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  • bbsailor
    replied
    Originally posted by Singer15 View Post

    I am not an expert, but going by an online coil inductance calculator (https://www.eeweb.com/tools/rectangle-loop-inductance) the inductance of a strat PU is supposedly 7.6H, twice my example. How low can the inductance be before it is too low for a regular guitar amp to work with?
    Singer15,

    Guitar pickups evolved to provide enough output voltage to drive the early Fender amp designs that had limited amplifier gain to reduce the tubes required to build the amp. It all gets down to the signal to noise ratio. If the pickup output is too low you need to raise the volume control to hear the pickup, but you also hear the hum and all the noise that the pickup coil is also picking up. The requirement to make a pickup produce enough voltage to drive the Fender guitar amps required about 6,000 to 7000 turns of very fine to produce about 80 milli- volts (mv) with the consequence or making a coil with inductance in the 2 to 4 Henry range.

    Your question is about if you can wind enough AWG 36 wire to produce a pickup that will fit your allotted space, match your amp input impedance and not affect to high frequency tonal quality that your ear would like.

    Here is how this all fits together.

    With inductance, when you double the turns you make the inductance (H) 4 times higher. If 1,000 turns makes a 1H pickup coil then a 2,000 turn pickup makes a 4H pickup. If I make a 1,000 turn pickup that is 1H then I add 100 turns, what is the new inductance? The 1,100 turn coil is 10 percent higher or 1.1. Now just square 1.1 to get 1.21 and that is your new coil inductance with 100 extra turns.

    All coils with inductance have a reactance or an AC resistance that is frequency dependent. Reactance is XL and is calculated as XL equals 2 pi (6.28) times the Frequency(F) times the inductance or L in Henries.

    Now we need to think about pickup loading to minimize how much the upper frequencies are being loaded down by these things.
    1. The on body volume control.
    2. The coax cable capacitance
    3. The amplifier input impedance typically rated at 1 million ohms but check your amps rated input impedance.
    4. The combined load of the on body volume control value in parallel with the amplifier input impedance which slightly lowers the combined value.

    Here is a rule of thumb that has been in effect but little has been said about this which I call "the rule of 40". Volume pot values are typically about 40 times higher than the DC resistance of the pickup to not load down the high frequencies of the pickup too much. If you use the 1000Hz frequency to do your XL calculations you will be within a plus or minus 5 percent value that is optimum. Typically low impedance pickup have about one tenth to one twelfth the number of turns compared to a high impedance pickup.

    A Shure A96U matching transformer has a bridging impedance (10 times higher than the source impedance) for low impedance microphones or pickups in the range of about 100 to 250 ohms. The turns ratio in 1 to 12 meaning that the voltage is boosted 12 times. If your high impedance pickup has 6000 turns the your low impedance pickup should have 500 turns so he output from the transformer would be near what a high impedance pickup would produce without a transformer. Here are the other things to consider. High impedance pickups will be more affected by coax cable capacitance of about 30 pf per foot of cable .A 10 foot cable is like putting a 300 pf capacitor across the pickup output. Only your ears will tell you if this affects your sound. Simple switch in a 300pf cap across the pickup and listen for any changes. Low impedance pickups bypass this effect as they are not as affected by cable capacitance at low impedance outputs. Since the Shure A95U is plugged in to the high impedance guitar amp input jack you get the voltage boot of the transformer without adding as much noise as if you added another amplifier stage. Low impedance pickups are less sensitive to hum and noise than a typical high impedance pickup would also pick up.

    If you use AWG 36 to make a high impedance pickup you will make a very large coil winding if you had 6,000 to 7,000 turns on the bobbin. You might want to make two 3,000 to 3,500 turn coils like a humbucker to keep the coil wires closer to the magnets and make the coils more efficient at sensing the vibrating strings but at a tonal cost of less high frequencies.

    All pickup designs are a matter of compromise, balancing the following things to consider.

