Pickup theory question

[big_teee]

Have guitars to build and pickups to wind. IRL

Are you building anything interesting?
Sounds like Fun!
Peace,
Terry

[David Schwab]

No, turns does not equal output; it is one factor.

We are talking about the difference between magnets, so we must assume the original question implies that all other factors on this pickup are the same. Just swap out the magnet.

Magnetic field strength at the string is another. A stronger magnetic field magnetizes the string more strongly and results in a higher output.

Try it a few times. You hear a different tone, but not a significant change in output. I've done this on a pickup where I swapped the magnet for an alnico V, a ceramic 8, and a neo. The ceramic and neo were a little louder than the alnico, but the neo was not louder than the ceramic. The ceramic was thinner sounding however.

So I find stronger magnets do give more output, but the tonal change is greater than the change in output.

This holds up to saturation, and it would take very strong magnet indeed to saturate the cores or the string. For example, using large neodymium magnets on the cores results in very high output, but terrible string pull.

That's not my experience at all.

It is only the field strength at the string that matters.

How do we know that? We can swap an alnico for an equal strength ceramic and they don't sound the same. I can also swap a ceramic for an equal strength* neo and they don't sound the same either. So there seems to be more to it than that.

(* measured at the pole)

It takes fewer turns of thinner wire to get a given resistance. You will also get a lower output and inductance. The resonance frequency of the whole circuit moves up and this usually results in a brighter sound.

Counting turns is better.

[Mike Sulzer]

David,

If we start off with a very weak magnet, we will have a very low output. Do you agree with this? As we increase the strength of the magnet, the output voltage increases linearly with the strength of the magnetic field at the string. If you continue to increase the strength of the field but observe less than a linear increase in the output, it is necessary to find out why. For example, are you really measuring the output of the pickup? Or, for example, are you listening to an amp with one or more stages overdriven? That would give the effect you describe, but there might be a lot of other things that could, too.

Strong neos in a humbucker: The test I am describing uses slugs in both coils with a 3/8" dia., 1/4" thick neo disk on the back of each slug. It is very difficult to assemble, but you can do it. This gives the effects I describe.

[David Schwab]

Mike, I make bass pickups. I plug the bass directly into the hi-z input on a digital mixer with no effects or gain stages or over drive. I can tell what the output is by looking at the meters, and also when I record samples into an audio editor. If I'm testing guitar pickups, I start off the same way.

I have been working on some guitar pickup prototypes, and one of them, a dual blade humbucker in the bridge position, seemed too harsh. So I swapped out the ceramic magnet for an alnico II. The pickup is smoother sounding now, but has about the same output. It does have less top end and attack however.

Now on testing neos, I'm taking about a neo bar magnet charging steel blades or poles. I took a dual blade bass pickup, and was able to switch magnets between an alnico 5, a C8 and an N42 neo. You don't want to use a big neo without having steel blades or poles. Measured directly at the blades I'm getting about 200G. I can't directly measure the magnet as it pouts my meter out of range. A Strat pickup with alnico rods reads about 600G at each pole. Is the Strat pickup louder? No.

So if you go back to the original over wound humbuckers, like a DiMarzio Super Distortion, what makes them loud? They are over wound. You can remove the ceramic magnet and replace it with an alnico, and then you have something like a Duncan JB, and it's still loud, just darker sounding.

Alembic used to sell these kits where you stuck a big ceramic magnet under your Strat pickups. (Rick, was that your idea?) Did it make them louder? Sure, but it also made them sound different. It wasn't as big a change in output as a DiMarzio FS-1, which is wound to 14K.

Try some magnets in a conventional pickup design.

[Mike Sulzer]

So how do you explain this? How can you do the best job of designing pickups if you do not understand why this is so? I am assuming that you agree that there must be a linear increase if you start off with very weak magnets. What is limiting the output level and changing the tone?

[Jeff Callahan]

No, turns does not equal output; it is one factor. Magnetic field strength at the string is another. A stronger magnetic field magnetizes the string more strongly and results in a higher output. This holds up to saturation, and it would take very strong magnet indeed to saturate the cores or the string. For example, using large neodymium magnets on the cores results in very high output, but terrible string pull.

How does changing the type of Alnico change the field geometry significantly? This is a subject for simulation with FEMM. But even if it did, so what? It is only the field strength at the string that matters.

I always thought of it as the strings vibration disrupting the magnet's field, then the magnet's fluctuating field induces current in the wire. Is this fundamentally wrong?

Thanks,
Jeff

[Mike Sulzer]

I always thought of it as the strings vibration disrupting the magnet's field, then the magnet's fluctuating field induces current in the wire. Is this fundamentally wrong?

