Mike Sulzer Posted October 8, 2007 Author Report Share Posted October 8, 2007 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 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. Quote Link to comment Share on other sites More sharing options...
JoeAArthur Posted October 8, 2007 Report Share Posted October 8, 2007 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 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. The coil of a pickup is wound perpendicular to the magnetic field of the pole piece. When the field of the pole piece changes due to the motion of the string, it induces a voltage in the coil. There is only one point of disagreement - where is the magnetic field. You insist it's in the string, I say it's in the pole piece. 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. Whatever tangent you want to go off into is fine with me. I really don't care what fantasy you care to believe. Quote Link to comment Share on other sites More sharing options...
Mike Sulzer Posted October 8, 2007 Author Report Share Posted October 8, 2007 "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? Quote Link to comment Share on other sites More sharing options...
erikbojerik Posted October 8, 2007 Report Share Posted October 8, 2007 (edited) It's a case of backward premises as I tried to point out. Mike's needs the magnet - as the string being a magnet, to be moving and therefore inducing the voltage. If that were correct, then yes, Mike would be right assuming that the field would fall off with distance. That is the only way that the field would be altering the field at the top of the pole - if the field were not "in" the pole to begin with. Joe, I had exactly this misconception at the beginning. What you are missing is an understanding of the assumptions going into this, and the phenomenon that Mike is trying to model in FEMM (keeping in mind there are some limitations of FEMM): #1 - Imagine the plot with (say) 1800 G on the pole piece, showing the intensity in color and the field lines. OK. #2 - Now imagine the plot for an identical situation (1800 G on the pole piece), with the only difference being the presence of a string vibrating above the pole so fast that it can be approximated by a solid object (a thin disc....because FEMM can deal only with fields frozen in time). If the string (disc) is ferromagnetic, do you agree that the string (disc) will have an induced field from the pole that will alter the magnetic field lines around the top of the pole piece? If you don't agree, consider this: pickup makers often use uncharged A5 rod, i.e. no magnetic field in it. They charge it by passing it between 1" neodymium magnets, which have a field so strong that it saturates the A5...even though the A5 never touches the neos. The A5 acquires an induced magnetic field so strong that it becomes permanent. Same thing happens to a guitar string, just at lower field intensity. #3 - Once you can accept that the string will become temporarily magnetized.....now take the field from #1 and subtract it from #2....you'll have something very close to Mike's plot. In FEMM the plots #1 and #2 will be barely different, difficult to see the differences due to how FEMM scales intensity, but they WILL be different, and the "extra" magnetic field due to the presence of the string is what Mike is modelling. But the magnetic field is in the pole and not the string. It is in both, as long as the string is made of ferromagnetic material. I the real world it would be ~1800 G in the pole and...something quite a bit less in the string. Instead of pickups, let's flip it around. Wind a small coil around your 12" pole - something shorter than 12". Put a current through it and create an electromagnet. Will the strength of the electromagnet be any difference depending on the position of the coil along the pole - as long as all of it remains on the 12" pole? No - because the primary poles of the electromagnet will be the ends of the pole. Depends. If the current through the coil is low, it would produce an induced field that is small compared with the rod, and you'll have little difference from a static rod no matter where you put the coil. But if you increase the current, eventually you will generate an induced field that ADDS significantly to the field put out by the rod, and the 2D distribution of this field WILL change with coil position, and WILL change the field strength at the ends if the coil is moved away from the center. Take this to its logical conclusion....pump so much current through the coil that the induced field is a million times greater than the 12" rod...so that the rod makes no difference...of course the strength (and distribution) of the field will change with coil position, because now the only significant source of magnetic field is the coil itself. You don't need a magnet in the middle of the coil to make an electromagnet...all you need is some ferromagnetic material, it needn't have any permanent field to begin with. Flip it back, and you have it - the magnetic lines of the pole must cut the coil regardless of the position of the coil on the pole. As long as the coil remains on the pole. That's why you can raise individual pole pieces on a pickup, without moving the coil, and still get a stronger signal. The lines of flux are moved closer to the string and are more easily disturbed by it. The lines of flux from the pole still cut the coil as long as it remains on the pole. The lines of flux are more dense as you get closer to the pole, and so the disturbance to the magnetic field by the string is of greater magnitude...that's why output increases, it has nothing to do with the fact that you've (in effect) moved the coil further away from the end of the pole. Edited October 8, 2007 by erikbojerik Quote Link to comment Share on other sites More sharing options...
