Talk:Pickup (music)

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January 19, 2004 Refreshing brilliant prose Not kept

Contents 1 Circuit Discussion 1.1 How Resonance Affects Sound 1.2 Altering Pickup Characteristics 2 Editing Discussion 3 FET preamps with piezo pickups 4 Multi transducer pickups 5 Piezo pickups: pressure or acceleration? 6 Magnetic Reluctance 6.1 Alfreds entry 6.2 LCs retort 6.3 Both descriptions? 6.4 My explanation of pickup operation. 6.5 Magnetic Pickups - Towards an improved wording 7 Interesting quote from book. 8 Magnectic circuits and grounding the magnet base 9 Output of a Pickup websurffATgmaildotcom 10 HSS vs SSH 11 Position or type? 12 Possible Error with Humbucker Description 13 Motherbucker 14 Single Coil vs. Humbucker

[edit] Circuit Discussion

Anyone know the equ cct of a variable reluctance pick-up?. Is the output inductive or resistive?--Light current 04:00, 13 November 2005 (UTC)

Try this link[1] I think this link answers my question very well. It states that guitar pick-ups( variluctance types) look like a voltage source in series with a large inductor (1 - 10 H) with series resistance of course, shunted by the self capacitance and the load capacitance ( which is usually much greater than the self cap). It is therefore not surprising that the sound obtainable from magnetic pick-ups varies so greatly with type loading etc. It seems amazing that guitarists have put up with such lousy transducers for so long!

Extract from above link:

A real coil can be described electrically as an ideal inductance L in series with an Ohmic resistance R, and parallel to both a winding capacity C. By far the most important quantity is the inductance, it depends on the number of windings, the magnetic material in the coil, and the geometry of the coil. The resistance and the capacitance don´t have much influence and can be neglected. When the strings are moving, an AC voltage is induced in the coil. So the pick-up acts like an AC source with some attached electric components (Fig. 2).

The external load consists of resistance (the volume and tone potentiometer in the guitar, and any resistance to ground at the amplifier input) and capacitance (due to the capacitance between the hot lead and shield in the guitar cable). The cable capacitance is significant and must not be neglected. This arrangement of passive components forms a so-called second-order low-pass filter (Fig. 3).

Thus, like any other similar filter, it has a cut-off frequency fg; this is where the response is down 3 dB (which means half power). Above fg, the response rolls off at a 12 dB per octave rate, and far below fg, the attenuation is zero. There is no low frequency rolloff; however, a little bit below fg there is an electrical resonance between the inductance of the pickup coil and the capacitance of the guitar cable. This frequency, called fmax, exhibits an amplitude peak. The passive low-pass filter works as a voltage amplifier here (but doesn't amplify power because the output current becomes correspondingly low, as with a transformer). Fig. 4 shows the typical contour of a pickup's frequency response.

If you know the resonant frequency and height of the resonant peak, you know about 90 percent of a pickup's transfer characteristics; these two parameters are the key to the "secret" of a pickup's sound (some other effects cannot be described using this model, but their influence is less important).

What all this means is that overtones in the range around the resonant frequency are amplified, overtones above the resonant frequency are progressively reduced, and the fundamental vibration and the overtones far below the resonant frequency are reproduced without alteration.

[edit] How Resonance Affects Sound

The resonant frequency of most available pickups in combination with normal guitar cables lies between 2,000 and 5,000 Hz. This is the range where the human ear has its highest sensitivity. A quick subjective correlation of frequency to sound is that at 2,000 Hz the sound is warm and mellow, at 3,000 Hz brilliant or present, at 4,000 Hz piercing, and at 5,000 Hz or more brittle and thin. The sound also depends on the height of the peak, of course. A high peak produces a powerful, characteristic sound; a low peak produces a weaker sound, especially with solid body guitars that have no acoustic body resonance. The height of the peak of most available pickups ranges between 1 and 4 (0 to 12 dB), it is dependent on the magnetic material in the coil, on the external resistive load , and on the metal case (without casing it is higher; many guitarists prefer this).

The resonant frequency depends on both the inductance L (with most available pickups, between 1 and 10 Henries) and the capacitance C. C is the sum of the winding capacitance of the coil (usually about 80 - 200 pF) and the cable capacitance (about 300 - 1,000 pF). Since different guitar cables have different amounts of capacitance, it is clear that using different guitar cables with an unbuffered pickup will change the resonant frequency and hence the overall sound.

