number of turns in a Coil

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.

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

Thank you

 

You need to know what its Inductance will be.

http://www.circuits.dk/calculator_multi_layer_aircore.htm

 

Let's say you don't care about inductance.

Mean diameter is 8.25mm. Mean length/turn is PI*8.25mm = 25.91mm.
AWG40 copper wire resistance per meter is 3.441R / m.

So for 55 Ohm, that's 617 turns.

Area: 4mm*2.5mm= 10mm2
Area of AWG40: 0.00501mm2
Therefore: 1996 turns
From experience about 75% of this value fit into the calculated space,i.e. abou 1500 turns.

In short: you can have 55Ohm with about 617 turns. Or you can have the dimensions you indicated with about 1500 turns.

 

AWG-40 has a resistance of 3.44 Ohms per meter.
Source: http://www.daycounter.com/Calculators/AWG.phtml
So, a length of AWG-40 that measured 55 Ohms would be 55/3.44 = 15.988 meters long, or about 52' 5-7/16" long.

You should get some kind of plastic tubing or round wood dowel to wind the coil on, that will make it easier. You'll need to make a couple of end plates (stiff cardboard would work fine) and space them exactly 4mm apart on the dowel or tube and glue them in place.

The difficult part will be winding on that much wire without breaking or stretching it.

 

I came up with somewhat different numbers.

599 turns, ~40 turns/layer, 15 layers, giving an OD of 10mm, wire length of 52.48 feet, inductance of 3.261mH, resistance of 55.05 Ohms.

 

Hand winding that narrow gauge wire will not give you a tight coil, except with great difficulty. It'll be a random wound coil. So you'll use more wire for a given number of turns, or in this case get less turns for a given length of wire, if you're trying to hit that DC resistance.

Somewhere I've seen the factor for estimating a hand winding but I can't recall where. Maybe in Wheeler's original work.

 

Wayneh,
If it's ~55-1/2 feet or ~16 meters of AWG-40, it will measure ~55 Ohms resistance.
Now, if they wind up with fewer turns, the impedance at a given frequency will be different; the L will be lower, and the parasitic C will also be lower. However, unless the wire gets stretched in the process of winding, the resistance will still be the same.

 

The last two weeks I made several small coils , AWG 22 to 30. For a given coil area (length * layer thickness) I get around 75% of the turns I calculate by dividing that area by the wire cross-sectional area.
So for 10sqmm and 0.00501sqmm wire cross-sectional area that's about 1500 turns that would fit in there.
That would be much more than 55Ohms.

 

I agree. That's why I said he'll "...get less turns for a given length of wire, if you're trying to hit that DC resistance."

 

That's pretty good; the theoretical maximum is just over 90%, with no deformation. Do you include the insulation thickness when you calculate the cross section?

Your wire gauges are much heavier wire than the OP's and are easier to hand wind cleanly. I'm pretty sure the hand winding percentage is in the 60s.

 

Of course! Not. I just needed an approximation.

 

The cross section of 40AWG magnet wire can be <60% copper, if the insulation is "heavy build". I'm sure this is not a concern to most people most of the time, but I was once trying to optimize the windings for a coil, and such details matter in that case.