A pickup physics questions

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So here we go, for anyone who can answer I've been curious for a while about this question.

So imagine a simple single coil pickup. When more wire is wound on to the bobbin, we get increased impedance and a higher output from the pickup, but on the other hand, the resonant peak in lowered. The question is, why does the resonant peak of the pickup lower? Or rather, what is the physical property of the coil that lowers the resonant peak?

Any help would be greatly appreciated.
 
Re: A pickup physics questions

man, this is a great question but since most of us aren't as "in deep" into pickups as some of the guys representing Seymour Duncan are, it's kind of hard to get you a good answer.

edit:

My girlfriend, a graduated mechanical engineering major, tells me that the relationship between impedance and resonance have an inverse mathematical relationship.

she said impedance is 1/resistance.

I know this doesn't explain why it happens, but it's all I got.
 
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Re: A pickup physics questions

...and this happens due to electromagnetism and the conductive properties of the material.
 
Re: A pickup physics questions

Stolen from the main SD website.

number of turns
The number of turns of copper wire wound around the bobbin greatly affects the performance of the pickup and the overall tone of your guitar. A pickup with lots and lots of turns will have high output but this output will come at the expense of dynamics and high end. A pickup with way too many turns will actually lose output due to resistive losses caused by excessive impedance. On the other hand, an under-wound pickup will come across sounding thin and weak. Getting the right number of turns for a great-sounding pickup is as much art as it is science.
 
Re: A pickup physics questions

The more wire, the higher the resistance. I think you already know that it is the higher resistance which affects frequency, first in the highs or treble. Impredance is just resistance for alternating currents. Now think of standing in water (ocean, lake or pool). Is it easier to walk slowly or run really fast? Slow is no big challenge.

Same thing in electronics. Impedance = restriction of movement (ie. current). So as your highs become affected, the resonant peak shifts down toward the lows. This has to do with the wavelength, or cycles per second, also called Hertz.

Sure, this might not be the technical answer you're looking for but it's true. I'll dig up the formulas if you really want but I have to find my textbooks first!
 
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Re: A pickup physics questions

astrozombie said:
My girlfriend, a graduated mechanical engineering major, tells me that the relationship between impedance and resonance have an inverse mathematical relationship.

she said impedance is 1/resistance.

I know this doesn't explain why it happens, but it's all I got.
Click to expand...

Um… no.

Source: Never send an ME to do a EE's job.
 
Re: A pickup physics questions

astrozombie said:
Well this is an awful amount of negativity towards a clear guess from someone who clearly stated he doesn't know the answer.
Click to expand...

What smiley can I use that will make it abundantly clear that I'm at least half-joking?

Edit: It's an engineering in-joke. Engineering students across different disciplines poke fun at each other.
 
Re: A pickup physics questions

STGFoley said:
So here we go, for anyone who can answer I've been curious for a while about this question.

So imagine a simple single coil pickup. When more wire is wound on to the bobbin, we get increased impedance and a higher output from the pickup, but on the other hand, the resonant peak in lowered. The question is, why does the resonant peak of the pickup lower? Or rather, what is the physical property of the coil that lowers the resonant peak?

Any help would be greatly appreciated.
Click to expand...

A guitar pickup is a second order low pass filter with resonance peak. The cutoff frequency and the amplitude of the resonance peak come out of the impedance, capacitance and resistance just like any other similar circuit.

The reason why you can't easily compute those cutoff and peak values is that you can't easily measure the impedance and capacitance in a pickup. There's constructions to do that, e.g. in Helmut Lemme's book.
 
Re: A pickup physics questions

Well to be exact in order to have a resonance peak you need at least three components: a coil (giving what is called inductance), a capacitor (giving capacitance), and a resistance. These three then form what is called and RLC circuit (see: wikipediafor the basic formula's).

In our case the coil obviously is the wire turned many times around a bobbin, also know as our single coil. Our capacitor is a little bit less obvious, but since all of us are tone feaks, we know cable's have capacitance, and specifically in our wound coil we have capacitance between the different wire turns, since what a capacitor is; is in its most basic form two conductors with some non conductive material in between. Finally we arrive at the most familiar component and the one we all know how to measure: the resistance, which basically arises from the thickness and length of wire we use (and its purity i.e. the main influence on its conductance).

