Does this make any difference?

[shadowfall]

I was searching for tutorials on guitar wiring and i saw some people use a switch for OOP on a single coil.Does this produce a different sound?I thought i that i had to have 2 pickups and then reverse the face of the one so when i combine the two ill get a differenct sound.
Also lets say i have 3 pickups N M B and i want to connect them in series.If i change the chain will i get a different effect? For example one way would be N-->M-->B .Another one would be B-->N-->M ,etc

[ashcatlt]

Mar 30, 2016 10:56:58 GMT -5 shadowfall said:

I was searching for tutorials on guitar wiring and i saw some people use a switch for OOP on a single coil.Does this produce a different sound?I thought i that i had to have 2 pickups and then reverse the face of the one so when i combine the two ill get a differenct sound.

Assuming that neither of the wires being switched also carry the "shield ground" for the pickup, it won't sound any different while playing. What it can do is if you're holding a note and it's got the amp reinforcing it in some controlled feedback, sometimes flipping the phase switch can force it to break to a harmonic, so it goes "ooooeeeeeeeeee". I've never really messed with it (probably should), but the technique is often attributed to Brian May, and used as an explanation for why he had a phase switch for each of his three pickups.

Also lets say i have 3 pickups N M B and i want to connect them in series.If i change the chain will i get a different effect? For example one way would be N-->M-->B .Another one would be B-->N-->M ,etc

Nope. Again, assuming that the bridge's bottom wire isn't also supposed to be its shield connection, there will be no noticeable difference. The order of series components doesn't really matter unless you start putting things in parallel between them, and even then it's kind of iffy. People will tell you this is because we're dealing with AC signals and the electrons go both ways, but it's the same with DC.

[newey]

Ashcatlt is right:

Assuming that neither of the wires being switched also carry the "shield ground" for the pickup, it won't sound any different while playing

To finish that thought, we should add that, if one of the wires is also the shield ground, it still won't give you a different tone while playing, but will likely give you some unwanted noise, whether playing or not playing.

If you do have one of the older style of pickups where the braided shield does double duty, those can sometimes be converted so as to have a separate signal wire. Often, it's easier to just find a more modern pickup, but if you have one of the old-style ones and insist on keeping it, write back and we'll point you in the right direction on possibly modding it.

[shadowfall]

Ok so lets say += parallel x=series

if i go N+(M+B) or (M+N)+B or (B+N)+M (and the same for x) there is no difference
I guess this (N+M)xB be different from this N+(MxB)

[col]

That's right, no difference.

N+(M+B) or (M+N)+B or (B+N)+M = N+M+B | M+N+B | B+N+M (and the other three possible orders). Like regular algebra, there is no difference at all.

With:

(N+M)xB and N+(MxB) (and each of the variants), there is a difference in tone. And, extrapolating from the above, there is no difference between (N+M)xB and (M+N)xB, or (NxM)+B and (MxN)+B

[ashcatlt]

Apr 2, 2016 6:58:07 GMT -5 shadowfall said:

Ok so lets say += parallel x=series

if i go N+(M+B) or (M+N)+B or (B+N)+M (and the same for x) there is no difference

Yes.

Think about how you'd wire those. Ultimately, you'd have the top wire of all of the pickups connected to the tip of the jack, and the all of their bottoms connected to the sleeve. Doesn't matter which ones you collect first, they all end up in the same place.

I guess this (N+M)xB be different from this N+(MxB)

1 + 1 = 0.5, 0.5 x 1 = 1.5.

1 x 1 = 2, 2 + 1 = 0.67.

So also yes.

[shadowfall]

And with the same logic i guess (N+M)xB = Bx(N+M) and (NxM)+B = B+(NxM)

I was trying to find which configuration apply to this : The order of series components doesn't really matter unless you start putting things in parallel between them, and even then it's kind of iffy.

