DC Resistance equals output???

[antigua]

Maybe so! Since this scatter plot analysis suggests that DC resistance is a reasonably close indicator of inductance when the wire gauge is normalized, and because DC resistance is directly related to wire length, it means that you can fairly reliably trace a line from one to the other. 

When people say a pickup is "hot", it means two things: louder and darker. The darker part comes from the inductance, because a pickup is an LC low pass filter, and a higher inductance "L" means a lower pass. As for louder, every turn of wire on the coil is like a single AC generator, so if you have 8,000 turns of wire, it's as though have have 8,000 AS generators producing a voltage in series, which all adds together, in the same respect that two 1.5 volt batteries in series gives you 3 volts, three give you 4.5 volts, etc. The number of little AC generators you have in series corresponds to the amount of wire, which corresponds to DC resistance (a critical caveat is that not every loop of wire produces the same voltage, the loops closer to the guitar string produce more voltage than those which are farther away).  

The catch is that the number of winds on the coil is not the only factor that determines voltage and inductance. The output voltage and inductance are also effected by the core materials, and the voltage specifically is effected by magnetic strength and coil geometry, but that only means we can't compare pickups that are of a significantly different types. If you have two Stratocaster single coils with AlNiCo 5 pole pieces, or two P.A.F. pickups with identical dimensions, then nearly all of those additional factors are the same, and therefore cancel out on both sides of the equation. Chances are, you're probably comparing pickups of a given type to being with. So if you have two Strat pickups, and one has a DC resistance of 6.0k, and the other 6.5k, the inductance, and therefor the "hotness", will track closely with the difference in DC resistance. Also note that while DC resistance is linear as more wire is added to the coil, inductance rises exponentially, so the "hotness" difference between a 5.5k and 6.5k Strat pickup is smaller than 6.5k up to 7.5k, which is smaller than 7.5k to 8.5k, etc. 

The major practical problem with associating DC resistance to the hotness of a pickup is that not all pickups are wound with the same wire gauge. Most all pickups are wound with 42AWG, but many pickups, such as Tele neck pickups, are wound with the finer 43AWG, and many high output humbuckers are wound with even finer 44AWG. The finer wire will show a higher DC resistance for the same length of wire. So if you're comparing two Strat pickups, and one is wound with 42AWG while the other is 43AWG, it's apples and oranges. Fortunately, that can be accounted for by converting the actual DC resistance of one wire gauge to the relative DC resistance of another. Then it's apples to apples again. 

This PDF offers values for resistance per foot of these different guages of copper wire mwswire.com/wp-content/uploads/2016/10/techbook2016.pdf

Here is the good part:




There's only three values here that are needed 1659, 2143 and 2593 ohms, the nominal resistance for 42, 43 and 44 AWG. The min and max suggest the margin for error, although it's not apparent here the degree to which copper wire deviates from the nominal value.

Therefore...


To convert 42 AWG to 43 AWG equivalent, you multiply the resistance by 1.29 ( 2143 / 1659 )
To convert 42 AWG to 44 AWG equivalent, you multiply the resistance by 1.56 ( 2593 / 1659 )

To convert 43 AWG to 42 AWG equivalent, you multiply the resistance by 0.77 ( 1659 / 2143 )
To convert 43 AWG to 44 AWG equivalent, you multiply the resistance by 1.21 ( 2593 / 2143 )

To convert 44 AWG to 42 AWG equivalent, you multiply the resistance by 0.64 ( 1659 / 2593 )
To convert 44 AWG to 43 AWG equivalent, you multiply the resistance by 0.83 ( 2143 / 2593 )


One practical use of these conversions if determining just how much hotter an "overwound" 43 AWG bridge pickup is than it's 42 AWG normal wind neighbors. Take the Antiquity II for example, the "hot bridge" has a DC resistance of 10.08k with 43 AWG whereas the neck and middle have DC resistance of 6.44k with 42AWG. So what would the DC resistance be for the hot bridge if it were 42 AWG? 10.08k * 0.77 = 7.76k. Could the hot bridge actually be wound with 44AWG? If the "hot bridge" were wound with 44AWG, the 42AWG equivalent would be 6.4k, which is the DC resistance of the neck and middle pickups, but because we know the "hot bridge" has a much higher inductance than the neck and middle, and is in fact "hotter", it's not possible that 44AWG has been used.  

The real tricky business might be determining what wire gauge a coil is wound with. The "default" is 42AWG for most all "vintage" Fender and Gibson pickups. In general, when a Fender single coil has a DC resistance over 8k ohms, the higher DC resistance suggests that it's bumped down to 43AWG. For a P.A.F. style humbucker, the cut off from 42AWG is somewhere between 9k and 10k ohms, as this is the point where the bobbin becomes "full" with 42 AWG, necessitating smaller 43AWG wire. The corresponding cutoff mark for 43AWG is harder to pin down, but if the DC resistance is over 15k ohms, odds are it's that even thinner  44AWG wire being used. In the case of P-90's, because they have such a large bobbin, most all are wound with 42AWG, despite reaching a DC resistances beyond 10k with just one coil.

