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Post by octupletnumber9 on Sept 9, 2010 3:00:40 GMT -5
I've read about this somewhere a long time ago. but can't find it now. Maybe this applies to parallel circuits and not series, but in one or the other there is a problem caused by the lack of a load on a circuit including two Humbuckers which lowers the overall output. I had a guitar that would get louder when the volume pot was turned down a bit. Instead of installing a pot in my new guitar, how do I determine the right value of resistor to install?
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Post by newey on Sept 9, 2010 9:29:39 GMT -5
Octo-
Hello and Welcome to Gnutz2!
I'm not clear on this. Wiring 2 HBs in series will give higher output than wiring the 2 in parallel, that's the nature of series wiring per Mr. Ohm's Law. But I'm not sure that's what you mean here.
If it's a guitar you no longer have, it's not really relevant to your new guitar, but it sounds like a problem. Was this one vol. pot or 2? Wired in series or parallel?
Again, not clear on what you're asking.
You have a new guitar; I assume it's 2 HBs based on the first part of this discussion. How is it wired, or how do you propose to wire it? Do you really not want "a pot" on it for some reason? If so, which pot don't you want? And why would you want to install a resistor in its place?
If you eliminate a pot from the circuit, you lower the overall resistance, since even at "10", the pot offers some resistance to the circuit. This is usually how one wires a "solo switch"- the Vol and Tone pots are taken out of the circuit with the flip of the switch, giving a "direct output" from the pups to the output jack, resulting in higher output for a boost to one's soloing.
Adding a fixed resistor in place of the pot will have the effect of having a pot permanently turned down by some amount. If the pot was 500KΩ, replacing it with a 500KΩ resistor will sound like the pot was turned to "0". Lower values will correspond to other knob settings. No resistor at all, as noted above, means it "goes to 11".
In short, we need more info on what you wish to accomplish here.
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Post by ashcatlt on Sept 9, 2010 13:56:08 GMT -5
Ah, newey, I'm sorry...
Removing a volume ( or tone) pot actually increases the resistance between the "loose ends" of the pickup wire. You are essentially replacing the 500K pot with an infinte resistance. Of course, this is parallel to (at least) the amp's input, but it still gets bigger: 500K||1M~350K, infinity||1M=1M.
This actually is not expected to change the overall output volume unless there was something wrong with the pot to beginwith. What it will do is increase the volume of the very highest frequencies which can be passed by the system. It won't actually change which frequency is highest, just makes that area a little louder relative to the rest. Removing a volume pot is more like turning a tone pot to 11 or 12.
Likewise, adding a single resistor, in series or parallel, won't be expected to have much impact on total volume - at least not with reasonable values. This, again, will impact tone more than volume. A volume decrease require a voltage divider, the reason we use all 3 lugs on the V, but only 2 on the T.
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Post by sumgai on Sept 9, 2010 16:03:53 GMT -5
Mein Gott, octo must be my age - I understood him perfectly fine. First things first....... octo, Hi, and to the NutzHouse! OK, so what the guys said above has value, but it didn't directly address, and answer, your question. You want to know what value resistor to insert into your current circuit such that it would cause a volume increase, right? Well, they did get to the correct answer eventually, even if by differing and round-about paths.... the proper answer is "None". Not knowing where you stand on the learning curve in electronics, I'll simply state this: in your previous guitar, you noted an increase in volume when "the volume control" was turned down. Fine. But let me ask you first, was this a single-volume-control guitar, or did it have two controls? And if it was a two control'er, did they both increase the apparent volume when turned down slightly? If it was a single pot (a Master volume), then it could well be that your particular pot had a quality control issue. It wouldn't be the first time we've seen a pot that registered less resistance as the wiper was moved away from the endstop, whereas we were expecting more resistance. If there were two pots involved, and they both exhibited the same behavior, then I'd have to go with a wildly wrong component being stuck in there. Ideally everything should work together, but if someone were to throw in a part that was way out of specification, then what you experienced might be possible. And if all of this (now it's my turn to conjecture.....) is merely an exercise to get more output from your current guitar, then I'd suggest that you consider either dropping in some hotter pickups, or else [url=http://guitarnuts2.proboards.com/index.cgi?action=display&board=schem&thread=3150&page=1 ]installing an on-board active buffer[/url] circuit. (Or what the hell, you could do both!) HTH sumgai
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Post by JohnH on Sept 9, 2010 17:17:57 GMT -5
The only time I could imagine that turning down a volume pot, or adding resistance, might actually increase volume is if you have two pickups with very different outputs or impedance. In that case, the output with both at full volume is mostly controlled by the weaker, lower resistance pup, and turning down its volume controll can increase volume as the higher output (and higher impedance) pickup starts to dominate.