[edit] Altering Pickup Characteristics

Basically, there are 2 different ways to change a guitar's sound as it relates to pickups:

1. Change the coil configuration of the built in pickups. This is possible with nearly all humbucking pickups. Normally, both coils are switched in series. Switching them in parallel cuts the inductance to a quarter of the initial value, so the resonant frequency (all other factors being equal) will be twice as high. Using only one of the coils halves the inductance, so the resonant frequency will increase by the factor of the square root of 2 (approximately 1.4). In both cases, the sound will have more treble than before. Many humbucking pickups have four output wires - two for each coil - so different coil combinations can be tried without having to open the pickup. Some single coil pickups have a coil tap to provide a similar flexibility.

2. Change the external load. This method is inexpensive but can be very effective. With only a little expense for electronic components, the sound can be shaped within wide limits. Standard tone controls lower the resonant frequency by connecting a capacitor in parallel with the pickup (usually through a variable resistor to give some control over how much the capacitor affects the pickup). Therefore, one way to change the sound is to replace the standard tone control potentiometer with a rotary switch that connects different capacitors across the pickup (a recommended range is 470 pF to 10 nF). This will give you much more sound variation than a standard tone control (Fig. 5).

Yes, Helmut Lemme's article is good. I do not agree with your comment about lousy pickups. Guitarists have gotten pretty much want they want. Since the inductance varies faster than linear with removal of turns, it is possible to increase the frequency response quite a bit without loosing too much output. But that resonant peak is a very important part of the sound when combined with fairly simple tone controls in the amplifier. Electric guitar is a mid range instrument, and it is especially important to limit the highs when playing with a lot of distortion.

[edit] Editing Discussion

Why was the right parenthesis I added removed? Wilson 16:43, 25 February 2006 (UTC)

Oh. Apparently when I edit this article, the edit box cuts it off at that "a". Can somebody else place that right parenthesis in there? Wilson 16:46, 25 February 2006 (UTC)

Strange coincidence. I upgraded to the latest version of the Linkification extension and I wasn't able to edit the article, either. Not the same problem as you, is it? — Omegatron 19:49, 25 February 2006 (UTC)

Hmm. I also updated Linkification. Probably has to do with it. I'll look into it. Wilson 21:53, 25 February 2006 (UTC)

What a bizarre coicidence.  :-) — Omegatron 22:32, 25 February 2006 (UTC)

Why are you removing all the Wikilinks, 'O'?--Light current 17:51, 25 February 2006 (UTC)

You removed them, and I put them back. Pay more attention to what you're doing, please. It's not the first time you've removed or mangled content accidentally. — Omegatron 19:52, 25 February 2006 (UTC)

I didnt remove them, I replaced deleted text that someone lese had removed. But sorry for accusing you! Please forgive!--Light current 21:04, 25 February 2006 (UTC)

Forgiven. :-) — Omegatron 22:32, 25 February 2006 (UTC)

Thanks 'O'!--Light current 22:35, 25 February 2006 (UTC)

[edit] FET preamps with piezo pickups

Do you have any details on how FET amps do not distort as sharply as bipolar transistor amps? I have not heard of this phenomenon. Doesnt it also depend on the actual circuit design?--Light current 15:26, 17 March 2006 (UTC)

Yeah, distortion and linearity depend on the amplifier topology much more than the type of device used. — Omegatron 15:51, 31 March 2006 (UTC)

Oh! I added that! How irresponsible of me. I was thinking of something specific. I'll fix it. — Omegatron 15:56, 31 March 2006 (UTC)

[edit] Multi transducer pickups

Does anyone know if the John Birch 'Full-range' pickups are multi-transducer? John's keeping tight lipped about them, but I recall when I first found him selling them back in the mid 90s he was claiming they were perfect for MIDI and the only thing I could think of that would justify this claim was if each string had a separate mag.

Incidentally, when is someone going to do something about that blatent ad for Roland's MIDI gear? 'Most famous' my a***!! Deke42 12:39, 31 March 2006 (UTC)

May be "most popular" would be better? We can't ignore it, it is a fact, not POV. Check out google or any specialized sites. Roland guitar synths became somewhat standard de facto, keeping dominant market share. All other products keep 1-2% at most. --GreyCat 14:54, 31 March 2006 (UTC)

Firstly 'Multi-transducer' and 'MIDI' aren't the same thing, secondly 'Dominant market share' and 'fame' don't have an equals sign between them either, and thirdly (?) 'Guitar synth' does not automatically indicate 'Pickup'.

Roland have their market share because the bottom fell out of the MIDI guitar market and they became the only game in town, but I'd lay odds that even guitarists who persevere with MIDI and who own Roland Guitar-MIDI units aren't aware that (If?) they're using Roland pickups. Indeed, of the only two players I know who still have MIDI units, neither of them could tell me what the pickups were (One turned out to be Yamaha, the other was Casio, I don't think either of them are in the MIDI market any more).

