i want to make same circuit of fm receiver.wanted to knw how many turns should be made of 29 guage wire.i was searching for how much guage.thx for d answr given by u in earlier posts.

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..

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.

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.

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.

active| newest | oldestHere are some results ;

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

how i build 1 milihenery inductor ..

i want to make same circuit of fm receiver.wanted to knw how many turns should be made of 29 guage wire.i was searching for how much guage.thx for d answr given by u in earlier posts.

n y 80 turns

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

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 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.

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

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 :-)

http://www.circuitstoday.com/single-transistor-radio

Thank u.

Otherwise, you need to find a small "ferrite toroid" and wind your own.

Are you sure that's an FM receiver ? It looks more AM to me.

And i have those circuits but they are of u knowor 5mH or

1.35mH

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