Question about tone control frequencies...

[aziltz]

hey guys, if you're interested, i started a new thread with some simulation data for the TS as a whole, comparing the clipping amp output to the tone control and so on.

http://www.diystompboxes.com/smfforum/index.php?topic=75775.0

[ScottB]

This is an awesome thread! Many thanks! I joined just so I could pursue an issue with something very similar that is not really addressed in plain terms anywhere that I can find.

Right now I am working on the guitar electronics only, and want to experiment with some on-board circuits. Maybe a box itself is in my future, but right now I am just trying to learn the intricacies.

I believe this is ultimately related to my quest to find a generalized way to map corner frequencies of the guitar's tone circuit. For simplicity's sake, this is in general what I am working with:



All I would like to do is figure out how to determine the -3db corner frequency on paper. Approximations are fine. I thought I knew what i was doing and then realized it did not work at all like I assumed it should. The corner frequency is inversely proportional to the value of C1, but seems proportional to the value of R2.

Since this seems to be very similar to the paradox that Projectile had questioned in this thread, can anyone give me insight on how I go about calculating The corner frequency of the circuit? I know it has to rely somehow on the rest of the circuit interacting, I just can't grasp how.

[johngreene]

R2 can be thought of as a 'shelving resistor'. It limits how much the circuit can attenuate the higher frequencies. You can think of the combination of R2 and C1 as a signal capacitor with an adjustable 'quality'. So your 3dB point is always the same and is a function of R1 (the series component) and C1 (the shunt component). The maximum amount of attenuation is a function of R1, R2, and C1 (near the 3dB point). As you go up in frequency it reduces to just the ratio of R1 and R2.
--john

[ScottB]

Thank you John, but I am still confused so let me break this down into digestible chunks.

chunk 1
If I understand you, and assuming we turn R2 completely off and remove it from the circuit, the Fc will follow the traditional rule of 1/(2*Pi*R1*C1). Is that correct?

chunk 2
Now, in the case of the value of R1 you mentioned the series component, did you mean only the top "half" of R1 depending on where the wiper is? Or did you mean the entire value of R1 top to bottom?

chunk 3
Does the Impedence of L1 (the guitar pickup) have any real effect? I assume not if the answer to the above is to only use the series portion (or top half) of R1.

chunk 4
Regardless of the above answers, I am still not clear in how R2 limits attenuation.

Thank you for the response

[johngreene]

Thank you John, but I am still confused so let me break this down into digestible chunks.

chunk 1
If I understand you, and assuming we turn R2 completely off and remove it from the circuit, the Fc will follow the traditional rule of 1/(2*Pi*R1*C1). Is that correct?

chunk 2
Now, in the case of the value of R1 you mentioned the series component, did you mean only the top "half" of R1 depending on where the wiper is? Or did you mean the entire value of R1 top to bottom?

chunk 3
Does the Impedence of L1 (the guitar pickup) have any real effect? I assume not if the answer to the above is to only use the series portion (or top half) of R1.

chunk 4
Regardless of the above answers, I am still not clear in how R2 limits attenuation.

Thank you for the response

Chunk 1. Yes
Chunk 2. Top half, depending on where the wiper is.
Chunk 3. Yes, it will. But will depending greatly on the pickup characteristics. For basic understanding of the circuit, let's just assume it''s a perfect voltage source.
Chunk 4. The capacitor will (idealy) reduce to zero impedance with high frequency. If R2 is zero then all the signal will be shorted to ground. With any non-zero setting of R2, there is a fix amount of resistance in that path so therefore it can not attenuate any further.

[ScottB]

Oh! Now I see! that turns out to be much simpler than I was making it out to be.

Thanks John, this is plenty to think about now. I'm not sure how exactly to calculate the limit of the attenuation due to the shelving resistor but I do see how that works. It clears up some major confusion I had.

[johngreene]

Oh! Now I see! that turns out to be much simpler than I was making it out to be.

Thanks John, this is plenty to think about now. I'm not sure how exactly to calculate the limit of the attenuation due to the shelving resistor but I do see how that works. It clears up some major confusion I had.

