Les Paul too bright

[JoeM]

You would have it in series with the signal.

OK, that makes sense.

[Glass Snuff]

True, but I still believe (and some may disagree with me) that simply having a capacitor in the circuit makes the whole circuit sound slightly different. I did a few tests a while ago between different capacitors and using jumpers and the capacitors always sounded warmer with the larger caps being a bit deeper or looser sounding and smaller caps were tighter in their response. This was with the tone pot full up, but I heard it. Not a huge difference, but enough of a difference that I could choose my favorite material and value out of about ten or twelve different caps.

The science behind this is simple: you will hear what you expect to hear. There's a reason for double blind testing.

Frequency dependent resistance, it's what they do. Resistors are frequency independent meaning resistance is even across all frequencies, capacitors are frequency dependent meaning the resistance at a given frequency is different than resistance at another frequency.

And if you calculate the resistive value of a cap in a guitar circuit, you'll find it's not going to add appreciably to the value of the tone pot on "10". Especially if you use 1 Meg pots.

The 'tone pot as an RC filter' model doesn't satisfy Occam's Razor as it doesn't explain the behavior of a tone pot. A 'tone pot as a Zobel filter' model does. To wit, if you have a .022 or better, a .010 cap, and you turn your tone knob from '2' down to '0', you'll hear the guitar get louder (It happens with a .047, as well, but the increase is slight and harder to hear.). If the tone control was an RC filter, this wouldn't happen.

In essence, the tone control acts as a pan pot between two resonances. There's a null in the middle (as predicted by Zobel) where there's a -12dB rolloff. For the two extremes of the pot's travel, it's simply damping these resonances. As the tone cap's resonance isn't excited until the resistive value drops below the null point, for tone pot settings above "5" the cap may be replaced with a straight wire.

Small value caps work well for bright guitars because they allow the tone control to be turned down enough to tame the brightness without producing mud.

(Yes, I'm repeating what I said in "two humbuckers - 2 vol - 1 tone issue", but that one's got sexy pictures. )

[Mike Sulzer]

The 'tone pot as an RC filter' model doesn't satisfy Occam's Razor...

Occam's razor has to do with the simplest explanation probably being the correct one. It is not about explanations that simply do not work.

Capacitors are not frequency dependent resistors. They have an impedance that varies with frequency, but the voltage lags the current by 90 degrees. Therefore they do not dissipate energy as resistors do.

Over most of the range from 10 down towards zero, the tone pot is lowering the resistance across the pickup-cable resonance. It gradually lowers the resonant peak, causing a loss of high frequencies. Panning from one resonant frequency to another is not an adequate explanation except as the control gets close to zero.

[David Schwab]

The 'tone pot as an RC filter' model doesn't satisfy Occam's Razor as it doesn't explain the behavior of a tone pot. A 'tone pot as a Zobel filter' model does. To wit, if you have a .022 or better, a .010 cap, and you turn your tone knob from '2' down to '0', you'll hear the guitar get louder (It happens with a .047, as well, but the increase is slight and harder to hear.). If the tone control was an RC filter, this wouldn't happen.

You are leaving the pickup coil and patch cord capacitance out of the equation. When you turn the tone control down all the way you now have a tuned resonant circuit. You get that resonant hump in the low frequencies which makes it sound louder, but you lost the resonant peak that was in the upper mids.

A Zobel network is a balanced bridge circuit. A tone control on a guitar is hardly that complex. It's also not exactly a RC filter, but it's a lot closer to that than a Zobel network or m-derived filter.

[Glass Snuff]

Occam's razor has to do with the simplest explanation probably being the correct one. It is not about explanations that simply do not work.

The simplest explanation that fits the facts. There's no satisfaction if observable phenomena are not explained.

Capacitors are not frequency dependent resistors. They have an impedance that varies with frequency, but the voltage lags the current by 90 degrees. Therefore they do not dissipate energy as resistors do.

True, but are you offering this as a defense of the "tonepot as an RC filter"?

Over most of the range from 10 down towards zero, the tone pot is lowering the resistance across the pickup-cable resonance. It gradually lowers the resonant peak, causing a loss of high frequencies.

Not over most of the range, as we're using audio taper pots. And I'd specify that we're losing high frequency energy, but not actually reducing the bandwidth.

Panning from one resonant frequency to another is not an adequate explanation except as the control gets close to zero.