Personally I think it's a pity 'The Ax' failed, but once it did MIDI guitar was dead in the water. But again, that's not fact, just an opinion. Now, about those John Birch pickups...Deke42 16:38, 31 March 2006 (UTC)

I'm pretty sure the AIX-101 is not a piezo pickup, it's a regular magnetic hexaphonic.

Another thought just struck me. Why have the multitransducers got their little advertising break? None of the other pickup paragraphs have them, and I didn't actually think they were allowed anyway. Deke42 16:42, 31 March 2006 (UTC)

Feel free to change it the way you see more acceptable :) I really have no idea how to really improve it. --GreyCat 16:41, 12 April 2006 (UTC)

[edit] Piezo pickups: pressure or acceleration?

Do piezo pickups measure the changing pressure between the strings and the body, or the vibration of the body itself (as pressure against the mass of the pickup)? The pictured one seems the latter, but built-in bridge piezos might measure the former. Does anybody have solid information on this? --Howdybob 08:56, 8 June 2006 (UTC)

I do not have solid info, but perhaps the following liquid may whet your appetite!I think you can do it either way!

For ex on my EUB, the piezo pu's are integrated into the string saddles on the bridge. Also, on some DBs, the piezo pu is squeezed into one of the little cutouts in the bridge. So these work mainly by pressure changes between string and body and give more of a raw unfiltered sound I would suppose.

On some guitars, a piezo pick up can just be stuck onto the body (inside or out), in which case it operates against its own mass only. One might conclude that this method would not give as much bass. Also the sound is going to have more of the characteristics of the gutar body than the strings.--Light current 12:01, 17 June 2006 (UTC)

[edit] Magnetic Reluctance

The idea of explaining magnetic pickups in terms of Magnetic Reluctance is fine technically, and allows for succint wording - but Magnetic Reluctance is quite a specialized concept. I think that it would be a lot clearer to the general reader to explain the operation in terms of (better known) basic physical concepts (Faraday's law etc - see, for example, my reverted changes). Alternatively, the page on Magnetic Reluctance could be expanded - at present it is also a little obscure. Also I think that it is important to mention that in the case of Magnetic Pickups, the strings need to contain a magnetic material.

Monsieurmagoo 11:27, 17 June 2006 (UTC)

Unfortunately your explanation was wrong. The strings do not get magnetised as such. see this lnk [2] There is no point in putting the worng explanation in just because its simpler. 8-( --Light current 11:42, 17 June 2006 (UTC)

My explanation was not wrong.

Firstly, the article that you cite does not say that the strings do not become magnetized. Secondly, there are many articles and online lecture notes which say that they do. See for example this article by Dr. Vassilis Lembessis in Europhysics News. In fact the Wikipedia article on electric guitars also makes this point. See also Journal of Applied Physics V 91 pp 7777, which explains that the strings need to be made from a soft magnetic material and gives an explanation for the pickup operation which is broadly similar to the one that one that I gave previously (which you reverted and seem to think is wrong).

The physics here is very straightforward and can be explained in a straightforward way. In my view invoking specialized concepts such as Magnetic Reluctance to explain such simple operation is poor style.

--Monsieurmagoo 12:02, 18 June 2006 (UTC)

Can you quote some (non WP) web refs that say the strings become magnetised and that this is necessary for the functioning of the pickup? Also as I dint have access to that vol JoAP, could you quote the relevant para her please ?8-|--Light current 16:50, 18 June 2006 (UTC)

I have never heard of the strings becoming permanently magnetized for correct function. The sound would then be changed if you twisted the string. It also seems like the waveform would be distorted due to the hysteresis of magnetization of the string. I thought it only required a conductive string for eddy currents, not a ferromagnetic string. — Omegatron 17:28, 18 June 2006 (UTC)

Nevermind. I'm reading around and everything says the string has to be ferromagnetic.