[sumgai]

Apr 2, 2016 6:58:07 GMT -5 shadowfall said:

if i go N+(M+B) or (M+N)+B or (B+N)+M (and the same for x) there is no difference
I guess this (N+M)xB be different from this N+(MxB)

Apr 2, 2016 13:43:38 GMT -5 col said:

That's right, no difference.

N+(M+B) or (M+N)+B or (B+N)+M = N+M+B | M+N+B | B+N+M (and the other three possible orders). Like regular algebra, there is no difference at all.

With:

(N+M)xB and N+(MxB) (and each of the variants), there is a difference in tone. And, extrapolating from the above, there is no difference between (N+M)xB and (M+N)xB, or (NxM)+B and (MxN)+B

I need to add some clarification here, even at the expense of not "keeping it simple, stupid".

In the main, the above statements are true, under one ideal condition - that all three pickups have the same electrical characteristics (specifically, the inductance value). If one of them is much hotter than the others, for instance in an S-S-H arrangement with a very hot Bridge humbucker, the equations will be more than a bit unbalanced.  Note that I'm talking about the combined tonalities of the pickups, not the overall volume differences between them. (Generally, if one pup is too loud, it gets adjusted (by a volume knob, or by lowering the pickup under the strings) until everything sounds "just right".)

The larger inductance of a hot humbucker (or even a very hot single coil) will cause the total amount of inductance to change, depending on whether it's in series or parallel. This affects the frequency response, better known as The Tone (aka The Mojo!).  However, while this difference might be large to a calculator's results, and it might be easily seen on a graph, in the real world  I believe it would be heard only if it's clean (unprocessed in any way), and in a quiet room. I expect that under most playing conditions, no one would ever notice it.

HTH





sumgai

[col]

sumgai,

I think you need to reread what I wrote (or maybe I wasn't very clear).

If you are referring to this:

'there is no difference between(N+M)xB and (M+N)xB, OR (NxM)+B and (MxN)+B', I am comparing arrangements within two possible scenarios.

[sumgai]

Apr 2, 2016 14:15:49 GMT -5 shadowfall said:

And with the same logic i guess (N+M)xB = Bx(N+M) and (NxM)+B = B+(NxM)

I was trying to find which configuration apply to this : The order of series components doesn't really matter unless you start putting things in parallel between them, and even then it's kind of iffy.

No, there's no "iffy" about it, you can put the series stuff first, or the parallel stuff first, it won't matter to the final calculations, nor to the final tonal output. However, you state something that concerns me: if we're dealing with only three pickups, then it will be kind of difficult to place a parallel "thing" between two series components. More than 3 pickups, yes, it can be done, but with only three?

But even then, the arithmetic and the calculations will still be the same, only expanded for the additional pickup(s) - you can still put them in any order.

HTH

sumgai

[sumgai]

col,

Yes, I took your meaning correctly. What I was pointing out was the potential difference that arises if the pickups are not of approximately equal value, particularly speaking of inductance.  Having one pup with a much larger inductance, as can happen when dropping in a very hot humbucker alongside of two single coils, which will change the overall calculations by no small amount.  I pointed out this potential "fly in the ointment" only because currently popular graphing tools will tend to show moderate differences in where a peak or a drop-off point might occur, especially when compared to a all-pups-the-same-value scenario.

But as I expressed earlier, that difference will most likely be unnoticible under most playing conditions.

sumgai

 

[col]

Hi Sumgai,

I think we must be speaking at cross purposes. Unless I'm mistaken, he was wondering if there is any difference in tone depending where in the series chain each of the pickups appear, relative to 'ground'. Of course, there is no difference. AND, this would be true irrespective of any differences between each pickup. The OP's brackets were superfluous too, just as they would be with regular algebra. I went on to mention that there is no difference in how the pickups are arranged in parallel in parallel (this may be obvious, but I covered it anyway, just in case).