    1. Magnet size and strength
    2. Pickup distance from strings
    3. Magnetic damping on the strings
    4. Noise that the pickup also picks up
    5. Coil size or compactness to fit allotted space, be visually appealing, or be efficient with the chosen magnet(s)
    6. Ability to properly feed the input impedance of your amp or transformer mounted on or near the amplifier.

    The best way to learn is to tinker with what you have and discover things that please you. I only offer the above things to help you better understand some fundamental ground rules that have governed pickup evolution since Leo Fender built his first electric guitars and amps. The best current transformer that I have discovered so far is using a Triad CSE186L with the a three turn primary winding of AWG16 removed leaving a .125 square inch space to mount an AWG8 string loop. An eighth inch round file will help makes space for the AWG8 wire to fit the space. Place electrical tape around the laminated frame in the wire window to prevent the AWG8 wire from being shorted out against the lamination frame. The output impedance is in the 100 to 150 ohms range and perfect for feeding an XLR low impedance mic input or a Shure A95U style mic matching transformer.

    Joseph J. Rogowski
    Last edited by bbsailor; 07-09-2020, 06:47 PM.

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  • Helmholtz
    replied
    Originally posted by Singer15 View Post

    I am not an expert, but going by an online coil inductance calculator (https://www.eeweb.com/tools/rectangle-loop-inductance) the inductance of a strat PU is supposedly 7.6H, twice my example.
    Typical strat PUs measure between 2H and 3H. Without poles, values would be even lower by around 30%.

    I tested this online calculator and found that it gives wrong (way too high) results. I don't think the formula it is based on is correct.
    There is a more complex formula in literature which gives quite correct results, but is a PITA to work with.
    And of course these theoretical formulas only really work for air-core coils.
    Last edited by Helmholtz; 07-09-2020, 05:28 PM.

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  • Singer15
    replied
    Originally posted by Helmholtz View Post

    Well, not lowish either. With an inductance of 3.8H your PU's impedance and resonant frequency would lie somewhere between a strat PU and a PAF type humbucker.
    I am not an expert, but going by an online coil inductance calculator (https://www.eeweb.com/tools/rectangle-loop-inductance) the inductance of a strat PU is supposedly 7.6H, twice my example. How low can the inductance be before it is too low for a regular guitar amp to work with?

    Leave a comment:


  • Helmholtz
    replied
    Originally posted by Singer15 View Post

    When I said LoZ I was thinking lowish!
    Well, not lowish either. With an inductance of 3.8H your PU's impedance and resonant frequency would lie somewhere between a strat PU and a PAF type humbucker.

    Leave a comment:


  • Singer15
    replied
    Originally posted by bbsailor View Post

    Singer15,
    The reason why I chose a 500 turn current transformer (CT) is to match the typical low impedance microphone input impedance rated at 150 to 250 ohms but typically has an actual input impedance of about 2400 ohms to act as a bridging impedance to not load down the mic too much, Now, if you want to target a high impedance guitar amp, do this.
    1. Do a web search for a current transformer with 5000 turns or more.
    2. Use the thickest string loop wire through the CT and a good low resistance string loop joint.
    3. You can use two CTs per pickup either with the CTs next to each other or with one CT on each end of the pickup string loop.
    4.You can wire or switch the CT outputs in series or parallel for different tones. Make sure the phase is correct or they will cancel out each other.

    Since the string loop is very thin compared to a typical pickup, no pickup cut out needs to be made in the guitar body. This makes the bridge to neck connection better since there are no pickup cutouts except about 1 square inch to accommodate the CT but not in line with the neck. .Also, the pickup output from the induced vibrating string is all very close to the string loop and not distributed up and down like in a high Z pickup.. You should get about 80 mv to 100 mv using a single 5000 turn CT.

    Helmholtz is correct.. If you use 5000 turns of AWG 36 you are making a bulky but traditional high Z type pickup that needs a body cutout. Order a few 5000 turn CTs and experiment!!!

