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I am making an inductor

  • Air core
  • Wire: 18 AWG guage
  • Turns: (I am not sure ) but could be about 6000 turns
  • The weight of the wire : 400 gram
  • The coil wrapped by machine
  • Coil is 5 cm
  • Diameter of core is 1.5 cm

I used an inductance meter to measure the inductance and I got a weird reading of this large inductor = 3.7mH.

The DC resistance also is wire = 3.5 ohm.

Could these measurements be correct although it is a large inductor?

If so, how can I increase the inductance to 150mH?

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Edit

Thanks guys for useful answers and comments

I just found i was wrong,

  • it is 5 cm lenght coil which is 50mm

  • the wire gauge is 18 AWG = 1mm

  • Flange of the coil is 1.5 cm = 15 mm ( maximum number of layers)

  • So one layer should be 50 turns (if roughly turned with no spacing)

So according to these numbers the turns should be 50 turn per layer * number of layer 15

50 *15 = 750 turns

So this inductor is about 650 to 750 turns not 6000 turns as i thought 😔

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  • \$\begingroup\$ To a first approximation, add more turns. Inductance is proportional to the square of the number of turns. So you need a total of 6000 * sqrt(150/3.7) = 38,000 total turns. There are also formulas for air-core inductors. electricaltechnology.org/2014/03/… \$\endgroup\$
    – mkeith
    Aug 14 at 23:59
  • \$\begingroup\$ The formulas are going to assume that each turn is tightly packed with the adjacent turns. If you want maximum inductance, you can't just randomly wind onto a spool. coil32.net/multi-layer-coil.html 66pacific.com/calculators/coil-inductance-calculator.aspx \$\endgroup\$
    – Mattman944
    Aug 15 at 0:05
  • \$\begingroup\$ You use the wire ineffectively if you aim to certain inductance (=150 mH) Try bigger coil diameter, say 10 cm and as tightly packed turns as possible. As few as 310 turns (about 100 meters of total wire length) can be enough - should be measured and fixed when done, because it's based on approximate formulas. An iron or ferrite core would reduce the needed turns radically, but a solid iron bar wouldn't work, it should be an eddy current free construction if you want to use it with AC or non-constant DC. \$\endgroup\$
    – user287001
    Aug 15 at 0:37
  • \$\begingroup\$ Tito, would you mind saying what is it for? There may different solutions. \$\endgroup\$ Aug 15 at 5:56
  • \$\begingroup\$ @aconcernedcitizen i am make solenoid lock (high pulling force) using 220v AC, so i need good resistance of the coil, the impedance of this now is 3.6 which means it output 61A!!!. I need it works on 3-5A \$\endgroup\$
    – Tito
    Aug 15 at 6:16

1 Answer 1

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In theory the inductance is proportional to the square of the number of turns, so 2x the turns would be 4x the inductance; however that only applies if all turns couple to all other turns. That doesn't occur in your structure for 3 reasons:

  1. all turns are not the same diameter -- so all the flux from a turn doesn't couple to a turn that is a different diameter
  2. When the spacing is similar to the diameter, the flux from each turn 'loops back' before reaching the other turn and doesn't couple.
  3. An iron core would improve each of the above issues.

However an iron core would limit the frequency response of the inductor. Basically, at higher frequencies (and sharp dV/dt edges), the apparent inductance would be smaller. You can partially mitigate this by using a ferrite core, but lacking that, a bunch of small nails (easiest way to get iron wire), with the heads cut off and bundled together (insulating if possible) would be better.

The aircore is reasonably independent of frequency; if you add an iron core, you might find that the "inductance" depends on the frequency at which it is being measured.

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  • \$\begingroup\$ This will be AC solenoid so i can not add iron core inside _____ Reason number 1, you mean each layer is not the same diameter of the next layer? If so how to beat it \$\endgroup\$
    – Tito
    Aug 15 at 7:02
  • \$\begingroup\$ @Tito measure the inductance on it is installed over the solenoid, as the solenoid increases the impedance \$\endgroup\$
    – Ferrybig
    Aug 15 at 12:08
  • \$\begingroup\$ @Ferrybig sorry did not understand you \$\endgroup\$
    – Tito
    Aug 15 at 12:36
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    \$\begingroup\$ Then it's not an air cored inductor -- the solenoid core will be iron and will increase the inductance. However the core will move, so the inductance will change. Be sure you measure the inductance in the 'open' position; but there will still be some iron in the core in that position. \$\endgroup\$
    – jp314
    Aug 15 at 15:20
  • \$\begingroup\$ "each layer is not the same diameter of the next layer" -- yes; you can't beat this without other compromises. Adding an iron core will 'beat' that because most of the flux will concentrate in the iron core --> turn diameter doesn't really matter. For a solenoid with 'pull' you need an inductance that is large and that changes with position. This really means that you need an iron core. \$\endgroup\$
    – jp314
    Aug 16 at 0:36

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