The inductance of a coil depends on its geometrical characteristics, the number of turns and the method of winding the coil. The larger the diameter, length, and the larger number of winding turns, the greater its inductance.
If the coil is tightly wound, turn to turn, then it will have more inductance than a not tightly wound coil, with gaps between the turns. Sometimes you need to wind a coil with a given geometry, and you don't have a wire with required diameter, then if use a thicker wire you should increase slightly number of turns, and if use a thinner wire it takes to reduce the number of turns of the coil to get the required inductance.
All of the above considerations are related to winding coils without ferrite cores.
Inductance of single-layer coils on cylindrical winding forms can be calculated by the formula:
L - inductance of the coil, μH;
D - diameter of the coil (diameter of the former), mm;
l - length of the coil, mm;
n - number of turns of windings.
There is could be two tasks in the calculation:
A. The geometry of the coil is given, find the inductance;
B. The inductance of the coil is given, calculate the number of turns and the diameter of the wire.
In the case "A" all data are given, it is easy to find the inductance.
Example 1. Let's calculate the inductance of the coil shown in the figure above. Put the values in the formula 1:
L=(18/10)2*202/(4.5*18+10*20) = 4.6 μH
In the second case the coil diameter and the length of the wound are known. The length of the wound depends on the number of turns and the wire diameter. Therefore, it is recommended to calculate in this order. Based on geometric considerations, determine the size of the coil, the diameter and the length of the wound, and then counting the number of turns by the formula:
After you have found the number of turns, determine the diameter of the wire with insulation according to the formula:
d - diameter of the wire, mm;
l - winding length, mm;
n - number of turns.
Example 2. We need to make a coil with a diameter of 10 mm and with a length the winding of 20 mm, the coil should have an inductance of 0,8 μH. The winding has one layer, turn to turn.
Put the values in the formula 2, we get:
n = 10*(5*0.8*(0.9*10+2*20))1/2/10 = 14
The diameter of the wire: d = 20/14 = 1.43 mm
To wind the coil with a wire of smaller diameter, it is necessary to place obtained by calculation 14 turns across the entire length of the coil (20 mm) with equal intervals between the turns (the step of winding). The inductance of the coil will be 1-2% less than the nominal value, it should be considered in the manufacture of these coils. To wind the coil with a thicker wire than 1.43 mm, the new calculation should be done with the increased diameter or length of the coil winding. You may also need to increase both the diameter and the length at the same time, until get the desired dimensions of the coil for a given inductance.
It should be noted that the above formulas is intended to calculate the coils with the length of winding l equal to or more than half of the diameter. If the length of winding is less than half the diameter of the winding D/2, the more accurate results can be obtained by using the formulas below:
L = (D/10)2*n2/((4D+11l)) (4)
n = (10L*(4D+11l))1/2/D (5)
Reference: "300 practical advices"