    Joseph J. Rogowski
    bbsailor & helmholtz,
    thanks for the replies. When I said LoZ I was thinking lowish! I am not concerned with the bulk of the pickup, because I have built some guitars with the idea in mind of being able to mount any size pickup without making holes in the guitar top. I recently built a pair of guitars that use a projected neck, with about 20mm clearance, enough for a very fat pickup. My true LowZ pickups sound great with this by the way, and into my amp I have no problems at all, since it can deal with HiZ or LoZ. I was just hoping to recreate something like those old Charlie Christian PUs that had 38ga wire, for my own amusement. So what I am unsure of is exactly where the crossover would be between LoZ and the need to have a mic channel, and a pickup with enough native winds to drive a standard guitar amp. I don't have any 38ga wire but I do have a spool of 36. Any suggestions as to the least number of winds that would get a useful signal?
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  • bbsailor
    replied
    Originally posted by Singer15 View Post
    I've had great success making lowZ PU's with what I learned on this forum. However, they all require a transformer or an amp with a mic channel. Now I'm wanting to make a low z pu that will not need a mic transformer, and I wonder if something akin to the old Gibson Charlie Christian would work with these specs: 5000 turns x 36ga. The DC resistance calculates at around 1200 Ohms. Inductance calculation says it would have about 3.8H. I would use 6 tiny neod mags 6mm diam x 3mm thick as the poles. Before I go ahead and waste a lot of wire on this idea it would be good to know if there is any hope of it working in a common guitar amp.
    Singer15,
    The reason why I chose a 500 turn current transformer (CT) is to match the typical low impedance microphone input impedance rated at 150 to 250 ohms but typically has an actual input impedance of about 2400 ohms to act as a bridging impedance to not load down the mic too much, Now, if you want to target a high impedance guitar amp, do this.
    1. Do a web search for a current transformer with 5000 turns or more.
    2. Use the thickest string loop wire through the CT and a good low resistance string loop joint.
    3. You can use two CTs per pickup either with the CTs next to each other or with one CT on each end of the pickup string loop.
    4.You can wire or switch the CT outputs in series or parallel for different tones. Make sure the phase is correct or they will cancel out each other.

    Since the string loop is very thin compared to a typical pickup, no pickup cut out needs to be made in the guitar body. This makes the bridge to neck connection better since there are no pickup cutouts except about 1 square inch to accommodate the CT but not in line with the neck. .Also, the pickup output from the induced vibrating string is all very close to the string loop and not distributed up and down like in a high Z pickup.. You should get about 80 mv to 100 mv using a single 5000 turn CT.

    Helmholtz is correct.. If you use 5000 turns of AWG 36 you are making a bulky but traditional high Z type pickup that needs a body cutout. Order a few 5000 turn CTs and experiment!!!

    Joseph J. Rogowski

    Leave a comment:


  • Helmholtz
    replied
    Originally posted by Singer15 View Post
    I've had great success making lowZ PU's with what I learned on this forum. However, they all require a transformer or an amp with a mic channel. Now I'm wanting to make a low z pu that will not need a mic transformer, and I wonder if something akin to the old Gibson Charlie Christian would work with these specs: 5000 turns x 36ga. The DC resistance calculates at around 1200 Ohms. Inductance calculation says it would have about 3.8H. I would use 6 tiny neod mags 6mm diam x 3mm thick as the poles.Before I go ahead and waste a lot of wire on this idea it would be good to know if there is any hope of it working in a common guitar amp.
    How did you determine inductance?

    A PU having an inductance of 3.8H definitely is a high impedance PU. In fact its impedance at 1kHz is around 24k. DCR is insignificant for impedance.

    Leave a comment:


  • Singer15
    replied
    I've had great success making lowZ PU's with what I learned on this forum. However, they all require a transformer or an amp with a mic channel. Now I'm wanting to make a low z pu that will not need a mic transformer, and I wonder if something akin to the old Gibson Charlie Christian would work with these specs: 5000 turns x 36ga. The DC resistance calculates at around 1200 Ohms. Inductance calculation says it would have about 3.8H. I would use 6 tiny neod mags 6mm diam x 3mm thick as the poles.Before I go ahead and waste a lot of wire on this idea it would be good to know if there is any hope of it working in a common guitar amp.