Nothing wrong with that, its just the way it is, and no point in adding extra resistances to address it IMO.
John
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Post by octupletnumber9 on Sept 10, 2010 0:39:44 GMT -5
Thank you, everybody, for the quick replies and interest.
I misled some of you concerning the pots. Let me offer some details.
I've got a relatively new Epi LP. I've replaced the factory PUs with a Gibson 500T for the neck and just installed a cheap Duncan Designed Basslines Detonator for the bridge I picked up for 24 bucks American, just to see what it sounded like. I figured it was worth 24 bones if only to appease my curiosity. BTW, the Detonator came as a factory bridge PU in the '06 Epi LP Studio. Its an Invader clone with less treble. On line, most people hate this PU. But, for the same reasons people dislike it, I was attracted to it.
Getting back on topic.
I have one vol pot for the bridge wired as a standard vol pot, one vol pot for the neck wired as a balance pot between the neck and the bridge, one pot (with a huge capacitor) for a master tone, and one master pot for an EMG Afterburner.* The PUs are in Series for a fat, full tone with higher output. Both vol pots are push-pull for shorting the circuit as an off switch.
Problem: the PUs both together are the same volume as just one PU. I wasn't surprised, as I already knew about this "load" problem. But, neither vol fixes the problem like it did for my old Guitar.
As for my old guitar, It had the Gibson 500T, a Peary Gates B, and the original three singles from a Squire Srat, all in series, three vol pots, three tones, a phase reverse toggle, the Afterburner, and a mini momentary off push button.* When I backed the vol off to 9, it got louder. Its been so long I can't remember if this happened for all vol pots, so it may have been a bad pot as mentioned already.
The thing is, I remember reading a forum, somewhere, where somebody wrote about two pickups not having a "load" causing a reduction in output. I'm not quite clear about what is meant by "load", I'm guessing it either means pulling current through a device or pushing it. And, that including a small amount of resistance after the component would solve the problem. I imagine it would be like a coil increasing voltage in the way a thumb over a garden hose increases the stream. In practice, it all seemed to work that way on the old guitar. But, it doesn't seem to be working on the new guitar.
In sum, I don't intend on removing pots, I just thought it would be better to put a small resistor after the PUs to provide that "load". I would need to know how much resistance would maximize the output as too much or too little would miss the point of maximization.
There is someone out there that knows what I'm talking about 'cause I read his post, so please don't dismiss me as schizoid. I know it all sounds counterintuitive, but that "someone" can explain this phenomenon much better than I.
Thanks again, everybody.
* I know most of you are thinking "What the hell? He needs an Afterburner with PUs that hot!? All I can say is: I have unusual (read immature) taste in tone. It's my guitar and that's what I like.
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Post by JohnH on Sept 10, 2010 19:12:00 GMT -5
very sorry but Im not getting this - I cant relate what you are describing to a known effect, although it sounds interesting indeed!
John
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Post by ashcatlt on Sept 11, 2010 0:09:56 GMT -5
I wouldn't call you schizoid, but you're at least a little bit Nutz! We like that around here.
