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Topic: Pickup engineering |
Patrick Thornhill
From:
Austin Texas, USA
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Posted 4 Dec 2018 4:12 pm |
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Hey y'all. I'm not sure if this is the correct place to pose this question. Probably not the correct forum, really, but I don't feel like joining any others. 
Why are pickups made with very long lengths of narrow wire? That is, why is magnet wire of very fine diameter and wound many times to create an electromagnet?
If, all else equal, the total amount of conductor (copper, in most cases) could be arrived at with shorter lengths of larger wire - or even a chunk of conductor with a slot for the magnet machined in it - why would the resulting pickup not work correctly? I assume that's the case at least since no one does it. What is it about the very long, narrow wire that makes it required?
Thanks! |
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Patrick Thornhill
From:
Austin Texas, USA
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Posted 4 Dec 2018 4:25 pm
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Just as an addendum, I presume it's a matter of surface area and not mass or volume, but I don't understand why.
Thanks! |
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Michael Brebes
From:
Northridge CA
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Posted 5 Dec 2018 6:44 am
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First thing, it does not become an electromagnet. That only occurs when you put a voltage through a coil of wire to excite a piece of regular iron to turn it into an "electromagnet". The magnet is there to create the magnetic field, which is disrupted by the movement of the strings that have been placed in that field. That field disruption will cause a very slight voltage potential to occur in a wire that is wrapped around the magnet. The more turns around the magnet the greater the voltage potential in the wire. The smaller gauge wire used the more turns can be wrapped around the magnet without getting the coil size too large. Hope that answers your question.
_________________
Michael Brebes
Instrument/amp/ pickup repair
MSA D10 Classic/Rickenbacher B6/
Dickerson MOTS/Dobro D32 Hawaiian/
Goldtone Paul Beard Reso
Mesa Boogie Studio Pre/Hafler 3000
RP1/MPX100 |
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Donny Hinson
From:
Glen Burnie, Md. U.S.A.
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Posted 5 Dec 2018 7:45 am
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| To simplify a little more: Each turn of wire on the pickup creates a minuscule voltage when the magnetic field around it is disturbed. (This is how an electrical generator works.) The more turns, the more voltage created - sorta like hooking batteries in series to get more voltage. You can even consider the pickup a transformer, of sorts. The string cutting through the magnetic field acts like the primary in the transformer. It induces a voltage in the secondary, which is the pickup coil, and that is then fed to an amplifier. The amplifier increases the voltage and current even more. And that, in turn, drives the speaker, which is like a pickup acting in reverse. Speakers turn a voltage into physical movement (of the speaker cone), which creates sound. Whereas, a pickup turns physical movement of the string into a voltage which drives the amplifier. |
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Jim Palenscar
From:
Oceanside, Calif, USA
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Posted 5 Dec 2018 8:25 am
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| The variables such as pickup height, number of turns of the wire around the magnet(s), distance from the wire to the magnet, thickness of the wire, type of magnet, type of wire, type of insulation, the method in which the wire is wound, composition of the plates above and below the coils, etc.- all make a difference in what you hear as an end result. |
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Fred
From:
Amesbury, MA
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Posted 5 Dec 2018 9:18 am Re: Pickup engineering
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| Patrick Thornhill wrote: |
If, all else equal, the total amount of conductor (copper, in most cases) could be arrived at with shorter lengths of larger wire - or even a chunk of conductor with a slot for the magnet machined in it - why would the resulting pickup not work correctly? I assume that's the case at least since no one does it. What is it about the very long, narrow wire that makes it required?
Thanks! |
Some one does do it. The frame of a Lace Alumitone is a single winding around the magnet. The tiny current is fed to a transformer under the frame to match the impedance and voltage requirements of instrument amplifiers.
Fred |
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Patrick Thornhill
From:
Austin Texas, USA
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Posted 5 Dec 2018 1:22 pm
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Thanks ya'll!
I understand at least in a basic way how it works, and that variations in winding etc. create different sounds. But what I'm asking is why thousands of turns of very thin wire is used rather than an equivalent amount of copper in the form of heavier wire (with the apparent exception of the Alumitone). My impression is that winding pickups is time consuming and costly, so if there were an alternative one would think pretty much any business would have pursued it. The fact that they didn't suggests to me that there really isn't a viable alternative.
Put it this way: Leo Fender's designs were extraordinarily economical. Everything he designed was an attempt to distill its function down to the easiest, most efficient, least costly form it could take. But he still spent the time and money winding a kilometer of hair-thin wire around magnets to make his pickups. That suggests to me that other, cheaper and easier approaches basically don't work (e.g a chunk of solid copper of equal mass to 8000 turns of 42 guage insulated wire can't be used to make a pickup for some reason). That "for some reason" is what I'm interested in. |
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Benjamin Davidson
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Posted 5 Dec 2018 2:16 pm
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The answer of why thousands of turns of very thin wire rather than a similar mass of copper can be found within Ohm's Law.
The electrical signal must interact with the pickup, or there will not be a change in the signal from the magnetic flux changes of the resonating strings. This is accomplished with small gauge wire, which inherently is more restrictive to electrical current, thus driving resistance up and creating the voltage signal we use to drive our amplifiers.
Now I want to go study this out so I can give you a more appropriate mathematical response. |
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Donny Hinson
From:
Glen Burnie, Md. U.S.A.
