I'm trying to design a 16uH 50Arms 4kHz air core inductor.

At first it sounded nice and simple. I found this formula L= (d^2 * n^2)/(18d+40l)


L is inductance in micro Henrys, d is coil diameter in inches, l is coil length in inches, and n is number of turns.

Everything seemed great. Then i came across that this formula is only used for radio frequency 1 - 30MHz or so.

How would i design an air core inductor at 4kHz?

  • \$\begingroup\$ Why do you want a 50A RMS air core inductor operating at 4kHz, with 0.8 milliWebers? (I don't think you will have to worry about parasitic capacitance.. but it still helps to know what you are trying to achieve.) \$\endgroup\$
    – jonk
    Commented Aug 4, 2016 at 18:01
  • \$\begingroup\$ @jonk trying to simulate the effects of a long run of wiring for equipment testing. That is the equipment will use a long run but to test the equipment a device will be built to simulate the long run. The spec is called out to us, by someone else. \$\endgroup\$
    – vini_i
    Commented Aug 4, 2016 at 18:17
  • \$\begingroup\$ @KingDuken Not sure that i understand what you mean by "make an oscillator"? \$\endgroup\$
    – vini_i
    Commented Aug 4, 2016 at 18:21
  • \$\begingroup\$ Have you read this: arrl.org/files/file/Technology/tis/info/pdf/9708033.pdf \$\endgroup\$
    – jonk
    Commented Aug 4, 2016 at 18:22
  • \$\begingroup\$ @jonk While i have not read that specific article the formula that is called out in the article is the same formula i have listed here. Is that formula still valid at 4kHz? \$\endgroup\$
    – vini_i
    Commented Aug 4, 2016 at 18:27

1 Answer 1


The equation is probably OK lower frequencies

Suppose you have a 2 inch diameter, 6 inch long coil, with 33 loops. The first equation would predict an inductance of

\$L= \frac{d^2 \cdot n^2}{(18d+40l)}\$

\$L= \frac{2^2 \cdot 33^2}{18 \cdot 2 + 40 \cdot 6} \approx15.8 \mu H \$

Using the second formula, 2 inch diameter is about 0.0508 m (radius 0.0254 m) and 6 inch length is about 0.1524 m

\$L = \frac{\mu_0 \cdot N^2 \cdot A}{l} = \frac{4\pi 10^{-7} \cdot 33^2 \cdot \pi \cdot 0.0254^2}{0.1524} \approx 18.2 \mu H\$

How tight is your design tolerance? Do you need it to be really close to 16 uH? If so, you can tune it by stretching (lower) or compressing (higher) to the extent the wire is flexible. You can also increase effective inductance by adding a small capacitor in parallel.

  • 1
    \$\begingroup\$ I might go with #6 house wire, insulated, at about 3 turns per inch, on a 4" PVC pipe (OD=4.5".) That would be about 4.75" coil length and an even number of 14 turns. See how that works and adjust the length. \$\endgroup\$
    – jonk
    Commented Aug 5, 2016 at 1:02
  • \$\begingroup\$ @jonk That's actually pretty close to what i was thinking. \$\endgroup\$
    – vini_i
    Commented Aug 5, 2016 at 15:47
  • \$\begingroup\$ Well, give that a shot then. Won't cost much, you can get everything from Lowes or Home Depot, and I think #6 will give you some breathing space regarding the current. Make it solid core. \$\endgroup\$
    – jonk
    Commented Aug 5, 2016 at 16:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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