I'm intrigued by the second circuit in Spehro's answer and/or it's precursor (FM Crystal Radio) https://electronics.stackexchange.com/a/217086/33670

FM crystal radio

But I don't happen to have any 18 Ga solid wire, while I have a vast excess of 14 Ga and 12 Ga solid wire as construction byproducts.

The two versions of this circuit at the linked page call for 4 or 5 turns of 18 Ga (1.02 mm dia.) wire with a 12mm inside diameter.

That calculates out to 343 nH or 537 nH when I find a coil calculator that takes wire diameter as an input. Changing to 12 Ga (2.05 mm dia.) lowers it to 285 nH to 445 nH.

Using 12 Ga, if I increase the diameter to 16.05 mm (i.e. a 14mm ID) this seems to restore (nearly) the original inductance values - but then I realize that the coil calculator I'm using has no entry for overall coil length, and thus is probably assuming a close-wound coil, which these are distinctly not.

I find other calculators that let me play with length of the coil (which I note is not specified in the source, unless I'm missing it, though it can be ballparked from the stated inside diameter and photographs) but they don't have wire diameter as a variable. Nor do many of the formulas I find given on various sites. I begin to plumb the depths of what I don't know about inductance, and what effect wire diameter has on it.

Going with one of the "no wire diameter" coil formulae, 86 nH to 135 nH at D = 13mm (the original, approximately) assuming the coil length is 25 mm. If it's 15mm, 128-200 nH. Formula I'm using is $$ I (uH) = {0.394*(Coil Radius In cm)^2*(Number of Turns)^2 \over (9*Coil Radius In cm)+(10*Coil Length In cm)} $$

Seems for 22 cm of straight wire (about enough to make the longer coil), "self-inductance" for 18 Ga is 264 nH and 12 Ga is 234 nH - not a lot of difference - 30 nH or 13%-ish. But either (rather than the difference between them) is also somewhat more than the "coil calculation" gives for the coil, in this instance. If "wire-diameter self inductance" for a given length of wire simply adds to coil inductance, it seems to be somewhere between "make small adjustment" or "lost in the noise" - if I get this correctly, which I very well might not. For self-inductance I'm simply using several web calculators that seem to agree, I have not transcribed that one into a local spreadsheet yet.

So, is that the way it works when changing wire sizes for a given widely-spaced coil - (simple addition of self-inductance for a given length of given size wire to the coil inductance for a given coil inductor with no wire diameter specified) or not?

  • \$\begingroup\$ Spent several frustrating hours messing with this circuit - located 5 miles from a 10,000 watt FM transmitter (normal receiver can pick up that station with no external antenna connected, and it doesn't have an internal FM antenna) - not a peep, not a whisper. \$\endgroup\$ – Ecnerwal Feb 19 '16 at 4:37

The AARL handbook indicates that the formulas are not terribly accurate for VHF small coils. They happen to have a graph showing measured inductance vs turns for AWG 12 wire. The handbook is a pretty useful guide to many things radio-related- even an old one such as the one I have (46th edition).

enter image description here

When there is some gap between the turns you can adjust the inductance up and down by compressing or expanding the length of the coil.


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.