erikbojerik Posted October 9, 2007 Report Share Posted October 9, 2007 (edited) I have a fair bit of experience with this business of magnetic fields induced both by current and by nearby magnetic fields, there's a lot of this going on at a pretty high level in the physics of the flight of charged particles through a mass spectrometer (one of the hats I wear at work). A mass spectrometer magnet IS two guitar pickups in reverse; coils wound around iron cores (both with special shapes), a pair of them with opposing poles facing each other and held apart by a yolk. In mass spectrometry, you want the field in between the poles (where the charged particles fly) to be as pure and homogeneous as possible, like to a few parts in 10,000 (0.01%). Electrically-induced fields become a problem due to induction from the EM coils producing a "fringe field" on the outside edges of the magnet; this is minimized by using a small number of turns of very heavy wire, and using a sh*tload of current (10s to 100s of amps). The coils get so hot you have to cool them with water...we have one 8-ton magnet where the turned "wire" is actually square copper tubing with water flowing through it...along with ~130 amps! Magnetically-induced fields are purposely introduced at the entrance and exit of this big EM by hanging shims of iron separated from the magnet core by strips of aluminum; the induced field changes with the shape of the shims, and the shims are machined so that they form a magnetic "lens" that helps to focuss the beam of charged particles as it exits the magnet. Joe, it is not fantasy. If you can accept that a guitar string can obtain an induced field from the nearby pole, then everything else falls into place. If you want to prove this with real-world experience, do this test for yourself: Take a neo and tape it to a wood table so the poles are oriented horizontally. Take a sewing needle and tape it to the table with the sharp tip pointing directly toward one of the poles; DON'T let the needle touch the neo. Leave it there for awhile, then remove the needle and pass the sharp tip in front of a compass. Let me know how it turns out. Edited October 9, 2007 by erikbojerik Quote Link to comment Share on other sites More sharing options...
SwedishLuthier Posted October 9, 2007 Report Share Posted October 9, 2007 OK guys, I have like 5 minutes online in a hotel lobby. I'll be on the road for the rest of the week. First impression: People have stopped listening to each other and just try to win a debate… Second impression: My experience wasn't that clearly explained as I thought... I took one single rod magnet as I wanted a completely symmetric coil to eliminate as many factors as possible. I then made a pair of single hole flanges that I attached to the rod (imagine a standard strat pickup when looking at it from the side). I then pushed the top flange down like half way. Wound the “pickup” to a couple of thousand turns. Placed the coil in my rig. Measured current and voltage feeding the rig with white noise. Flipped the coil. Repeated. The current was spot on regarding current. The voltage deviated a bit but only a few percent. Together the output was very close. Causes to doubt the experience: - It was not a real string vibrating (might be able to do exactly that when I get home this weekend) - It was not a “true” pickup (don’t realy know what that woul change) - I might have done something really weird winding the coil… Now some input. The speaker/headphone is calibrated to as close as possible mimic the electrical input to the microphone amplifier of the sound card used, compared to the input I get when playing the same pickup used in a real guitar played with real strings. The distance from the generator (headphone speaker) is a bit greater that the normal distance from pickup to strings (about double). How fast does the magnetic field fade away? Compared to that of a vibrating string? The idea of a 12” long magnetic rod with a coil at one end and testing the output of that would really be interesting. Unfortunately my supplier doesn’t carry that long magnets… Quote Link to comment Share on other sites More sharing options...
