# Will an inductor wound using wire of 0.6438 mm (22 AWG) be close enough to 0.508 mm (25 SWG) for a neglectable difference?

Degi-Key has 25SWG as an obsolete item. While the instructions for a project say to use, "25cm length of 25SWG wire. Wrap the wire around a cylindrical object of 6mm diameter and take it out after eight turns." http://www.electronicshub.org/fm-bugger-circuit/

Additionally, is there a general rule (such as 0.2 mm tolerance) for small inductors (less than 20 turns ~ maybe) for most gauges of wire?

• Wow, how old are these instructions? – Ignacio Vazquez-Abrams Jul 18 '16 at 23:08
• But in general, wire thickness usually only affects its ampacity, not its magnetic properties. – Ignacio Vazquez-Abrams Jul 18 '16 at 23:09
• I edited the title, to show that an inductor is being "wound", not just the inductance of the wire. – Marla Jul 18 '16 at 23:45
• Why not use 24AWG wire? – brhans Jul 19 '16 at 1:35
• Adam, I appreciate the vote of confidence. However, if you wait before selecting one answer as "correct", you may get a wider diversity of answers. You can still "unclick" your acceptance of answer and wait until later to decide. Sometimes the engineers here stop supplying answers if one answer has been accepted – Marla Jul 19 '16 at 1:38

The inductor wound with 0.508 wire would have a length of 4.064 (perfectly wound). The inductor wound with 0.643 wire would have a length of 5.15 (8 turn coil)

Difference in length is 1.09 mm.

This is about a 27% difference in length. Inductance therefore is going to be reduced by about 27%. Not a trvial amount. Increasing inductor length using same amount of turns (wider copper) results in reduced inductance (Inductance inversely proportional to Length).

Inductance would equal 0.789 of the original designed inductance.

One could try using 9 turns of the larger diameter wire, attempting to get inductance back to normal. Inductance is proportional to the number of turns squared.

$(9^2/8^2)*0.789$ = 0.99 of original designed inductance.

Advanced users : I have kept it simple here. Of course for 8 turns one uses 9 times the copper diameters for coil length. And obviously the "perfect" winding with no space between turns.

It doesn't often turn out this simple to change turns to accommodate change in copper width (coil length). Just fortunate that one additional turn (in this case) can restore original inductance.

Accurately calculating the inductance of a coil is not as easy as you might think, particularly if it is short. Rather than trying to do it myself, I used the online calculator at Bob’s Electron Bunker. For 8 turns of 0.508mm wire on a 6mm former at 50MHz it returned 0.3349uH, and for 0.6438mm wire the result was 0.2954uH (11.8% less).

Now the critical question:- will this difference be 'neglectable'? The answer depends on what tolerance is acceptable in the particular circuit. In this FM 'bugger' project the coil is part of a tuned circuit which determines the operating frequency. The other part is a 50pF variable capacitor which is used to tune the transmitter to a frequency somewhere in the 88-108MHz FM broadcast band.

The resonant frequency of a tuned circuit is inversely proportional to the square root of its inductance. This halves the effect of inductance variation. 11.8% less inductance only increases the frequency by about 6.5%, so an 88-108MHz tuning range would become 94-115MHz. If you are not concerned about covering the entire FM broadcast band then this difference may indeed be 'neglectable'.

Also the variable capacitor may be able to tune a wider range than necessary, so the slightly lower inductance could possibly be 'tuned out' to still cover the full FM broadcast range. If so then the larger wire size would make no practical difference.