Humbuckers- highs lost from the volume pot. Should I hack this 355' some more?

M

mixsit

Well-known member
I guess it's pretty common that as you go off of wide open on the pickup volume these things drop some high end. I've sort of wanted to find a way to get around this off and on for years but never pursued it. Mostly I'm back at it again drifting more and more to the clean side of rock styles.
The guitar is a 70's ES355 that I have long since swapped in split-coil Semore Duncan's and tossed the Gibson vibrato bar for a solid post tail. Presumably any 'collector' status is already dead so that's not an issue.

So I guess the question would be is active' the way to do this?
I don't know if I'd regret it and completely lose it on the tone end?
 
You actually don't have to go active at all, theres a way of wiring the pots so that you don't get the treble bleed through the pot. I'll see if I can pull up a schematic or something, while I do that see if you can find something by doing a search in the Seymour Duncan forums which has a ton of wiring tips.
 
If you want to attenuate the highs, a 0.02uF capacitor can be wired across the output. This can even be done inside a lead if you don't want to hack the guitar.

If you have *lost* highs and want to regain them, check for dry joints, change the strings, or consider a maple instead of rosewood fretboard - all these change the tonal characteristics to reinforce higher frequencies.

Failing that, a multi-band EQ is your friend.
 
One possibility is a high pass cap on the volume pot. It goes between the pickup lug and the output lug on the volume pot. Most commonly, it uses a 1000pf (or .001 mf) cap. Personally, I don't like them, as they tend to make the whole guitar brighter, though I've used them from time to time by going to an even lower cap. I don't mind a 500 pf cap, but lower is better in my mind.

You could also try using a 1 meg cap. But it has the same problem, it makes the guitar brighter when the volume is at full.

Another possibility is to use a buffer amp of some sort (assuming you like the sound of your current pickups). There are a lot of them on the market, but I'd suggest a Bartolini, just because they make the best internal electronics out there. If you have it before the volume control, it would bring the impedence down to the point where you won't notice the impedence loading of the volume control, which of course is what your having problems with.

Then, of course, there is the option of going with a whole set of low impedence pickups. They may or may not be active, but low impedence none the less. Bartolini makes some very nice low impedence passive pickups, or at least they used to. I'm not completely sure about their new guitar pickups, but I would imagine they are still low impedence. And of course you could always go with active pickups, along the lines of EMG's.

The problem with the active options, at least from my point of view, is that you loose some of the interaction you get when you drive a tube amp with high impedence pickups. HOWEVER, if you are mostly playing through a clean amp, or if you usually have one or more pedals between you and your amp, then it isn't really an issue anyway, so do whatever you need to do. For me, I never play completely clean anyway (well, rarely), and I would HATE to lose the dynamic interaction with the amp. For you, it sounds like the issue is different from mine.


Light

"Cowards can never be moral."
M.K. Gandhi
 
Light said:
One possibility is a high pass cap on the volume pot. It goes between the pickup lug and the output lug on the volume pot. Most commonly, it uses a 1000pf (or .001 mf) cap. Personally, I don't like them, as they tend to make the whole guitar brighter, though I've used them from time to time by going to an even lower cap. I don't mind a 500 pf cap, but lower is better in my mind.

You could also try using a 1 meg cap. But it has the same problem, it makes the guitar brighter when the volume is at full.
Click to expand...

How could that be? When the volume is all the way up, the pickup lug is shorted to the output, so the bypass cap would be shorted as well.
 
I found a section in Semore's Q&A, #116. I still need to digest this stuff! :) Lot's of good info.
Thanks a bunch for getting me started guys.
Wayne
 
ggunn said:
How could that be? When the volume is all the way up, the pickup lug is shorted to the output, so the bypass cap would be shorted as well.
Click to expand...
That's only true for a pot thats set up to completely short to the hot of the output when turned all the way up.

Normally when the volume is all the way up the resistance to the ground is the same as the resistance to the hot if the volume was all the way down. This means with the volume all the way up theres 1 meg ohms of resistance for the highs to bleed to the ground.
 
Light said:
One possibility is a high pass cap on the volume pot. It goes between the pickup lug and the output lug on the volume pot. Most commonly, it uses a 1000pf (or .001 mf) cap. Personally, I don't like them, as they tend to make the whole guitar brighter, though I've used them from time to time by going to an even lower cap. I don't mind a 500 pf cap, but lower is better in my mind.

