Italian Made Gaglio Pickups And The Fantasma System

[Joe Gwinn]

The bifilar winding means that the flux linking turn n of winding 1 and winding 2 is the same. But is this enough to imply large M?

To the first order, yes.

For a large mutual inductance M = k sqrt(L1*L2), k (which varies between 0 and 1) is 1 (or nearly so). From Close, The analysis of linear circuits, page 335, we have "Physically, this corresponds to the case where all the flux links all the turns of both windings."

Exactly.

The bifilar winding does not mean that turn n of winding 1 has the same linking flux as turn m of winding 2. With the open core, we do not expect this to be the case. I think we have to determine, rather than assume, M.

If the two wires of the bifilar pair are physically close to one another, most flux that links one will of necessity link the other. How can it be otherwise? There is little space between the wires, and flux that links only one wire must pass between the wires.

[Mike Sulzer]

If the two wires of the bifilar pair are physically close to one another, most flux that links one will of necessity link the other. How can it be otherwise? There is little space between the wires, and flux that links only one wire must pass between the wires.

Consider an air core solenoidal coil with one winding. Pass a current through it. The flux generated by turn 1 at one end clearly does not all pass through turn n at the other end. (It diverges.) Then if there are two windings, bifilar, all the flux from turn 1, winding 1 does not pass through through turn n, winding 2. To make this approximately so, one must use a high permeability core that "confines" the flux, and then extend this core out one end of the coil around back into the other end.

[Joe Gwinn]

Consider an air core solenoidal coil with one winding. Pass a current through it. The flux generated by turn 1 at one end clearly does not all pass through turn n at the other end. (It diverges.) Then if there are two windings, bifilar, all the flux from turn 1, winding 1 does not pass through through turn n, winding 2. To make this approximately so, one must use a high permeability core that "confines" the flux, and then extend this core out one end of the coil around back into the other end.

The definition of mutual inductance is in effect self scaling. The effect of the leakage flux is to reduce the inductance of short solenoidal coils. A bifilar solenoidal coil will show this identically in both windings. The mutual inductance is due to the flux of either coil that links both coils. So, even if some of the flux generated by one coil fails to fully link all turns of that coil, what flux does link the coil also links the other coil in the same proportion.

As a thought experiment, consider four identical torioidal coils in two series strings. In each string, one coil is wound on its own core while the other is cowound on the same core as used by the other string.

[Mike Sulzer]

The definition of mutual inductance is in effect self scaling.

Let's see what this equation: M = k sqrt(L1*L2) has to say about that. Suppose you insert a core into a bifilar wound air core solenoid. Suppose L1 and L2 double in value. Then in the equation for M the sqrt term doubles. The core has altered the flux linkage, so k increases, too. So M is rising faster than L1 or L2. It does not scale as you say.

Consider this also. Equations for finding inductance contain n^2. This is because each turn couples to every other turn, not equally, but in an amount accounted for by the constant in each equation. Mutual inductance also must consider the coupling of each turn of winding 1 with every turn of winding 2. Above you have stated that one considers each turn with the corresponding turn of the other coil. That neglects all the other combinations, the ones for which the flux linkage is not complete in the case we are discussing.

One might also consider the model for a transformer. One starts with an ideal transformer. One adds an L in parallel with either the primary or secondary. This is the magnetizing inductance. Then one adds an L in series with either the primary or secondaary. This accounts for the imperfect coupling, that is, k less than 1. These represent two degrees of freedom; their values are computed separately.

Finally, one might consider why all power transformers use an enclosed core. If one could use an open core and keep k up by using a bifilar winding, it would be done.

At rf where one might use a transformer with a k significantly less than one, it is because one uses the leakage inductance as part of tuned circuit to make a bandpass filter. You tune it out.

[Joe Gwinn]

Let's see what this equation: M = k sqrt(L1*L2) has to say about that. Suppose you insert a core into a bifilar wound air core solenoid. Suppose L1 and L2 double in value. Then in the equation for M the sqrt term doubles. The core has altered the flux linkage, so k increases, too. So M is rising faster than L1 or L2. It does not scale as you say.

We've slid off the original question to a similar looking but unrelated question, obscured by a cloud of math.

The question before us is how large the mutual inductance between the windings of a bifilar wound coil will be, relative to the inductance of each of the two windings taken alone.


We can make the following observations:

1. Given that the winding is bifilar, the two windings are identical. Therefore, they have the same number of turns, so N1=N2=N; and because these two windings have exactly the same environment, the same inductance, so L1=L2=L. The details of this environment are not relevant; it's enough that that both windings see the same environment.

2. If we connect the two windings in series aiding, the overall inductance will be 4*L. This is true because inductance varies with the square of the turns count. The inductance of either coil is proportional to N^2, while the inductance of the series-aiding double coil is proportional to (N1+N2)^2= (2*N)^2= 4*N^2. The N^2 term cancels, leaving the ratio 4.