Unless I'm mistaken, I don't believe that the OP was asking if there is difference between, say, B+(MxN) and N+(BxM) - of course there is a differnce. In this case, if the three pickups were identical, there would be no difference in overall impedance, but there would still be a tonal difference. Just as there is a tonal difference between MxN and BxM, even with theoretically identical pickups.

[sumgai]

Apr 2, 2016 17:52:01 GMT -5 col said:

Hi Sumgai,

I think we must be speaking at cross purposes. Unless I'm mistaken, he was wondering if there is any difference in tone depending where in the series chain each of the pickups appear, relative to 'ground'. Of course, there is no difference. AND, this would be true irrespective of any differences between each pickup. The OP's brackets were superfluous too, just as they would be with regular algebra. I went on to mention that there is no difference in how the pickups are arranged in parallel in parallel (this may be obvious, but I covered it anyway, just in case).

100% agreement. But here's my baseline: in almost every scenario we discuss, particularly with new people, the unspoken assumption is that all of the pickups are of roughly the same value, both in inductance and impedance. I'm simply trying to point out that this assumption isn't always correct, that's all.

Also, referring to what I bolded above... the OP has expressed interest in combination series and parallel together, not just series alone.  This is why I raised the whole point in the first place.  I just didn't want anyone to expect XYZ results, and end up with ABC results instead.

Unless I'm mistaken, I don't believe that the OP was asking if there is difference between, say, B+(MxN) and N+(BxM) - of course there is a differnce. In this case, if the three pickups were identical, there would be no difference in overall impedance, but there would still be a tonal difference. Just as there is a tonal difference between MxN and BxM, even with theoretically identical pickups.

Hmmmm... B+(MxN) and N+(BxM) each have two (supposedly identical) pickups in series, and a third one in parallel to that combo - electrically the same results, and in theory they both should yield the same tonal results, if all other factors are accounted for.  BxM and NxM give different results because they use two pickups in difference positions along the string, which means that we now have to talk about harmonics and other fun stuff.  When all three pups are in play, no matter how they're combined, string harmonics are no longer a topic of interest.

HTH

sumgai

[col]

Hi sumgai,

I'm not sure if we are talking about the same thing here. Are you suggesting that with three pickups (B, M & N), assuming they are electrically identical, they will combine to create an identical tone, so long as the arrangement (irrespective of the specific pickup chosen for each position) is the same? So, B+(MxN) would sound the same as M+(BxN); and B(M+N) would sound the same as M(B+N)? I don't think that is what you are suggesting, but it reads like that.

I'll reread the earlier posts, just in case I'm missing something.

[reTrEaD]

Apr 2, 2016 14:15:49 GMT -5 shadowfall said:

I was trying to find which configuration apply to this : The order of series components doesn't really matter unless you start putting things in parallel between them, and even then it's kind of iffy.

If the "things in parallel" remain in parallel with the same things, the order of series components still doesn't matter.

Let's say we connect the Neck, Middle, and Bridge pickups all in series.
And connect a capacitor in parallel with the Middle and Bridge series pair.
It won't matter if we make the series connection as:

Neck
Middle
Bridge
--or--
Neck
Bridge
Middle
--or--
Bridge
Middle
Neck
--or--
Middle
Bridge
Neck

As long as the capacitor is in parallel with the Middle and Bridge pair, these will all sound identical.

Of course it won't be possible to have:

Middle
Neck
Bridge
--or--
Bridge
Neck
Bridge

Because the Middle and Bridge are separated in the string
So you can't put the capacitor in parallel with the Middle and Bridge pair.

Not sure if that clarified things or made it more confusing...

[shadowfall]

Apr 2, 2016 15:15:04 GMT -5 sumgai said:

col,

Yes, I took your meaning correctly. What I was pointing out was the potential difference that arises if the pickups are not of approximately equal value, particularly speaking of inductance.  Having one pup with a much larger inductance, as can happen when dropping in a very hot humbucker alongside of two single coils, which will change the overall calculations by no small amount.  I pointed out this potential "fly in the ointment" only because currently popular graphing tools will tend to show moderate differences in where a peak or a drop-off point might occur, especially when compared to a all-pups-the-same-value scenario.

