# Two tone controls in parallel?

Discussion in 'Pickups & Electronics [BG]' started by DrVenkman, Nov 1, 2010.

1. ### DrVenkman

Jan 22, 2010
Pacific NW
This is more an exercise in me trying to understand things than anything else; bear with me please.

Suppose we had two tone circuits set up in parallel (see the attached diagram - hopefully I did that right). Now suppose that the two caps have the same value.

If I set the pot on cap 1 at about the middle, it bleeds of a certain amount of "sound" at and above a given frequency (and correct me here and anywhere else if I'm wrong). Now as I turn the pot on cap 2 up, it starts to bleed even more of that given frequence off, right? So I could get the same effect just by turning pot 1 a little more, at least to a point.

But if I have both pots turned to zero resistance, then I've just got two caps in parallel. These add, so now I'm bleeding off a different set of frequencies than before, right? So first question, at what point does that happen? If I've got pot 1 set to zero, is it a really abrupt transition that won't take effect until pot 2 is really close to zero too? Or is it more gradual? Or is there a way to make it gradual?

Second question is what if the two caps have different values? My weak understanding is one would bleed off higher frequencies, and the other bleed of lower frequencies (plus even more of the higher frequencies being bled off by the first cap). Is this correct? And then as both pots reach zero, would it bleed off even lower frequencies?

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2. ### DrVenkman

Jan 22, 2010
Pacific NW
Thread bump in the hopes someone can enlighten me.

3. ### FunkMetalBass

Aug 5, 2005
Phoenix, Arizona 85029
Endorsing Artist: J.C. Basses
In parallel your potentiometers add reciprocally (two 500k pots full-on act as one 250k pot full-on). Both of your capacitors are in parallel, so they add (two .01 uF caps act as one .02 uF cap).

Now, passive tone controls are the same as a simple low-pass filter (AKA: RC circuit), in that they bleed off the higher frequencies as determined by the capacitor value and pot value + pickup inductance.

First question: Yes - it's a different frequency since you have doubled your capacitance. If both of your pots are turned to 100%, nothing is happening. Very little bleeding signal is being sent through the capacitor to the ground. As you start to turn one down, you are increasing the signal to ground, and your single pot acts like a tone pot normally should - how quickly it takes effect is dependent upon its taper.

Second question(ish): If they have different values, your overall capacitance changes. It could shift the cut-off frequency up or down, depending on the capacitance (increased capacitance = lower cut-off frequency). Here's what's going on with a tone pot:

On the left is when your tone pot is full-on (i.e., this bypasses the signal going to the ground). There's a tiny bump at the cut-off frequency (red), but other than that, not much is happening. On the right is when you slowly start turning your tone knob off (i.e., increasing signal to ground). The more signal you send through the capacitor and to the ground, the steeper the slope after the cutoff frequency.

The cut-off frequency is determined by the overall capacitance and overall resistance. So, even if both pots have wildly different capacitor values, each one will simply react as though it has the sum of those pots on each one. If you have two 500k pots with .01uF caps and the pots are in parallel, you can reasonably expect that the cut-off frequency will be the same as one 250k pot with a .02uF cap. Turning each pot full-off will be about half as much effect as turning the 250k pot to its halfway point.

4. ### DrVenkman

Jan 22, 2010
Pacific NW
Now that you've explained it, that makes perfect sense. Thanks!