I don't want to sound like we've got the be-all-end-all of guitar electronics knowledge around here, but we have (and have had) a number of quite competent EEs around, and I've never seen anybody mention anything like this. I'm not an EE, but it doesn't make sense within my (limited) understanding of the way these things work.
I wish you could point us toward the original post. If you do find anything on this, please bring it around.
Two humbuckers in series should certainly be noticeably louder than either one on its own. My LP gets louder (and darker) when I switch to series mode, even with both Vs at 10. My mini-strat (SHS) gets louder when I put all 4 coils in series and it has absolutely 0 resistors (variable or fixed) internal to it.
Is it safe to assume that you're not testing this through a compressor or so extreme distortion?
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Post by octupletnumber9 on Sept 11, 2010 0:24:44 GMT -5
I wish I could find the original post, too. Yes, I am not testing through compression or distortion. I'll search again for the original post. If I find it I'll post a link. Thank you for the help.
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Post by octupletnumber9 on Sept 11, 2010 2:06:38 GMT -5
AHAA! I found it. It turns out this phenomenon I described applies to parallel PUs, not Series. I'm kind of surprised you guys didn't know about this because it's from the original guitarnuts site eight years ago. This is the answer to my question. It's a quote from John S. Atchley and I hope I'm not violating any copyright laws. . Enjoy. "Recently (June 2002) I received this email... Mayby you can help me with my little problem: I have a gibson es345 ('69) with original PAF humbuckers. When I put the switch in the middle position the guitar sound out of phase, which is very annoying. When I bring down the level of one of the pickups, the sound fattens up, clearly a phase problem. I allready had two repairmen check the guitar but none could solve the problem. The wiring seems to be OK. My repairman told me it has to be a problem with the humbuckers. Have you heard of this problem before, and do you think It could be a problem with the humbuckers? ...to which I sent this response... Does the sound fatten up even if you only back either volume control off just a little bit, like by 10% or so? If so, this is normal for vintage Gibsons. With both volume controls at maximum the two pickups are wired directly in parallel and each applies a fairly high load to the other. As soon as you back either volume control off even just a little you throw 100k or so of resistance between the two pickups which isolates them so they no longer load each other. The 100k of resistance isn't that significant to the high impedance amp, so you get almost full volume from both pickups and the sound is fat. A "cure" for this is to put a 47k resistor between each volume pot and the pickup switch. This amount of resistance will barely reduce your volume at all, but there will always be at least 94k of resistance between the two pickups, preventing them from loading each other. Not all vintage gibsons do this because often the cheap pots they used didn't go all the way to the end of their electrical travel anyway, leaving a little resistance between the pickups even at full volume. I'm glad you brought this up, I need to make mention of this "fix" on the website because I've not seen anyone else note it. Finally, I should add that a similar dynamic is at work in Fender style guitars. When you select the neck and middle pickups together, each is loading the other, likewise when you choose the mid and bridge pickups together (or the neck and bridge on a Tele style guitar). That is why the signal usually gets a little quieter, instead of louder, when you select the combination positions on the selector switch. If you want a different sound from your Strat or Tele try putting about a 20k resistor between the hot lead of each pickup and the selector switch. You'll sacrifice a bit of overall volume (not very noticeable at all if you're feeding a hi impedance amp or effect) but will get a fatter sound in the mixed positions. I'm a little surpised that aftermarket pickup manufacturers don't seem to have tumbled to this. It would be so easy to include a little series chip resistor inside the pickup. I guess it's hard to buck tradition and they have to figure that a lot of people like the Strat mixed positions the way they are. Once upon a time I loved the neck/mid position until I built a guitar that blended the middle pickup and discovered just how thin the mixed positions are compared to what they can be with just a bit of resistance between the pickups."