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Posted 5 Dec 2018 4:38 pm
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| Patrick Thornhill wrote: |
| ...But he still spent the time and money winding a kilometer of hair-thin wire around magnets to make his pickups. That suggests to me that other, cheaper and easier approaches basically don't work (e.g a chunk of solid copper of equal mass to 8000 turns of 42 guage insulated wire can't be used to make a pickup for some reason). That "for some reason" is what I'm interested in. |
Patrick, for a magnetic transducer (what we call a "pickup") to work, you a property in the conductor (the copper part) called "inductance". Now, we get that inductive property from coils of wire which are set up in a magnetic field to construct a pickup. A chunk of solid copper has lots of mass, but literally zero inductance at audio frequencies. You can't just make a pickup from a chunk of copper and a magnet. There's no inductance there, so it won't work. The inductance we need for a pickup to function needs a coil of wire (a conductor), and the inductance of that coil comes from the configuration of that conductor (the size, shape, and number of turns in the coil), not from it's mass.
Meanwhile...
Want to improve a pickup? Make it from silver wire instead of copper! |
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Dave Mudgett
From:
Central Pennsylvania and Gallatin, Tennessee
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Posted 5 Dec 2018 6:14 pm
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OK, you asked.
With a typical guitar pickup, there's a magnet (with or without magnetic loading) surrounded by a coil. The guitar strings perturb the magnetic field, thus creating a change in the magnetic flux, and thus inducing the EMF in the windings, which is transmitted through the guitar's electric circuit.
Everything you need to know (quantitatively) about this is embodied in Maxwell's 3rd equation of classical electromagnetic theory (from http://hyperphysics.phy-astr.gsu.edu/hbase/electric/maxeq2.html), which is commonly known as Faraday's Law of Induction:
where E is the electromagnetic force (EMF) induced across a surface represented by the surface integral, ΦB is the magnetic flux across that surface, d/dt is its time rate of change, and B is the surrounding magnetic field.
But for a tightly wound coil of wire, composed of N identical turns, each surrounded by a uniform magnetic field, Faraday's law of induction specializes to
E = -N ∂ΦB/∂t
where E is the electromagnetic force induced in the coil, N is the number of turns of wire, and ΦB is the magnetic flux through a single loop.
In other words, the EMF is produced by the changing magnetic flux, which is perturbed by the moving steel string - the iron in a steel string is a magnetic material. All things being equal, the EMF is proportional to the number of winds. But all things are not equal, and the construction of the coil affects how that EMF is transmitted through the guitar's electric circuit. That is what accounts for the differences between how pickups are wound. Everything impacts this: the strength and geometry of the magnets, the gauge of the coil wire, the number of turns, and so on. There is a several-dimensional space of things that can be varied infinitely.
Edited to remove the link to the 2nd image, which was being disallowed.
Last edited by Dave Mudgett on 7 Dec 2018 9:43 am; edited 1 time in total |
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Patrick Thornhill
From:
Austin Texas, USA
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Posted 6 Dec 2018 1:58 am
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Ah ha!! It’s the Ohm’s law bit that I was unclear on, but make perfect sense - the narrow wire presents more resistance which in turn increases the voltage produced (per Faraday’s law) when the magnetic field is perturbed by the motion of the strings.
My hypothetical solid chunk of copper presents basically zero resistance, hence no inductance. |
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Peter Harris
From:
South Australia, Australia
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Posted 6 Dec 2018 4:28 am
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OK Donny, now explain that bit about the SILVER wire.... 
_________________
If my wife is reading this, I don't have much stuff....really! |
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Donny Hinson
From:
Glen Burnie, Md. U.S.A.
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Posted 6 Dec 2018 7:25 am
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| Patrick Thornhill wrote: |
Ah ha!!...
My hypothetical solid chunk of copper presents basically zero resistance, hence no inductance. |
So then Patrick, using that logic, let's just use a big chunk of carbon, which is a poor conductor and has lots of resistance. Do you think that would give us the necessary inductance?  |
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Georg Sørtun
From:
Mandal, Agder, Norway
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Posted 6 Dec 2018 9:16 am
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Hmm, why not put in some magnets and amplify the current induced in the vibrating strings? Only need an enormously efficient and noiseless pre-pre amp to get the output up to usable levels … worked in the -80s so why not now?
Oh, works best with non-conductive bars. |
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Patrick Thornhill
From:
Austin Texas, USA
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Posted 6 Dec 2018 6:07 pm
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| Quote: |
| So then Patrick, using that logic, let's just use a big chunk of carbon, which is a poor conductor and has lots of resistance. Do you think that would give us the necessary inductance? Winking |
Touché, Danny Hinson, challenge accepted!
My understanding is that the woven graphite we call carbon fiber is actually a pretty good electrical conductor (never heard anything about resistance, so I’ll research). I don’t actually know how that stuff is made and how it differs from pure, elemental carbon - am I allowed to use that as my “big chunk of carbon�
Peter Harris, I think Donny’s test-question to me relates to the statement about silver: the material in question must be a good conductor, and silver, along with gold, are about the only elements that conduct better than copper (or are at least the best that are commonly available and not radioactive). |
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J Fletcher
From:
London,Ont,Canada
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Posted 7 Dec 2018 9:19 am
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| As far as I know , it is the coil that creates the inductance , resistance is just a by product of the metal wire . The inductance is the important quality , not the resistance . Well you need the magnet too , and the vibrating string . |
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