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Is there any formula to calculate number of windings to get a 50 Watts voice coil of a speaker?

let's say, I want to wind a 20mm diameter voice coil for a speaker. And my coil is made out of copper wire which is 1 mm thick (diameter) and has a resistance of 13 ohms per 1000 meters. And I want the voice coil to use 50 Watts. How do I calculate the number of windings required?

P.S: I'm developing a new speaker design, so, I'll experiment with overheating issues. Just ignore all other parameters and let me understand the relationships between given parameters to initially wind the voice coil. I'm stuck in finding out the number of windings to experiment with, so if some formula gives 50 turns, I'll initially wind it and experiment with all other parameters one by one (eg: overheating).

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  • \$\begingroup\$ Problem with that is 13 Ohms/km = resistivity of 1.02x10^-8, but the actual resistivity of copper is 1.68x10^-8. Perhaps if your wire had a square section? \$\endgroup\$ Jan 19 '18 at 9:23
  • \$\begingroup\$ Square or hexagonal wire is not unknown in speaker design, for exactly this reason. I have no idea if that's what the questioner is using though. \$\endgroup\$ Jan 19 '18 at 11:42
  • \$\begingroup\$ just take it as an example :) I need to know the relationships between parameters to calculate how much turns does it require to withstand 50 Watts from the power amp. \$\endgroup\$ Jan 19 '18 at 13:17
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Calculate the length of the wire required first by its resistance. If 1000 m per 13 ohm, that simply means 500 m per 6.5 ohm or 76.92 m per ohm. Now if you are wiring a coil for 4 ohms, you need 307 m wire on the coil. Now calculate the length of one turn of wire on the coil. If one turn is 2 inches, for example, then you need 2000 turns for 4 ohms impedance.

The coil is designed on ohms not watts. Watts indicate the capacity or load a coil can handle, which depends on the gauge of wire. And before wiring, you can check gauge and watt chart for adequate watt calculation.

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No, you cannot calculate that with only these parameters.

Think about the relations between the parameters of the voice coil you list:

  • coil diameter
  • wire diameter
  • wire resistance
  • number of turns

And how that relates to the maximum power handling capability.

Suppose a certain voice coil can handle 50 W, what happens when I put 100 W into it? I think it will get hotter compared to the 50 W situation.

But how hot is OK? That isn't listed anywhere in the parameters for the voice coil. The coil itself is OK until the copper starts to melt. But I'm sure that at much lower temperatures other things had heat issues already like the material we're winding the coil onto. Some voice coils are wound on an aluminium tube which will help in dissipating the heat. That would increase the power handling capability of voice coil. But hey, you didn't list which material is used for winding the voice coil onto.

Also, if the voice coil is allowed some movement (like in long-throw speakers) that moves the air around the speaker which also helps in cooling.

If you're using the speaker in a free air, a closed enclosure or a bass reflex enclosure that will also have a significant impact on voice coil movement and therefore cooling and therefore power handling.

See how I got from just the voice coil to the complete design of a speaker?

That is because it is not only the voice coil which determines the maximum power handling capability, it is the complete design.

If you look up datasheets of loudspeaker drivers (so the speaker themselves, not a finished box with a speaker in it) then in a proper datasheet you should find the maximum power handling capability listed with a note saying how that's measured like what volume of enclosed box the driver was mounted in.

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  • \$\begingroup\$ At high temperatures you'd likely also damage the insulation on the coil, causing adjacent turns to short together. \$\endgroup\$
    – user253751
    Jan 19 '18 at 9:37
  • \$\begingroup\$ Speaker assembly can use insulation and adhesive selected to survive 240C - (in the 1980s, maybe higher today). \$\endgroup\$ Jan 19 '18 at 11:46
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    \$\begingroup\$ If you put 50W into a speaker it'll use 50W regardless of the number of turns. \$\endgroup\$
    – Finbarr
    Jan 19 '18 at 14:18
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    \$\begingroup\$ I just want to make the coil use the 50W power, so how many turns does it requires to use all 50W power? Hmm, so please explain to me, after reading my answer, how you can still think that anyone can tell you how many turns you need for that 50W. \$\endgroup\$ Jan 19 '18 at 14:46
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    \$\begingroup\$ @Pretty_Girl100 This is the message: That is because it is not only the voice coil which determines the maximum power handling capability, it is the complete design of the speaker + speaker enclosure which matters. If you do not understand my answer than you have a lot to learn about loudspeaker design. \$\endgroup\$ Jan 20 '18 at 15:37
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Well, your exact question can't be answered as it stands - or rather, the answer is, you won't be able to fit that much wire on a voice coil that needs the correct impedance. The wattage question is moot because your wire can handle not only 50 Watts, but around 5000 Watts if you keep the Amperage around 9 Amps continuous.. And it can handle about 15,000 Watt spikes... But that's not really what you want to know probably.