Now we can see adding more wire influences all three the components of the equation. First we increase the inductance by adding more turns of wire. Yet this also increases the capacitance because there is more wire to interact and most obviously it also increases the resistance.

Now for our configuration the most basic approximation of our circuit is this one: link. It is the capacitance of the coil in parrallel with the resitance and inductance combination. This would give us a resonant frequency of edb36e676badaaec3722f5a411a39a26.png.

Yet this is a simplification of what really happens, but it gives you an idea. When both the inductance (L) and the capacitance (C) become larger 1/LC becomes smaller and also R is increasing, so the total under the square root is becoming smaller, therefore its resonase goes down. In reality there is a way more complicated circuit that needs to be used as it is very crude to only use one capacitance for the entire pickup system, all dependent on the wire thickness, isolation thickness, the winding pattern etc. That's why making pickup is indeed as much art as it is science.

To be complete indeed the total amplitude of the signal increases with more windings since the induced current (due to the vibrating strings, altering the magnetic field) in the coil increases. Yet this comes at the expence of a lower resonant peak and thus less high end, as all our ears already knew ;).
 
Re: A pickup physics questions

STGFoley said:
So here we go, for anyone who can answer I've been curious for a while about this question.

So imagine a simple single coil pickup. When more wire is wound on to the bobbin, we get increased impedance and a higher output from the pickup, but on the other hand, the resonant peak in lowered. The question is, why does the resonant peak of the pickup lower? Or rather, what is the physical property of the coil that lowers the resonant peak?

Any help would be greatly appreciated.
Click to expand...

Well to be exact in order to have a resonance peak you need at least three components: a coil (giving what is called inductance), a capacitor (giving capacitance), and a resistance. These three then form what is called and RLC circuit (see: wikipediafor the basic formula's).

In our case the coil obviously is the wire turned many times around a bobbin, also know as our single coil. Our capacitor is a little bit less obvious, but since all of us are tone feaks, we know cable's have capacitance, and specifically in our wound coil we have capacitance between the different wire turns, since what a capacitor is; is in its most basic form two conductors with some non conductive material in between. Finally we arrive at the most familiar component and the one we all know how to measure: the resistance, which basically arises from the thickness and length of wire we use (and its purity i.e. the main influence on its conductance).

Now we can see adding more wire influences all three the components of the equation. First we increase the inductance by adding more turns of wire. Yet this also increases the capacitance because there is more wire to interact and most obviously it also increases the resistance.

Now for our configuration the most basic approximation of our circuit is this one: link. It is the capacitance of the coil in parrallel with the resitance and inductance combination. This would give us a resonant frequency of View attachment 40031.

Yet this is a simplification of what really happens, but it gives you an idea. When both the inductance (L) and the capacitance (C) become larger 1/LC becomes smaller and also R is increasing, so the total under the square root is becoming smaller, therefore its resonase goes down. In reality there is a way more complicated circuit that needs to be used as it is very crude to only use one capacitance for the entire pickup system, all dependent on the wire thickness, isolation thickness, the winding pattern etc. That's why making pickup is indeed as much art as it is science.

To be complete indeed the total amplitude of the signal increases with more windings since the induced current (due to the vibrating strings, altering the magnetic field) in the coil increases. Yet this comes at the expence of a lower resonant peak and thus less high end, as all our ears already knew ;).
 
Re: A pickup physics questions

Thanks for the replies everyone. So as I understand it from having a look at these answers and then following these points around the internet, is that impedance is in AC what resistance is in DC, but impedance is more
complex, because in AC there is reactance to be considered; reactance being the property of capacitor or inductor that means they more strongly oppose current at higher frequencies.

The next logical question I suppose, is why reactance should oppose current depending on frequency? I guess explaining that means getting into some pretty gnarly theory, but if anyone wants to have a go at explaining, it'd be very welcome.
 
Re: A pickup physics questions

The behavior of an inductance or capacitance in a circuit is frequency dependent. The particulars of this are explained in electromagnetics; it has to do with charge separation across a capacitor, and the relationship between the current through and the magnetic field around the wire of an inductor.

