Les Paul Lo-Z pickups

[Louie Seven]

hi

2.5K is correct.... my mistake...the diagram I have is very blurred..but upon close examination, there is indeed a decimal point.

I surrender to superior forces!

NOW, I have a theory question...it's a general question about stacked coils. Nothing really to do with Lp Recording guitars or low impedance....

IF a person took a standard humbucker and cut it in half...(seperating the coils, as it were) leaving all the wiring and connections intact, and then positioned one coil atop the other (assuming that all other practical considerations such as cavity depth, pickup mounting, etc, were dealt with properly)...would this arrangement work?

Opinions ?

Louie

[David Schwab]

From my limited experience with stacked coils... the problem is either the bottom coil is a dummy, and that makes the sound weak in the upper coil, or the bottom coil senses the strings, but then it's out of phase, so it cancels low end.

Duncan stacks have the magnets going through both coils, so the bottom coil is active. I made a Tele lead pickup like this. Very bright sounding, but in a good way. I could have wound a lot more wire to help this. You may notice that stacked pickups usually are wound pretty hot.

Dimarzio and Kinman have a dummy coil on the bottom, with a magnetic shield between them, and the bottom coil has no magnets, but does have metal slugs for increased inductance.

I'm not sure how EMG and Bartolini have the coils/magnets set up... But EMG gets around the dummy coil issue by using a summing amp, and thus isolating the active coil from the load of the dummy. Alembic does the same thing.

[Louie Seven]

It's clear that I'm going to be doing a bit more research before I tackle this.

Yep!

But hey, a thousand feet of 28 or 30 will allow me two or three shots at it.

eh ?



Thanks everyone.

Louie Seven

[dpm]

Sorry for the misinformation is my previous post. The guitar I was thinking of was a Les Paul Signature. Those pickups (I'm pretty sure) were either 200 or 500 ohms.... I think more likely 200. Those 2 numbers look so similar on a meter lol

[Louie Seven]

I remember when those Lp Signature guitars first appeared.

It was, in the early 80's and we were walking down up Van Ness Avenue in the peoples republic of San Francisco.

Suddenly, there was a music store I'd never seen before. I don't know if it was new, or if I'd just overlooked it....

At any rate, we strolled in and there on the floor was two blonde Gibsons that looked like strange Rickenbackers. A guitar and a matching bass.

The headstocks said 'Les Paul Signature'. I just marveled, scratched my head and left.

L7

[David Schwab]

Of course now we have the Epiphone Jack Casady Bass

Which is a remake of the LP Signature bass.

Supposedly they didn't get the pickup quite right.

[Soulweb]

New Member...help with LP recording bass pot.

Sorry to be such a newb. I am 38 and have been playing for 25 years. I play jazz almost exclusively. So when I saw the PBS special with ol Les playing his Recording I jumped in and bought one. I LOVE it. But I have one small problem, the bass pot seems shot. I have asked countless online vintage salespeople if they have this part, and of course they do not.

in my searching I came across this thread and was excited. Not only do you guys know what I'm talking about, you seem to know in depth what I'm talking about.

Please allow me to explain my problem. I'm not a technical guy at all so again, forgive my ignorance.

The guitar I bought looks virtually mint. I assume it say under someone's bed for the past decade. As a result, the bass pot only works on "10". If I turn it below 10 the signal seems to cut drastically and immediately. In other words, from 9 to 1 the sounds is the same...as though it's cut in half. if i turn the knob back to "10" it sounds like you would expect the guitar to sound when you turn the bass knob to "10".

I have tried the usual mickey-mouse DIY fix-its of turning the pot back and forth quickly as well as shooting a few bursts of Radio Shack tuner cleaner into the pot to no avail. There's no scratching and the knob turns freely. It just doesn't work. Maybe it's oxidized or corroded? I don't know.

But in lieu of trying to fix this, can any of you steer me in the right direction to finding a replacement? I have read about "Deoxit", but will this be worth trying if the Radio Shack tuner cleaner didn't do anything?

This forum is my last hope. The whole thread has been a huge eye-opener for me. Now I hope someone can help me get this beautiful guitar to sing again. It wants to play some Jim Hall or some Ed Bickert!

Thanks so much for your patience in advance.

[David Schwab]

First try some WD-40. It works a lot better than the stuff made for pots!

You can probably get a replacement from an online electronics supply, like Mouser, Digikey, etc.

It's not a standard guitar pot value, but it is a standard value.

[R.G.]

