How to make inductors?

[legend killer]

Hi everyone,
Is it possible to make inductors of desired values by just winding few turns of wire on a pencil.Basically the problem is that the inductors are not available in the city or anyother nearby.So whenever I want to make a college project or something like that,the circuit is never completed and it just stays on paper.
Some of my collegues told me that if you just wind few turns of wire on a pencil,and then take the pencil out of the coil I will get an inductor,whereas some told me that it not possible to make an inductor by just winding few turns and the circuit with these kinds of coils never work.
I am really confused and I need your expertise on this.Is there a formula relating type of wire,number of turns and the inductance in Henrys?

 

[#12]

Whoa...this takes me back.
How far back?
Notebook one, page 4!
L in uH= [MeanRadius x Nturns]^2 / [9MeanRadius + 10 length]
You have to consider the current carrying ability of the wire, but air core inductors don't saturate magnetically.
Mean Radius = [I.D. + O.D.]/2

 

[ericgibbs]

Hi everyone,
Is it possible to make inductors of desired values by just winding few turns of wire on a pencil.Basically the problem is that the inductors are not available in the city or anyother nearby.So whenever I want to make a college project or something like that,the circuit is never completed and it just stays on paper.
Some of my collegues told me that if you just wind few turns of wire on a pencil,and then take the pencil out of the coil I will get an inductor,whereas some told me that it not possible to make an inductor by just winding few turns and the circuit with these kinds of coils never work.
I am really confused and I need your expertise on this.Is there a formula relating type of wire,number of turns and the inductance in Henrys?

hi,
Look at this link.
Air Core Inductor Coil Inductance Calculator

Use a high value resistor, say a 1megohm, and wind the enamelled copper wire onto the resistor body.
Wrap a couple of turns of the copper wire around the resistor wire at each end of the coil and solder the copper and resistor wire together at the junction.

This inductor can be inserted into a pcb or project board.

 

[legend killer]

Thank you very much for replying Eric and #12.
I know little about inductors but still I want to confirm that does the inductance depends on the spacing between the turns?

 

[ericgibbs]

Thank you very much for replying Eric and #12.
I know little about inductors but still I want to confirm that does the inductance depends on the spacing between the turns?

hi,
The formula states length and number of turns are required as values for finding the inductance, so the answer to you question is there...

length and number of turns

 

[Pommie]

Mean Radius = [I.D. + O.D.]/2

Isn't that the mean diameter?

Mike.

 

[#12]

pommie...I think you're right. Should be [ID + OD]/4

 

[legend killer]

hi,
The formula states length and number of turns are required as values for finding the inductance, so the answer to you question is there...

length and number of turns

Oh yes,I didnt realise that,I am really glad to discuss the problems with genius like you.
One more think I would like to ask you is that where should I start making inductor.Should I first wind the turns on the pencil and then take out the pencil,then connect the two ends of coil with the two ends of resistors?Why do I need to connect the resistor?

 

[ericgibbs]

Oh yes,I didnt realise that,I am really glad to discuss the problems with genius like you.
One more think I would like to ask you is that where should I start making inductor.Should I first wind the turns on the pencil and then take out the pencil,then connect the two ends of coil with the two ends of resistors?Why do I need to connect the resistor?

The resistor replaces the pencil and it remains in position to support the coil...

 

[legend killer]

I guess the value of resistor is so high so as to make sure the resistor does not interfere with the operation of inductor.No current drawn and hence high Q.

 

[ericgibbs]

I guess the value of resistor is so high so as to make sure the resistor does not interfere with the operation of inductor.No current drawn and hence high Q.

Not really, the 'Q' value is the ratio of the inductive reactance and the resistance of the coil winding , not the coil former.

 

[#12]

Ah yes. coil resistance is in series with the inductance. The coil former is in parallel with the inductance.

 

[legend killer]

Not really, the 'Q' value is the ratio of the inductive reactance and the resistance of the coil winding , not the coil former.

I am getting this a little bit.If I am not acting like asking for spoon feeding,sir can you plz explain explain me little technically,because I am really curious to learn it.Intutively I have the idea that if two resistive components are connected in parallel,and if one is very much greater than other,then it is as good as saying that that component does not exist.

 

[ericgibbs]

I am getting this a little bit.If I am not acting like asking for spoon feeding,sir can you plz explain explain me little technically,because I am really curious to learn it.Intutively I have the idea that if two resistive components are connected in parallel,and if one is very much greater than other,then it is as good as saying that that component does not exist.

You asked about 'Q', the 'R' value is the resistance of the coil.
The resistance of the coil is distributed along the length of the wire used to wind the coil so its considered to be effectively in series with the inductive reactance of the coil.

Q = (2pi * F * L)/R

 

[legend killer]

I now understand the concept about Q.But I think there has been a miscommunication from my side .I was asking about the resistor which is attached to the coil.I think you already gave the answer,that is ,it supports the coil.

 

[ericgibbs]

I now understand the concept about Q.But I think there has been a miscommunication from my side .I was asking about the resistor which is attached to the coil.I think you already gave the answer,that is ,it supports the coil.

hi,
If a 1 megohm is chosen as the coil former by comparison the resistance of the coil is negligible, so the effect of the 1 meg upon the circuit can be ignored.
If you ever wind RF type narrow gauge wire air cored coils, you dont want them flapping around in the breeze as the inductance/tuning will be affected.
The higher gauge wire coils can be self supporting.

 

[legend killer]

thank you very much for the help

 

[BrownOut]

One more thing. Instead of a pencil, use a machine screw with a pitch about right for your wire. Wind the wire on the screw, then rotate the screw out of the coil. The windings are prefectly uniform, and spring out a little so that no windings end up shorted. As long as there is space between coils, you don't need to use coated wire. I've used this method with 22ga hookup wire ( insulation removed ) to make self supporting RF coils.