But as I expressed earlier, that difference will most likely be unnoticible under most playing conditions.

sumgai

 

3 pickups but humbuckers with parallel/series/split configurations


reTrEaD i want seperate V/T for every pickup so there will be a cap for each one

[col]

sumgai ,

Rereading the above, (assuming three identical pups) it seems that you are indeed suggesting that there would be no difference in the overall output, no matter how the three individual pickups are placed in a PUx(PU+PU) or PU+(PUxPU) arrangement. I think there would be; here's my reasoning on the matter.

If, say, the bridge pickup is placed thusly:

Bx(M+N)

How will this single pickup have its output filtered by the other two pickups compared with:

B+(MxN) ?

(I'll refer to these has 'hybrid' arrangements).

Assume no signal is generated by the other two pups (threat them purely as filters), then the output from the bridge pup will not be the same in the two arrangements. Now, consider the other two pups (now generating signals), although they might be (in theory) electrically identical, they each will produce a different individual output because of their different locations between the neck and bridge. Then, they each will have their own output filtered depending upon their (electrical) relationship to the other two pickups. The individual output of each pup in each of the hybrid arrangements will be affected both by their (electrical) position and by the signal they generate because of their physical location.

Now, there must be a way of representing the output of each pup mathematically, and then combine those outputs into a single expression for the end output, but I am clearly not the man to do that. I appreciate that I am treating each pup as an individual entity (and that might be a mistake), but I don't know how else to explain what I mean. Hopefully there is someone here (maybe you, sumgai) who can express this mathematically, or at least better express the problem/reality than me.

One final thought on the matter. The output of an individual pup is not dependent upon the output of the other pups. Each pup is only affected by the filtering arrangement. Just because the output of the other two pups are affected by their electrical location, it does not follow that the signal of the third pup would somehow change in some reciprocal manner, equalising the overall output of the system to some nominal signal, based purely upon the number of (identical) pickups in a particular arrangement. I hope I'm managing to get across my thinking on this.

It has just occurred to me that we (the collective 'we') have actually discussed this problem before. There was no definitive outcome:

guitarnuts2.proboards.com/thread/3239/coil-combinations

[JohnH]

I've mostly lost track of what is being argued for here. But lets say we are talking about series/parallel combos of three similar single coils, A,B and C.

The only difference between them is where there are placed; B , M or N positions

Let's say each has individual impedance of '1' and output of '1', (in arbitrary relative units)

There are two basic configurations:
A + (BxC) has output 1.333 and impedance 0.667.
A x (BxC) has output 2 and impedance 1.5

So these two types of configuration are clearly different to each other.

Further, taking the first type A + (BxC), of its total relative output of 1.333, 0.667 of that comes from the 'A' pickup, and 0.333 from each of 'B' and 'C'. So given that the three pickups are placed differently and so pick uo different signals, it does indeed matter which one is in the dominant 'A' position. B and C can be swapped with each other however.

Similarly, with A x (B+C) arrangements, half the total relative output comes from 'A', so it matters which one it is.

[sumgai]

col,

I see the problem here.... in your post (5 messages above this one), you used the same formula for both equations - A + (B x C). In your most recent post, you did indeed use different formulas. I'd like to believe that you suffered a case of "fat fingers" in your earlier post, thus prompting my (hopefully) gentle chiding. I stand corrected, per your newest message.

---

(col, you're familiar with the following, but I'm spelling it out for others who are not yet experienced in this stuff.)

In point of fact, the simple formula for two pups is easy to say out loud:

When they're in series, the impedance is doubled, and when they're in parallel, the impedance is cut in half.

