4
\$\begingroup\$

I'm trying to wrap my own inductors (air coil) for an RF circuit that I am making and I am having a little trouble solving for the number of turns I need to reach the desired inductance.

\$L=\dfrac{D^2N^2}{18D+40l} \$

  • L = inductance in uH
  • D = coil diameter in inches
  • l = length of the coil
  • N = number of turns

I'm wondering if there is a way to negate l from the equation using the gauge of the wire. I'm using 32AWG magnet wire and have found that 113 turns is equivalent to 1 inch of l. Can I just replace l with N/113?

My goal is to be able to measure the diameter of the non ferrite core I am going to wrap around and plug in the desired inductance. It shouldn't matter how long the coil is in the end unless I reach the end of the core and wrap another layer on top.

If possible, is there an online calculator that will do this for me?

Improve this question
\$\endgroup\$
1
  • \$\begingroup\$ You are assuming what is called a close wound coil. That is one possibility, but not the only one - spacing the turns can be one of the easiest ways to adjust the value of the result. \$\endgroup\$ – Chris Stratton Mar 22 '13 at 4:46
2
\$\begingroup\$

You have experimentally derived the equivalence that 113 turns yields one inch of coil length, yielding the conversion factor (1 inch / 113 turns). Obviously this tacitly depends on the coil diameter, but is a sound figure for the diameter under which you established the measure.

If you have 113 turns that is one inch. If you have 226 turns that is two inches. and so forth. So Length definitely equals number of turns / 113. Looks to me like you can safely substitute N/113 for l. Just keep in mind this doesn't generalize because of the implicit dependence on diameter built into the 113.

You should probably just ask the second part of your question as a new question. It's way more interesting than the part I'm answering here :-).

Improve this answer
\$\endgroup\$
2
  • \$\begingroup\$ cool! to make this useful for more people (people using non-32AWG wire) how about the following formula: L= d^2*n^2/(18d+40(N/(T))) L = inductance in uH D = coil diameter in inches l = length of the coil N = number of turns T = 1/(diameter of the wire) \$\endgroup\$ – OhmArchitect Mar 22 '13 at 2:59
  • \$\begingroup\$ @DannyKmack you know that the diameter in these equations has nothing to do with the diameter of the wire right? It's the diameter of the inductor cyclinder... \$\endgroup\$ – vicatcu Mar 22 '13 at 3:07

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