ampage Tube Amps / Music Electronics 
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6/2/2005 9:17 PM  
Joe Gwinn 
Iteration is not required to find turns count I finally got around to mathematically analyzing (using calculus as opposed to computing numerical examples) the winding of a pickup. It turns out that for all coils where the length of a turn varies linearly with height above the core (all pickups with bobbins comply), mean turn length is a complete replacement for quadratics and iterative solutions: both approaches give exactly the same answer. Consider a pancake coil (space between bobbin flanges equals the width of the wire). In such a coil, the winding is a flat spiral. Right on the core, the length of a turn is Lmin. At the top of the winding, the length of a turn is Lmax. Between these two extremes, the winding length varies linearly with height above the core. This is a consequence of simple geometry. In the mathematical limit of an infinite number of turns using wire of infinitesmal diameter, the total length is the integral of this linear function. (Clashing infinity against infinitesmal to yield a sensible and correct answer is exactly what integral calculus does, so the above statement isn't nonsense.) This integral (and thus the corresponding sums of layers and turns) is a quadratic, which forced the numerical solution of quadratic equations by various methods, including iteration. Mathematical evaluation of the integral yielded a very simple result, that the total length (of infinite the winding) was simply the average of Lmin and Lmax multiplied by the depth of the winding. So it turns out that, whatever the size of wire used (so long as it's reasonably small compared to the bobbin dimensions), whatever the fill factor, whatever the values of Lmin and Lmax, the total length is the average of Lmin and Lmax times the number of turns. No quadratics or iteration is needed. So, to solve Problem #2 (given measured Rdc, core and bobbin dimensions, and wire data, estimate number of turns), use the core and bobbin dimensions to compute the mean turn length; use Rdc and the wire data to compute total wire length; divide total wire length by mean turn length to yield the turns count. The resulting turns count will be in error to the degree that the actual wire as wound differs from the nominal wire data, and due to errors in the bobbin and core dimensional data. 

6/2/2005 11:28 PM  
Andrew C 
Joe, I don't mean to party poop but I don't think you have an accurate reading of Lmax to play with, certainly not in the terms the problem was set. You define Lmax as a bobbin dimension, but that would only work if the bobbin was wound full. I believe Lmax should be the outer dimensions of the actual wound coil (or an average of several dimensions in the case the coil has any taper or other extrusions). I guess the problem solver wants to estimate coil windings from published specs, which often leave out Lmax. Lmin and bobbin height can often be deduced from the general style of the pickup. Unfortunately your method required that the coil be in the near vicinity of someone with a careful pair of hands and a set of calipers. To remedy the situation we could apply a function to represent the relative fullness of the bobbin, but its probably going to lead back into the quagmire. Andrew C. 

6/3/2005 7:12 AM  
Joe Gwinn 
On 6/3/2005 5:28 AM, Andrew C said: [QUOTE]I don't mean to party poop but I don't think you have an accurate reading of Lmax to play with, certainly not in the terms the problem was set. You define Lmax as a bobbin dimension, but that would only work if the bobbin was wound full. I believe Lmax should be the outer dimensions of the actual wound coil (or an average of several dimensions in the case the coil has any taper or other extrusions).[/QUOTE]I guess I didn't say it clearly, but Lmax is the length of the longest turn, which is determined by the dimensions of the coil winding, not the bobbin and core. The bobbin and core constrain the bottom and two sides of the winding, but not the top. So one would determine Lmax by measuring the dimensions of the outside of the winding itself. This will be somewhat less than the flange dimensions.


6/3/2005 7:28 AM  
Dr. Strangelove 
Tape thickness? Joe Gwinn wrote:
Plastic film tapes are usually around .005" thick. I've only a rough idea about the fabric (.010.014") or the paper ones (.005".010"), although adhesive seems to account for .002" of the total thickness. drh  

6/3/2005 8:10 AM  
Joe Gwinn 
On 6/3/2005 1:28 PM, Dr. Strangelove said:


6/3/2005 12:19 AM  
Dr. Strangelove 
Re: Iteration is not required to find turns count I borrowed another Barden Tele Bridge, took pertinent measurements again. They are different from the last set. See what you get with these.
drh  

6/3/2005 8:04 AM  
Joe Gwinn 
On 6/3/2005 6:19 AM, Dr. Strangelove said: [QUOTE]I borrowed another Barden Tele Bridge, took pertinent measurements again. They are different from the last set. See what you get with these.
However, just for fun, I'll compute a turns count from the above table. The error shouldn't be too large. This constitutes Homework Problem #2. Compute Lmin: The core is a slender rectangle 2.375 by 0.0625 inches, with the thin ends rounded. We will ignore the thickness of the wire. The perimeter is therefore Pi*0.0625+2(2.3750.0625)= 4.8214, so Lmin= 4.8214 inches. Compute Lmax: The width of the bobbin flange is 0.2925 inches, and the core width is 0.0625 inches, so the maximum winding height above the core is therefore (0.29250.0625)/2= 0.115, so Hmax= 0.115 inches. This inflates the dimensions of the outer turn to 2.375+2*0.115= 2.605 inches by 0.2925 inches, the outer dimensions of the flange. This will be slightly larger than the true outer turn, but we will use the flange dimensions anyway. The perimeter of this outer turn is Pi*0.2925+2(2.6050.2925)= 5.5439 inches Compute Lmean: Lmin= 4.8214 inches and Lmax= 5.5439 inches, so their average is (4.8214+5.5439)/2= 5.1827 inches. So, Lmean= 5.1827 inches, or 0.4319 feet. Compute total length of wire used: The measured DC resistance is 2,230 ohms, and unstretched AWG #43 wire is nominally 2,143 ohms per 1000 feet, to the total wire length is (2230/2143)*1000=1,040.6 feet of #43 wire. Compute number of turns: The total length of wire is 1,040.6 feet and the length of the average turn is 0.4319 feet, so there are something like 1040.6/0.4319= 2,409 turns on this pickup. This would be the high end, as the actual coil will be slightly smaller. But, 2409 is just within the range given in the above table, "24002600?".


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