Mike Sulzer

I did this test: I had available this bobbin with pole pieces: GuitarUSA strat type bobbin with high permeability (non-permanent magnet) ferrite pole pieces, 7/8 inches long, flush with the top of the bobbin, and sticking out the bottom. There was about 3/8 inches of pole piece below the bottom of the coil. The test is to position this over the pole pieces of a pickup in the guitar. Thus the string is already magnetized. Do this with top or bottom facing the strings and compare the voltage. I accepted the inaccuracy of picking and casual positioning, used the #1 string open, and looked at the second harmonic on the FFT analyzer. I got about three times the voltage with the top of the bobbin (pole pieces flush with the top) facing the strings than with the bottom of the bobbin (with the pole pieces extending from the bobbin) facing the strings.

Mike, do you have the full Jackson ref? If you could do a Stokes-Maxwell tutorial, that would be great and probably educational for all the people following this discussion (the "Stoke's" that I have in my head is a sphere sinking through a fluid). Sure, the first part is easy: Jackson, John David, Classical Electrodynamics, Wiley (1999, 3rd Edition) ISBN 0-471-30932-X. This classic on the subject is not easy. I am sure there is something as good for magnetic induction, but easier to read and more up to date. I am not familiar with what it might be, but a look at some E&M 101 type course at your favorite university would probably give a title. Let me think about the second part. Mike

"Here is where I disagree; in order for the changing magnetic flux to generate current in the coil, the relevant surface is the surface of the wire itself that makes up the coil." No, it certainly is not. See pages 208-209 of the latest version of Jackson, or apply Stoke's theorem to the differential version of the relevant one of Maxwell's equations.

I do not think we are in agreement yet. For flux, think of dividing the surface inside the loop into small patches. Multiply the area of each patch by the strength of the component of the field pointing perpendicular to the area and add them all up. If the field is changing through the area; the largest contribution to the changing flux is where the field is changing the most. If we move the string magnet closer to or further from the pickup, the relevant component of the field strength changes most through the pole piece. So it makes the most contribution. Everywhere on the surface counts; the major contribution comes from where the field changes most rapidly, and that is generally where the field is strongest. Your "at home" experiment is not completely specified until you describe the field. If you specify a field that does not change in space where the loop moves, then you get no voltage no matter how you move the loop. This is because the total flux through the loop does not change. If the field does change significantly with location, then you do get a voltage when the loop moves because the total flux changes. (integral of B(dot)da is different at diferent times) "Here's the key...wherever those moving contours cross the turns of the coil, you have a changing magnetic flux across a conductor, and you'll make current." How can you predict a numerical value from that? It does not work. You must look at the change in the integral of B(dot)da. You can have contours moving across conductors and get nothing; the effects on different parts of the surface can cancel.

Suppose you have one pickup switched on and its volume and tone all the way up. These pots reduce the brightness of the sound a bit. If you add a master volume control (assume it is all the way up), it decreases the brightness a bit more. You have to be the judge of whether it is a big effect or not. Things are more complicated with both pickups switched on, or with varioius combinations of levels of the volume controls, but in general the effect of the mv is to lose some brighntess. The higher the value value of the mv, the less loss of brightness when it is all the way up, but the more it loses treble when it is in an intermediate position. (Cable capacitance effect) But that is just like any other pot. By the way, where do you get those bypass pots?

Let's go back and look at the law of induction, or the relevant one of Maxwell's equations. 1. We have a magnetic field that is changing in time. 2. We draw a closed path somewhere in the space the field occupies. 3. There is, in general, is a voltage around this path. 4. We compute this voltage by defining a surface; the loop lies on this surface. 5. The part of the surface inside the loop is what counts. 6. We get the voltage by adding up the effects of the time-varying field at each point on the surface. 7. It is the field component perpendicular to the surface at each point that counts. To apply this to a coil: 1. Define a loop as above coincident with a turn; do for each turn of the coil. 2. The total voltage across the coil is the sum of the voltages around all the turns. (They are in series.) 3. The simplest surface to use for each turn is the plane it lies in if the turn come back on itself. If the turn meanders across the coil, this harder to think about, but it still works. 4. So we have to add up the voltages determined by the fields passing through the surfaces defined for each turn. Now here's how I would apply this to a coil wound around the pole piece: we have to look at the field passing through the entire turn. (The donut has no physical significance.) Whether a particular turn is wound close to the core, or further away, the plot shows that the field that is large and perpendicular to the turn is in the pole piece. There is very little contribution outside the pole piece because: 1. The field is small. 2. It is mostly parallel to the turn. We have a difference of view here, but please consider that the analysis above is based on interpretating the general equation for this specific case.

