# Does anyone calculate cutoff freq of their low pass (tone) controls?

Discussion in 'Tele-Technical' started by JD0x0, Jan 27, 2015.

1. ### JD0x0Poster Extraordinaire

Age:
28
Feb 22, 2009
New York
I was wondering if any other members here, actually calculate the cutoff frequency, when they're selecting a tone cap, for a low pass filter in their guitar(s)
Those of you, who do, what do you generally like the corner frequency to be at, and also, why was that your choice? Do you like different roll off points, for different guitars?

I have a feeling most members just throw a cap in, because it 'works' in another guitar or a similar setup. Is anyone as picky as me, when it comes to their filter? I've found, where the cutoff frequency is, makes a pretty distinct difference in the rolled off tone.

R-C filter calc here...
http://www.muzique.com/schem/filter.htm

2. ### waparker4Doctor of TeleocityAd Free Member

Nov 9, 2011
I don't think that R-C filter applies to passive guitar tone controls.

For one thing notice that the filter in the calculator has either the cap or the resisitor in series with the signal. The tone control is a cap and a resistor together, and the whole unit is in parallel w/ the signal.

Plus a pickup is a high impedance device, and then there is the pickup's RLC circuit complicating matters.

From my experience, a tone control works like this:
At max, it is almost like it's not there at all, and 500k vs 250k doesn't make much of a difference.
As you roll it off, the resonant peak of the pickup diminishes.
Then you roll it off more, the treble cutoff point starts to go down in frequency.
At minimum, the resonant peak comes back, but it is at a lower frequency.

3. ### dsutton24Poster ExtraordinaireGold Supporter

Dec 29, 2010
Illinois
That calculator is of no use in a guitar tone control. If you plug in .047µF and 250K you get a cutoff frequency of 13 hz, well below the threshold of hearing. 500 ohms and .047µf yields about 7 KHz, which is 3.something octaves above middle C.

I doesn't work because that simple formula disregards the huge inductance of the pickup, as well as the capacitance and resistance of the pickup itself, and accordingly the pickup itself has a resonant frequency. And, that's even assuming that you're dealing with only one pickup.

There was a gentleman who used to hang out here and pick apart all technical threads without regard to things like facts... Well, anyway I set out to calculate the knee frequency of a guitar passive tone control and quickly discovered that it's a much more complex problem than you'd think. After a long session of nodal analysis and self flagellation I think the answer ended up being something around 700 hz with 500 ohms dialed in, and about 2500 hz with 250K . I looked around and can't find it now, if you search long enough, you should be able to find it.

As for most people 'throwing a cap in', there's no substitute for experimentation, especially in a case where somebody might be advocating the use of calculations that is meaningless. It's not like anyone is starting from scratch and just going through their entire collection of caps, there are plenty of well established values that work, and they're a good place to start.

4. ### JD0x0Poster Extraordinaire

Age:
28
Feb 22, 2009
New York
That's because the POT isn't the resistor. The PICKUP is the resistor. Plug in a .022uF cap with a 7000ohm resistor. Cutoff, looks about right. Same goes for the other caps, when you use the pickup's DCR as the resistor. NOT the pot's value.

here's an example. We'll use 6000 ohms as our resistor.
.015uF cornerfreq=~1770hz
.022uf =~1206hz
.033uF=~804hz

I mean, I could be wrong, but seems to fit in right around where it should be. I guess it could be a coincidence, though.

5. ### dsutton24Poster ExtraordinaireGold Supporter

Dec 29, 2010
Illinois
Huh? So what did you do with the inductance of the pickup? Inductance has an impact on filter response, it's typically around 7 Hy, that's relatively huge. Inductive and capacitive reactance are out of phase with each other, and that phase angle is affected by resistance.

I'm telling you that a simple RC calculation does not give you any meaningful picture of what a guitar tone circuit does. Look at your numbers, the .033µf (typically you'd use a .047 with single coils) yields a knee at something above C above middle C. Does that even make sense from a musical standpoint? No, of course it doesn't.

You're trying to make a technical argument that doesn't apply to the situation. You might as well be trying to write a budget while ignoring what your expenses are.

It's deja vu all over again.

RLee77 likes this.

6. ### luapselTDPRI Member

Oct 6, 2011
New haven CT
JD0x0 - you said, "maybe I'm wrong"- you were. The pickup is NOT a 7000 ohm resistor. When they say "7kohm" for a pickup that's DC resistance, not AC reactance, or impedance. AC acts differently toward a capacitor or inductor(coil) than DC does.
Basically, it's like waparker4 said: Those "plug-in formula" pages are NOT for a guitar circuit.
If you want to draw a guitar circuit, you have to start with an inductor (coil)- That's the PICKUP. Then, IN PARALLEL with the pickup, you must draw a 250k variable pot, in parallel with a .022 if capacitor.
THEN you must calculate the inductive reactance ( XL) , of the Pickup, (2 pi x FxL), to determine what band of frequencies will pass through the pot and cap "RC" circuit. THEN you can determine what capacitor- will give -what "cutoff frequency". That's why AC is so much more complicated than DC.
P.S. - don't bother. If you want full power, take the tone cap out. If you really like playing with tone, get a TBX bass-treble cutoff circuit...
Otherwise, stick with 250k pot, .022 cap. Good luck!

Last edited: Sep 10, 2017

7. ### dsutton24Poster ExtraordinaireGold Supporter

Dec 29, 2010
Illinois
You resurrected a two year old, 5 post thread just to tell a long time member he was wrong about a minor technical point? Not a great way to get off to a good start.

Relax a bit.

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