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Post by borsanova on Sept 11, 2010 5:41:43 GMT -5
I have a gibson es345 ('69) with original PAF humbuckers. When I put the switch in the middle position the guitar sound out of phase, which is very annoying. When I bring down the level of one of the pickups, the sound fattens up, clearly a phase problem. I allready had two repairmen check the guitar but none could solve the problem. The wiring seems to be OK. My repairman told me it has to be a problem with the humbuckers. Have you heard of this problem before, and do you think It could be a problem with the humbuckers? What is described here is absolutely correct, it is not a bug and the repairman has no idea of guitar wiring. It happens because in an out-of-phase config the two pickups are cancelling each other out (at least partially and mostly for the lower frequencies). The effect is most prominent (a very sharp and twangy sound) when both pickups are about the same volume. Now if you lower one of the independent volume controls the cancellation diminishes since the other pickup starts to prevail. Thus the sound gets fatter and louder. This is possible only in a Les Paul config with independent volume pots for each pickup. It is not a wiring problem and can be very useful to achieve very different sounds without switching, just by turning your volume knobs. I use it for switching between rhythm and solo parts. I use the 10/10 out of phase for rhythm and when it comes to the solo I simply turn down the neck volume to 8 or 7 in order to have a full humbucker bridge tone. All the other suppositions like pickups loading each other are AFAIK not involved. Btw: The same principle is used in modern astronomy: When you want to observe a faint planet close to a much more brilliant sun you use an out-of-phase config of your observatories to cancel the sun out and maybe get a glimpse at the planet.
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Post by ashcatlt on Sept 11, 2010 9:40:03 GMT -5
I vaguely recall having read that somewhere in the past. I don't think he's actually talking about what we normally call OoP here, though it would work there for the same reason. Bor is right that an out of phase combo is thinnest when both pickups are close to the same volume.
I wonder about the actual explanation, there, though. I always thought that it was the parallelization (reduction) of the overall resistance which caused two coils in parallel to be quieter. But what do I know.
It still doesn't adress your issue, though.
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Post by borsanova on Sept 11, 2010 11:04:48 GMT -5
I don't know if octu is actually talking about oop, but the source he cites certainly does. I have a 345 copy with oop switch and it does the same thing. I wonder about the actual explanation, there, though. I always thought that it was the parallelization (reduction) of the overall resistance which caused two coils in parallel to be quieter. But what do I know. As for the explanation, you can imagine your string signal like a diagram curve with peaks and downs much like you can observe on an equalizer. When you put two pickups in parallel, their curves (signals) mix, if you put them in series they add. But if you put one of them out of phase, it is as if one of your curves was flipped upside down, that is the two signals subtract. It is a wonder that there is still so much signal left. Probably it is due to the fact that pickups in different positions pick up quite different string signals. This seems to be more true for the higher frequencies and that is the reason why the higher frequencies are retained, while the lower ones are almost completely canceled out.
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Post by ashcatlt on Sept 11, 2010 14:07:14 GMT -5
Nah, the dude that wrote the original question (which Mr Atchley attempts to answer) said it was almost like the pups were OoP. If this was actually the case, though, it'd be much more noticeable than a little bit of apparent tone loss.
As I understand it, two coils in parallel add the same as in series, but the lower overall resistance means that for a given current output, the voltage must be less.
Any time you've got two coils sensing different parts of the string, you are more likely than not to have at least some of the harmonics out of phase between the two. Some harmonics will interfere constructively (add) and some destructively (subtract). The fundamental is always moving the same direction all the way along the vibrating length of the string. When we say that one pickup is Out of Phase with the other, we mean that the fundamental is inducing a positves voltage in one while it's negative in the other.
Lower harmonics propagate along longer stretches of the string, so it's sort of more likely that they will be moving the same direction above both pickups. That's a function of the vibrating length of the string. That is, it depends where on the neck one is fretting, and is difficult to predict.
Anyway, it's more likely that the fundamental and lower harmonics will cancel in an OoP config, which makes it thin. The higher harmonics add or cancel in a more scattershot way, creating something like a comb filter. This happens to some extent, though, even when it's wired "in phase" as any standard strat will demonstrate.