Thanks,
Jeff

Many different viewpoints are correct since they predict what happens and are logically consistent. However, it is often most useful to stick to the actual physical principles. What does one mean by "disrupting the magneitc field"? Currents produce a magnetic field, but we do not have significant macroscopic currents in the string in a normal pickup. The currents that count are microscopic ones: each atom can be thought of as a small loop of current. These currents are always there, but in the absence of an applied field they are randomly oriented. Only when they tend to line up do they make a significant magnetic effect. The static field from the pickup partially lines them up, so they make a magnetic field. Fields add, and so you can look at this as disrupting the total field if you want. However, it is a time-varying field that induces a voltage in the coil. When the string vibrates it is the part of the total field due to the lining up of the currents (magnetization) that is time varying. Thus you can ignore the large static field from the pickup and just consider the small time varying part from the vibrating string. There are a lot of things you can figure out about pickups intuitively if you use this viewpoint.

For example, is the pickup output linearly related to the string motion? No, or at least not quite. This is because: 1. For a constant level of string magnetization it would be very close to linear. 2. But as the string moves farther away from the pole piece it becomes less magnetized because the static field falls off with distance. So the nearly linear relationship is modified.

[David Schwab]

Is this fundamentally wrong?

Nope. That's the standard variable reluctance transducer model.

The permanent magnet's field is always there, but is static. If you don't disturb it, no current flows through the coil. You can either move the coil through the lines of flux, or move some other magnetic object through the field. It doesn't matter if you have strings or not, because you will always have the magnetic field around the pickup. The strings become magnetized in the field, but the field was already there.

Pickup (music technology) - Wikipedia, the free encyclopedia

The vibration of the nearby soft-magnetic strings modulates the magnetic flux linking the coil, thereby inducing an alternating current through the coil of wire.
...
More generally, the pickup operation can be described using the concept of a magnetic circuit, in which the motion of the string varies the magnetic reluctance in the circuit created by the permanent magnet.

Another definition:

magnetic transducer:

McGraw-Hill Science & Technology Dictionary:
magnetic transducer

(electromagnetism) A device for transforming mechanical into electrical energy, which consists of a magnetic field including a variable-reluctance path and a coil surrounding all or a part of this path, so that variation in reluctance leads to a variation in the magnetic flux through the coil and a corresponding induced emf (electromotive force).

[David Schwab]

Currents produce a magnetic field, but we do not have significant macroscopic currents in the string in a normal pickup.

But in this case we have the energy coming from the permeant magnet. We have no current flowing in the coil without disrupting the magnet's field. While that current produces its own magnetic field, that's not a major part of how a pickup works, and in fact probably works against it, as do eddy currents.

[FunkyKikuchiyo]

No energy ever comes from a passive pickup. Magnets don't have energy, they just have polarity. The energy comes from the string which is why it is almost completely pointless to talk about "output". As I said earlier, it is a relative, emotional term. I don't see how it could be scientifically justified that having a stronger magnetic field wouldn't increase the efficiency of a pickup, provided that the core is up to the task (as is the case with most pickups). As the field gets VERY weak then the signal does weaken and eventually disappears. I don't see how it could be simply binary; saying either there is a magnetic field or there isn't. Chances are it isn't a linear scale. More magnetic field means more eddy currents (yes, I said it) and such, so it becomes more open to inefficiencies.

Move a pickup farther away, the signal decreases. You've effectively decreased the magnet strength that way, too. So, if you use a stronger magnet and adjust it lower, you have a net change of zero in terms of "output", though the pickup itself can be said to be much stronger. If you decide that pickups with different magnets need to be at different distances from the strings, then you've destroyed your ability to make empirical statements about power. If one major variable is wide open to your interpretation, then it just isn't empirical. I'm not saying that it isn't useful to think about output, but again... I strongly think it needs to be considered an emotional, intuitive trait, like "warmth", "jangly" or "sweet".

[bbsailor]

Many different viewpoints are correct since they predict what happens and are logically consistent. However, it is often most useful to stick to the actual physical principles. What does one mean by "disrupting the magneitc field"? Currents produce a magnetic field, but we do not have significant macroscopic currents in the string in a normal pickup. The currents that count are microscopic ones: each atom can be thought of as a small loop of current. These currents are always there, but in the absence of an applied field they are randomly oriented. Only when they tend to line up do they make a significant magnetic effect. The static field from the pickup partially lines them up, so they make a magnetic field. Fields add, and so you can look at this as disrupting the total field if you want. However, it is a time-varying field that induces a voltage in the coil. When the string vibrates it is the part of the total field due to the lining up of the currents (magnetization) that is time varying. Thus you can ignore the large static field from the pickup and just consider the small time varying part from the vibrating string. There are a lot of things you can figure out about pickups intuitively if you use this viewpoint.

For example, is the pickup output linearly related to the string motion? No, or at least not quite. This is because: 1. For a constant level of string magnetization it would be very close to linear. 2. But as the string moves farther away from the pole piece it becomes less magnetized because the static field falls off with distance. So the nearly linear relationship is modified.

Mike, David and all

This discussion relates to the diffrences between the art, science and engineering of guitar pickups.