erikbojerik Posted October 10, 2007 Report Share Posted October 10, 2007 Well Peter, Joe may have stopped listening, but I'm still quite interested in how this really works! 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. If the headphone N-S is perpendicular to the pickup's N-S, then you'd get similar output flipped/unflipped. There might also be something I don't understand going on with the norths and souths in the pole piece and headphone...in one orientation you'll have like poles facing each other, in the other orientation you'll have opposite poles facing each other...I would have expected differences just from that alone...or maybe you get some strange compensation going on that somehow equalizes the two tests....don't know. I think you need to make either a 1-pole pickup or full 6-pole pickup and test it in an actual guitar. FWIW I got the message below from Rick Turner, who spent many a year working for Gibson on their pickups including coming up with new designs (some of which never made it to market); he now runs his own company. First of all, I consider aperture to be the length of the string sensed, not the depth of the coil. Secondly, magnetic phenomena drops off as the square of the distance. To think that the bottom turns of wire...the ones farthest away from the strings contribute as much to the voltage as the closest ones is simply to not understand how any of this works. There are things you can do to make the lower turns more efficient, and that's one of the "secrets" of string surrounding pickups like the Rickenbacher horseshoe pickups or some of the National/Valco lap steel pickups. To test this, wind about 1000 turns of wire on a Strat or Tele style bobbin with the two bobbin plates spaced about 3/16" apart. Use all the same length magnets. Put one bobbin at the very end of the magnets, and the other bobbin at that 3/16" distance. Now try the pickup with the coil closest to the strings. Now flip it upside down so the magnets are at the same distance, but the coil is down at the bottom. Tell me which is louder... No, don't bother; I already know. So now show that to someone who thinks that distance doesn't matter... A magnetic pickup works in a 3D matrix of flux, coil positioning, and string motion. Everything counts. "Pancake" coils are inherently more efficient...but such a pickup will sound different. As I said, everything counts... Note, you will find more induced current in more distant turns with an "iron load" coil core as opposed to a coil that has a ceramic magnet core. String motion will vary the flux at the bottom of a pole magnet...a bit...but won't do that with a ceramic magnet. Like I said, everything changes everything with mag pickups. __________________ Rick Turner www.renaissanceguitars.com www.d-tar.com Quote Link to comment Share on other sites More sharing options...
Mike Sulzer Posted October 10, 2007 Author Report Share Posted October 10, 2007 "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. Quote Link to comment Share on other sites More sharing options...
Mike Sulzer Posted October 11, 2007 Author Report Share Posted October 11, 2007 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 Quote Link to comment Share on other sites More sharing options...
erikbojerik Posted October 11, 2007 Report Share Posted October 11, 2007 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 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. Quote Link to comment Share on other sites More sharing options...
SwedishLuthier Posted October 11, 2007 Report Share Posted October 11, 2007 Just another quickie from a hotel lobby: My headphone speaker’s magnetic field is oriented in the same direction as the magnetic pole. I use the headphone rig to eliminate all human factors like picking variations. I can test it with a real string/guitar but that have to wait to the weekend. Rick Turner is a well respected guitar builder. But maybe (just maybe) the key lies in the phrase “Tell me which is louder... No, don't bother; I already know” . Did he actually make that coil, or did he just assumed that that he know what the result would be? Did he do it with 1000 turns or the more appropriate 3-4000 turns (half a traditional SC)? Would that change anything? Dunno… And what about the “pancake pickup is more efficient” business? What does that mean in the context of the discussion we have had here? The opinion were raised that the sound of a short wide coil mainly differed from a tall thin because the difference in inductance. But if Rick Turner is right there must be something more than that to it. Quote Link to comment Share on other sites More sharing options...
Mike Sulzer Posted October 11, 2007 Author Report Share Posted October 11, 2007 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 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. Quote Link to comment Share on other sites More sharing options...
