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If you have to possibility to built yourself a headphone's driver, what will look like? - Page 2

post #16 of 18
8/2/10 at 4:50pm
Thread Starter 

hi deltaydeltax,

 

well, you are not discouraing me.......well not yet but your comments are having a real good influence to make me thinking of changing my mind about making them, probably should i buy them?......the problem is where, nobody wants to deal will small volume and custom sizes.

 

Well, back to making the drive, I've was planning to first reverse engineering a driver dismounted from audio technica headphone and try to achieve something similar with similar specs.

 

Like you said, I realized that the best thing is starting by the magnet, then.......making a coil and a cone. Cone is not a big deal for me, was planning to use some mylar film 0.0005" and press it into heated molds. I have access to cnc machines so making a mold isn't to complex for me, well hope fully the machine will work soon, another project on the way!! http://picasaweb.google.ca/plduchesne/CNC#

 

The baskets.......well, i am not very familiar with magnet field and all speaker stuff, will learn on the terrain as the project goes on. for the basket, will machine some to have something similar to the actual drivers that i have from audio technica.

 

for impedance analysis, well again, not a pro but was thinking using a software to control quality.

I do have wt3 software  http://www.daytonaudio.com/wt3.html

  it's the easiest way for me now to mesure it. It's probably not the best way but as a start, It should work.

 

Thanks for digikey info, didn't know that they were selling magnet too, might be the best bet as a start to do some testing,

 

petecul


Edited by petecul - 8/2/10 at 4:52pm
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post #17 of 18
8/2/10 at 7:40pm

 

You can use the simple DC resistance equation to find the length of magnet wire you'll need for various coil resistances:

http://en.wikipedia.org/wiki/Electrical_resistance#DC_resistance

 

The area of the wire you choose (A in the equation), is easier looked at in circular mils:

http://en.wikipedia.org/wiki/Circular_mil

 

I'm not sure of any equations which will predict the impedance in AC. I'll have to go digging through an energy conversion book and see if a formula exists there. You may have to hook up the coil you wind to a series resistor, drive this network with a signal generator putting out a 1V sinewave and increase the frequency starting from the lowest frequency available, and look for the 0.707V on an oscilloscope. Then, work backwards using the lo-pass or high-pass equation, depending on how you have the two components wired (the resistor going to ground with give a low pass response, while the inductor grounded will give a high pass response). This will give you the coil value in Henries once run through the equation.

 

I'm not doing a very good job of explaining how to do this. If you follow what I said, then all is well. If not, then let me know and I'll draw a diagram and step by step directions on how this is done. I had one of the students working for me do this for a coil we were using. She really didn't like doing it, but she did find the value of an unknown coil, and what she calculated ended up being very valuable.

 

I'm also not sure how much excursion you'll get based on the b-field of the magnet, and the b-field generated by the coil. The weight of your .5mil thick cone, the adhesive, your wire, and the bobbin will all have an effect on this.

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post #18 of 18
8/5/10 at 8:05pm
Thread Starter 

Thanks for those maths......really!

 

I am still figuring out how to coil up to 46awg but soon will be doing electrical stuff

 

thanks deltaydeltax

 

petecul

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