U.S. Patent 4,524,667  and [3] — Omegatron 17:59, 18 June 2006 (UTC)

Im sure you didnt intend to usurp my post 'O' so im moving it into chron order for you 8-). Now Ive had a quick look at that ref[4] And Im very worried 8-( that its not IMO correct. It says the wire is magnetised to work the pickup. Now its true if you did magnetise the string,(it would only be a weak magnet) it would indeed induce some emf in the coil. But thats not how mag pickups work. THey will work with completely demagnetised strings! Is this University of Delaware 'Kosher'? Idont think this chap 'knows his onions'!--Light current 18:22, 18 June 2006 (UTC)

The ferromagnetic string becomes magnetic in the presence of a magnetic field, just like a paper clip stuck to a magnet can pick up more paperclips. So it's a "temporary magnetism" as opposed to a "permanent magnet". Of course, the difference is subtle, and I think is always just a less strong version of the other, depending on magnetization and coercivity. — Omegatron 21:19, 18 June 2006 (UTC)

Well I dont really think thats a helpful way to think about things. For instance, when the string goes outside the influence of the permanent magnet, is the string magnetised or not? Its much easier to think of this as a variable reluctance circuit! Like varying a series resistor in a seies circuit and noting the voltage change across the load resistor. Do you see?--Light current 21:25, 18 June 2006 (UTC)

Easier for whom? Magnetic circuits are a very obscure concept. The typical description of a magnetic wire moving around near a coil is correct and understandable by the typical person. — Omegatron 00:54, 19 June 2006 (UTC)

Easier for everyone (and more correct). I dont agree that they are obscure. They should have been covered on all undergrad EE courses. They are quite simple really: completely analogous to a series circuit with a voltage source and resistors in. The voltage source represents the MMF, and the reluctances of the cct are represented by resistors. The magnetic flux equates to electric current and of course flux passing thro' a reluctance gives a drop in MMF. But you wouldnt say that the MMF difference across a reluctance is is source of MMF would you? No more would you say the pd across a resistor is a voltage source!

BTW since you have been interjecting your posts at seemingly random points, these posts do not now make logical sense when you read from top to bottom.--Light current 01:32, 19 June 2006 (UTC)

No it has to be ferromagnetic to affect the flux density in the airgap and thus cause the flux to change and cut the coils giving an induced voltage. But the string is 'soft' magnetic material. Thts why it doesnt remain magnetised! Unfortunately Im having trouble finding a ref for how mag pickups work. I know how they work, but I cant prove it by refs!--Light current 17:40, 18 June 2006 (UTC)

I have to say that the physics and experimental techniques this chap uses seem very dubious to me 8-(. I think we may need a third opinion from someone like User:Alfred Centauri!--Light current 18:34, 18 June 2006 (UTC)

Heres an extract from that patent in which the author recognises that the operation depends upon magnetic reluctance:

The elongated permanent magnet has a width which is less than the width of the elongated passageway in each coil defining an elongated cavity between the elongated permanent magnet and the exterior of each coil. A pair of magnetic pole pieces are disposed in each elongated cavity having an end piece adjacent the permanent magnet and enclosed by a portion of the coil and wherein the other opposite end face of one of the pair of magnetic pole pieces is adapted to be positioned in a spaced relationship to the ferromagnetic strings to define a magnetic reluctance path between each of the ferromagnetic strings to the permanent magnet wherein the reluctance path includes the predetermined air gap, the magnetic pole piece having a preselected permeability and a portion of the permanent magnet member.

--Light current 18:39, 18 June 2006 (UTC)

Brief (but correct) explanation of operation of mag pu's. [5]

[edit] Alfreds entry

Two things seem clear to me: (1) The guitar string material must have a high permeability in order to have any significant affect on the magnetic circuit and (2) this means the string will be magnetized in the presence of the pickup magnet. It occurs to me then that describing the operation of the guitar pickup using the magnetic circuit approach or the 'vibrating magnet near a coil' may be equivalent - just different perspectives. Alfred Centauri 02:00, 19 June 2006 (UTC)

What do you mean by the term 'magnetised'?--Light current 02:06, 19 June 2006 (UTC)

Magnetized in the sense that the magnetic domains of the steel in the guitar string align with the magnetic field of the magnet in the pickup while in the presence of that magnetic field. Alfred Centauri 02:25, 19 June 2006 (UTC)

Yeah OK. Just has some mag flux thro' it but not permanently magnetised tho' as in a magnet! FYI, the string cores are made from steel. The windings could be made from brass, nickel or who knows what, so I think the pickup effect must rely on the core. I dont know what the type of steel is from a magnetic pov.

Just one little rider, to tax your brain a little: Does a resistor passing current act as a source of emf? And does this have anything to do with the subject in question? 8-)--Light current 02:35, 19 June 2006 (UTC)

Resistors are linear; ferromagnets are not. — Omegatron 03:04, 19 June 2006 (UTC)

You are correct again 'O'. Ferromagnetic materials are not linear (especially near the origin!) and they also have Hysteresis. And you know what that means for magnetic pickups dont you! But lets assume they are linear(ish) for the purposes of this discussion. 8-)--Light current 03:33, 19 June 2006 (UTC)

A resistor, in a purely circuit context is a 'sink' for emf. Of course, a real resistor will have some inductance and so will be a source (but not a net source) for emf when there is a changing current through it. I'll have to think about whether this could have anything to do with this particular discussion. Hmmm.... Alfred Centauri 03:16, 19 June 2006 (UTC)

1. For further links which say that the strings become magnetized, see, for example: [6] and [7]

2. I assumed that it would be obvious that when I said "magnetized", I did not mean "permanently magnetized". Perhaps I should have made this clearer.

