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    How to calculate wire length needed on a coil for a given impedance?

    Let's say you have a 3" diameter voice coil, using 30 gauge copper round wire. How long does this wire need to be to give you a 2 ohm impedance? I am talking about a single coil here, no dual voice coil configurations.

    I know there are programs like SpeaD from Redrock Acoustics that will tell you this, but what is the formula for doing this by hand?

    Thanks



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    Re: How to calculate wire length needed on a coil for a given impedance?

    Let me know if this is right. I used this page for the following calculations. I'm going to use 24 gauge wire in this scenario.

    24 gauge copper wire has a resistance of 25.67 ohms per 1000', which comes out to .02567 ohms per foot. I want 2 ohms, so by doing some math I should need about 77.91 feet (~935") to get a 2 ohm resistance through a 24 gauge piece of copper wire.

    The circumference of a 3" circle (our voice coil) is around 9", so each turn of the voice coil will take 9" off the length we need to get 2 ohms. So to figure out how many turns I need I will divide 935" by 9" and get ~104 turns. The diameter of 24 gauge copper wire is .0201", so multiply .0201" by 104 and we get a winding width of ~2.1 "

    Is all this correct or did I mess up in my math or figures somewhere?



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    Re: How to calculate wire length needed on a coil for a given impedance?

    Ok.....Lmao. now I'm kinda drunk but I'm gonna try. The voice coil ( thats already made ) won't change they way you think it will. I've taken a 4ohm woofer and ran it over 200ft and it didn't change. I guess I need more info because the circumferemce of the voice coil doesn't matter. Do the math all you want but its gonna take a lot more then that to change it




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    Re: How to calculate wire length needed on a coil for a given impedance?

    Quote Originally Posted by PocketRocket View Post
    Ok.....Lmao. now I'm kinda drunk but I'm gonna try. The voice coil ( thats already made ) won't change they way you think it will. I've taken a 4ohm woofer and ran it over 200ft and it didn't change. I guess I need more info because the circumferemce of the voice coil doesn't matter. Do the math all you want but its gonna take a lot more then that to change it
    Maybe it's because you really are drunk (no offense meant lol ), but I don't think you understand my question. I'm talking about if you wanted to design your own coil from scratch and have another company make it for you (such as Precision Econowind)

    Think of a 3" copper voice coil. That coil consists of a piece of copper wire with a given gauge wrapped around a former SEVERAL times. One wrap counts as one turn. Now, the copper wire that the 3" coil is made out of has a certain diameter. The diameter is going to depend on the gauge of the wire. We're using 24 gauge in this scenario. In order for a piece of 24 gauge copper wire to measure 2 ohms, it has to be a certain length. A shorter wire will have less resistance than a longer wire, thus the shorter wire will have a lower impedance.

    And the circumference of the voice coil DOES matter in a way. If I want 30mm of xmax on a driver, and I have a .75" top plate, my winding width needs to be a certain height. Given the same gauge wire, a 2.5" voice coil will have a longer winding width than a 3" voice coil. So for the 3" coil to get the same winding width of the 2.5" coil, you will have to change the wire gauge on the 3" coil.

    My question is how long does this 24 gauge wire have to be until we measure a resistance of 2 ohms. Let me know if this makes more sense or not



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    Re: How to calculate wire length needed on a coil for a given impedance?

    Ha, yeah it makes more sense....but due to my hangover at this point I can't answer the question.




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    Re: How to calculate wire length needed on a coil for a given impedance?

    Does anyone know? I'm thinking my explanation is correct, but I'm not sure if I can scale the resistance the way I am doing. For instance, a piece of 24 gauge copper wire has a resistance of 25.67 ohms per 1000'. Is it correct to say if I divide the length by two, I will also reduce the resistance by two?

    @sundownz ;



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    Re: How to calculate wire length needed on a coil for a given impedance?

    Ok I found a calculator online to do this and it seems like my results were pretty close to what the calculator gave me so I think I've answered my question

    Coil Physical Properties Calculator



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    Re: How to calculate wire length needed on a coil for a given impedance?

    Also... sounds like you are calculating single layer coils -- most coils you see are 4 layers... at the minimum 2 layers typically.

    One-layer is pretty rare other than edge-wound PA drivers.



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    Re: How to calculate wire length needed on a coil for a given impedance?

    Quote Originally Posted by sundownz View Post
    Also... sounds like you are calculating single layer coils -- most coils you see are 4 layers... at the minimum 2 layers typically.

    One-layer is pretty rare other than edge-wound PA drivers.
    Yeah but this was for simplicity's sake. I did some for multiple layers since this thread. These are the lengths needed to make a 1.5ohm coil using multiple layers. Anyone please let me know if you see mistakes

    Re = 1.5ohm
    Coild Windings = 18 gauge copper
    Coil Size = 2.5"
    2.5" Circumference = 7.85"
    18 gauge diameter = .0403"

    ohms per foot = .006385
    foot need for 1.5ohm = 234.9256ft, or 2819.1072"
    turns = 359
    Winding Width for 1 layer = 14.46"single coil
    Winding Width for 2 layer = 7.23" single coil
    Winding Width for 4 layer = 3.62" single coil
    Winding Width for 8 layer = 1.81" single coil
    So for a single layer we just multiply the number of turns by the wire diameter and that will give us the winding width for the single layer. For the dual layer, I assume we will divide the number of turns by two (since I'd guess both layers have the same amount of windings so that they are even), then multiply that answer by the wire diameter to give us the winding width of each layer. Does that sound right?
    Last edited by *Ace*; 03-27-2013 at 04:34 PM.



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