I'm not by any means a pickup expert, but I've been doing electronics, including power and signal transformers for about 30 years. Here are a couple of things I think will add to the discussion here.

The size of the signal a coil makes is proportional to the number of turns and the strength of the field it moves across. Field strength*turns ~signal level.

The inductance of a coil is proportional to the number of turns squared if nothing else about the inductor changes. Cut turns by a factor of two, cut inductance by a factor of four. And also cut signal level by only a factor of two.

The inductance of a coil is proportional to the area of the coil. Make the inner, non-turns-containing area bigger, make the inductance bigger.

The self capacitance of a coil is complexly dependent on exactly how the wires are wound, as is common "wisdom" for pickup studiers. But obviously, the fewer turns, the lower capacitance, because capacitance is a surface-area effect and fewer turns is lower surface area.

The groove that ordinary pickups are running in is several thousand turns. In the various winding configurations, this leads to a self-resonance and cutoff frequency somewhere between 1kHz and 7kHz. There are probably other burbles I don't know about but this is the synopsis of what I read.

If you cut the turns by a factor of four, filling the same volume of the coil with bigger wire, you ought to cut the inductance by sixteen and the capacitance by some unknown amount. The signal level goes down linearly with the turns, a factor of four.

The high frequency cutoff goes up by the square root of the LC product. You only need to get a factor of maybe four to get you out to the very high end of the audio range, 10k to 16K for the high frequency cutoff.

So a modest reduction in turns - 1/4 to 1/10 the turns - cuts inductance by 16 to 100. It extends the fundamental high frequency cutoff out to the top end of the audio range or above. It cuts signal by an amount which only needs a modest boost, 4 to 10, to restore signal level.

And while we're on that signal boost, there are some VERY good amplifiers for magnetic coil pickups that were developed for phono cartridges and tape players - remember those? Like, before CDs?

Something like the National LM387 is so quiet that it is quite difficult to tell it's there, even at high gains.

[Joe Gwinn]

I'm not by any means a pickup expert, but I've been doing electronics, including power and signal transformers for about 30 years. Here are a couple of things I think will add to the discussion here.

The size of the signal a coil makes is proportional to the number of turns and the strength of the field it moves across. Field strength*turns ~signal level.

Don't forget the area of the turns. The presence of permeable material increases the effective area. Effective area will affect both music voltage and coil inductance. And hum voltage.

The inductance of a coil is proportional to the number of turns squared if nothing else about the inductor changes. Cut turns by a factor of two, cut inductance by a factor of four. And also cut signal level by only a factor of two.

The inductance of a coil is proportional to the area of the coil. Make the inner, non-turns-containing area bigger, make the inductance bigger.

The self capacitance of a coil is complexly dependent on exactly how the wires are wound, as is common "wisdom" for pickup studiers. But obviously, the fewer turns, the lower capacitance, because capacitance is a surface-area effect and fewer turns is lower surface area.

Self-capacitance varies mostly with the physical size and shape of the coil, and won't reduce that much.

The groove that ordinary pickups are running in is several thousand turns. In the various winding configurations, this leads to a self-resonance and cutoff frequency somewhere between 1kHz and 7kHz. There are probably other burbles I don't know about but this is the synopsis of what I read.

If you cut the turns by a factor of four, filling the same volume of the coil with bigger wire, you ought to cut the inductance by sixteen and the capacitance by some unknown amount. The signal level goes down linearly with the turns, a factor of four.

The high frequency cutoff goes up by the square root of the LC product. You only need to get a factor of maybe four to get you out to the very high end of the audio range, 10k to 16K for the high frequency cutoff.

So a modest reduction in turns - 1/4 to 1/10 the turns - cuts inductance by 16 to 100. It extends the fundamental high frequency cutoff out to the top end of the audio range or above. It cuts signal by an amount which only needs a modest boost, 4 to 10, to restore signal level.

And while we're on that signal boost, there are some VERY good amplifiers for magnetic coil pickups that were developed for phono cartridges and tape players - remember those? Like, before CDs?

Something like the National LM387 is so quiet that it is quite difficult to tell it's there, even at high gains.

Yep. You have captured the essence of active pickups, such as those from EMG.

[David Schwab]

You have captured the essence of active pickups, such as those from EMG.

EMG's actually are high impedance coils (~8K) wired in parallel to the differential inputs of an op amp.

Alembic pickups are low Z and boosted up to line level.

[R.G.]