3. As I've said at the start of this discussion, I do think that the explanation in terms of variable magnetic reluctance is sound. I think that the concept of Magnetic Reluctance was developed to give a convenient analogy between magnetic and electrical circuits. This kind of analogy is clearly of benefit to engineers working on complex magnetic systems. However, the operation of the magnetic pickup is very simple, and can be explained in terms of more fundamental (and much more well known) concepts. So what I am saying is that the two explanations are essentially the same, but the one that I gave is simpler and closer to the fundamentals, and therefore in my view, more appropriate for an encyclopedia.

4. As regards the argument that Magnetic circuits are familiar to EE undergrads, while this may be true, I learned about magnetization and induction in school a couple of years before I went to college - I think that most people do. We're not just writing for electronic engineers here, I really think that the simpler explanation is more accessible.

5. I was asked to quote from the Applied Physics article which I referenced above, (Lenssen et al, Applied Physics V 91 pp 7777) so here it is: "Conventional pickup elements for guitars consist of a permanent magnet and a coil of isolated wire. The amount of flux through the coil is modulated by the vibration of the nearby, soft-magnetic string and therefore induces an alternating current in the coil."

6. Finally, I think that most of the non-patent articles that we have cited in this discussion have explained the operation of the pickup in a similar way to the way that I described it (including the peer reviewed Applied Physics article above). Does Light Current still believe that my explanation was wrong?

--Monsieurmagoo 11:19, 19 June 2006 (UTC)

""Conventional pickup elements for guitars consist of a permanent magnet and a coil of isolated wire. The amount of flux through the coil is modulated by the vibration of the nearby, soft-magnetic string and therefore induces an alternating current in the coil." Isn't this precisely the magnetic reluctance description of operation (without using the word reluctance)? Alfred Centauri 12:21, 19 June 2006 (UTC)

It doesn't mention the word reluctance because there is no need to introduce the concept to explain the pickup operation. Also, this description doesn't talk about altering "the reluctance of the magnetic path seen by the magnet" (which is what this page currently says). Finally it doesn't attempt to explain the operation in terms of analogy with an electrical circuit (which I think is the point of Magnetic Reluctance). Instead the emphasis is on "inducing an alternating current in the coil". The understanding here is clearly in terms of Faraday's law. --Monsieurmagoo 12:51, 19 June 2006 (UTC)

Regarding your first two points: What you say is true but that's not necessarily 'good'. In other words, the description of operation from the journal is good - as far as it goes. Sure, Faraday's law tells us that changing the magnetic flux linking a coil produces a voltage between the ends of the coil. However, the more curious reader of this article may ask "how does the vibrating string modulate the flux through the coil?" Faraday's law doesn't address this. Thus, Faraday's law alone cannot completely describe the operation of the pickup. Alfred Centauri 13:13, 19 June 2006 (UTC)

[edit] LCs retort

One of the difficulties I have with the 'strings become magnets' idea is why the pickup manufacturers have not found this out yet. After all, if this idea was correct, they could leave out the permanent magnets in their pickups and supply a small stand alone magnet with which to magnetise the strings by stroking etc. Im sure this would save the makers pots of money!--Light current 13:33, 19 June 2006 (UTC)

The strings would then need to be made out of a hard magnetic material, like neodymium, and if the string were twisted, it would change the sound, so the strings would have to have two marked sides and be installed in a certain orientation and constantly maintained, and do long, thin permanent magnets like to be arranged with the poles perpendicular to the axis of the magnet? I'm sure the manufacturers, who have been doing this every day for decades, know more about the function of the pickups than you do, and choose these designs for a good reason.

