's why the Benny is useful; not for the reason he describes, but to make the load appear more resistive.

Interesting stuff.

Perhaps you replaced a bit of equipment that had a crappy wall wart? Or a neighbour did!

Clifford Heath.

Agreed, you could simulate it, and it isn't addressed in the articles I read, but does it do much other than add a bit of harmonic distortion? I don't see how.

That's always a danger in a crowded band, yes.

Not me, it was far too powerful, and stronger outside than inside. I suspect an aging line and pole replacement program in progress likely removed a HV leaker, somewhere in the neighborhood.

RFI was so bad I hadn't bothered checking the AM band in perhaps a year or two--it had been like listening to a Tesla coil or an EDM machine.

Cheers, James Arthur

A place I worked at had a hash in the AM band, I walked around with my am radio and traced it to a power pole about 35ft from our building. We called the power company and told them they had an arcing power pole at the corner of... They responded immediately and by the end of the day our noise was gone. Looking back saying the pole was arcing was maybe just a little over kill, but it sure got action. Mikek

I walked around outside a bit with an AM radio when we were having the RFI, but it was everywhere--I couldn't get a bearing on it, other than it was on the powerlines.

I probably should've done that. We have some hams in the area; I'm surprised none of them complained. Until yesterday I'd assumed it was one of my neighbors arc-welding 24/7, or trying to reanimate Frankenstein perhaps.

Nice work, Sherlock!

Cheers, James Arthur