For the limit, assume the capacitor is a short and then you have a simple voltage divider (ignoring other factors like source and load impedance). So when R2 is equal to the setting of R1 (between the wiper and top) the attenuation is 1/2.

[ScottB]

This is excellent stuff. I've scoured the web and asked in various forums, it turns out I should have just come here first. Thanks!

When the shelving resistor ratio (thanks for that, it opened my search up) limits the attenuation aka a simple voltage divider, obviously it has no effect on lows because the cap looks like an open to them. The very highs obviously work the opposite, so at 1/2 ratio half of the highs are missing the cap completely.

Now, thinking of a simple bode plot, assuming an ideal class 1 on paper, where does this ratio start its effect? Is it dependent on the corner frequency and simply reduces the slope (from -6db/octave to say -3db/octave) or is it independent and only related to the cap properties itself (basically leveling off the slope at some higher frequency, creating a sort of S curve on the bode plot)?

[johngreene]

This is excellent stuff. I've scoured the web and asked in various forums, it turns out I should have just come here first. Thanks!

When the shelving resistor ratio (thanks for that, it opened my search up) limits the attenuation aka a simple voltage divider, obviously it has no effect on lows because the cap looks like an open to them. The very highs obviously work the opposite, so at 1/2 ratio half of the highs are missing the cap completely.

Now, thinking of a simple bode plot, assuming an ideal class 1 on paper, where does this ratio start its effect? Is it dependent on the corner frequency and simply reduces the slope (from -6db/octave to say -3db/octave) or is it independent and only related to the cap properties itself (basically leveling off the slope at some higher frequency, creating a sort of S curve on the bode plot)?

Just think of the capacitor as a frequency-dependent resistor. Add it's reactance to R2 and compute the voltage divider. It sounds like you have the correct idea (S-curve).
Here's the TS tone control section, including the 1k/.22uF cap response:


Here's what happens when you put a 100 ohm resistor in series with the .22uF cap. You can see the 'shelving'.

[ScottB]

That is exactly what I envisioned, excellent!

Is there any general formula for the "shelf"? What I mean is there any rough guide to help envision what frequency the shelf "corners" in a similar way that you can roughly calculate the initial corner frequency of the filter circuit?

And in my simplified diagram, if R1 is turned all the way up, ie: having no series component, the tone control still works. Is that because the impedence of the pickup itself is still present and actively determining a filter circuit corner frequency (for a humbucker that would be roughly 10kohms on average I guess).

[johngreene]

That is exactly what I envisioned, excellent!

Is there any general formula for the "shelf"? What I mean is there any rough guide to help envision what frequency the shelf "corners" in a similar way that you can roughly calculate the initial corner frequency of the filter circuit?

And in my simplified diagram, if R1 is turned all the way up, ie: having no series component, the tone control still works. Is that because the impedence of the pickup itself is still present and actively determining a filter circuit corner frequency (for a humbucker that would be roughly 10kohms on average I guess).

Yes, now you start getting into the second order effects. If the pickup was a perfect source, the cap would have no effect. 

[ScottB]

Man this is so cool!

Iafter more research, especially here, I decided I needed to bite the bullet and learn one of these tools. I was a little intimidated at first but I went ahead and ripped LTSPICE, played with it a little while and before long I was graphing frequency response curves all over the place!

I highly recommend this to anyone who seems to need some other way to understand what is going on. I now see the relationship between the shelf resistor and the tank resistor. Different diagrams behave slightly different or almost not at all, but in each case I couldn't guesstimate what the differences would be. Now I see it visually in a graph.

I'm not sure how to model the pickup exactly; I used a generic AC source and a 1M load to model the marshall amp. I tried a parallel resistor load to simulate the pickup impedence but it seems to make little difference if any regarding the tank resonance. Doesn't matter too much, the concepts are now much clearer. There are a few things I question but that may be the limitations of the simulator, or it may be the limitations of my model.

Thanks!

[Projectile]

It is cool.

I enrolled in college for an electronic engineering tech degree, all because of this forum.  It starts in a few weeks. I'm excited!

[aziltz]

It is cool.

I enrolled in college for an electronic engineering tech degree, all because of this forum.  It starts in a few weeks. I'm excited!

sounds like a good investment, I hope you enjoy it!