You could also try using a 1 meg cap. But it has the same problem, it makes the guitar brighter when the volume is at full.
Click to expand...
Hmm. I sure do not want for more high end wide open. If anything I could actually go for a bit more weight on the bridge p/u in the dual coil mode, but that probably gets into more of the p/u that I stuck in there.
Does a capacitor like this make a high pass filter, thus less lows overall? Would the highest (1 meg) roll the most off or least?
Wayne
 
mixsit said:
Hmm. I sure do not want for more high end wide open. If anything I could actually go for a bit more weight on the bridge p/u in the dual coil mode, but that probably gets into more of the p/u that I stuck in there.
Does a capacitor like this make a high pass filter, thus less lows overall? Would the highest (1 meg) roll the most off or least?
Wayne
Click to expand...

No, it's a bypass, allowing the highs to bypass the effects of the volume pot. The lower the capacitance, the higher the lowest frequencies that get through it.
 
ggunn said:
How could that be? When the volume is all the way up, the pickup lug is shorted to the output, so the bypass cap would be shorted as well.
Click to expand...



You still get some high freqencies bleading to ground. I mean, there is still a path to ground there, and you must remember that impedence varies with frequency. The impedence of the pot is lower for high frequencies than it is for low frequencies, so when you get to frequencies which are high enough, the pot may as well have an impedence of zero.



Light

"Cowards can never be moral."
M.K. Gandhi
 
mixsit said:
Hmm. I sure do not want for more high end wide open. If anything I could actually go for a bit more weight on the bridge p/u in the dual coil mode, but that probably gets into more of the p/u that I stuck in there.
Does a capacitor like this make a high pass filter, thus less lows overall? Would the highest (1 meg) roll the most off or least?
Wayne
Click to expand...


As ggun said, it just allows some of the highs to bypass the volume pot.

Honestly, I think the solution which you are likely to be happiest with is a buffer amp in your guitar. Unfortunatly, if you want to keep your guitar's two volume setup, you would need two buffers, one for each pickup, because you want it before the volume controls.

I'd recomend the Bartolini AGDB dual buffer IF it has dual outputs (which I'm not sure about, I'd have to ask my rep). If it doesn't, than two AGB buffers would be ideal. There is one problem with this recomendation - these are not "short list" products from Bartolini, so they can take up to six months or even a year to get them. Special orders from them just take a long time, and they can be difficult to get ahold of, but I'm reasonably certain that you would be very happy with the improvement in your sound.

Or, if you know how to use a soldering iron, you could always MAKE YOUR OWN.



Light

"Cowards can never be moral."
M.K. Gandhi
 
Light said:
You still get some high freqencies bleading to ground. I mean, there is still a path to ground there, and you must remember that impedence varies with frequency. The impedence of the pot is lower for high frequencies than it is for low frequencies, so when you get to frequencies which are high enough, the pot may as well have an impedence of zero.
Click to expand...

Resistance is frequency independent. Reactive impedance, capacitance and inductance, varies with frequency - as frequency goes up, capacitive impedance goes down and inductive impedance goes up - but resistive impedance does not change.
 
ggunn said:
Resistance is frequency independent.
Click to expand...


Of course it is, resistance is a DC measurement, which has no frequency. When you are dealing with an AC signal - such as, oh, for instance, an audio signal - you are dealing with impedence, which is absolutely frequency depentdent.


Light

"Cowards can never be moral."
M.K. Gandhi
 
Light said:
Of course it is, resistance is a DC measurement, which has no frequency. When you are dealing with an AC signal - such as, oh, for instance, an audio signal - you are dealing with impedence, which is absolutely frequency depentdent.
Click to expand...

It is true that you cannot measure capacitance and inductance by DC methods, but you can most assuredly measure resistance by either DC or AC methods. Put another way, resistance is not a DC measurement, it is a frequency independent measurement.

Impedance is frequency dependent, but impedance is a combination of resistive and reactive components, the reactive components being capacitance and inductance. The reactive components of impedance are frequency dependent, the resistive component of impedance is not. Resistance is not a separate quantity from impedance, it is a component of impedance.

A pure resistor exhibits the same resistance (i.e., impedance with zero value reactive components) to both AC and DC voltages. For example, a voltage divider circuit made from carbon resistors will split up an AC voltage, no matter what the frequency, in precisely the same proportions as it will a DC voltage.

While it is true that every real world passive electrical component has resistive, capacitive and inductive components, I have trouble believing that a 1 meg pot has a big enough capacitive component in parallel with its resistive component that it would appear as anything close to a short to ground at any frequency in the audio range.
 
ggunn said:
It is true that you cannot measure capacitance and inductance by DC methods, but you can most assuredly measure resistance by either DC or AC methods. Put another way, resistance is not a DC measurement, it is a frequency independent measurement.
Click to expand...