To generate current in the coil, you need magnetic flux through the turns of the coil itself, i.e. outside the rod magnet (not inside). Just outside the rod, the field lines are so much more dense at the top that I can't help but think that the output at the top will be stronger than the output at the bottom. Try making the same plot, but make the rod Alnico5 at ~1800 G, I'd be interested to see what that looks like. OK, sorry, there is a real problem with that plot. There are two objects; the one on top is a neo magnet; the one below was supposed to be a soft steel pole piece. Somehow it got changed to neo. There is a new plot below. It shows more change from top to bottom, and so I agree, it is indicating more than the 2% that Peter measured. I believe that it is just the field in the pole piece that counts. The coil is wound right around the pole piece, and the only significant vertical component is inside. (It is the whole area inside a turn that counts, and only the component perpendicular to the turn.) http://www.naic.edu/~sulzer/bigSource2.png

That is exactly what should happen. If it is wired according to the EMG diagram for the 81, the slider goes to the output jack and off to your 1 megohm guitar amp input. Only if the amp impedance were much lower would the taper be affected.

OK on the 1/d^3 dependency...but still the field falls off by quite a lot. If I zoom in and try to read the intensity from the colors, at a position of 1 rod radius outside the ends of the poles, I get these fields: top of pole = 1.52e-4 (hard because of the concentration of field lines, but not far off) bottom of pole = 2.768e-5 That's a difference of a factor of 5.5...while Peter is measuring 2% difference. That is true, but the interesting thing is that the fields inside the pole piece at the two ends appear to differ by less. Those are the fields that matter since the coil encloses them.

I think that shield is a high permeability material like permalloy, not steel. To get effective shielding into a small space, it is necessary to use special materials.

No, you want to use the same number of turns, not wind for the same resistance. The output is not proportional to the resistance, it is proportional to the number of turns. If everyone uses #42, then you can use the resistance as an alias for the number of turns. You will need to wind for a higher resistance to get enough turns.

Then there is something funny going on here. If the slider of the pot goes to the output jack which then runs to a standard high impedance (1 Meg ohm) load in the amp how could the pot behave in that way? On at least one model (RG-20) the slider appears to run back into the pickup. It could be internally loaded there, causing funny taper effects with high resistance pots. Other than that, it is hard to see how the taper could be affected.

The number of turns would have a very small effect. I suspect that he has tried the coil experiment. The pancake thing does not sound right. A standard humbucker consists of two non-pancake coils, and it has high output.

True that...but outside the pole (where the coil is) the source itself still falls off as 1/d^3, and so should the test perturbation. That's why I'm surprised at Peter's result. No, the field only falls of as (1/d)^3 far from the source. Peter's result is in agreement with this simulation, with the limitaion that we do not really know how big his source is.

The question regarding Peter's test is what happens when the source is large like a dynamic driver, rather than small like a string. Here is a simulation like the one above (with the weaker coercitivity), but with a large magnet instead of a string. Note that the field at both ends of the pole piece is very similar instead of there being a fall off from one end to the other. http://www.naic.edu/~sulzer/bigSource.png

Pots load passive pickups and so 250 vs 500K makes a diference for the treble. An active pickup has a preamp that isolates the pickup from the pots and cable. Using a 25K pots means that the cable no longer cuts treble when the pot is part way down.

Why would it act like a switch? I might be missing something, but the EMG 81 circuit has the slider of the pot going to the output jack just like with a passive pickup. If you use a large value pot you would have the same disadvantage, that is, loss of treble at intermediate volumes. With a 25K pot there will be no treble loss.

"Peter, is your headphone source shooting down right over the top of the pole? If the headphone is oriented so that its magnet has a N-S orientation parallel to the pole piece like the disc in Mike's FEMM diagrams, i.e. either N or S is pointing at the top of the pole, then the field should still fall off with distance, and I don't understand how you get the results you get." I will have some time to respond in more detail later, but the answer is probably the size of the headpone driver coil/polepiece. The (1/r)^3 fall off is not observed close up, where the relevant scale size is the size of the source.

