.

Remember Me

### Create an account

Fields marked with an asterisk (*) are required.
Maximum length of name and username fields is 16 symbols
Name *
Email *
Verify email *
Rating 5.00 (1 Vote)

# Online calculators

## Multilayer air core inductor on a rectangular former:

The calculator allows calculating the number of windings of the multilayer air core coil on a rectangular former at the design stage. You can select measurement units - mm or inches, and also AWG or SWG wire number - the wire diameter will be determined automatically. This calculator based on Rosa's straight wire self-inductance formula (formula 9 on p.305) and formula for mutual-inductance between two unequal straight wires from American National Bureau of Standards circular C74 (formula 182 p.273 with error correction). The coil is represented as many parts of straight wires. Calculating the self-inductance of each part, and mutual inductance of each possible pair of straight wire parts and summing it all, we get a self-inductance of the multilayer coil. This is the "virtual winding numerical method" that is the same as in the calculation of multilayer coil with round former. The accuracy of the calculation is to 5%. At a very high value of inductance, it's possible the small hang. See more about source formulas. Many thanks to Robert Weaver for his great help.

##### Calculate number of turns

ENTER THE INPUT DATA:
 Select units: mm/cmmil/inch AWG → 000000000012345678910111213141516171819202122232425262728293031323334353637383940 SWG → 7/06/05/04/03/02/001234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950

 L = mHμHnH – Required inductance a – Width of former cross-section b – Height of former cross-section l – Length of winding (former) d – Diameter of wire without insulation k – Diameter of wire with insulation Automatic

RESULT:
 N = – Number of turns n = – Number of Layers c – Winding thickness Lw – Required length of wire Ω =  Ohm – DC Resistance of coil

Another useful calculator:

0 #1 naman 2016-02-28 19:00
hi i calculated the number of turns = 230 for a 0.5mH inductor having a=13 b=15.2 and l=54 , 18 SWG wire. when i wound the coil using my cnc , the actual inductance came out to be .335mH .
can you help me as to what may have gone wrong here
0 #2 Coil32 admin 2016-02-28 20:55
I can surmise that the winding has been imperfect (not dense) or wire number is not 18 SWG.

Сomments from anonymous guests are enabled with moderation.