Hi

Im making an LM3876 based amplifier, but needs an inductor. (0.7uH)
Iv'e bought some from futurlec (part no IND00068) on 0.68uH, don't know if I can use them?

I've read some different guides on LM3876 building, and I can read that I can make one myself with some 0.40mm coated wire wounded 10 rounds around a 10 ohm resistor.

I only have 0.20mm wire, can I use that, but eg. with another amouth of rounds? I can only find a "air inductor" calculator, but you can not set other wire diameters in.

This is my first amp, so I got a lot of questions.

 

Hi

Im making an LM3876 based amplifier, but needs an inductor. (0.7uH)
Iv'e bought some from futurlec (part no IND00068) on 0.68uH, don't know if I can use them?

I've read some different guides on LM3876 building, and I can read that I can make one myself with some 0.40mm coated wire wounded 10 rounds around a 10 ohm resistor.

I only have 0.20mm wire, can I use that, but eg. with another amouth of rounds? I can only find a "air inductor" calculator, but you can not set other wire diameters in.

This is my first amp, so I got a lot of questions.

Greetings.

I am a master DIY inductor maker. It all started when I built my first crystal radio at the age of 8. I won't tell you how long ago that was!

.68 uH is a small inductance, so it WILL be an air inductor. If the resistor value is LARGE enough that it won't have any effective shunt effect across the coil, you can do this (I wouldn't use anything LESS than a 1 megohm resistor for this in ANY application). However there are better ways. You can buy toroid cores with a permeability of 1...the same as air..but a lot more solid!

The GENERAL formula for an air wound, single layer solenoid is found in the ARRL Handbook. It is:

L= d^2*n^2/(18d+40L, where d=coil diameter in inches (wire center to wire center), L=coil length in inches, and n= number of turns. Answer in microhenries.

It's not as straightforward for iron or ferrite cores. You will want to look at the charts supplied with the core material if you go that route. Amidon Enterprises has a lot of good data sheets online.

Hope this gets you started. It's a really really sad addiction!

eric

 

0.4mm wire is roughly equivalent to AWG 26, with a resistance of 40.8 Ohms per 1000 feet.
0.2mm wire is roughly equivalent to AWG 32, with a resistance of 184.1 Ohms per 1000 feet.

While you could wind a coil using the .2mm wire, it may have too high of a resistance for your circuit.

8 turns of 0.4mm wire wrapped tightly on a Bic pen barrel (0.33" diameter) would give you a coil of roughly 0.67uH.
7 turns of 0.2mm wire wrapped tightly on the same form would give you a coil of roughly 0.66uH.

(eta)
Eric, that formula you posted is also good for multilayer air coils.

There's also Wheeler's formula for single-layer air coils only; not quite as accurate, but sometimes it's easier to use, and plenty good enough for such a small inductor:
L(uH) = (r^2) * (N^2) / (9*r + 10*h)
where:
r = coil radius (inches)
N = number of turns
h = coil height in inches

 

Ok thanks.

I got a lot of air-inductors laying around (multilayer) But I don't have any equiptment to measure them with. (They are all from a amplifier)

Here is a picture of them I got from futurlec.com - Is it that type I could use?

 

You can read their values by the same color codes as the resistors.

 

Jep, it's 0.68uH, but is it usable for my project?

 

Hello,

Here is a page from the EDUCYPEDIA on component codes.
http://www.educypedia.be/electronics/datacomponent.htm
Inductors and chokes are a little bit down the page.

Greetings,
Bertus

 

Jep, it's 0.68uH, but is it usable for my project?

What is its DC resistance? Be careful about measuring with a DMM. Short the leads of your DMM while on the ohms range, and get the resistance of the leads. Subtract that value from the measured resistance of the inductor.

 

Greetings.

I am a master DIY inductor maker. It all started when I built my first crystal radio at the age of 8. I won't tell you how long ago that was!

.68 uH is a small inductance, so it WILL be an air inductor. If the resistor value is LARGE enough that it won't have any effective shunt effect across the coil, you can do this (I wouldn't use anything LESS than a 1 megohm resistor for this in ANY application). However there are better ways. You can buy toroid cores with a permeability of 1...the same as air..but a lot more solid!

The GENERAL formula for an air wound, single layer solenoid is found in the ARRL Handbook. It is:

L= d^2*n^2/(18d+40L, where d=coil diameter in inches (wire center to wire center), L=coil length in inches, and n= number of turns. Answer in microhenries.

It's not as straightforward for iron or ferrite cores. You will want to look at the charts supplied with the core material if you go that route. Amidon Enterprises has a lot of good data sheets online.

Hope this gets you started. It's a really really sad addiction!

eric

Great info !!! how can I calculate for iron core inductor ? is this formula practical enought? L=N^2 x Al

Thank you in advance

 

Why would you be thinking of iron core when an air core is only 8 turns?

Bob

 

Im building an 8H choke , I need to calculate number of turns.

 

Ok thanks.

I got a lot of air-inductors laying around (multilayer) But I don't have any equiptment to measure them with. (They are all from a amplifier)

Here is a picture of them I got from futurlec.com - Is it that type I could use?

If it says 0.68 uH why not?
They are a bit uncommon though.
Wire resistance even for small RF inductor will be small at 0.68uH

Its not air coil. These are called color wheel RF inductor or just RF inductor.
There are 4 different sizes

1/8W quite uncommon
1/4W more common
1/2W mostly this kind is found
1W very uncommon for DIY but used in commercial appliances.

 

Hello,

Did you notice the date of the post with the picture and the date of the post of kia2094?

This is a very old thead with a necropost.

Bertus

 

Preferably the component supplier is named and even a link and the poster total noon.