    Leave a comment:


  • Nickdino
    replied
    Hi, new to the forum. Does anyone have soundclips of their low-z pickups? Anyone selling them?

    Leave a comment:


  • bbsailor
    replied
    Originally posted by bbsailor View Post
    Bea,

    Try using solid core magnet wire with a thin insulation.
    Four strands of AWG11 will equal the size of AWG 8 and offer about 52 micro ohms per inch for the 4 parallel AWG11 strands. Make sure you make a low resistance joint on each of the strands. I find that using thin wall copper tubing that closely matches the wire diameter makes a good low resistance joint. Use silver solder.

    I hope this helps.

    Joseph J. Rogowski
    I need to correct my post

    Two equal length wire strands of the same size will equal a strand that is three sizes larger. Two strands of AWG 11 at about 104 micro ohms per inch will equal a single strand of AWG 8 at about 52 micro ohms per inch.
    When I used 4 strands of AWG 11 this resistance should have been cut in half again to be about 26 micro ohms per inch or equal to a single strand of AWG 5 wire. This technique can work with pairing any number of AWG solid wire strands.

    Sorry, for my mistake, I just returned form a vacation in Europe.

    Here is the way I have measured various wire sizes to get a visual graph of wire size and string loop resistance effects.

    I use a Velleman PCSGU250 that is a 2 channel digital scope and a 50 ohm output signal generator. Here is how it is set up.

    I put a BNC Tee connector on the graphics generator 50 ohm output and a short BNC coax to channel 2 scope input.

    The open end of the BNC Tee connector end is terminated with a red and black screw terminal to attach to the low impedance pickup stimulation coil.

    I use a low impedance pickup stimulation coil wire size that is 10 times larger area than the high impedance pickup wire size. My AWG 32 stimulation coil has about 500 turns and I put a 500 ohm resistor in series to minimize the loading effects on the stimulation coil by being directly connected to the 50 generator output source resistance.

    The Velleman tester comes with a software package to use the signal generator alone or the oscilloscope alone or a combination of both to see the signal frequency graph driving the stimulation coil and the output of the low impedance pickup directly from the current transformer output feeding the oscilloscope channel 1. The graphs can be labeled and saved. The good part is what you can learn by using a variety of string loop wire sizes, number of thicker or smaller strands, and output level all graphed from low to high frequency low impedance pickup output.

    When your ear tells you something is different with this wire string loop or current transformer number of secondary turns being in the range of 500 to 100 turns. The typical mic input is about 2400 ohms to act as a bridging impedance (10 times higher than the pickup output impedance) for rated 150 ohm microphones which have a real output impedance ranging from about 75 ohms to 250 ohms.

    The largest wire I can fit in a Triad CSE186L current transformer is one strand of AWG 8. But first the three strands of the CSE186L AWG 16 wire must be removed to expose a single square opening to allow a snug fit of one turn of AWG 8 solid core wire. You need to wrap a turn of insulating tape around the laminated transformer core on each side to prevent direct contact with the exposed transformer laminations that might short out the string loop. Once you make a good string loop joint (low resistance) the resistance of the string loop will depend on its length. A minimal string loop of about 7 inches will span a 2 inch string width, the bend and string joint and the amount of string loop wire going through the current transformer. Making a 7 inch string loop of AWG 8 is 52 micro ohms times 7 or 364 microohms. I then add 10 percent to account for typical leakage inductance in this type of pickup now 364 plus 36 is now about 400 ohms. Since a 500 turn current transformer has an output impedance change of turns ratio squared. Each additional inch of AWG 8 wire loop only adds 52/4 or 13 micro ohms to the output impedance. The 500 turn transformer is now the DC Resistance of the string loop times 250,000 or one quarter million. Since the string loop is measured in micro ohms resistance, the actual output impedance is very close to one quarter the string loop resistance plus 10 percent to come very, very close to the actual output impedance measured by an LCR meter at 120 HZ. If you use a 1000 turn current transformer you now need to use a string loop with 4 times that area because a 1000 turn current transformer is now turns ratio squared or 1000 times 1000 or 1,000,000 (1 million). Since the out is based on turns ratio, a 1000 turn current transformer only gets about twice the output level but the size of the string loop wire is now very, very large and harder to work with. That is why I have focused on a 500 turn current transformer but you can use what you have in stock but understand the consequences and the output impedance limitations bases on the actual input impedance of your amplifier device.