Mike Sulzer Posted October 11, 2007 Author Report Share Posted October 11, 2007 Just another quickie from a hotel lobby: My headphone speaker’s magnetic field is oriented in the same direction as the magnetic pole. I use the headphone rig to eliminate all human factors like picking variations. I can test it with a real string/guitar but that have to wait to the weekend. Rick Turner is a well respected guitar builder. But maybe (just maybe) the key lies in the phrase “Tell me which is louder... No, don't bother; I already know” . Did he actually make that coil, or did he just assumed that that he know what the result would be? Did he do it with 1000 turns or the more appropriate 3-4000 turns (half a traditional SC)? Would that change anything? Dunno… And what about the “pancake pickup is more efficient” business? What does that mean in the context of the discussion we have had here? The opinion were raised that the sound of a short wide coil mainly differed from a tall thin because the difference in inductance. But if Rick Turner is right there must be something more than that to it. 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. Quote Link to comment Share on other sites More sharing options...
psw Posted October 11, 2007 Report Share Posted October 11, 2007 FWIW I got the message below from Rick Turner, who spent many a year working for Gibson on their pickups including coming up with new designs (some of which never made it to market); he now runs his own company. First of all, I consider aperture to be the length of the string sensed, not the depth of the coil. Secondly, magnetic phenomena drops off as the square of the distance. To think that the bottom turns of wire...the ones farthest away from the strings contribute as much to the voltage as the closest ones is simply to not understand how any of this works. There are things you can do to make the lower turns more efficient, and that's one of the "secrets" of string surrounding pickups like the Rickenbacher horseshoe pickups or some of the National/Valco lap steel pickups. To test this, wind about 1000 turns of wire on a Strat or Tele style bobbin with the two bobbin plates spaced about 3/16" apart. Use all the same length magnets. Put one bobbin at the very end of the magnets, and the other bobbin at that 3/16" distance. Now try the pickup with the coil closest to the strings. Now flip it upside down so the magnets are at the same distance, but the coil is down at the bottom. Tell me which is louder... No, don't bother; I already know. So now show that to someone who thinks that distance doesn't matter... A magnetic pickup works in a 3D matrix of flux, coil positioning, and string motion. Everything counts. "Pancake" coils are inherently more efficient...but such a pickup will sound different. As I said, everything counts... Note, you will find more induced current in more distant turns with an "iron load" coil core as opposed to a coil that has a ceramic magnet core. String motion will vary the flux at the bottom of a pole magnet...a bit...but won't do that with a ceramic magnet. Like I said, everything changes everything with mag pickups. __________________ Rick Turner www.renaissanceguitars.com www.d-tar.com This is very interesting for the kind of thing I am making and experienced with my "sustainer" designs which work a little like a reverse pickup. My coils are very much of the "pancake" variety. I think when he says "don't bother" we do know that adjusting the coil further from the string result in a volume drop, it is hardly worth testing. New noiseless designs like this from fender (pictured with one of my new coils) show how far they are prepared to separate the coils and put a lot of substantial magnets to separate the upper sensing coil from the lower noise canceling coil. Stacked pickups have been less successful in the past because the upper and lower coils did cancel out some of the signals...these new designs have gone to great lengths to try to separate the sensing and canceling coils. The kinman has mag poles only in the top coil (more steel poles below) and a steel shield to stop the string being sensed by the lower coil and detracting from the single coil tone. ...pete Quote Link to comment Share on other sites More sharing options...
Mike Sulzer Posted October 11, 2007 Author Report Share Posted October 11, 2007 Stacked pickups have been less successful in the past because the upper and lower coils did cancel out some of the signals...these new designs have gone to great lengths to try to separate the sensing and canceling coils. The kinman has mag poles only in the top coil (more steel poles below) and a steel shield to stop the string being sensed by the lower coil and detracting from the single coil tone. ...pete 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. Quote Link to comment Share on other sites More sharing options...
erikbojerik Posted October 11, 2007 Report Share Posted October 11, 2007 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. 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. Quote Link to comment Share on other sites More sharing options...
Mike Sulzer Posted October 11, 2007 Author Report Share Posted October 11, 2007 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. 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. Quote Link to comment Share on other sites More sharing options...
erikbojerik Posted October 11, 2007 Report Share Posted October 11, 2007 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. 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. Quote Link to comment Share on other sites More sharing options...