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Post by borsanova on Sept 11, 2010 15:48:25 GMT -5
The guy said his 345 sounded like out of phase only when the two pickups were used together. Together with the description of what happened when one of the volumes was turned down to 9 or 8 this is the proof that they were actually parallel out of phase. Probably a previous owner wired her up in the well-known B.B. King manner.
As for the rest we're saying pretty much the same things with different analogies. But you are talking about string phases and I'm talking about electrical current phases. Since the first translate into the latter the difference it not so big, but it exists. In particular, only in an electric current the oop-cancellation could, at least in theory, completely cancel the signal.
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Post by JohnH on Sept 11, 2010 15:49:16 GMT -5
I tried adding those 47k resistors to an LP a few years ago, for a different reason, in an attempt to control the often described issue of one volume pot turning off both pickups. Later I realised that this was a problem that did not need solving.
It had two effects that I did not like:
Although it did not reduce the overall output by much, it started to dull the tone by reducing the resonant peak in the high treble.
Also, relevant to this thread, it does indeed stop the interaction between the two pickups and you just seemed to get an average of the two signals, like they had been blended in a mixer. It loses the very interactions that make these direct parallel settings interesting and unique, different from just an in-between version of the two pickups. That effect is what Ash describes above.
But try it, YMMV! - comparitive sound clips would be interesting
cheers John
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Post by ashcatlt on Sept 11, 2010 16:58:36 GMT -5
Well bors, I'm not trying to argue. Just discussing/trying to illuminate a little.
The thing is that it doesn't really matter whether we're talking OoP or not. The two coils in parallel interact - either by loading each other or by reducing the overall resistance (or both) - in such a way that the output is not louder, and may be quieter than either pickup by itself.
What you seem to be talking about is possibly more correctly termed absolute polarity. When the string moves toward the pickup does it create a positive or negative voltage?
Phase is a frequency dependent phenomenon. Two signals of opposite absolute polarity will cancel one another to the extent that:
A - the various frequencies are phase coherent (moving the same direction at the same time) between the two sources, and
2 - the various frequencies are close to the same volume.
When we're talking about two (or more) coils on a guitar, it's very rare for either - let alone both - of these conditions to be met. Flipping the phase on one coil from an HB - especially an SC sized HB - will get you very close to 0 output, but even then it's not complete cancellation. My mini-strat will do this, and I've tried it, and it ain't silent, though it's mostly unusable.
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Post by borsanova on Sept 11, 2010 19:13:54 GMT -5
I'm not an expert on interaction between two parallel pickups, and I admit that I don't really understand how the "loading" effect works.
But I have listened very carefully to all kinds of pickup combinations on my 22 Dual and this is what I found.
First, you are right by saying that 2 pus in parallel may appear to have less volume than each one of them alone. Certainly their output will be smoother, which may contribute to the impression. If there is really a loss in overall output (and there is no reason why not believe it), it is however much less perceivable than in any oop configuration.
Since we agree on what happens with out of phase, I want to investigate a bit more into what happens between two pickups in parallel (in phase), exposing what I found on my guitar.
My Twenty Dual offers pickups with very different impedance, especially in the bridge position where I have the humbucker with about 14 k, the single coil with 7 k and the dual with overall 3,5 k. If I put these in parallel with the neck pickup (7,5 k), their interaction varies noticeably. What happens is that the pickup with lower impedance prevails in the mix. Thus with the humbucker I hear much more of the neck pickup's signal and with the dual I hear more of the bridge pickup, while with the single coil both pickups seem to be evenly matched. Note especially that in the parallel mix with the neck pickup my bridge dual pickup results not only louder than the single coil, but even much louder than my humbucker. These things get even more evident if I use the neck coil tap and combine for instance the neck single coil (4k) with the bridge humbucker (14k). In this case the bridge humbucker is almost completely overshadowed by the neck pu.