Don't forget the area of the turns. The presence of permeable material increases the effective area. Effective area will affect both music voltage and coil inductance. And hum voltage.

Actually, the signal size is dependent on the size of the field change across the conductor times the number of turns it cuts, directly from the differential form of Faraday's law. For everything else equal, the number of turns is what determines that. Area of the coil doesn't really enter into it, except as you pick up more or less of the field change by changing the fill of the surrounding space with permeable material. Yes, if you dramatically change the structure of the coil, all bets are off because you've changed the M-field.

But if you're trying to understand what affects what and how, you might take an example pickup setup, with a fixed-size bobbin, fixed magnetic and mechanical/material structure and fixed amount of strings dragging the M-field through the coils. In this test case of trying to understand what number of turns does, changing turns only causes signal strength to change proportional to the number of turns.

I think you're referring to the integral form of Faraday's law the line integral of E being equal to the surface integral of B. The answers come out the same (they HAVE TO!) but the differential form is more accessible for people who haven't had a fields course in a few years.

For changing the surrounding space's fill with permeable material, yes, you can arrange for the string movement (which we'll assume is the same, trying to limit conditions so we understand the contribution of one thing at a time) to drag more or less of the available M-field through a higher or lower percentage of the coil. For setups like guitar pickups, most of the M-field volume is leakage, as there is little of it focussed to lie where the moving string can drag the field over the coil and make signal.

So yeah, area of the coil, height of the coil, build if the coil, fill of the 3-space around the coil with which materials all affect signal. But to understand, you need to separate the effects and understand them one at a time, then combine them. Or at least I do - that's how I understand it best. Maybe I'm just slow.

I develop an understanding by holding everything constant (mentally at least) and changing one facet. When I understand what changing turns does, I'll change something else. M-field density is another good one. Magnetic cross sectional area of the inside of the coil is another one; so is permeability of the fill. So is the reluctance of the external path and where any high permeability materials lie in the external flux path.

I know, it's a plodding approach, but it looks like I'm going to have to teach myself. I have two speeds - slow and stop...

[Joe Gwinn]

Actually, the signal size is dependent on the size of the field change across the conductor times the number of turns it cuts, directly from the differential form of Faraday's law. For everything else equal, the number of turns is what determines that. Area of the coil doesn't really enter into it, except as you pick up more or less of the field change by changing the fill of the surrounding space with permeable material. Yes, if you dramatically change the structure of the coil, all bets are off because you've changed the M-field.

...

I think you're referring to the integral form of Faraday's law the line integral of E being equal to the surface integral of B. The answers come out the same (they HAVE TO!) but the differential form is more accessible for people who haven't had a fields course in a few years.

Well, the differential form is easier mathematically, but harder to understand in a qualitative way. The original experiments by Faraday were implicitly integral form.


For math: http://scienceworld.wolfram.com/phys...radaysLaw.html.

What the first equation says is that the voltage per turn is proportional to the time rate of change of the total flux passing through the surface whose border is the wire loop.

What the second equation says is that the voltage per turn is proportional to the time rate of change of the surface integral of the magnetic flux passing through the wire loop.

These are different ways to say the same thing. If the magnetic field is uniform, the total flux is simply the field strength times the area of the loop.


For a simple physical description, without integrals or differentials: http://hyperphysics.phy-astr.gsu.edu...ic/farlaw.html.

[tbryanh]

Because low-Z pickups have fewer turns, the voltage they produce is lower, but is this the real issue? Probably not. We know we can use a transformer to increase the voltage.

Probably the real issue is how much power does the pickup produce? When the impedance of the pickup is the same as the amp, the pickup transfers the most power to the amp and drives the amp hard. This is usually desirable. If not, we turn the volume pot down on the guitar.

The low-Z pickup probably does not produce quite as much power as a high-Z pickup. Is this bad? Not with all the high gain amps and foot pedals available today. Was this bad back in the day when the Les Paul Recording was made. Probably.

Distortion was already popular when the Recording was made, and it became more and more popular as time passed. But the Recording probably didn't drive the amp as hard as other guitars, so it did not become a popular guitar and was discontinued.

[David Schwab]

First, look how well low Z microphones work. They do have a transformer, but they also need very small coils to keep them light. But they work just fine, and have good output.

And a Mic pre.

So enter the onboard preamp. Sure, you can use a transformer with a Low Z guitar, but I'd bet an active preamp would get a better tone.

My bass has line level output. It it were a guitar, I'd have no problem over driving an amp.