I think the description that a high schooler can understand is superior to the magnetic circuit/reluctance description that only engineers would understand. They are equivalent. — Omegatron 13:54, 19 June 2006 (UTC)

Well 'O' I must say Im disappointed in your apparent laxness on this matter! 8-( You, like MrMagoo are suggesting that we enter inaccurate information, just because its easier to understand 8-? A little thought will show you that if this idea was viable, the strings would not need to be made from a 'hard' magnetic material as the small amount of coercivity present in normal steels would perform the function in exactly the same way as they do in your explanation. ie the strings can only form a relatively weak 'magnet' due to the flux generated by the pu magnet if the permeability is high (B=uH). ie H is low for this 'magnet'. As regards the orientation of the string wrt to the pickup: you will recall that guitar strings vibrate in an elliptical manner and so some induction would still be present. It has yet to be shown that the two ideas are equivalent. I doubt it - but as always Im prepared to be proved wrong 8-|--Light current 14:40, 19 June 2006 (UTC)

Alfred: Because the strings are magnetized by the pickup magnet, they have their own magnetic field. That field varies in space. Because the strings are vibrating, the field and therefore the flux linking the nearby coil changes with time. And therefore an EMF is set up in the coil through Faraday's Law. The explanation given in the paper is complete, it is just succint and assumes some knowledge of the physics. So all you need is magnetized strings and Faraday's law to describe the operation completely. --Monsieurmagoo 13:42, 19 June 2006 (UTC)

The 'magnetization' of the string varies with its position relative to the magnet as do the magnetic field lines of the system composed of the magnet, string, and coil. Claiming that the magnetic field of the string is a separate entity ('they have their own magnetic field') that is solely responsible for the induced voltage in the coil implies that the magnet has its own magnetic field that doesn't change and thus doesn't participate in the induction of voltage in the coil. Is this really what you wish to claim? Alfred Centauri 18:52, 20 June 2006 (UTC)

I'm not an expert on Magnetism by any means but here is what I was thinking: The permanent magnet in the pickup is a hard ferromagnet, and therefore its magnetization is essentially independent of applied fields for moderate field strengths. We can therefore, to a first approximation, think in terms of the superposition principle. Thus it is reasonable to mentally separate out the fields from the string and from the magnet. From this perspective, since the magnet is fixed relative to the coil, its contribution to the flux is constant and so it does not contribute to the EMF. Thus, to a first approximation the only function of the magnet in the pickup is to magnetize the strings. Here are some references which support this: [8] (problem 2) and [9]

That said, to be completely comprehensive and general about the pickup operation, you would need to take into account higher order effects such as the possibility of additional induced eddy currents in the Magnet. Clearly their contribution to the EMF would be smaller than that from the magnetized strings, and may be negligible or may be made negligible by design (the calculation of this would probably take me much more time than I really should be spending on this..). I agree with the idea of Omegatron, viz that we should give the vibrating magnetized string description but say that it is an approximation and that a more general description is provided by variable Magnetic Reluctance. --Monsieurmagoo 00:09, 21 June 2006 (UTC)

[edit] Both descriptions?

Can we include both descriptions? The induced magnetism version is arguably correct, though misleading when simplified. The magnetic reluctance version is more correct, but much more difficult to understand. Let's use the simpler one first, and then say "in more detail..." and explain magnetic reluctance. Those who don't care to learn more will still understand the basic idea. — Omegatron 19:01, 20 June 2006 (UTC)

Seems like an entirely too reasonable suggestion to me. Alfred Centauri 21:18, 20 June 2006 (UTC)

I dont agree we include both descriptions until both descriptions are shown to be correct.--Light current 22:44, 20 June 2006 (UTC)

What, specifically do you think is incorrect about it? The permanent magnet induces a magnetic field in the ferromagnetic string, which changes as the string moves, but is roughly the same and in the same direction. The magnetic field of the string and permanent magnet are superimposed on each other. The moving magnetic string induces a current in the coil. The field of the permanent magnet doesn't induce a current in the coil since they are motionless relative to each other. — Omegatron 23:51, 20 June 2006 (UTC)

I don't believe superposition is valid here due to the non-linearity of the B-H characteristics of the magnet. However, what Monsieurmagoo appears to be suggesting is, like any good engineer, we linearize the system via the 'small-signal approximation'. Still, there's something 'fishy' about all this that I can't quite put my finger on. For example, what would happen if the magnet is removed from the pickup and placed on the opposite side of the string? I suppose, we would need to insert a core of some type in the pickup. But then, wouldn't the core become magnetized? Crap, I'm going in circles - time to sleep on this problem. Alfred Centauri 03:25, 21 June 2006 (UTC)

Alfred, maybe you ought to consider the magnetic circuit and its equivalent electric circuit and cast your mind back to my earlier question about emf and resistors? 8-)--Light current 11:44, 21 June 2006 (UTC)

The magnetic circuit is non-linear too, so the reluctance will not behave like a resistor. — Omegatron 12:24, 21 June 2006 (UTC)

THe reluctance of non ferromagnetic materials like air is linear because the BH curve is linear. Most of the reluctance in the magnetic circuit is provided by the path of flux thro' the air. This reluctance will behave as a resistor in the equ. cct. When the string moves within the flux generated by the field of the magnet, it alters the total reluctance of the magnetic path. Alteration of the total reluctance causes the magnitude of B to change (H provided by the permanent magnet is constant). Changes in B induce an emf in the coil! Simple!--Light current 13:30, 21 June 2006 (UTC)

[edit] My explanation of pickup operation.

