Tone Pot Frequency Question

[ahren707]

Hey guys,

Maybe a silly question here but I am trying to understand and be able to visualize the frequency change that is happening when I turn my tone pot from 10 to 0. I understand there are different pot cuttoffs, I'm pretty sure mine is a 500k. What I can't seem to visualize in my head is what is actually happening frequency wise as I turn from 10 to 0. Is everything above 500k being attenuated more and more at an equal amount or is the angle of the attenuation increasing thereby cutting off the higher highs more and more relative to the frequencies closer to 500k? Does that make sense?

Thanks
Ahren

[ehdwuld]

the pot is a variable resistor

the cap is a filter


Think of it as a water hydrant and a bucket

As you turn the pot it adds resistance
Forcing the signal through the filter

The filter removes more and more of the signal
By shunting filtered frequency to ground

A bucket with a row of 1/4 inch (5.5mm) holes

The size of the holes is the value of the cap
if the faucet it all the way open

The more signal comes into the bucket than can come out the first hole at the bottom of the bucket

More pot equals more flow through outer filter the bucket

And more tiny holes start sqirting water

That may be confusing

I apologize

[eclecticsynergy]

Hey guys,

Maybe a silly question here but I am trying to understand and be able to visualize the frequency change that is happening when I turn my tone pot from 10 to 0. I understand there are different pot cuttoffs, I'm pretty sure mine is a 500k. What I can't seem to visualize in my head is what is actually happening frequency wise as I turn from 10 to 0. Is everything above 500k being attenuated more and more at an equal amount or is the angle of the attenuation increasing thereby cutting off the higher highs more and more relative to the frequencies closer to 500k? Does that make sense?

Thanks
Ahren

The pot changes only the amount of rolloff as you turn it.*
It's the cap which determines the cutoff frequency.

A large cap like 1μF will cut off a lot of midrange along with the treble. Smaller caps leave more midrange and definition, while still rolling off the highs.
A very small cap will roll of just the air & sparkle while lowering the resonant peak. This can be useful in certain guitars because it can seem like a midrange boost.

*EDIT: People sometimes choose a lower value pot, but that doesn't change the cutoff frequency.
A smaller value pot like 250K sounds darker than a 500K because it allows a bit more rolloff even when it's wide open.

[freefrog]

Hey guys,

Maybe a silly question here but I am trying to understand and be able to visualize the frequency change that is happening when I turn my tone pot from 10 to 0. I understand there are different pot cuttoffs, I'm pretty sure mine is a 500k. What I can't seem to visualize in my head is what is actually happening frequency wise as I turn from 10 to 0. Is everything above 500k being attenuated more and more at an equal amount or is the angle of the attenuation increasing thereby cutting off the higher highs more and more relative to the frequencies closer to 500k? Does that make sense?

Thanks
Ahren

See the following pic (I think it was initially posted on the Duncan forum):

https://nextgenguitars.ca/product_images/uploaded_images/tonepotfreq.png

What it shows it that a typical tone cap really reshapes the resonant peak of a guitar pickup when set very low .

During 66% of its curve or more, the tone pot flattens the resonance mostly because of its decreasing resistance. So much that a tone pot set @ 3.5/10 wouldn't behave much differently with or without tone capacitor...

Counter-intuitively, the caps doing the most difference when a tone control is full up are those with a very low capacitance (a few nanofarads)... It's not to say that "normal" caps of 100nF / 47nF / 33nF / 22nF don't influence the tone at 6.5/10: they do, by filtering differently the harmonics. But the whole "EQing curve" of the pickup doesn't really change whatever is the cap used, during most of the taper.

Also, the corner / cut-off / roll-off frequency is not fixed with a tone pot: it depends partly on the inductance of the pickup used. The highest the inductance, the darkest the tone, when the tone control is full up but also when it's at 0.

IOW, a 500k Ohm pot it not what will set the resonance at 500hz: if this happens, it's due to the value of the tone cap finally cooperating with the inductance of the pickup to shift down the resonance (instead of simply flattening it as it does for the most part).


Below is a botched 5spice sim allowing to visualize all this. For those accustomed to my pics: this time, I've not "normalized" (flattened) the response . I wanted it to look more like the curve in the other web page (which is also how the "crude" resonant peak of a pickup looks when it's electrically induced).

[Teleplayer]

It’s not easy to visualise even if you understand the theory of passive filters. It’s because the tone control has two different mechanisms at play.

1) For the top part of the tone control, typically positions 6 - 10 on an “audio” pot, it works as adjustable Q filter to control the LC resonance of the pickup and guitar cable combined. It’s this resonance that gives a pickup it’s characteristic brightness, so changing the Q is a powerful effect.

2) In the bottom half of the tone control the tone cap value has a bigger influence on the tone. It becomes effectively a RLC low-pass filter where the “C”element is the tone cap.

Edit: I wrote this a couple of hours ago but I was slow to post it, while freefrog has written practically the same thing.

[freefrog]

Teleplayer : thx for sharing. Spot on sum up IMHO and you evoke log / audio pots, that I had not mentioned. Not to mention that it's nicer not to talk alone in the lounge... ;-)


As a footnote to my previous answer: a simple way to check which frequencies are affected is to play a pickup direct to the mixing board, to record it and to pass the whole track through a frequency analyzer stacking the results. Should make obvious what happens exactly.

