Being that thin I'd have thought it would be quite resistive.
Being that thin I'd have thought it would be quite resistive.
As for a broken coil wire, I'd have thought that would put the resistance up rather than down.I must be confused. Two times I have fixed a single coil pickup, that the coil wire was broken away from where it is soldered to the lead on the bottom bobbin. I don't recall them reading a higher DCR than they were supposed to. IIRC both read 0 ohms.Yes, a broken coil wire would give an open circuit.
Am I imagining that, or is it different in a humbucker? Please explain.
Sincerely,
Electronically Challenged
Making the world a better place; one guitar at a time...
You must be confusing an open circuit/overload reading with a short/circuit 0 ohms reading. If the wire has broken away, how can you have a circuit? And if there's no circuit, you've got an infinite resistance reading. I don't know exactly where you were measuring across, but if you were measuring across the min lead connection solder joints on the pickup, you would have got an open circuit reading.Originally Posted by McCreed
Now if someone had been passing sufficient voltage/current through the coil to burn the insulation off, then you could get a near 0 ohms reading and no output, but that is very far from a broken wire and very unlikely to ever happen short of extreme stupidity like trying to plug the guitar lead into the mains.
Well, looking up AWG tables gives a diameter of 0.15mm as probably 34 or 35 AWG.
Interestingly, if you assume an average 6" length per wind (the table I used only had inches), then 4400 winds is about 2200 feet. That equates to a resistance of 272ohms per 1000 feet, which is basically the 34 AWG wire range.
So I would guess they have used 34 AWG wire to wind it, which will definitely give some output from the pickup with 4400 turns per coil, but certainly not the output that 30k would have given if that was 44 AWG. From 'resistance per 1000 feet' tables, 15k per coil would indicate 12700 windings (which is a lot!). A typical standard guitar humbucker has around 5000 windings per coil.
Less windings = less output and also less inductance, which moves the resonant peak higher and reduces mids and increases treble.
So it will probably sound relatively quiet and clear.
But it does seem like someone picked up the wrong reel of wire for these particular pickups.
From what I can gather, the mudbuckers used 42awg. , i think that's about 5.5 ohms / m
Well, the reason for using thinner wire is to get more turns on in the same space, but it would appear Gibson didn't.
Using feet again, at an average 1.659 ohms per foot resistance, then 15k ohms would require about 9000 feet of wire or 18,000 turns per coil! (Maybe slightly less as the length per turn increases as the effective coil radius goes up as the wire gets wound on). That's about double the number of turns on a P90. I can see from photos that there's a lot of wire there!
So with a real one (or good copy) you are going to get an awful lot of output, with almost no treble, all bass and mid. Instant overdrive into most amps unless you turn the bass's volume control down. The Artec pickups also use ceramic, not alnico magnets like Gibson did, which given 42AWG wire wound to 30k overall, would give even more output!
well I think the the point of Cliff throwing a mudbucker in there was how it handles distortion. I believe the lack of treble is why they are referred to as "mud"buckers.
Well, your current pickup will sound a long way from that!
Well yeh! Especially considering one coil is in the bin in 100 wind pieces
If the rewind goes well, then I’d suggest getting some alnico magnets as well.
Here’s a documented rebuild of an Artec pickup. https://music-electronics-forum.com/...eb-0+mudbucker
(I’ve only partially read it so I hope it makes sense)
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