multi-gauge wire humbucker question

[Joe Gwinn]

Joe,
Is there some mathematical shortcut to derive the extra turns needed of a smaller gauge wire onto two identical bobbins once we know the proportion of the cross-sectional areas of the two wires in question? I bet the ratios of the different wire's cross-sectional areas and the extra turns needed to equalize the turns area products will have a fixed coefficient.

I suppose scatter winding may throw off the calculations a bit.

I'm keeping this strictly air-core coils for now.

If the bobbins are identical, wind the same number of turns on each, as David Schwab said. Scatter winding will have minimal effect. Nor will different wire sizes much matter, so long as the coils have roughly the same physical dimensions (so the typical turn is about the same for both coils).

If the bobbins differ, figure out the area of a typical turn on each bobbin, and adjust the number of turns so their area-turns products are equal.

Let's try an extreme example, two circular coils, one being three times the diameter of the other. Assume the smaller coil has 1000 turns of thin wire wound on a form 1" in diameter. The area-turns product will be 1000 (Pi/4) 1^2. The other coil is 3" in diameter. How many turns must it have to have equal area-turns product? We will call this unknown turns count "N" and solve, so:

N (Pi/4) 3^2 = 1000 (Pi/4) 1^2

The (Pi/4) terms appear on both sides, and so will cancel, yielding:

N 3^2 = 1000 1^2

Dividing both sides by 3^2 yields:

N = 1000 (1/3)^2, which equals 1000/9, or 111.1 turns.

Now, just for amusement, compute the inductance, using the "single-layer solenoid" and "circular loop" formulas from Inductance - Wikipedia, the free encyclopedia. Each turn is a circular loop. The smaller coil has 1000 such loops, while the larger coil has only 111 turns. Being lazy, I am going to simplify by taking the N^2 term from the w>>1 case of the solenoid formula and apply it to a collection of circular loops; this is in effect a very short solenoid.

The inductance of a 1000 turn coil is (1000/(1000/9))^2=9^2= 81 times greater than that of a 111 turn coil, if they were the same diameter. But they are not the same diameter, so we are not done. Applying the circular loop formula, after some math, a 3" loop has 1.852 times the inductance of a 1" loop, assuming use of #42 AWG wire, with radius 0.00125". So, the net effect is that the smaller coil has 81/1.852= 43.7 times the inductance of the larger coil. This is precisely why the Shur humbucking coil works.

[Mike Sulzer]

Well... I guess I like a less partial shield in general.

So, what do you think are the primary sources of problematic magnetic fields (besides transformers?)

Bob Palmieri

Perhaps large switched currents such as occur when using light dimmers.