How many turns do I need for a compact air-core coil with specific dimensions?

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To calculate the number of turns for a compact air-core coil with specified dimensions, the turns per layer can be determined by dividing the height (4mm) by the wire thickness (0.08mm), resulting in 50 turns per layer. The number of layers is calculated by the formula (OD - ID) / (2 * wire thickness), yielding approximately 31 layers. Therefore, the total turns amount to 1550, assuming rectangular packing; for triangular packing, this should be adjusted by multiplying by 0.6. The discussion highlights the challenges of working with 40 AWG wire, which is fragile and difficult to handle, suggesting that achieving the desired turns may require specialized techniques or training. Ultimately, a practical approach involves accepting a lower fill factor to avoid wire breakage while still achieving a functional coil.
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Hello there

Its a air core coil I am not too sure how to calculate the
number of turns I have to wind.

Specs are below

Height : 4mm
Outside D: 9.5mm
Inside D :7mm
Coil wire thickness : 40 AWG / 0.08mm
Coil resistance : 55 Ohms
Coil weight : 0.7g

Important figures highlighted. I have made a detachable bobbin.U sing self bonding wire.
coil is air -core.

How can I calculate the number of turns that i would require to make this coil

Thank you
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Turns = turns per layer * number of layers

turns per layer = height / wire thickness = 4 mm/0.08 mm = 50

number of layers = (OD-ID)/(2*wire thickness) = (9.5-7)/(2*0.08 ) = 31.25 = 31 layers

therefore, Turns = 31 * 50 = 1550
 
oh btw that's assuming rectangular packing, if its triangular packing then divide the result by 0.78
 
Having been in the transformer business, I'm very pessimistic.
Phelphs dodge gives for #40 Bondeze wire a nominal diameter of 0.0037 inches.
This works out to a diameter of 0.0939 mm.

When hand winding you usually get maybe 60% fill.

#40 wire is THIN, so is extremely difficult to work with.

Good Luck
 
Carl is right, AWG 40 breaks incredibly easily. 60% is also a more practical number, so multiply the result I originally gave you by 0.6 (or if using triangular packing, which I suspect is what it would come down to, divide by 0.78 then multiply by 0.6)
 
You need 15.4m of wire to achieve the resistance. 0.08mm needs care.

With a setup that rotates the coil former you can still make a perfectly compact coil of 0.08mm by hand but only with training and good hands.

Then each layer packs nearly 50 turns, I estimate 47 (again under good conditions...) so the first layer packs 1.0m and the latest 1.4m with 1.2m as a mean, so you need 13 layers.

I could never pack properly successive layers in the same direction (wire returns makes a mess), nor did I see a coil made this way, so each layer takes >0.08mm and the 13 perfect layers fit well in OD=9.5 ID=4

BUT

you'd go crazy before finishing the 13 layers. With said training and good hands you achieve 4 layers or 50 turns with this tiny wire, then you throw all through the window.

Next solution: pack the wire disordered, taking advantage of the available 50% filling factor. For 650 turns of tiny diameter you DO need to rotate the coil former or you'll break the wire. At 50% filling you can't make much of a mess, the winding must already look good.
 
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