That's a 4:1 ratio, and if one double checks his math, and the two answers don't come out to that ratio, then something went bump in the night.

When dealing with three pups, we break it down into pieces, like so:

For A x (B + C):

Two units are in parallel, giving us half of one unit.  That is then placed in series with the remaining unit. Presto, that's a value of 1.5 times a single unit.

Similarly, for A + (B x C):

Two in series yields twice the value of one unit.  Putting that in parallel with the remaining unit gives us something less than the value of a single unit.  In point of fact, it will be 0.75 times the value of that single unit.

It doesn't take too long to see that the ratio between these two lash-ups is 2:1.  Not as much difference in tone between the two, compared to just a pair of pickups with their 4:1 ratio, but still, it's quite noticible - the frequency peak will move up or down the chart.

I hope that everyone can understand this, I realize that I broke it down like a story problem instead of using arcane math formulas.  If anyone wishes to see the proofs, in math form, just ask.

HTH

sumgai

[col]

Hi sumgai,

I think I know from the confusion arises. The OP seemed to be concerned about the order of the pups in series. Of course, the order makes no difference. And, just

Apr 2, 2016 6:58:07 GMT -5 shadowfall said:

Ok so lets say += parallel x=series

if i go N+(M+B) or (M+N)+B or (B+N)+M (and the same for x) there is no difference
I guess this (N+M)xB be different from this N+(MxB)

I applied belt and braces and made clear that the same was true for parallel pups, preempting this:

Apr 2, 2016 14:15:49 GMT -5 shadowfall said:

And with the same logic i guess (N+M)xB = Bx(N+M) and (NxM)+B = B+(NxM)

I was trying to find which configuration apply to this : The order of series components doesn't really matter unless you start putting things in parallel between them, and even then it's kind of iffy.

I'm sure we've been posting at cross purposes - is the above from where it stems? I think we have confused the hell out of everyone. Or maybe it was just me who that!

[shadowfall]

Just to make things more clear. Do these combinations produce the same sound? N+(MxB),N+(BxM),(MxB)+N,(BxM)+N .And the same thing if we change +,x and even OOP combinations. Also the pickups are all different .You can have SSH,HSH,HHH,etc

[newey]

Do these combinations produce the same sound? N+(MxB),N+(BxM),(MxB)+N,(BxM)+N

Yes.

And the same thing if we change +,x and even OOP combinations.

Maybe- it depends on what you mean by "change x,y". (MXB)+N won't sound the same as (M+B)xN. (MxB)+N(OOP) will be different than the other two.

And remember, these differences are theoretical. In reality, and depending on the pickups involved, the tonal distinctions may be so subtle as to be indistinguishable.

[shadowfall]

Apr 4, 2016 5:16:34 GMT -5 newey said:

Do these combinations produce the same sound? N+(MxB),N+(BxM),(MxB)+N,(BxM)+N

Yes.

And the same thing if we change +,x and even OOP combinations.

Maybe- it depends on what you mean by "change x,y". (MXB)+N won't sound the same as (M+B)xN. (MxB)+N(OOP) will be different than the other two.

And remember, these differences are theoretical. In reality, and depending on the pickups involved, the tonal distinctions may be so subtle as to be indistinguishable.

By changing i meant that if i take N+(MxB),change to +for x and get Nx(M+B) and then apply the same thing (4 combinations in total) i will still get the same sound as the first type of connection.Ofcourse (MXB)+N won't sound the same as (M+B)xN

[ashcatlt]

Apr 4, 2016 7:41:34 GMT -5 shadowfall said:

By changing i meant that if i take N+(MxB),change to +for x and get Nx(M+B) and then apply the same thing (4 combinations in total) i will still get the same sound as the first type of connection.Ofcourse (MXB)+N won't sound the same as (M+B)xN

The answer is still yes.

I guess I stirred a pot when I started talking about "sticking things in parallel between them". I used the world "iffy" because it's just more complicated a subject than I wanted to get into at the time, and didn't actually apply to the OP*. Probably should have left it out. What I meant was something like what ReaTread said.