Bass strings are difficult to pull on, and so the obvious problems of using neo that b_l referred to do not apply so much as with a guitar. Those magnets are too big and too strong for guitar; I have never tried with a bass. There is another issue that is rarely mentioned. The pole pieces magnetize the string, but also they serve to amplify the the fluctuating field from the vibrating string. Neo has a relative permeability of 1.04, which means it does very little amplification. Alnico or steel have higher permeability, and so even though the field is weaker, you still get good signal. I do not see any benefit to using neo as the pole pieces rather than neo on the bottom (or even very small neo on the top) of a pole piece with higher permeability.

"Most people know where it is, and do not have to pretend it goes into the string and somehow produces a primary magnetic field stronger than the pole piece." No, I am saying it is much weaker than the field of the pole piece. Where did you get the idea that I thought it was stronger? Inside the pole piece the field from the string is far wekaer than the field of the pole piece. That is why we need to get rid of it to easily see the pattern of the field from the string. But you agree with the interpretation of the law of induction in my previous post?

Mike, No. Read again - Erik is the one that claimed the string was the conductor. In a pickup, the coil would be the conductor. The string just disturbs the magnetic field of the pole piece. By subtracting out the magnetic field of the pole piece there would be no magnetic field to cut the wires of the coil and generate a voltage. Sure, you can do it with a model, but it's ignoring reality. Look at it this way - you are eliminating the cause and claiming that the effect still exists independently without it. Joe, you are ignoring the physics. The field of the pole piece induces magnetization in the part of the string that is over it. This magnetization produces a magnetic field. When this field changes from the motion of the string, it induces a voltage in the coil. This is one way of analyzing how the original field is disturbed. There are some limitations to the concept of "cut the magnetic field lines, get a voltage". Suppose we have a loop of wire with a volt meter in it. Suppose we put this loop in a magnetic field pointing through the loop. Assume that the field is constant in space. No matter how you move the loop there is no voltage induced around it as long as the orientation of the loop remains constant. This is because the magnetic flux remains the same. If you rotate the loop about an axis perpendicular to the field, you do get a voltage. In this case the magnetic flux through the loop changes. Suppose the field is not constrant in space. Now if you move the loop, even keping the orientation the same, you do get a voltage because the flux changes. This is a direct consequence of the law of induction; this is also one of Maxwell's equations. If you do not agree with this, please refer to a source that explains the physics as you see it.

Joe, Erik's responses reminded me of something else. Are you thinking that the conductivity of the string is an essential part of how the pickup works? It is not. In principle the string could be made of non-conducting ferrite. As Erik pointed out, it is the magnetizability of the string that counts

"...producing it's own unique magnetic field instead of merely influencing the field of the pole piece." Those two things are the same thing because the influence results from magnetization. A ferromagnetic material material that is not really "hard" (such as neo) can be magnetized in quite complicated ways with an external field. Making those pole patterns is not a problem.

Joe, Didn't you like Peter's example of how a magnetized nail picks up other nails? Can you offer any physical explanation of how this happens unless the nail becomes magnetized? (Magnetization means that the current domains tend to line up. What other physical change in the nail could cause it to pick up other nails?) Of course the string is not magnetized the same along its whole length. Why would it be?

"I'm missing something...what about the stacked single doesn't work according to Peter's test? Are you saying that (for a strong uniform perturbation of the pickup's field) the signal itself should also cancel if the two coils are in opposite polarity and have equal output?" The result of Peter's test implies that the two coils of the SHB would sense the same signal. Then the the out-of-phase connection would give nothing across the two coils in series. This is indeed what does happen for signal far from the coil, one that changes litte over the length of the pole piece. The SHB can only produce output from a signal that changes in amplitude or phase over the length of the pole piece. (Or it works in some other way that I do not understand.) So the conclusion is: if Peter's test did use a source that simulates the string and given that the result is that the signal is the same whether or not the coil is closer to the source, then the SHB does not work as described above. I think it is more likely that Peter's source does not properly simulate the string, but we do not know until we hear from Peter, and that might be a few days.