When I first noticed these interactions, I thought about using resistors in order to have my pickups evenly matched, but then I saw that it would have been far too complicated. So I gave up on it and accepted the situation. In some case it is possible to use one of the volume knobs to correct a bit, especially when I'm looking for the thinnest sound in out of phase mode, but usually I won't take the time, except sometimes when I'm recording a reggae rhythm or something similar.
I remember that, time ago when I first exposed them, these results of my studies were criticised by an electronics expert among the board users (maybe not on this board, in any case I don't remember the name), but they are empirical findings and as that they cannot adapt to theory, but theory must adapt, trying to explain the evidence.
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Post by sumgai on Sept 12, 2010 3:38:59 GMT -5
OK, that's it, the kid gloves are comin' off. Besides, I can hear Chris "Pinwheel" Kikta clear over on this side o' the continent! ash, you, out of the whole bunch here, should remember one word - filtering. borsa, you should read on for a mini-course in inductive interaction (two pickups in parallel or in series) John, you yourself should have remembered your most informative post, linked here: Modern and Vintage wiringocto, Man, if you werent a newbie here, I'd turn you over to newey for some well deserved lashes with a wet e-string! In your first post, where you alluded to having a previous guitar that worked better by turning down one of the volumes, you specifically failed to mention "vintage wiring". Now you may not have been knowledgable enough at that point, but you are now (I hope! ), and therein lays all of the reason for everything said since Reply #1. Yes, much of it was informative, and some of it was germane, but it all missed the mark until you finally quoted the original post (from guitarnuts.com). I'll simply state that if you had said "vintage" early on, at least one of us, and probably most of us long-timers, would have tumbled immediately to what you were talking about. ,,,,,,,,,,,,, Now, having dished out the brickbats, let's get back on track. The first problem here is easier to deal with. When volume pots are wired "backwards", they tend to load a pickup (or several of 'em) in an ever-changing manner. That is, when you rotate the wiper, the load on the pickup changes. Accordingly, this does indeed affect the tone. Not so much for "standard" pickups, but hotter pups are more affected, thus there is a more noticble change of tone as the wiper is rotated. Now, if the pots (controls) are wired "forwards" or normally, the load on each pot is unchanged, as the wiper is rotated. This leads us to the fact that if a pickup's load is changed, and therefore the tone is altered, then if the load is unchanged, it must follow that the tone is unaltered. From that, we see that this is not an effect that can happen only in Les Pauls - any multi-volume-control guitar can suffer this malady, provided that it is wired "backwards". Let me go two steps further here...... In series - the impedance of each coil is added together. Now, since coils have a tendency to pass low frequencies and cut down on high frequencies, it's apparent that the net effect will be that we hear less highs (than with just one pup). We tend to think "more chunk", or "fatter tone", etc. This is because the mids appear more prominent to us, since we are now missing some of the highs. So what happened to the volume? Simple - more impedance means less load (we're closer to having no load) on the pups, which in turn means greater output.* Slightly more, but enough for most players to be able to tell the difference. In a parallel connection, impedances divide down algebraically. This lower impedance works to allow more high frequencies passing through, thus we hear "more sparkle" or some such. Sadly, since we're also putting more load on the pickups, the output level also declines a small but noticable amount. The other factor I alluded to is that of filtering. When a coil is in the circuit, there is a certain amount of filtering - the frequency response is no longer flat (linear). (We already know this from our studies of how a capacitor operates, so much of this shouldn't be new.) What happens when two coils are in a circuit? Well, in a perfect world (easily purchased with a mere 10lbs of Unobtanium), the two would react the exact same way to whatever frequency was present, and depending on series or parallel, the net effect would be increased at the appropriate freqencies. However, no guitar is perfect (else how could we come to be Nutz, eh?). The very reason we have more than one pickup is to capture, and transmit onwards, the string's vibrations occuring at differing points (nodes). These nodes have different harmonic strengths