What you need to know, is you need thinner speaker wire.

Your ~17 Gauge wire, to get it down to the correct ohmage would be about - 1000m / 13 Ohms = x/8 Ohms (if it is an 8 ohm speaker system (generally home systems)- or 4 if it is a 4 Ohm (generally, a car speaker system)).

Thus, you'd need 615m (8 ohm) or 307m (4 Ohm)..

Extremely large, It's pointless to calculate because the number of turns in the thousands and with the thickness of the wire, you won't be able to fit it in your magnet.

You can save your time winding, and money, buy using thinner copper. each halving of diameter will increase the ohmage by 4 times. So if you use 0.25mm solid wire, you'll be happy to use 16 times less length of wire. Or about 17m for 4 ohms, or 34m for 8 Ohms. And this is about what most woofer speakers use - 28 Gauge sold copper wire, about 120 feet of it, give or take.

Once you know the length of wire, and I'm just giving approximations here, you can calculate the humber of turns (and layers) required. You have to know things like, how long is the throw if your woofer (as you will need to make sure your VC length is as short or shorter than this. Also, how many turns do you need on the VC to make it fit perfectly in your magnet?

So calculating the number of turns is the least of all the problems here. Making a speaker can be fun, but if you are doing experiments - you don't need to reinvent the wheel - there are many people who have done this - and far more - and compiled these into beginner guides. Q factors, capacitance, inductance, etc. will all be necessary to understanding making a speaker and it's enclosure.

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    \$\begingroup\$ Please read about quantities and units and consider using the scientific quantities: resistance instead of ohmage and current instead of Amperage, etc \$\endgroup\$
    – Huisman
    Feb 17 '20 at 21:07
  • \$\begingroup\$ Is this a sight requirement, or are you just trying to RESIST the evolution of the English language? :) \$\endgroup\$ Feb 25 '20 at 0:09
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    \$\begingroup\$ Yes, it's a sight [as you put it] requirement. "amperage" hurts my eyes. Let's leave the evolution of the English language to the English.SE stack. \$\endgroup\$ Feb 25 '20 at 4:09
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    \$\begingroup\$ I'm trying to persuade you to use the scientific language instead of the English language. Unless you want to write a novel about exciting electrons frolicking happily through a wire until they met a resistance and started to sing hommage (excuse my French) to their amperage. \$\endgroup\$
    – Huisman
    Feb 25 '20 at 6:46
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interesting question... first you need to decide what tupe of amp you will use. then you will know 4 or 8 ohm speaker it will be. normal 100 watt amp will give you 20 volts into 4 ohms. to do that you firstly need to make coil that can withstand that conditions. mainly it will be coil the same 4 ohms in free air (in speaker it will change depending on magnetic system parameters). to make it just search for inductance coil calculating program. remember P=IU so as example for 4 ohm speaker ordinary will be used low woltage (+/-24V power supply) amp that can give about 30V max with no load. you need about 15V. to get 50W now you know that you need wire that can handle ~3.33Amps. wire for transformer windings has secure working point about 2.5 A per aquare millimeter. during fact that speaker has poor but better cooling than transformer 1 mm2 will be okay. remember 8 ohm version will requre 2 times smaller wire. now you can calculate the first coil with software and start experiments.

next one 4 OHM will be the lowest point of impedance of speaker. read about Thiele and Small parameters and you will understand difference between DC resistance and impedance - its not the same thing. so in resonance point impedance will rise up to several dozens ohm. of course in resonanse point speaker will give maximum output at the same conditions. But if you in surprising manner (industrial amp in supermarkets or similar) will feed real 50W at resonance poin (that can be 100 V or more) you will blow up the speaker :) to try out physics of T-S parameters just grab some speaker and amp give a sweep signal from 20Hz to 20 KHz (frequency generator software) and see how changes voltage in multimeter or what you have from measuring devices

P.S. remember this is calculation for sine wave signal. audio signal is mixed type signal and 50W RMS are huge power for audio system. as example many of high grade home stereo systems have just 10-20W RMS and sounds enough for large room.

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    \$\begingroup\$ Welcome to EE.SE. I tried to read your answer but gave up due to lack of capitalisation and punctuation. Please see Write to the best of your ability. \$\endgroup\$
    – Transistor
    Dec 5 '18 at 21:49

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