Below is the spectrum of a bridge Bill Lawrence HB recorded like that (played in chords from unfretted strings to 12th fret).
Volume full up. Tone pot full up on the upper left, then tone pot @ 0/10 with a kind of Q filter (bottom left pic), with a non typical low value 2.2nF tone cap (upper right screenshot) and with a regular 22nF cap (bottom right). Should put my 5spice sims in perspective...

[Chistopher]

Also worth noting that if you put a resistor in series you can decrease the lower range of your tone control. I have a 4k7 resistor on my Strat's tone control with a 10 nF cap which get's rid of the strange behavior that usually happens on settings less than 3.

[hamerfan]

Also worth noting that if you put a resistor in series you can decrease the lower range of your tone control. I have a 4k7 resistor on my Strat's tone control with a 10 nF cap which get's rid of the strange behavior that usually happens on settings less than 3.

New to me, but I never use the tone pot lower than 7.

[ahren707]

freefrog Thank you for the photo from the Duncan Forum. That is really what I was looking for. I had no idea that the "natural" resonance peak of the pickup was so huge. Thats a really great way to visualize it for me. Thank you.

[ahren707]

Just wanted to thank everyone for their responses. Really helped me understand a lot of things.

I am going to try to steelman what I think I understood and you guys can tell me if I got it right or not.

Here goes...

The cap is the actual filter that sets the cutoff frequency. The tone pot controls how much of the signal goes through that filter. The actual end result is a combination of the two signals, the one going through the filter and one not going through the filter. (I am picturing this like two parallel busses in a mix that are then summed into one mix bus.) At tone control 10 all of the signal is unfiltered and at tone control 0 it is 100% filtered.

Is this correct?

[Chistopher]

Not exactly.

The cutoff frequency is calculated as f_c = 1 / (2πRC) where R is resistance and C is capacitance. As you rotate the tone control you change the result of this equation. The cutoff frequency is the frequency in which the signal is -3db from the unfiltered value. In short, the potentiometer doesn't just control how much signal gets through the cap, it adjusts how low the bar is for what bass frequencies can get through.

I am blanking on a metaphor for this behavior, but let's say you have a pipe with water in it and that water has certain impurities. smaller impurities are bass frequencies and larger impurities are treble. The way you described it would be to have a valve that splits between two paths, one path is unfiltered and the other path filters out particles larger than 1 inch. No matter what the valve is set to, all particles under 1 inch will get through. The way it actually works, and this is where it gets a little hard to describe, would be like if adjusting the value also adjusted how strict the filter is. With the tone at 10, all signal passes through the unfiltered pipe, with the tone at 5 half the signal passes through unfiltered and the other half filters out particles larger than one inch, with the tone at 0 all the signal passes through the filtered pipe with no particles larger than 1/2 inch getting through.

If someone can find a better metaphor, please elaborate

[ahren707]

Chistopher Ok, so it changes how much is going through the filter and the cutoff frequency of the filter. And is this because the more signal you send into the filter the more resistance there is?

[ahren707]

From what I'm reading it is basically an EQ sweep, with an attenuation of 10 DB per Decade or a 10 DB reduction for every 10x in frequency above cutoff? When tone knob is at 10 is let's all signal through unaffected and when tone pot gets turned down it slowly lowers the cutoff frequency until it reaches is lowest possible frequency which is determined by the size of the capacitor?

does this sound right?

[freefrog]

When tone knob is at 10 is let's all signal through unaffected and when tone pot gets turned down it slowly lowers the cutoff frequency until it reaches is lowest possible frequency which is determined by the size of the capacitor?

does this sound right?

No it doesn't work like that concretely.

At 10, the resistance of your tone pot still affects the signal unless you use a no-load tone pot.

From 10 to 0, the resonant peak of your pickup(s) is firstly dampened by the resistance of the tone pot (and again, it wouldn't act differently without capacitor when it comes to the shape of the resonance and therefore to the related sound. If you want to check that by yourself, set a tone pot at 6 then put a jumper across its cap and listen: even with the cap shorted by the jumper, you should hear almost identical sounds except some "nuances" in the harmonics, beyond the resonant frequency). THEN, when the resistance bettween pickup and cap reaches a very low value (because the pot is lowered), another resonant peak appears, much lower pitched, and due to the value of the cap + the inductance of the pickup.

See the screenshots that I've shared about a simulation posted once on the Duncan forum, about my own simulations, and about a pickup played direct to the board with pot full up or @ 0/10. If it's not enough, there are similar explanations and pics on other forums.
For instance, see the post 67 here:

https://www.talkbass.com/threads/my-1970-fender-jazz-022-capacitor-compared-to-047.1145058/post-18265479

[EDIT - I had shared another link, with an interesting screenshot but what appeared to me as a misleading comment... Might make more sense to share the following link, giving access to a useful tool. I don't know if it still works but when it does, it's really helpful :
https://guitarnuts2.proboards.com/th...nse-calculator ]

Unlike mines, these curves are flat until resonance because they are based on other theoretical models of pickups than my pics above. But they all show the same thing about how tone controls work.

[uOpt]

There is a fundamental difference between a tone pot all the way closed and not.

If the pot is all the way closed the capacitor is simply in parallel with the pickup coil. Since the pickup coil has its own capacitance, and since that capacitance is part of a second-class low pass filter with resonance peak you simply lower the frequency (but not the amplitude) of the resonance peak. Typical tone pot capacitors are way too high value to make this a universally useful sound.

A partially closed or all open tone pot dampens the amplitude (but doesn't lower the frequency) of the resonance peak.

That is why there is this sudden change in behavior once you close the tone port all the way.