It's not an explanation, so much as a metaphor. Since a tone pot is a log pot, it's close to zero over much of its lower range. The actual Zobel resistance value is about 75K, give or take a bit depending on the L value of the pickup.

[Glass Snuff]

You are leaving the pickup coil and patch cord capacitance out of the equation.

Actually, I feel it is those that claim the tone control is an RC filter that are leaving the coil and cord out.

A Zobel network is a balanced bridge circuit. A tone control on a guitar is hardly that complex. It's also not exactly a RC filter, but it's a lot closer to that than a Zobel network or m-derived filter.

A Zobel for a speaker crossover generally consists of a resistor and capacitor, calculated to offset the rising impedance of the voice coil. Really quite similar, albeit reversed. If we dial in the proper value on the tone pot, we get a very well behaved -12dB rolloff, suggesting the L has indeed been damped.

I'm open to other suggestions, of course, but so far this model of a pickup creating an output that depends on its load explains the observed behavior much better than the more common idea that a pickup provides a fixed output and the highs get "bled to ground" though an RC filter.

[David Schwab]

I'm open to other suggestions, of course, but so far this model of a pickup creating an output that depends on its load explains the observed behavior much better than the more common idea that a pickup provides a fixed output and the highs get "bled to ground" though an RC filter.

Caps pass AC current. So when the tone control is on zero it is indeed shunting the signal above a certain fgrequency to ground, determined partly by the value of the cap, and also the other parts in the circuit like the pickup. But when it's not on zero it's doing something different. It's mostly flattening the resonant peak due to the resistance of the pot.

Mike Sulzer explained that very well not long ago in another thread. Maybe he will chime in again.

The cable capacitance works much like the tone control, but with a much smaller value cap.

The reason I said a tone control is not exactly like a RC filter is there is no resistance in series with the signal.

[jakeac5253]

Well since this concept seems to be up for interpretation as opposed to set in stone, may I interject my theory. Since the tone pot setup is in parallel, when the tone pot is full up there is full resistance between the rest of the circuit and the cap. Very little signal goes to the cap, but some still does because of the nature of a circuit. When you roll the pot down you decrease the resistance to the third lug thus increasing the likelihood that the signal will pass through the cap. Since the cap is connected to the bottom of the tone pot which is connected to nothing else in the circuit, this effectively opens anything that passes through it. Now what would get opened when you roll the tone pot down? Anything that passes through the cap. Since we know that caps leave the treble alone and cut the bass, the bass isn't passing through the cap but getting halted at the third lug and getting shorted back to the rest of the circuit while the highs pass to open circuit. Whether the correct terminology is 'pass to open circuit' or 'decreases the pickups inductance at those frequencies' I'm not exactly sure. Changing the value of the cap changes how much of the original signal is left in tact versus how much passes the cap.

[JoeM]

Well since this concept seems to be up for interpretation as opposed to set in stone, may I interject my theory. Since the tone pot setup is in parallel, when the tone pot is full up there is full resistance between the rest of the circuit and the cap. Very little signal goes to the cap, but some still does because of the nature of a circuit. When you roll the pot down you decrease the resistance to the third lug thus increasing the likelihood that the signal will pass through the cap. Since the cap is connected to the bottom of the tone pot which is connected to nothing else in the circuit, this effectively opens anything that passes through it. Now what would get opened when you roll the tone pot down? Anything that passes through the cap. Since we know that caps leave the treble alone and cut the bass, the bass isn't passing through the cap but getting halted at the third lug and getting shorted back to the rest of the circuit while the highs pass to open circuit. Whether the correct terminology is 'pass to open circuit' or 'decreases the pickups inductance at those frequencies' I'm not exactly sure. Changing the value of the cap changes how much of the original signal is left in tact versus how much passes the cap.

Go back and and read the explanations here - these guys know what they're talking about.

[Glass Snuff]

Since the cap is connected to the bottom of the tone pot which is connected to nothing else in the circuit, this effectively opens anything that passes through it.

Fail. The cap is connected across the output of the pickup. Its value becomes part of the LCR of the pickup.

To quote Lemme:
"The integral "heliocentric" view on pickups: Pickup, pots in the guitar, cable capacitance, and amp input impedance are an interactive system that must not be split up into its parts. If you analyze the properties of the parts separately you will never understand how the system works as a whole. The sound material a pickup receives from the strings is not flavoured by the pickup alone but by the complete system. This includes the guitar cable."

