I use three values here, namely 0.3, 0.7 and 1 mH (These are the values used in the Ariel loudspeakers). Here's how to get from inductance value to actual-coil-in-your-hot-grubby-little-paws.
|Wire size||Area||Diameter (mil)||Diameter (mm)||Resistance|
|8 AWG||8.336mm^2||128.5 mil||3.264mm||0.00206 Ohm/m|
|10 AWG||5.261mm^2||101.9 mil||2.588mm||0.00328 Ohm/m|
|12 AWG||3.309mm^2||80.8 mil||2.053mm||0.00521 Ohm/m|
|14 AWG||2.081mm^2||64.1 mil||1.628mm||0.00829 Ohm/m|
|Coil Inductance||10 Gauge R||Equivalent length||12 Gauge R||Equivalent length|
|0.3 mH||0.04 Ohm||12.195m||0.07 Ohm||13.436m|
|0.7 mH||0.06 Ohm||18.293m||0.11 Ohm||21.113m|
|1.0 mH||0.09 Ohm||27.439m||0.14 Ohm||26.871m|
You can use the calculator at Shavano Music Online, which gives
|1 mH||0.11 ohm||30.83 mm||61.66 mm||75 turns|
|0.7 mH||0.09 ohm||28.52 mm||57.04 mm||65 turns|
for 10 gauge and
|1 mH||0.16 ohm||25.56 mm||51.13 mm||82 turns|
|0.7 mH||0.13 ohm||23.73 mm||47.46 mm||71 turns|
|0.3 mH||0.08 ohm||19.8 mm||39.6 mm||51 turns|
for 12 gauge wire.
Alternatively, the calculator at Lalena (using Shavano's dimensions) gives
|1 mH||0.1 ohm||30.83 mm||61.66 mm||73 turns|
|0.7 mH||0.08 ohm||28.52 mm||57.04 mm||63 turns|
for 10 gauge and
|1 mH||0.15 ohm||25.56 mm||51.13 mm||79 turns|
|0.7 mH||0.12 ohm||23.73 mm||47.46 mm||69 turns|
|0.3 mH||0.07 ohm||19.8 mm||39.6 mm||50 turns|
for 12 gauge wire.
formula they use is
L=((D^2 * T^2)/(1000*(18*D+40*C)))
Where D is the average turn diameter, C is the coil length, and T is the number of turns.
There's also a calculator on Peter Ortner's site which is based on A.N. Thiele's paper Air Cored Inductors for Audio. To summarise the summary :-) Thiele derives a time constant K = (L/R) (with L in uH, R in ohm) which he uses to get the optimum dimensions for an inductor. The inner radius, thickness and width of the coil are all equal, namely C = sqrt( K / 8.66).
Apparently, the length of the wire required is W = 0.1873 * sqrt(L*C)
The number of turns is N = 19.88 * sqrt(L/C)
The wire diameter is D = 0.841*C/sqrt(N)
The mass of the wire is (C^3)/21.4.
For the 1mH 10 gauge coil above, L = 1000, R = 0.11
K = 9090.9, C = 32.4.
W = 33.7 m
N = 110.4 turns (Lalena gives 103 turns for a coil with these dimensions)
D = 2.59 mm
and you will need 1589 grams of wire (which is 36 metres long according to WST, so things seem to check out).
Looking at North Creek's resistance values, and the ones obtained by the calculators and Thiele's formula, I must assume that they have a kick-ass formula that tells them how to wind coils with such a low resistance... (1.0 mH and 0.09 ohm can be achieved using 2.93 mm diameter (about 9 gauge) wire with Thiele's formula)
The classical coil inductance formula is L=(r^2 n^2)/(9r+10l) with the dimensions in inches, or L=(r^2 n^2)/(25.4*(9r+10l)) with r and l in mm, but I doubt if that would be at all relevant here :-)
The impedance of a woofer will show a peak at resonance, and then a slow rise from the nominal impedance with increasing frequency.
A Zobel network consists of a series network of a capacitor and a resistor (a high wattage unit) connected in parallel with the woofer.
The purpose of the Zobel network is to cancel the rise in impedance, which is caused by the driver's voice coil impedance. In effect, the voice coil impedance is resonanted out using a capacitor.
With R the nominal impedance of the loudspeaker (4 or 8 ohms) and Le the voice coil impedance, C=Le/(R^2). Other sources suggest that the resistor should be 1.25 times larger than the voice coil resistance.
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