    Tinker with different types and numbers of string loop wire and let your ear tell you what sounds best. The theory that I offer only tells you how to target you input device impedance but high impedance pickups have evolved since about 1935 when the first electric guitars allowed the guitarist stand up, solo, and be heard while sitting in the rhythm section. All guitar pickups have resonant peaks due to targeting high impedance amplifiers, many thousand turns of very fine wire to generate enough voltage to drive these amps, 100pf of coil wire capacitance, coax cable capacitance of about 10 pf per foot to feed the high impedance amplifier. The consequence of this is the evolution of the typical "electric guitar sound" is due to the resonant point of most high impedance pickups being in about the 2 to 4 KHZ peak resonance range. Even high impedance magnetic pickups designed for acoustic guitars tend to have some of that typical "electric guitar sound". The low impedance pickups that I describe have a relatively flat response, allowing you to electrically alter the sound to match you desired sound.

    Keep on tinkering!

    Joseph J. Rogowski

    Leave a comment:


  • Joe Gwinn
    replied
    Originally posted by bea View Post
    practical question: how to insulate these?
    Source for the copper will problably normal NYM-Cable where i have a lot of spare material. Remove the thick PVC insulation and then apply something thin. But what?

    Insulation does obviously not need to be perfect, just good enough to insulate these tiny currents over most of the distance in order to avoid the skin effect of the large cross section.
    If this is a one-turn coil, the voltages will be tiny, so any kind of good-quality varnish intended for wood will work. Wood? Yes, so the cured varnish film is stretchy enough that it won't crack and fall off the copper. The metal must be very clean (no flux or oils), or adhesion will fail. Cleaning in an ordinary domestic dishwasher could do it, followed by drying the water off, dipping in varnish, and hanging out to dry in a warm clean place.

    Leave a comment:


  • micha70
    replied
    Originally posted by Helmholtz View Post
    Not necessarily. What matters is the total current loop resistance (determined by cross section times length). So if the cross section is lower only for a fraction of the loop length, its contribution to total resistance doesn't necessarily dominate. But even a continuous cross section of 19mm˛ should be more than enough for excellent bass response.
    Great! Lesson learned.

    Originally posted by Helmholtz View Post
    When you use rivets, their unpredictable contact resistance may significantly increase the total loop resistance.
    Absolutely true. I had a hard time finding the right rivets and „process“ to get a fairly consistent results. Meanwhile I use the solid 5mm bar and drill the holes for the rivets rather than using tubing. Before pulling them in, they are pressed in first. Still a risk. I‘ll have to see how tone will change over time, too.

    Originally posted by Helmholtz View Post
    By principle these "transformer PUs" tend to have a more uneven audio frequency response than a directly wound multi-turn PU of same(!) impedance.
    OK. Need to dig into that. Don‘t understand all of that.

    Bests,
    Micha
    Last edited by micha70; 10-28-2019, 05:40 PM.

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  • Helmholtz
    replied
    The rest of the string loop it's just "show".
    Not necessarily. What matters is the total current loop resistance (determined by cross section times length). So if the cross section is lower only for a fraction of the loop length, its contribution to total resistance doesn't necessarily dominate. But even a continuous cross section of 19mm˛ should be more than enough for excellent bass response.
    When you use rivets, their unpredictable contact resistance may significantly increase the total loop resistance.

    By principle these "transformer PUs" tend to have a more uneven audio frequency response than a directly wound multi-turn PU of same(!) impedance.

    Leave a comment:


  • micha70
    replied
    Hi there!