3. If we connect the two windings in series opposing, the fields exactly cancel one another, and the inductance is zero.

4. The formula for the inductance of two inductors having flux linkage (and thus mutual inductance M) is Ls=L1+L2+2*M for series aiding, and Ls=L1+L2-2*M for series opposed. Only the sign before the 2*M term varies.


We can now plug and grind to determine the size of M, given the above two data points:

4 L = L + L +2*M and 0 = L + L -2*M

4L -2L = 2M and 2L = 2M

Both equations independently lead to the conclusion that L=M, and therefore that all flux linking one coil also links the other coil.

By the equation M=k Sqrt[L1*L2], we can now compute the coupling coefficient k:

k=M/Sqrt[L1*L2]=L/Sqrt[L^2]=1

Saying that k=1 is another way to say that all flux that links one coil must link both coils, and that L=M. This is true regardless of the size and construction details of the bifilar coil. Nor does the nature of the core, if any, matter.

[David King]

OK then, I'm glad I waited to wind this puppy.

[Mike Sulzer]

Joe, you are right; I am sorry to be so dense and put you to so much trouble explaining this.

So then it it true that this pickup cannot work at all like the maker says because magnetic coupling between the windings is important at all guitar frequencies.

[Joe Gwinn]

Joe, you are right; I am sorry to be so dense and put you to so much trouble explaining this.

OK. I knew that L and M had to be equal because I could not see how the two windings of a bifilar coil could see different flux. The math came later. I find that I get lost if I cannot draw the physical picture.

So then it it true that this pickup cannot work at all like the maker says because magnetic coupling between the windings is important at all guitar frequencies.

Yes. Gaglio's understanding of how his pickup works is at least incomplete. Which isn't to say that the pickup doesn't work. Many inventors have crazy ideas about how and why their invention works, but life goes on.

[David King]

The idea seems pretty obvious to me and I wonder if there is prior art that wasn't included in the patent search. Do you guys think this patent would stand a chance in an actual dispute against a Fender or Gibson? I'm sure a clever lawyer could breeze around the convoluted English. I don't get the impression that Gaglio put a lot of thought into achieving the broadest possible interpretation. It may be more of a vanity patent than an actual legal protection.
That said I, would license the idea if I wanted to manufacture for profit.

[David Schwab]

David, when has Fender or Gibson ever done anything like this?

I have a large number of pickup patents on PDF going way back to the very first pickups, and I haven't seen this technique used before.

The closest thing would be a couple of Bill Lawrence patents, but he didn't use bifilar wire. It was concentric coils, and the main coil had both ends connected to the amp.

[Mike Sulzer]

OK. I knew that L and M had to be equal because I could not see how the two windings of a bifilar coil could see different flux. The math came later. I find that I get lost if I cannot draw the physical picture.

Yes. Gaglio's understanding of how his pickup works is at least incomplete. Which isn't to say that the pickup doesn't work. Many inventors have crazy ideas about how and why their invention works, but life goes on.

The pickup works, but does it really do anything that requires it to be made as it is? For example, what is its output level? If the level is lower than he is implying, well, we already know how to make a pickup with low output that has flat response to high frequencies. Just use fewer turns.

[Joe Gwinn]

The pickup works, but does it really do anything that requires it to be made as it is? For example, what is its output level? If the level is lower than he is implying, well, we already know how to make a pickup with low output that has flat response to high frequencies. Just use fewer turns.

I have no idea. We are now beyond theory.

It's time to build one and see. It's quicker and surer to build and test than to argue about what it ought to be able to do or not do.

[David King]

Well, on to the brass tacks then. If we take his assertion that it's wound hotter than standard pickups, first, how do we count the turns, does each turn count twice? I wouldn't think so. We'd need to figure out approximately what gauge will take up less than half the space of AWG 42. Would that be 44 or 45? I don't have either so I'm stuck with 43 and a taller bobbin.

[David King]

David, when has Fender or Gibson ever done anything like this?

I have a large number of pickup patents on PDF going way back to the very first pickups, and I haven't seen this technique used before.

The closest thing would be a couple of Bill Lawrence patents, but he didn't use bifilar wire. It was concentric coils, and the main coil had both ends connected to the amp.

David, I didn't mean to imply that Gibson or Fender or anyone else had ever done anything like this but I'm just wondering what would happen if a big company was attracted to an idea that was poorly protected and they just went for it knowing that their legal budget was orders of magnitude larger than Gaglios and that they could spend him into the ground. (Kind of like what the windshield washer timer guy had to go through with the auto makers)...

Re the prior Art, I was thinking "art" outside musical instrument pickups end of the coil winding spectrum. Like some sensor array for subatomic particles or something -you know?

[David Schwab]

David, I didn't mean to imply that Gibson or Fender or anyone else had ever done anything like this but I'm just wondering what would happen if a big company was attracted to an idea that was poorly protected and they just went for it knowing that their legal budget was orders of magnitude larger than...

Like DiMarzio getting a patent on basically the same patent as Kinman's?

Fender and Gibson seem to play fair and buy or license products.