Mike Sulzer Posted October 11, 2007 Author Report Share Posted October 11, 2007 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. 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 Quote Link to comment Share on other sites More sharing options...
erikbojerik Posted October 12, 2007 Report Share Posted October 12, 2007 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 Hehehe...looking at the field just outside the pole, to me this last diagram looks MORE consistent with Peter's result. But as the last 2 diagrams show, a LOT depends on the relative strengths of the pole's field and the test magnet's field. The field inside the pole obviously affects what's going on outside the pole...but as far as current generation goes, all the action is outside the pole. Here is how I think about it; in order to generate current in a coil, you have to have magnetic flux across a conductor. Magnetic flux is field flowing across a surface (gauss per cm2). So in 3D the relevant surface is the one enclosing the coil, think of it as a torus (a donut, the kind with a hole in the middle, not one of those jelly-filled monstrosities... ). Magnetic flux has to pass through that surface in order for there to be any current movement in the coil....and that surface is outside the poles. Quote Link to comment Share on other sites More sharing options...
Mike Sulzer Posted October 12, 2007 Author Report Share Posted October 12, 2007 ...but as far as current generation goes, all the action is outside the pole. Here is how I think about it; in order to generate current in a coil, you have to have magnetic flux across a conductor. Magnetic flux is field flowing across a surface (gauss per cm2). So in 3D the relevant surface is the one enclosing the coil, think of it as a torus (a donut, the kind with a hole in the middle, not one of those jelly-filled monstrosities... ). Magnetic flux has to pass through that surface in order for there to be any current movement in the coil....and that surface is outside the poles. 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. Quote Link to comment Share on other sites More sharing options...
erikbojerik Posted October 13, 2007 Report Share Posted October 13, 2007 (edited) Two ways of saying the same thing I think. I'm good up to this point: 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. From the source of all things true and proven, Wikipedia: Faraday's law of induction (more generally, the law of electromagnetic induction) states that the induced emf (electromotive force) in a closed loop equals the negative of the time rate of change of magnetic flux through the loop. This simply means that the induced emf is proportional to the rate of change of the magnetic flux through a coil. The relevant "at home" experiment is to take a loop of conducing wire and move it through the static field of a permanent magnet sitting on the table, you'll see current moving through the loop as long as you're moving it...stop, and the current stops. But this doesn't work for guitar pickups, you need to move the field across the conductor, and it is the string vibration that does this. If I understand FEMM correctly, the boundaries between different colors are lines of constant field strength, whereas the contour lines are showing the direction of magnetic flux. And the closer together the contours are, the higher the flux. In all of these plots there is magnetic flux, but no current through the coil...because the flux is not changing. So we need to imagine what a time-variable field will look like on such a plot. With a changing flux (superimposed on the static field) induced by the moving string, the color boundaries will expand & contract like baloons, and the contours will move and sweep through the 2D space accordingly. 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. Sure the flux will change inside the pole as well, and the field is strong there, but the contours inside the pole do not intersect the coil at all...so when they move they don't do work on the coil and don't contribute to the current. In a time-varying flux (only a variable flux will generate current), the horizontal contours outside the coil will move up and down perpendicular to the turns of the coil...current. You'll have variable flux not only in the up-down direction across the coil, but also in the radial direction across the coil. Edited October 13, 2007 by erikbojerik Quote Link to comment Share on other sites More sharing options...