The Secrets of Electric Guitar Pickups

[Glass Snuff]

Caps pass AC current. So when the tone control is on zero it is indeed shunting the signal above a certain fgrequency to ground, determined partly by the value of the cap, and also the other parts in the circuit like the pickup. But when it's not on zero it's doing something different. It's mostly flattening the resonant peak due to the resistance of the pot.

With no resistance, the cap is shorting the output of the pickup at certain frequencies. With resistance, a load is provided so the pickup can do some work (again, at certain frequencies).

The "something else" is reducing the resonance of the system until it's quelled, then with further reduction undamping the resonance of the pickup/tone cap combination. I'll admit, in the 4 years I've been aware of this it hasn't really done me much good, but I'm presenting the idea in the hopes that someone more clever than I can conceive of a better tone control.

[David Schwab]

With no resistance, the cap is shorting the output of the pickup at certain frequencies. With resistance, a load is provided so the pickup can do some work (again, at certain frequencies).

The "something else" is reducing the resonance of the system until it's quelled, then with further reduction undamping the resonance of the pickup/tone cap combination.

That's what I just said in the paragraph you quoted.

I'll admit, in the 4 years I've been aware of this it hasn't really done me much good, but I'm presenting the idea in the hopes that someone more clever than I can conceive of a better tone control.

A "better" tone control is easy. Make it active. Every fancy filter circuit will cause too much insertion loss in a passive system.

I happen to like passive tone controls for what they do. Some attempts to make them "better", like the Greasebucket make them worse in my opinion.

I use .02 caps in many of my basses. When turned all the way down they produce a really nice resonance.

[Steve Conner]

Lemme's explanation of how the pickup/controls/cord system works is about the best I've ever seen. And I hadn't seen it before, did he just write it recently?

The passive volume/tone circuit in a guitar or bass is, well to me, an example of one of these classic circuits that "just works". It's a simple circuit that has a musically useful effect. But because it's passive, every part interacts with every other, and when you try to think about it too hard, your head hurts. Is it a RC filter? Kinda. Is it a resonant low-pass filter like a wah pedal? Sort of. It has characteristics of both types of filters, and the blend of the two depends on both the volume and tone pots. And changing your guitar cord has every bit as much effect on your tone as changing your pickups!

You can't use the standard tactic taught in EE school of designing the circuit in stages, all buffered from each other, so that you can design each stage separately without having to bother about the interactions between them. (Well, it probably wasn't a standard tactic back in the tube era, when real EEs walked the earth. )

[David Schwab]

Right, look at a passive tone stack in a tube amp. You can wire one of those up in a guitar, and it will work, but you need a recovery gain stage to get the levels back up, and you will never get the tone from the pickup you had without the tone stack.

So the usual method is to buffer the pickup, and then do your tone stack and recovery amp. Buffered pickups can sound a bit different and some people don't care for that tone, but you can tune the buffer with caps and resistors to get the same kind of response as you get with a passive system.

[Mike Sulzer]

The simplest explanation that fits the facts. There's no satisfaction if observable phenomena are not explained.

Yes, of course it has to fit the facts; we are not discussing religion or politics.

True, but are you offering this as a defense of the "tonepot as an RC filter"?

Of course not. Most folks here are proabably tired of me pointing out that the function of the tone pot is to damp the pickup-cable resonant circuit, with the tone capacitor playing a role only near the bottom of the (resistive) range.

Not over most of the range, as we're using audio taper pots. And I'd specify that we're losing high frequency energy, but not actually reducing the bandwidth.

Over most of the resistive range. When you damp a resonant circuit, you increase the bandwidth, not reduce it; sorry if I was unclear. What you are doing is removing the emphasis on a band of frequencies that we perceive as "high", thus making the bandwidth closer to flat.

It's not an explanation, so much as a metaphor. Since a tone pot is a log pot, it's close to zero over much of its lower range. The actual Zobel resistance value is about 75K, give or take a bit depending on the L value of the pickup.

Let's leave the metaphors to those two topics I mentioned above. This is engineering.

At zero, the tone circuit is indeed again a resonant circuit. But it functions as a low pass filter since the resonnat frequency is low enough so that the major perceived effect is the loss of highs, not lows.