Here are a couple of practical examples:

Cheap low-pass crossover for a tweeter in a particle-board home stereo speaker: a capacitor in series with the tweeter. Low bass doesn't get through. DC doesn't get through. The highest treble gets through. At some point in between these extremes, there's a spot called the "crossover region" where the capacitor becomes "visible" to the input signal as it goes down in frequency. The response of the tweeter -- with the capacitor -- will drop over a predictable slope when you look at a plot of output level vs. frequency. As you go down, down, down in frequency, eventually you hit 0 Hz, or DC, in which case an ideal capacitor will look like an open circuit (theoretically, infinite impedance), and nothing will get through. Go back up in frequency, and the capacitor looks less and less like an open circuit, until eventually it looks like the opposite -- a short circuit -- and for practical purposes the tweeter behaves like the capacitor isn't even there, at the highest audio frequencies.

An inductor is kind of the same in a simple series circuit, only the opposite. It looks like a short circuit (well, a simple wire with some resistance is a better model) to DC and to the lowest bass frequencies, but to very high frequencies, that same inductor looks like an open circuit. That's why you can put an inductor in series with a bass driver to roll the highs off. It's just a shallow, first-order crossover (like the high-pass in the paragraph above), but it illustrates the frequency-dependent behavior of an inductance in basic circuit theory.

In reality, pickups, cables, amps, pedals -- many of the things we deal with as guitarists -- have the qualities of inductance and capacitance. Sometimes on purpose, sometimes from non-idealities of real, physical products. Sometimes they're in parallel, sometimes in series. Sometimes they're combined in elaborate ways that yield complicated behavior.
 
Re: A pickup physics questions

STGFoley said:
Thanks for the replies everyone. So as I understand it from having a look at these answers and then following these points around the internet, is that impedance is in AC what resistance is in DC, but impedance is more
complex, because in AC there is reactance to be considered; reactance being the property of capacitor or inductor that means they more strongly oppose current at higher frequencies.

The next logical question I suppose, is why reactance should oppose current depending on frequency? I guess explaining that means getting into some pretty gnarly theory, but if anyone wants to have a go at explaining, it'd be very welcome.
Click to expand...

You are now getting north of the required complications. The parameters you are looking for come out of the very basic properties of the second order low pass filter. There is no difference in how the pickup does it or any normal combination of coil, capacitor and resistor.

There is a big difference in how you can measure these values but really all you need is learn basic electronics to the low pass filters. There's no magic in the pickup.
 
Re: A pickup physics questions

uOpt said:
There is a big difference in how you can measure these values but really all you need is learn basic electronics to the low pass filters. There's no magic in the pickup.
Click to expand...

The magic is all Faraday and Maxwell.
 
Re: A pickup physics questions

STGFoley said:
Thanks for the replies everyone. So as I understand it from having a look at these answers and then following these points around the internet, is that impedance is in AC what resistance is in DC, but impedance is more
complex, because in AC there is reactance to be considered; reactance being the property of capacitor or inductor that means they more strongly oppose current at higher frequencies.

The next logical question I suppose, is why reactance should oppose current depending on frequency? I guess explaining that means getting into some pretty gnarly theory, but if anyone wants to have a go at explaining, it'd be very welcome.
Click to expand...


Ok, impedance is NOT more complex than resistance, it IS resistance. You might be confusing inductance. That is definitely more complex.
Reactance is its own thing and should NEVER be considered in guitar circuits. It is a reality, yes but it is a very theoretical value which will not play much of a role in small signal electronics. Your statement is right, but just try to think of capacitance on its own and save yourself the headaches. In the majority of cases, it is a minimal loss. In 3-phase power electrical machines it becomes a big deal.

Reactance is the result of the time-delay in charging up a capacitance which creates phase issues. You will temporarily cut parts of a wave signal, affecting its harmonics as well. When the capacitor discharges, it is providing more energy/power than the signal originally had but for a very short time (a pulse really) and this sets up a new, different set of harmonics. Because wave signals are additive, you will emphasize some frequencies and attenuate others. This is how the sound gets affected when you fool around with all the components AND you can arrange the components in any number of ways (series, parallel, loops). Usually we limit ourselves to very simple circuits in guitars and we obtain plenty of artistic colors for making music. If you really wanted to get scientific about it, you could open up an infinite range of tones which in the end would not be much more useful than what's available to us now. That's partly why so little has changed in electric guitars over the past 60 years or so.

If you want to know why it doesn't matter much in guitars, consider how the frequency of a simple machine runs of 60 Hz (or 50 Hz in other parts of the world) coming out of your wall. Doesn't matter what it is; a coffee maker, pump, iron, heater, etc. In a guitar playing music, the frequency changes CONSTANTLY. Are you going to try and calculate each and every frequency? What good will it do? None because you could never make a circuit which would be "right" for all frequencies. You can try but it becomes pretty silly pretty fast. That's why some basic tone shaping tools are handy and we use them without worrying about the calculations for this vast array of parameters which doesn't amount to a hill of beans.