Nice in theory, completely untrue in reality. There is simply no way around it, volume pots blead highs, even a 1 meg pot. In the real world, DC resistance is nothing more than a convenent way to judge impedence.



Light

"Cowards can never be moral."
M.K. Gandhi
 
Light said:
Nice in theory, completely untrue in reality. There is simply no way around it, volume pots blead highs, even a 1 meg pot. In the real world, DC resistance is nothing more than a convenent way to judge impedence.
Click to expand...

Whatever. I guess my EE profs were all wrong, as well as the engineering text I consulted last night to make sure I was being as accurate as I could.... ;^)

For the record, I never said that a 1 meg volume pot won't bleed highs when turned all the way up, although the next time I have one of my guitars apart, I'll clip a lead and see for myself; that's easy enough to do. I am taking issue, however, with your assertion that a 1 meg pot has zero or near zero impedance to high frequencies in the audio range. Show me the numbers. What's the single pole rolloff point for that high pass filter to ground? Do you postulate a parasitic capacitance in parallel with the resistance of the pot to ground to explain what you observe? What do you surmise is the value of that capacitance?

I really don't mean to be pedantic, but the theory that you so easily discount as "completely untrue" is every bit as "real world" as any tinkering with the innards of a guitar by a tech with no formal training in electrical engineering. That self same theory is behind the design of all our electronic toys. No offense intended, but I respectfully submit the possibility that the reason the theory seems wrong to you is that you do not understand the theory.

The shoe is on the other foot. Muttley, I feel your pain... ;^)
 
ggunn said:
Whatever. I guess my EE profs were all wrong, as well as the engineering text I consulted last night to make sure I was being as accurate as I could.... ;^)
Click to expand...



First of all, I never said that the zero impedence range would be within human hearing. It would likely be FAR above the range of human hearing, if it is there at all. Second of all, my propensity for dramatic exageration should, by this point, be well established.

Third, your EE professors are not wrong, but they DO simplify things by imagining theoretically perfect parts, as do most electrical designers. Why not? I mean, for practical purposes, it doesn't matter, things still work the way they should; but when it comes to the actual sound of an instrument, we do have to deal with the realities of the components we use. Carbon resistors, including those in the track of a pot, are quite frequency specific. Granted, it is less of an issue with low voltage electronics, such as the inside of a guitar, but it is there.


Light

"Cowards can never be moral."
M.K. Gandhi
 
ibanezrocks said:
That's only true for a pot thats set up to completely short to the hot of the output when turned all the way up.

Normally when the volume is all the way up the resistance to the ground is the same as the resistance to the hot if the volume was all the way down. This means with the volume all the way up theres 1 meg ohms of resistance for the highs to bleed to the ground.
Click to expand...

Uh... AFAIK, guitars all either use either an audio or linear taper pot. Both of those are dead shorts when turned all the way up. What you describe would cut the guitar's output in half, which would be a pretty odd thing for someone to do when designing a guitar, IMHO. :)

Pots lose HF depending largely on the value of the pot. By changing the maximum resistance value of the pot, you're raising or lowering the input impedance of the "fader box" inside the guitar relative to the output level of the pickups. It's basically just like changing the impedance of your preamp. Effectively, you're changing how much the pickup output is loaded down. Since audio only cares about voltage, not about current, optimal signal transfer occurs when you have as little current drain as possible to avoid a voltage drop on the line.

Once you have appreciable voltage drop on the line, everything changes, though. Since you effectively have capacitance to ground by virtue of having a shielded cable, the result is that the resistance between the pickups and the output coupled with this capacitance turns into an RC low-pass filter.

Here's where I'm confused: the cutoff frequency should be proportional to the inverse of the resistance. Thus, in theory, lower resistance should be brighter. In reality, though, the reverse is true. I'm still trying to figure that out.


Some related links:

http://buildyourguitar.com/resources/lemme/
http://www.kpsec.freeuk.com/imped.htm
http://en.wikipedia.org/wiki/Voltage_drop
http://www.play-hookey.com/ac_theory/filter_basics.html
http://www.play-hookey.com/ac_theory/lo_pass_filters.html
 
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Oh, son of a... I do get it. Within a certain range, increasing the resistance increases the brightness because it pulls the resonance peak (just below the cutoff frequency) LOWER, putting it in a range that your ear perceives as useful HF sound rather than airy goodness.

At least I -think- based on what info I can gather that this is what's happening here.
 
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