The string concentrates the flux lines through itself because of its high permability cf the surrounding air). When the string is sitting directly above the pole piece centre line, the high permability of the string lowers the total effective reluctance of the magnetic circuit (cf what it would be with no string). When the string is displaced to either side of the pole piece centre line (or further away in a vertical direction), it is easily seen that the path length the flux has to travel to reach the low reluctance string has been increased slightly. This increased path length neccesarily increases the total reluctance of the magnetic circuit slightly. The small change in the total flux in the magnetic circuit thus caused is converted to an emf by the massive number of turns of wire in the pickup. --Light current 14:02, 21 June 2006 (UTC)

[edit] Magnetic Pickups - Towards an improved wording

On the basis of our discussions and the references we have gathered (in particular the Applied Physics paper), can I propose the following revised wording:

“A magnetic pickup comprises of a permanent magnet and a coil of wire. The vibration of the nearby soft-magnetic strings modulates the magnetic flux through the coil and therefore induces an alternating current through Faraday’s law. The signal created is then carried to amplification or recording equipment via a cable (or by radio transmission). More generally, the pickup operation can be described by the principle of variable magnetic reluctance.”

I think that the second sentence is fact, as is established by many references. I accept that there is still some debate as to whether this sentence describes the operation completely, given our discussions over the extent of secondary contributions to the induced EMF from the Magnet. Nonetheless it provides a much clearer basic understanding for the general reader and if someone is interested in the details they have access to the more general description.

Note that this description (unlike the current description) mentions that the strings are made from a soft magnetic material, so it provides a technical improvement as well an improvement in clarity. Monsieurmagoo 15:55, 21 June 2006 (UTC)

I would suggest some modifications ie:

“A magnetic pickup comprises a permanent magnet and a coil of wire wrapped around it. The vibration of the nearby soft-magnetic strings modulates the magnetic flux linking the coil and therefore induces an alternating emf through Faraday’s law. The signal created is then carried to amplification or recording equipment via a cable (or by radio transmission). More generally, the pickup operation can be described by the principle of Magnetic reluctance.”

my mods are in bold. Yes, this now seems to describe a variable reluctance circuit reasonably well!So I :suppose I will reluctantly ;-) have to agree with it! 8-)--Light current 16:18, 21 June 2006 (UTC)

8-) Cool. --Monsieurmagoo 16:31, 21 June 2006 (UTC)

If you hadn't noticed, I updated the actual article a few minutes ago and added my desired changes. Since we all agree on the concept, we can work on the exact wording just by editing it. — Omegatron 17:38, 21 June 2006 (UTC)

Oops. We forgot to include references in the artile. — Omegatron 02:57, 23 June 2006 (UTC)

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Here is something to consider in improving the definition of magnetic pickup.

Only about half the guitar pickups in existence actually have wire wrapped around permanent magnets. In the others, pole pieces (usually soft steel) are magnetized (not permanently) by magnets behind the coil. Then these cores magnetize the strings (again, not permanently).

Also, I think this type of approach is more helpful:

A magnetic pickup for a steel stringed instrument consists of two things.

1. A coil of wire with a ferromagnetic core. This coil is sensitive to the total time varying magnetic field. The sensitivity depends on many things, but two important ones are the permeability of the coil, that is, its ability to amplify the magnetic field, and the number of turns.

2. A method for magnetizing the parts of the strings near the core of the coil.

Then one can proceed to describe how it works. Thus we can view those pickups that use permanent magnets as the cores as cleverly using the magnet to perform both functions, and, in practice, with some limitations to how well both are performed. It also is apparent why the pickup is sensitive to other magnetic fields in the environment, that is magnetic hum and buzz. Mike Sulzer 17:53, 31 October 2006 (UTC)

[edit] Interesting quote from book.

