Calling Lex@Lace -Alumitone Question

[MR COFFEE]

Lex,
A question on the Alumitone design...

They read over 5K ohm resistance on my ohmmeter, and I just can't believe that is all wire turns resistance, no matter how thin the windings are. The coils are just too small to have that much wire resistance.

You put an RC (resistor-capacitor) filter on the output to kill RFI (radio frequency interference) up inside that shrink wrap over the C-core coils, right?

Maybe 5K and 820 pf or thereabouts?

I am all for y'all working on the double transformer idea that Joe advocates for - I think it's a beatiful can of worms is just squirming to get out!!

And welcome to the forum!

mr coffee

[salvarsan]

The Lace patent application US 20110048215 A1 shows the Alumitones.

In all drawings, there are two series connected coils.

<a href="http://www.google.com/patents/US20110048215"><b>Sensor assembly for stringed musical instruments</b></a>

As I guesstimate, the target DC resistance (2500 ohms per coil) needs about 10k winds of #46.

-hizself

[MR COFFEE]

Hi Salvarsan,

Check out my math, and tell me if I'm mistaken that there is a resistor in there. I've read the patents, and you are correct, they don't say anything about a resistor. But I wouldn't assume product development ended with the patent.

My math:
4800 turns on each coil (per Lex's post)
2500 ohms resistance per coil if it's all wire resistance (5K ohms divided by 2)
Therefore, each turns resistance is 2500 divided by 4800 turns equals 0.52 ohms per turn.
AWG 46 is 14 ohms per meter, therefore .52 ohms divided by 14 ohms equals .037 meters (the length of each turn)

This would mean that each turn around the teensy weensy coils takes an average of 3.7 centimeters or 37 mm. of wire, which is way longer than the circumference of the coil forms. So there HAS to be a resistor in there, right?

I think we are so accustomed to pickup windings running in the neighborhood of 5000 ohms that our brains turn off when we read 5K on an ohmmeter. Maybe it's not a RFI\EMI stop, maybe it's for voicing. Maybe the capacitor is in parallel with the resistor to equalize the loss of highs. I dunno, that's why I asked.

I've bought several Alumitones and I like 'em... I'm just curious, like I would assume all the folks around the Pickup Makers forum are.

Don't we all like to think about how our stuff works and how it might be taken further?

Let's give Lex a chance to give us all the straight scoop. It's patented, and his openness in the introductory thread suggests Lace isn't paranoid and that they are open to sharing ideas and knowledge just like Joe (bbsailor) has.

Lex?

[madzub]

Nah, i've cut open a few alumitones and chipped away the epoxy and cut open the coils and there aint no resistors or caps in there, just really thin wire on plastic bobbins over the laminated C cores. The humbuckers i had were 3.7k ohms, forget what the singles measured.

[MR COFFEE]

Hi madzub,

Thanks for chiming in with your findings!

So your alumitone humbuckers measured 3.7K ohms? I picked the lowest number of turns Lex reported for my calculations - 4800 turns. He said #46 AWG was the smallest wire, too...

So back to my math: (and I don't claim to be great at math, either...)

4800 turns on each coil (the least number of turns on each coil per Lex's post)

1850 ohms resistance per coil if it's all wire resistance (3.7K ohms divided by 2)

Therefore, each turns resistance is 1850 divided by 4800 turns equals 0.385 ohms per turn.

AWG 46 is 14 ohms per meter, therefore .385 ohms divided by 14 ohms equals .027 meters (the length of each turn)

This would mean that each turn around the teensy weensy coils takes an average of 2.7 centimeters or 27 mm. of wire, over an INCH long, which I believe is STILL way longer than the circumference of the coil forms.

So if there isn't a resistor in there, where is the extra resistance coming from?

If my math is f**ked, please explain it to me. I'm open to being corrected!!! Or other explanations I haven't thought of...

Maybe it isn't an obvious resistor, SMD is pretty damn small. So are SMD capacitors.

So if my math is right and I'm still way off, let's wait for Lex's explanation of what is the real deal...

Hey Lex, chime in when you're ready...

[MR COFFEE]

OK, Madzub,

I redid the math using your resistance measurement (3.7K) and the maximum number of turns and smallest wire guage (#47 AWG) Lex gave us…

7400 turns on each coil (the greatest number of turns on each coil per Lex's post)

1850 ohms resistance per coil if it's all wire resistance (3.7K ohms divided by 2)

Therefore, each turns resistance is 1850 divided by 7400 turns equals 0.25 ohms per turn.

AWG 47 is 17.4 ohms per meter, therefore .25 ohms divided by 17.4 ohms equals .014 meters (the length of each turn)

This would mean that each turn around the teensy weensy coils takes an average of 1.4 centimeters or 14 mm. of wire, about a half inch long, which does jive with a realistic guess of the avearage length of a turn around the coil forms.

So all coil resistance for the humbucker you dissected does sound plausible.

Was this a “deathbucker” you dissected?

And I'm still scratchin my head about my 5K bass bar. Defective?? Stretched the wire too tight while winding?

[David Schwab]

From what I understand, they wind different amounts of wire on the transformer coils to voice the pickups. Not all the Alumitones read the same. So to get a fuller tone the Bass Bar probably has more wire wound on it.

I make some pickups with individual coils per string. On one of them, 10,000 turns of 42 gives me 2.2k. So I'd imagine thinner wire would read a lot higher.