The art of designing a pickup relates to "trial and error" while seeking to find a combination of parts that sounds good. These parts are the (1) magnet types; (2), magnet strength; (3) number of winds, (4) coil shape being tall and narrow or short and wide; (5) amount of ferrous metal near the coil and magnetic path and (6) the interaction of all of the above with the pot loading, coax loading and amplifier input impedance loading.

The science of pickup design seeks to explain general concepts of physics related to having moving strings generate a voltage in a coil while interupting a magnetic field. The science also seeks to explain how all of the above six parts interact and uses the laws of physics to define or help explain the interaction and relationships of those parts.

The engineering of pickups uses knowledge of the science to replicate the art to obtain a repeatable product that satisfys the ear to produce a desireable sound. The engineering also seeks to measure critical pickup parameters (inductance, resistance, impedance, wire size, insulation thickness, and loading) to achieve a consistent replicated product.

A guitar pickup is a generator and a tuned circuit that changes when any of the parts mentioned above are altered. Increasing the turns on a pickup will generally increase the output but other things happen when turns are added. The added wire is in a weaker magnetic field so the extra turns do not increase the output as much as if they were closer to the magnet. The the output increases but not in linear way. Also, the extra turns add resistance and capacitance. Additional turns also lower the coil resonant frequency. Coils that have more turns have a higher impedance and are more affected by the pot value, coax capacitance and amplifier input impedance loading.

Over time with enough pickups having been made, several standard designs (form factors) have evolved. These include the (1) standard Strat single coil in various positions; (2) the standard Tele single coil; (3) standard humbucker and (4) hot humbucker (5) P90 low and wide. Most other pickups are variations of these and/or reference these pickups. The only new design to evolve is the Lace Alumitone but these still reference the older physical form factors.

With enough people skilled in the art, science and engineering of guitar pickups, several rules of thumb have evolved. I'll kick it off and offer one that I find useful.
1. Make total pickup loading 35 to 40 times the DC resistance of the pickup coil.

Hopefully, others can offer other useful rules of thumb.

Happy Holidays

Joseph Rogowski

[Jeff Callahan]

What does one mean by "disrupting the magneitc field"?

In Milan's book, "Pickups, Windings and Magnets..." he states, "The ocillations of the string disturb the the magnetic field which induces the coil to produce a feeble electric current. The voltage of this current is proportional to the number of turns in the coil and can be about 60mV in most single coil pickups and up to 500 mV in the most powerful humbucking models."

Back to the original question, if one places identical pickups (other than the magnets) in an identical spot on the guitar, (i.e. same distance from the stings, and bridge), would the weaker magnet produce a weaker current? If so, how much more would one have to wind on the weaker magnet to match the voltage? If an A2 bobbin creates say 30mV at a given setup and an A5 creates 60mV at the same setup, could you wind the the A2 x% more to get closer to the 60mV level?

After all the responses, I realize this not too important of a question, so I'll shut up after this.

Thanks,
Jeff

[David Schwab]

No energy ever comes from a passive pickup. Magnets don't have energy, they just have polarity.

So does a battery! Magnets do indeed generate energy from the atomic spin.

So define polarity.

The energy comes from the string which is why it is almost completely pointless to talk about "output".

Really? So remove the magnet from the pickup. Does the string produce any energy or current in the coil now? No.

More magnetic field means more eddy currents (yes, I said it) and such, so it becomes more open to inefficiencies.

The magnetic field only creates eddy currents in the presence of a conductor such as a pickup cover. The coil wire is probably too thin to worry about, but then the current in the coil produces its own magnetic field.

Move a pickup farther away, the signal decreases.

No surprise there. Same thing happens with a microphone.

[David Schwab]

In Milan's book, "Pickups, Windings and Magnets..." he states, "The ocillations of the string disturb the the magnetic field which induces the coil to produce a feeble electric current. The voltage of this current is proportional to the number of turns in the coil and can be about 60mV in most single coil pickups and up to 500 mV in the most powerful humbucking models."

Mike doesn't recognize the variable reluctance model of pickups.

Back to the original question, if one places identical pickups (other than the magnets) in an identical spot on the guitar, (i.e. same distance from the stings, and bridge), would the weaker magnet produce a weaker current? If so, how much more would one have to wind on the weaker magnet to match the voltage? If an A2 bobbin creates say 30mV at a given setup and an A5 creates 60mV at the same setup, could you wind the the A2 x% more to get closer to the 60mV level?

You can wind more wire on, but now you are increasing the inductance and lowering the resonant peak. It will be louder, but will also sound darker.

So, why use an A2 in a pickup instead of an A5? Generally it's for a softer or warmer tone, with less high end. If you wind more wire, you might get too muddy, and then you'll need a stringer magnet. Or it might be just what you are looking for. But you can't wind to a certain output, because that depends on factors like the strings used, and how hard someone picks, etc.

[Mike Sulzer]

Magnets do indeed generate energy from the atomic spin.

No. The energy dissipated in the load on the pickup and in the pickup coil comes from the vibrating string. The permanent magnetic field does not contribute energy to the process.