and timings, as I'm sure you'll all agree, yes? Then it's "merely" a matter of determining how Coil B is affecting the frequencies generated by the string at the position of Coil A, and adding that determination to how the frequencies found at Coil B are affected by Coil A, right? Pretty much. In a nutshell, to be sure. Not for the faint of heart, but if one studies some basic AC theory, it all starts to make sense. Trust me on this. The SummaryA pair of coils in series will cut more highs, but present less load to the pickups, thus giving a slightly higher output. (Because of greater inductance/impedance.) A pair of coils in parallel will allow more highs to pass through, but load down the pickups a bit more heavily, thus reducing the output some small amount. (Because of less inductance/impedance.) There are probably some other items in this thread that should be addressed, but I think I've done enough damage for awhile. ;D HTH sumgai * The word "load" can have two ways of looking at it. For our purposes here, I'm using the classic Engineering definition. The best example of how this works would be to think of using a 250KΩ pot versus a 500kΩ one - the larger value tends to yield more highs and greater output, yes? That larger value is said to have put less of a load on the pickup. Here's another way to think of it - as you reduce the load towards zero ohms (get closer to a direct short), you make the pup work harder and harder. (This is exactly like you shouldering a sack of cement (a load) and carrying it across the room - the greater the load, the harder you have to work, right?) As you increase the impedance away from zero ohms, that's less of a load - like carrying a candy bar instead of a bag of cement. When you get to the point of no load at all.... think "blast switch" here, and everything should start making sense.
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Post by ashcatlt on Sept 12, 2010 4:01:15 GMT -5
Or, think of Old Mr. Ohm and his law. V = I*R. I = V/R. If voltage remains constant, a greater resistance reduces the amount of current demanded of the pickup. A lower resistance demands more current. In this case, that current demand is the load.
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Post by borsanova on Sept 12, 2010 4:37:13 GMT -5
The SummaryA pair of coils in series will cut more highs, but present less load to the pickups, thus giving a slightly higher output. (Because of greater inductance/impedance.) A pair of coils in parallel will allow more highs to pass through, but load down the pickups a bit more heavily, thus reducing the output some small amount. (Because of less inductance/impedance.) Or, think of Old Mr. Ohm and his law. V = I*R. I = V/R. If voltage remains constant, a greater resistance reduces the amount of current demanded of the pickup. A lower resistance demands more current. In this case, that current demand is the load. Thank you both. These words make sense and got me clarified a few concepts. Now this also explains why, with an unevenly matched pair of pickups in parallel, the one with lower impedance prevails over the hotter one.
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weelamm
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Post by weelamm on Jul 19, 2011 15:42:05 GMT -5
Sorry for being late in the thread, but I have a newbie question.
In OP's second post, where they finally release the details of their setup, they define their 2nd vol-pot as "one vol pot for the neck wired as a balance pot between the neck and the bridge".
Does this not imply it is a blender pot, and not a true volume control? Certainly this would behave as a load-shifter and not contribute directly to the desired effect of having an increased output by backing off slightly on 1 of the volume controls. In this setup, there is only 1 true volume control (ignoring the Afterburner of course0.
Imo, in a normal 2 volume control rig, slightly increasing the resistance in one of them would encourage more current to travel through the control with less resistance which would indeed show a small increase in volume, at least when both were nearly at the same potential.
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Post by newey on Jul 19, 2011 18:31:56 GMT -5
Weelam-
Hello and Welcome!
I didn't read the statement that way, although you're right, it's ambiguous. I thought they meant that it was a std pot wired to blend the neck into the bridge.
But I also think that you're right, either way. Whether it's a true blend pot or a std. pot, it's not exactly the same scenario as would be with an LP with individual volume controls in parallel. When one wires a regular pot as a blend, that's like a series resistance with the other pickup, at least as I understand it (which, admittedly, may be part of the problem . . .)
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