# I NEED HELP MAKING A 0.35mH INDUCTOR? Answered

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RE-POSTED by  Request :-O)

Here are some results ;
• L inductance = 350 uh
• Diameter of coil = 2.05"
• Number of turns = 80
• Gauge of magnet wire = #29
• length of coil = 1"
See the graphic table..
BTW the wire gauge can be a larger number = smaller diameter.
A gauge of #29 single layer coils of wire will be an inch long.

A

I am trying to make a FM receiver i needed it for that . Here is the schematic.

OK, now I can help, this is a bare wire air core and will probably be just
a few loops hand turned and tuned by pulling them apart of squeezing
the loops together.

BTW this is a high frequency tuning inductor.

I will try to get you a wire size, a loop diameter and the number of turns
to make this coil inductor soon.

Can you provide the article where you found this schematic ?
It may save me a bunch of tables work by referring to an available
RF coil or mention how to make that..

A

Here are some results ;
• L inductance = 350 uh
• Diameter of coil = 2.05"
• Number of turns = 80
• Gauge of magnet wire = #29
• length of coil = 1"
See the graphic table..
BTW the wire gauge can be a larger number = smaller diameter.
A gauge of #29 single layer coils of wire will be an inch long.

A

Hey ! iceng "wire gauge can be a larger number"
i am pretty confused from this word . what i know the small the diameter of the wire the larger is the gauge (in numbers) am i right?
and can you please tell me what will be the diameter of the coil if i use a #18 gauge wire? Thank's in advance for your help.

Yes you are right !  Look here.

18 gauge wire is 1.024 mm thick.
80 turns will be 80×1.024×1.1slop = 90 mm = 3.54" minimum coil length
but the monograph has determined a 1" length

So lets try 70 turns × 1.024 wire dia × 10% = 78 mm = 3.1" coil length
K is now 3
coil length = coil dia / K = 2.05/3 = 0.68"
This is a worse diverging solution :-(

Next try a line through 350 uH to a 3" dia coil intersecting the Axis at 4.5
Using 80 turns through the 4.5 axis results in a K of 1.1
Coil length = coil dia / K  = 3 / 1.1 = 2.72"
this solution is converging to 3.54" which is the actual physical windings.

Now try a line through 350 uH to a 4" dia coil intersecting the Axis at 4.9
Using 80 turns through the 4.9 axis results in a K of 0.83
Coil length = coil dia / K  = 4 / 0.83 = 4.8"
this solution could work by spacing the actual physical windings but is too large.

Zero in with  a line through 350 uH to a 3.5" dia coil intersecting the Axis at 4.8
Using 80 turns through the 4.8 axis results in a K of 0.9
Coil length = coil dia / K  = 3.5 / 0.9 = 3.8"
This solution will work by loose spacing the actual physical #18 windings.

COIL  ( Diameter = 3.5"  Length = 3.8"  Gauge = 18  Inductance = 350 uh )

This is a single layer coil solution.
There are monographs for multiple layer coils that require machine
winding precision  and are smaller.

A

Good to hear.

You need to hire  EE teachers  to start exposing you to DC and AC circuit theory
and the math is simple.
Power semiconductors SCR, TRIAC, MOSFET,  BIPOLAR,
VACUUM TUBES are making a comeback, Li battery theory,
DC motors, Stepper, Induction, Synchronous, Traction and Homopolar motors.

Your probably do are not challenged by your environment except by females :-)

Diameter of wire here doesn't really matter.
to use a toroid, you have to know its "Al" value, which isn't something you'd generally get at the local shop !

The simplest thing to do would be to wire, say 10 turns on it, and make it oscillate with a signal generator against a capacitor, the inductance is then Al x n^2, where n is the number of turns, and you can deduce Al.

Steve