SwedishLuthier Posted October 13, 2007 Report Share Posted October 13, 2007 Hi guys, bat at home again. I have repeated the experiment but this time I wanted to see if there were some type of frequency dependency that played a trick on me. The idea I hade mentioned earlier in this thread regarding why a Tele pickup sound so different with and w/o the steel base plate was centred on the fact that higher frequencies will produce a higher output (discussed in an earlier thread). So I repeated the experiment but with specific frequencies instead of white noise. The result (in Volts, peak value): Frequency (Hz) Coil upwards Coil downwards 100 .50 .53 250 .75 .69 500 .67 .64 1000 .62 .62 2500 .64 .65 5000 .76 .72 10000 .90 .81 So a bit more differences that the former test but pretty consistent with the conclusion. I will continue with another test ASAP. To get rid of picking variations I will borrow an E-bow and repeat the test with real strings. I’ll report back. And what about the “pancake pickup is more efficient” business? What does that mean in the context of the discussion we have had here? The opinion were raised that the sound of a short wide coil mainly differed from a tall thin because the difference in inductance. But if Rick Turner is right there must be something more than that to it. 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. My main interest was the claim that they differed in sound, not output. I think when he says "don't bother" we do know that adjusting the coil further from the string result in a volume drop, it is hardly worth testing. Yeah, but in that case you are also lowering the magnet. Not really applicable to this discussion. Both of my experiments contradict Mr Turner completely. If I understand FEMM correctly, the boundaries between different colors are lines of constant field strength, whereas the contour lines are showing the direction of magnetic flux. And the closer together the contours are, the higher the flux. In all of these plots there is magnetic flux, but no current through the coil...because the flux is not changing. So we need to imagine what a time-variable field will look like on such a plot. With a changing flux (superimposed on the static field) induced by the moving string, the color boundaries will expand & contract like baloons, and the contours will move and sweep through the 2D space accordingly. 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. Sure the flux will change inside the pole as well, and the field is strong there, but the contours inside the pole do not intersect the coil at all...so when they move they don't do work on the coil and don't contribute to the current. In a time-varying flux (only a variable flux will generate current), the horizontal contours outside the coil will move up and down perpendicular to the turns of the coil...current. You'll have variable flux not only in the up-down direction across the coil, but also in the radial direction across the coil. And that is why I early on suggested two similar plots, one with a rod magnet (the pickup magnet) with a magnetic dist (string) in close proximity to each other, and one with a greater distance between the two magnets. Those two plots put close to each other would give us a very simplified picture of the changes in the magnetic filed (flux) due to the string moving. I can’t do it so if some one would step up to the challenge I would be grateful. Quote Link to comment Share on other sites More sharing options...
psw Posted October 13, 2007 Report Share Posted October 13, 2007 QUOTE(psw @ Oct 11 2007, 06:44 AM) * I think when he says "don't bother" we do know that adjusting the coil further from the string result in a volume drop, it is hardly worth testing. Yeah, but in that case you are also lowering the magnet. Not really applicable to this discussion. Both of my experiments contradict Mr Turner completely. Yes...well I take your point. I am not completely following this thread...I have my own self interest. Why this comment is of particular interest is that my unique sustainer "pancake coils" as shown in the picture, do have some interesting qualities, as far as efficiency driving the string and speed at which it can change magnetic states. We have never really been sure why, or even if it is imperially true, however it does seem to work and certainly has some different qualities from other coil forms tried. Now...I am not suggesting that this would make a good model for a pickup coil you understand...but it is of interest for my particular application. I wonder what exactly was meant by that comment. One member "Col" in the sustainer thread that has been doing some research of late into this, seems to suggest that a coil of roughly square profile, not dissimilar to pickup coils, where all the windings are surrounded as close to the others as much as possible. My gut feeling in designing my coils was that the "pancake coil" maximizes the number of overlapping winds. A wide coil would have more overlapping winds than a tall coil. The idea stemed from my feeling that each wind in the coil is a conductor that generates a magnetic or electrical response to a magnetic field (depending on if a electro-magnetic driver or sensor) and each overlapping wind is wound around a magnet increasing the power in a coil with a lot of overlapping winds... Of course...although I have read a bit and done a lot of experimental work with these "reverse pickups" (i.e. sustainer string drivers) this is all very much a gut feeling that I would welcome clarified. It is really an intriguing comment that must have some basis I suspect, but I have never come across it, even with my self interest in the subject... Would love to hear some ideas on it...or some genuine theory along these lines. cheers... pete Quote Link to comment Share on other sites More sharing options...
Mike Sulzer Posted October 13, 2007 Author Report Share Posted October 13, 2007 Two ways of saying the same thing I think. 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. Quote Link to comment Share on other sites More sharing options...
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