Most broadcasting and recording studios as well as many high fidelity systems use variable reluctance pickup units. THese posess a fixed magnetic field produced by a permanent magnet. The stylus is mechanically attached to an iron vane or armature located in an air gap between the poles of the magnet. Movement of the armature, created by the stylus, changes the reluctance of the magnetic path and creates a changing flux through the magnet. A coil of wire is wrapped around the magnet and the changing flux in the system generates a varying voltage across the coil terminals. Electrical output from such a pickup is on the order of 10-20mV

Julian L. Bernstein (Associate Dean of the DaySchool , RCA Institutes INc), Audio Systems, publ John Wiley and Sons, 1966, Lib of Congress No:66-17326

If you were to replace the words 'stylus mechanically attached to iron vane or armature' by the words 'guitar string' and the word 'armature' by 'string', you have a description of the guitar pick up, as I initially said! Voila! 8-)--Light current 17:34, 21 June 2006 (UTC)

Ahh well, thats another interesting discussion over with!--Light current 17:43, 21 June 2006 (UTC)

[edit] Magnectic circuits and grounding the magnet base

Hello, folks. Sorry for starting this new topic, but I didn't want my question to get involved on the discussion about wether is "best" to see the pickup effect in one way or another. I'm actually "pro-circuit" AND "pro-Faraday" myself. Both views are needed to understand the phenomenon, and they are the same thing... And the pickup is actually a difficult thing to understand...

Well, it was thinking about this magnectic circuit model that the question struck me: how actually is the circuit?

There is the magnet section, with low reluctance, and and air section, with high air reluctance, and then a ferromagnectic string section, again with low reluctance... But then how does the circuit closes? Trough air and wood, or can we ground the base of the magnet to the strings, creating a low-reluctance return path?

I've read about placing a metal plate on the magnet base and grounding it, but just for the purpose of shielding the pickup. Doesn't it also affects the intensity of the induction, making the magnectic reluctance much smaller, and changing much more?? What happens if we do this? (I'm still afraid to open up my beautiful guitar... :> )

I'm studying the electric guitar in my EE masters, and I would like very much making contact with other people who have studied the subject, and I would also appreciate some references about this!... Please, get in touch!...

Thank you all... -- NIC1138 22:56, 31 August 2006 (UTC)

[edit] Output of a Pickup websurffATgmaildotcom

It is well known that a plucked string has different maximum displacement as time goes by.

So the pickup's output changes with time ( follows string's ADSR curve) .

Hence we need to clear up how pickup output mV was measured and specified in the main article.

[edit] HSS vs SSH

Well, I'm not sure on the arguments for SSH, but I do know that Fender and Squier uses HSS. The same goes on SOFT (a music forum) and another guitar forum. If we take this argument qualitatively, it should be HSS, given that this is a kind of "official" designation, and has also drifted into popular use.

In any case, I think this little dispute needs to be resolved ASAP. Ariedartin JECJY ^Talk 10:37, 24 October 2006 (UTC)

[edit] Position or type?

Let's say you've got a guitar with a single coil next to the bridge and then a humbucker half way between the single coil and the neck. Will the position or the type of pickup make more of a difference in the tone? Gopherbassist 18:01, 9 December 2006 (UTC)

The type of pickup will most likely make the most difference. There is, however, a difference in sound due to position. The closer to the bridge your pickup is the more treble response you'll get (that is a generalization, but it gets the point across).

[edit] Possible Error with Humbucker Description

Dear Wiki users, I'm currently reading "Guitar Electronics: A Workbook" by Donald Brosnac (Brosnac, Donald. Guitar Electronics: a workbook. Ojai, Calif.: d.B. Music Co., 1980.). On Page 21, Brosnac discusses the operation of the humbucker pickup in terms of its parts, and I noticed that the Wikipedia article and the book do not agree completely on how the humbucker operates.

The possible error might be in the statement that BOTH the coils AND magnet are out of phase with each other. This book does not mention the polarity reversal of the magnets; rather, the reverse polarity of the coils affects the magnetic fields of the magnets. Thus, the magnets do not need to be inversed - the coils take care of that. Finally, the magnetic induction from the string movement are automatically flipped to be in phase with each other when the induced currents are created in the coil (Brosnac 21). Cmsalmeron 01:55, 20 February 2007 (UTC)cmsalmeron

Check my edit. Danilo Roascio 23:17, 1 May 2007 (UTC)

[edit] Motherbucker

I removed the section on Motherbuckers because it seemed out-of-place in the section on pickup construction and looked like a possible product advertisement to me. Does anyone feel that this is a radically different pickup design that deserves special mention in this article? Steve Carlson^Talk 19:48, 17 March 2008 (UTC)

[edit] Single Coil vs. Humbucker

I wondered whether anyone had found audio clips that would provide an example of the difference in tone/sound of single coil pickups compared to humbuckers? If anyone does I would like to know. Nowiky (talk) 00:28, 7 April 2008 (UTC)