If you're shopping for pickups, and all the pickup maker is offering are DC resistance values, these considerations and calculations can help you determine how their product compares to others, despite of the scant technical information provided. 

[reTrEaD]

xeroks: Resistance is useless
Chris: No, it's just futile.




You might enjoy this: guitarnuts2.proboards.com/post/37769/thread

[ms]

Assuming that you have allowed for various factors as you have described, I would expect resistance to be a better indicator of output than inductance.  This is because resistance increases almost linearly with the number of turns, as does the output.  (The linear relationship is not exact because additional turns are longer than the initial turns.)  On the other hand, inductance increases faster than the number turns.  If you put on more turns with the same coil geometry, the inductance increases with the square of the number of turns, but usually the coil is getting bigger as you add more turns and so the increase is not as fast as the square.  Inductance has to do with the flux generated by each turn passing through every turn, and so as you increase the number of turns, the number of combinations rises with the square.  (one turn: 11; two turns: 11, 22, 12, 21; etc.

[antigua]

Jan 17, 2018 11:31:26 GMT -5 ms said:

Assuming that you have allowed for various factors as you have described, I would expect resistance to be a better indicator of output than inductance.  This is because resistance increases almost linearly with the number of turns, as does the output.  (The linear relationship is not exact because additional turns are longer than the initial turns.)  On the other hand, inductance increases faster than the number turns.  If you put on more turns with the same coil geometry, the inductance increases with the square of the number of turns, but usually the coil is getting bigger as you add more turns and so the increase is not as fast as the square.  Inductance has to do with the flux generated by each turn passing through every turn, and so as you increase the number of turns, the number of combinations rises with the square.  (one turn: 11; two turns: 11, 22, 12, 21; etc.

That's a good point. Each turn of wire more akin to a single battery, and so for 8000 turns you're really just putting 8000 batteries in series, which is just a sum, not a square. And the turns closest to the string are stronger "batteries" than those near the bottom. I'll have to rework the paragraph on inductance to get that point across. 

As far as non comparing non-similar core pickups, one virtue of the inductance is that it takes into consideration the core material, which also plays a role in the voltage output. For example, based on this test  guitarnuts2.proboards.com/post/80432/thread , the "Texas Hot Bridge" has a higher wind count than the Mexican steel/ceramic Strat pickup, but the steel pole pieces of the Mexican pickup endows it with over 4dB additional output at 500Hz. In this case, the higher inductance of the Mexican pickup is also reflective of its higher voltage output. 

OK, I think I'm clear on this now. The inductance is only as important as the capacitance, in that both determine the resonant peak, but neither is really relevant to the overall output voltage, except at resonance. I need to go back and correct some statements where I've asserted that inductance and output are more closely related than they really are.

So what mostly makes a passive pickup a "high impedance" pickup, the high DC resistance, or the inductance?

[ms]

Jan 17, 2018 14:00:58 GMT -5 antigua said:

Jan 17, 2018 11:31:26 GMT -5 ms said:

Assuming that you have allowed for various factors as you have described, I would expect resistance to be a better indicator of output than inductance.  This is because resistance increases almost linearly with the number of turns, as does the output.  (The linear relationship is not exact because additional turns are longer than the initial turns.)  On the other hand, inductance increases faster than the number turns.  If you put on more turns with the same coil geometry, the inductance increases with the square of the number of turns, but usually the coil is getting bigger as you add more turns and so the increase is not as fast as the square.  Inductance has to do with the flux generated by each turn passing through every turn, and so as you increase the number of turns, the number of combinations rises with the square.  (one turn: 11; two turns: 11, 22, 12, 21; etc.

That's a good point. Each turn of wire more akin to a single battery, and so for 8000 turns you're really just putting 8000 batteries in series, which is just a sum, not a square. And the turns closest to the string are stronger "batteries" than those near the bottom. I'll have to rework the paragraph on inductance to get that point across. 

As far as non comparing non-similar core pickups, one virtue of the inductance is that it takes into consideration the core material, which also plays a role in the voltage output. For example, based on this test  guitarnuts2.proboards.com/post/80432/thread , the "Texas Hot Bridge" has a higher wind count than the Mexican steel/ceramic Strat pickup, but the steel pole pieces of the Mexican pickup endows it with over 4dB additional output at 500Hz. In this case, the higher inductance of the Mexican pickup is also reflective of its higher voltage output. 

OK, I think I'm clear on this now. The inductance is only as important as the capacitance, in that both determine the resonant peak, but neither is really relevant to the overall output voltage, except at resonance. I need to go back and correct some statements where I've asserted that inductance and output are more closely related than they really are.

So what mostly makes a passive pickup a "high impedance" pickup, the high DC resistance, or the inductance?

The highest impedance is at the resonance, and that is a result of the inductance, cable capacitance, and coil capacitance.