    Originally posted by David King View Post
    Micha, great work on these, they are really well executed and imaginative designs.
    Thanks a lot! As said, still learning.

    Originally posted by bea View Post
    @micha: welcome - but why didn't You show the last one in GBB? And what's the design? low output wide bandwith or high output smaller bandwidth?
    Thank you and sorry for the late reply. I don't follow this forum on a daily basis. The last design is now on the GBB forum as well as this one. https://www.gitarrebassbau.de/downlo...0043&mode=view
    I don't know why, but I can't upload files, so I linked to the picture of the GBB forum. Sorry!!!
    My basic idea is always to get low output / high bandwith pickups and color the sound with active electronics. I don't know if this will work out, but I'll keep trying. As you might have seen it, there are also some frequency plots of the last two designs take by a friendly forum menber from the GBB forum and they came fairly close to what I was looking for. First the "MW" pickup, second the "D" pickup
    https://www.gitarrebassbau.de/downlo...0383&mode=view
    https://www.gitarrebassbau.de/downlo...0384&mode=view
    Not suprising, since the current sensors are exactly doing what they are designed for.

    Originally posted by bbsailor View Post
    Micha,

    When making low impedance pickups with thick string loop wire, look up on the web the "skin effect" This skin effect says that higher frequencies do not penetrate fully to the core of thicker wire diameters. http://diyaudioprojects.com/Technical/American-Wire-Gauge/
    Here are a few wires that I have used.
    AWG 4 .2043 inches diameter 5.189mm diameter 650Hz for full current depth penetration
    6 .162 4.1148mm 1100Hz
    8 .1285 3.2639mm 1650Hz
    24 .0201 .51059mm 68Khz

    When I used very thick wire for the string loop I noticed that the output level increased due to lower resistance of the string loop with more current being generated but I also noticed that the lower frequency response was being more emphasized. In your experiments, try using multiple parallel strands of thinner wire but with the same area/resistance as thicker wire and listen for the tonal changes due to the skin effect.

    You did a nice job on your pickups. Thanks for sharing with us.

    Joseph J. Rogowski
    Thanks Joseph! Very useful information and exactly right as you can see. Thumbs up!

    I did some experiments with multiple strands and found the sound a little to focused on the higher frequencies and too weak on the low end. In the two designs I want to use on my current guitar project, the string loop is always 14mm˛ times two, but it is reduced due to the available diameter in the current sensors to max. 19mm˛. So, this 19mm˛ is the effective cross section. The rest of the string loop it's just "show".
    The difference between the "MW" and the "D" pickup is that the "MW" uses a copper tube with 5mm outer dia and 3,2mm inner dia, while the "D" uses a solid 5mm dia bar. This may explain the tonal differences you can see in the frequency plots. I thought that this is more, because of the differences in orientation of the magnetic field. But this is far beyond my knowledge. For the time being, I keep trying building and listening to the result.

    Bests,
    Micha

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  • bbsailor
    replied
    Originally posted by bea View Post
    Joseph,

    exactly that was the impression from my prototype with the thick wire. But nevertheless the 6 mm˛ loop of my prototype produced an at least reasonably balanced signal. Despite of that i do consider playing with 4x1.5 mm˛ next time.

    practical question: how to insulate these?
    Source for the copper will problably normal NYM-Cable where i have a lot of spare material. Remove the thick PVC insulation and then apply something thin. But what?

    Insulation does obviously not need to be perfect, just good enough to insulate these tiny currents over most of the distance in order to avoid the skin effect of the large cross section.

    @micha: welcome - but why didn't You show the last one in GBB? And what's the design? low output wide bandwith or high output smaller bandwidth?
    Bea,

    Try using solid core magnet wire with a thin insulation.
    Four strands of AWG11 will equal the size of AWG 8 and offer about 52 micro ohms per inch for the 4 parallel AWG11 strands. Make sure you make a low resistance joint on each of the strands. I find that using thin wall copper tubing that closely matches the wire diameter makes a good low resistance joint. Use silver solder.

    I hope this helps.

    Joseph J. Rogowski

    Leave a comment:

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