"Did he play good?"
"Ya man, you should have heard the reactance of his bridge pickup during the second solo. He really needs another 2pi/5 of phase correction there. I think his tubes were exhibiting sag in the mu value because he didn't bias right and he should have at least another 1,200 ohms on the primary of his output transformer!" I don't think so...

You seem REALLY interested in the physics of all this, you should buy a textbook or take a course. It's fascinating but I find wiring my own guitars and amps plenty rewarding without becoming an electrical engineer. Better to use your ears.

In case you're wondering I'm a materials engineer and had to take a bunch of electrical courses for my degree but I was guitar player first, so I sat in those courses filtering everything through my sick guitar gear mind. I really enjoyed learning how all my gear worked but the math was brutal and exams were bloody hard.
 
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Re: A pickup physics questions

L, C, and R interact to determine which frequencies are present at any given point in a circuit. Since a pickup "has" all three of these characteristics, and they are all affected by the specs of the coil windings (among other things), adding winds affects (increases) all three, hence a tonal change.

When the inductance goes up, more current moves, and the frequencies generated also change. That is one reason why tone is affected: your pre-amp is hit harder, and with different frequencies.

When resistance goes up along with inductance, there is some increased "push back." In an ideal resistor, this would cut all frequencies equally, but in reality, resistors have an easier time dissipating the highs, so there goes a little treble.

Capacitance, which acts as frequency-specific resistance, also goes up with more winds. More capacitance means that the frequency cutoff shifts downward, blocking the high end starting at a lower frequency than before.

It is somewhat analogous to fluid flow. Turning up a pump increases pressure, thus increases flow. However, the mechanical parts of the pump react to the increased flow, imparting their own changes to the character of the flow. I.e. you get more flow, but it's a more turbulent flow.

Similarly, changing the values of pots and caps, or even changing where they are placed in the circuit, also affects what makes it out of the guitar.
 
Re: A pickup physics questions

Natman said:
Ok, impedance is NOT more complex than resistance, it IS resistance.
Click to expand...

Resistance is a limiting case of impedance in which the reactive component is zero. Another way of looking at it is, "What is the impedance of the circuit element at DC?" That's resistance. It's "simple". In ideal circuit elements, the impedance of an inductor, or the impedance of a capacitor, is an imaginary quantity. In a circuit with both resistance and an inductance or capacitance, we say that the impedance is "complex" because it is the sum of a real (resistive) and an imaginary (reactive) component.

This web page here explains the details pretty well:

http://www.ampbooks.com/home/tutorials/lesson-009/
 
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Re: A pickup physics questions

Sure, but it is the resistance or losses due to the stuff in the circuit (whatever it happens to be). It is measured in ohms because it is virtually the same thing even if it originates from inductive and capacitive losses. The imaginary part (j) is where reactance comes in and affects phase.
 
Re: A pickup physics questions

Natman said:
"Ya man, you should have heard the reactance of his bridge pickup during the second solo. He really needs another 2pi/5 of phase correction there. I think his tubes were exhibiting sag in the mu value because he didn't bias right and he should have at least another 1,200 ohms on the primary of his output transformer!" I don't think so...
Click to expand...

This is dead right. This thread could be criticised for getting into territory which has less to do with playing guitars, than it has to do with electronics. However, while mucking around with the innards of my guitar, these questions do arise, and although I have looked up textbooks, it's much more understandable for me to get these questions answered in the context of guitar tone.

I think there is something magic about guitar pickups, in that it is the limitations imposed by physics that give us the unique tone. As you say, this is the partly the reason pickup design hasn't changed a great deal in 60 years. Presumably, if the problem of amplifying a guitar was floated today, the solution that engineers would strike on would be quite different from the pickup, as the frequency response would be considered too irregular to be an optimal reproduction of an acoustic instrument.

Nevertheless, the pickup was invented, and evolved the electric guitar: a musical instrument in its own right. As we all on this forum have quite a bit of affection for electric guitars, then I think it is a natural that we are curious about how they work.

Thanks for taking the time to answer everyone. It certainly gives me a very good base for further research.
 
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