JohnH did a fine job of explaining why it matters which pickup is "outside the parentheses" even when all the coils are identical, but especially when they're different.

Doesn't GuitarFreak do some of this now?



* It kind of does when we start talking about individual V and T pots, though...

[shadowfall]

So to sum things up the only thing that matters is which pickup is outside the parenthesis.If i choose what goes in and out the parenthesis i can do any of the 4 different combinations and i will get the same sound

[JohnH]

Apr 4, 2016 13:28:48 GMT -5 shadowfall said:

So to sum things up the only thing that matters is which pickup is outside the parenthesis.If i choose what goes in and out the parenthesis i can do any of the 4 different combinations and i will get the same sound

Yes I think that that is a true statement.


Ash - to date, GuitarFreak only deals with simple series or parallel combos, in or out of phase.

[sumgai]

Apr 4, 2016 13:28:48 GMT -5 shadowfall said:

So to sum things up the only thing that matters is which pickup is outside the parenthesis.If i choose what goes in and out the parenthesis i can do any of the 4 different combinations and i will get the same sound

Yes, with an important qualification: So long as the operators (the x and + signs) remain in their same positions respective of the parentheses.

BTW, I can find only 3 possible positions for each "2 and 1" configuration.  Considering that we have one pickup outside of the parens, and we have only three pups, it stands to reason there can be only three possible combos:

(n m) b

(n b) m

(m b) n

Given that we use the same operator within the parens above, and the opposite operator for the outside pup, all three combos will sound exactly the same.

Swapping the operators within and outside of the parens, we can have 6 combos that have "2 and 1" configurations, but we now know that three of those are duplicates, as are the remaining three.  Adding the all-in-series and all-in-parallel combos gives us 4 different tones when all three pups are in play*.  Of course, that's excluding OOP, HOOP, blending, and any other modifiers.

HTH

sumgai

* Why is my brain whispering "Mike Richardson, Mike Richardson" here?

[JohnH]

BTW, I can find only 3 possible positions for each "2 and 1" configuration.  Considering that we have one pickup outside of the parens, and we have only three pups, it stands to reason there can be only three possible combos

(n m) b

(n b) m

(m b) n

Given that we use the same operator within the parens above, and the opposite operator for the outside pup, all three combos will sound exactly the same.

Swapping the operators within and outside of the parens, we can have 6 combos that have "2 and 1" configurations, but we now know that three of those are duplicates, as are the remaining three

WARNING! GN2 error detection routines have detected conceptual inversion. Reboot after reinstalling factoids about pickups not contribting equally in series/parallel combinations.

[sumgai]

Well, John, I reverted to our "standard" assumption of all three pups having roughly equal values, particularly inductance and impedance. I know, after all the diatribe I posted above, I should know better, but for the sake of simplicity....

How will pickups contribute differently, depending on whether they are connected in series or parallel?  We've been saying for just about all of this thread that the order of the succession doesn't matter, for any given arrangement of three pickups.  Or have I mistaken your meaning here?

sumgai

[col]

Hi sumgai,

My assertion is that the order does not matter except in what I have referred to as 'hybrid' arrangements.

So, BxMxN is electrically identical to MxNxB, etc. Obviously, parallel is parallel, so no order issue there. (I know you know all this). The only apparent disagreement is with PUx(PU+PU) and PU+(PUxPU) arrangements. I say (and I think John agrees) that there is a difference in output in each arrangement of the following two examples:

1) Nx(M+B) and Bx(M+N) and Mx(B+N)

2) N+(MxB) and B+(MxN) and M+(BxN)

Even assuming identical pups, since the signal generated by each pup is different (because of their differing locations), the combined output of all three pups would be different in each specific arrangement. I make this assertion more from gut feeling than anything else. John would be the one to explain it properly.