[antigua]

Jan 17, 2018 14:33:23 GMT -5 ms said:

Jan 17, 2018 14:00:58 GMT -5 antigua said:

That's a good point. Each turn of wire more akin to a single battery, and so for 8000 turns you're really just putting 8000 batteries in series, which is just a sum, not a square. And the turns closest to the string are stronger "batteries" than those near the bottom. I'll have to rework the paragraph on inductance to get that point across. 

As far as non comparing non-similar core pickups, one virtue of the inductance is that it takes into consideration the core material, which also plays a role in the voltage output. For example, based on this test  guitarnuts2.proboards.com/post/80432/thread , the "Texas Hot Bridge" has a higher wind count than the Mexican steel/ceramic Strat pickup, but the steel pole pieces of the Mexican pickup endows it with over 4dB additional output at 500Hz. In this case, the higher inductance of the Mexican pickup is also reflective of its higher voltage output. 

OK, I think I'm clear on this now. The inductance is only as important as the capacitance, in that both determine the resonant peak, but neither is really relevant to the overall output voltage, except at resonance. I need to go back and correct some statements where I've asserted that inductance and output are more closely related than they really are.

So what mostly makes a passive pickup a "high impedance" pickup, the high DC resistance, or the inductance?

The highest impedance is at the resonance, and that is a result of the inductance, cable capacitance, and coil capacitance.

I get that, but is the inductance contributing one way or another to the output voltage at, say, 500Hz, or is it irrelevant at that frequency?

LTSpice modelling makes it look as though the inductance is irrelevant below the LC resonance, because you see a flat line, with or without the inductor in circuit. 

[pablogilberto]

Hi ms and antigua

Really enjoying this stuff!

Some follow-up questions:

1. Where do we attribute the output voltage now? Is it in the DCR (which correlates to the number of turns) or in the Inductance?

2. I agree that the windings closer to the strings will generate more voltage because they are nearer to the magnetic source (strings), what I want to understand is the tonal consequence of this.

Let's say I wound two typical Strat AlNiCo V. Both with the same exact bobbin and materials.
Assume that my goal is to do 8,000 turns on both of them BUT what I will do is complete 6,000 equal windings first (same pattern and tension) and then for the remaining 2,000 windings, this is what I will do:
Coil A - wind the last 2,000 windings strictly in the upper portion of the coil (near the strings)
Coil B - wind the last 2,000 windings strictly in the lower portion of the coil (far from the strings)

I understand the Coil A will produce more output BUT what will be their tonal difference?

I believe that the R and C parameters will be the same, how about the Inductace L?
Will they differ?

Thanks!

[antigua]

Feb 5, 2021 21:27:31 GMT -5 pablogilberto said:

Hi ms and antigua

Really enjoying this stuff!

Some follow-up questions:

1. Where do we attribute the output voltage now? Is it in the DCR (which correlates to the number of turns) or in the Inductance?

2. I agree that the windings closer to the strings will generate more voltage because they are nearer to the magnetic source (strings), what I want to understand is the tonal consequence of this.

Let's say I wound two typical Strat AlNiCo V. Both with the same exact bobbin and materials.
Assume that my goal is to do 8,000 turns on both of them BUT what I will do is complete 6,000 equal windings first (same pattern and tension) and then for the remaining 2,000 windings, this is what I will do:
Coil A - wind the last 2,000 windings strictly in the upper portion of the coil (near the strings)
Coil B - wind the last 2,000 windings strictly in the lower portion of the coil (far from the strings)

I understand the Coil A will produce more output BUT what will be their tonal difference?

I believe that the R and C parameters will be the same, how about the Inductace L?
Will they differ?

Thanks!

1) The overall voltage output is mostly a product of the magnetic circuit, and that includes how close the pickup is to the guitar strings, but also how much of the coil is close to the guitar strings, and how much permeable metal is involved in the construction. 

Each turn of wire generates a tiny voltage, and the purpose of having 8,000 turns is in order to get strong series voltage of about 100mV, and the inductance and DC resistance that come as a result of 8,000 turns of wire is by itself unimportant, it has the unwanted side effect of loading the pickup and decreasing treble response. But because inductance and DC resistance both vicariously describe how many turns of wire are on the coil, and the inductance also factors in the permeability of the metal parts, they can be used to guess what the voltage output will be. In other words, they are corollaries, but not causes. 

2) Supposing the distance between the pole pieces and the string are the same, and only the coil is moved up or down, per your example, there will be no difference in sound, because the way the strings are being magnetized remains the same, and the coil is remaining on the same axis. The only difference is that the magnetic field field from the string is more dispersed at distance, so the voltage is lower. 

If you lower the entire pickup, then the string gets magnetized differently. When the pickup is farther from the strings, the pickup is less able to magnetically pull on the strings, and so they vibrate differently, and that's the #1 cause for pickup height, or magnetic strength in general, to affect the sound. Setting that aside, if the magnetic is farther from the strings, a wider area of the guitar string will be magnetized, and that will decrease the output of high frequencies, as it's analogous to having poor focus on the strings.

In summary, it's the way the string is magnetized that will affect the sound, not the position of the coil in relation to the strings.