Calculating Limit Frequency

[fuzzy645]

According to the General Guitar Gadgets site the limit frequency of a filter can be calculated as:

Where R1 is expressed in ohms and C1 expressed in farads.

So when I plugged this into excel for the following values:
R1 = 220K (therefore 220,000 ohms)
C1 = .022 microfarads ( therefore 22000 farads)

The formula is 1 / (3.14 * 2 * 220000 * 22000)

The result came out to be 0.000000000033 which is not what I expected.

Any thoughts on my math here, or perhaps I need very different values.

[CynicalMan]

Look up the SI prefixes. Mega means 10^6. Micro means 10^-6, or 0.000 001. So 0.022 microfarads means 0.000 000 022 farads.

[slacker]

I find it easier to shift the units to Meg Ohms and uF so 1 Meg Ohm = 1 and 1uF = 1. 

So in your example R1 = 0.22 and C1 = 0.022

[fuzzy645]

ahhhh, thanks so my math was wrong 

So if we take a simple tone pot in a guitar with a 250K pot and .047 uf cap (functioning as a low pass filter) the limit frequency would then be:

1 / (PI * 2 * 250000 * 0.000000047 ) which is 13.54 hz which would allow frequencies lower than the limit frequency to pass and higher frequencies are rolled off.

Taking a 2nd example of a typical "treble bleed" circuit in a guitar (functioning as a high pass filter) which is often a 1000 pf cap in series with the same 250K pot, the limit frequency would be

1 / (PI * 2 * 250000 * 0.000000001 ) which is 636 hz which would allow frequencies higherr than the limit frequency to pass and lower frequencies are rolled off.

Sound right??

[ashcat_lt]

Nope.  The Tone pot at max resistance allows the most treble out, right?  But you've got it rolling off the entire audible spectrum.  Can't be right can it?

A passive guitar circuit is a lot more complex then that. 

[fuzzy645]

Nope.  The Tone pot at max resistance allows the most treble out, right?  But you've got it rolling off the entire audible spectrum.  Can't be right can it?

A passive guitar circuit is a lot more complex then that.  

Interesting point. A tone pot is (often) wired with the input into the wiper and the outer lug gets the cap to ground.  If you measure the pot (wiper to outer lug) when the pot is fully CW (on 10) you will get the full 250K (I just verified this) so the circuit at that point is a 250K pot (resistor) in series with the cap.....but as you said we are hearing full treble at that point.

Now, as you roll the pot CCW (down towards 0), the resistance decreases (so w/the pot on 0 fully CCW there is in effect almost no resistance between input signal and the cap) and you then hear the treble roll off maxed out.  Truthfully, you will hear the treble rolloff with JUST the cap only if you took the pot entirely out of the picture (I have tested this eg a fixed treble rolloff),  so in reality there probably needs to be a way to revise this formula for R being non-existent (I think).  You certainly can't set R=0 in the forumula as this will be a divide by zero error.  Maybe we need to say as R approaches 0 or something. 

Of course, maybe I'm totally wrong and the R is the resistance of the pickup input signal (usually around 6k or so)....so if you set the R=6K you get a limit frequency of 564 hz via the formula.

Thoughts??

[R.G.]

We simplify networks (and indeed everything) down to simpler models that let us do fast calculations. As long as the simplifying assumptions do not cause the answers to be too different from what happens in the real world, we are happy with the results. A whole lot of electronics relies on what assumptions you make.

It is a common mistake to neglect the effect of source impedance and load impedance when calculating frequency responses.

A single R-C network has a characteristic frequency of F = 1/(2*pi*R*C); but this holds true only under the assumption that the signal source that drives the network has a source impedance  (internal series resistance is a good first simplifying assumption for this) of so small that is it insignificant compared to the impedances of the R and C at any frequency of interest. It also only holds true to a good enough approximation if the load impedance driven by the R-C network is either dramatically different than the impedance of the R and C.

If the source and load impedances don't match these simplifying assumptions, then to get an answer that matches the real world better, you have to include the values of the source and load impedance into your network to get answers that will match the real world.

One problem with a tone pot is that the source impedance is a magnetic pickup. This is a coil of wire that all by itself has a complex R-L-C impedance. This complicates the simple view so that you have to include the source impedance of the pickup to get good matching of the real world. And the volume control after the tone pot has an impedance near the tone pot's value, it it interacts with the tone pot setting as well.

[ashcat_lt]

...and you've got that big long capacitor between the guitar and amp/first pedal...

[DougH]

Which is why math, simulations, etc can be great for comparing two things to each other and getting the jist of the overall effect of a circuit change, but can be pretty lousy at predicting real-world absolute values of things. Unless you take into account everything in the environment- source impedance of preceding stage, load impedance of next stage and everything between that and your ears- it's not really going to tell you the absolute frequency you're going to hear with a particular filter for example. But who cares, as that frequency is just a number and our ears hear sounds, not numbers. And even if we do associate a particular sound with a number, that association can get clouded by the complexity of a guitar pickup signal waveform.

But the math can be real useful for understanding the relative effect of a circuit change: i.e. If I change my tone control this way does it make it more bright or more dull? For that it is very useful and can help steer you in the right direction if you're trying to make a change to something. With audio "quality", in the end your ears will make the final decision.

[R.G.]

The real world is just a special case, albeit an important one.

[DougH]

The real world is just a special case, albeit an important one.

[greaser_au]

Which is why math, simulations, etc can be great for comparing two things to each other and getting the jist of the overall effect of a circuit change, but can be pretty lousy at predicting real-world absolute values of things.

The math & the simulations are only modeling based on a series of expected behaviours, after all. Depending on what you need, the model can be more or less rigorous.

When I did my trade level certificate courses a series of basic idealised assumptions were made, such as rule-of-thumb Av=Rc/Re for an unbypassed transistor stage, it pretty much ignored the effects of the next stage load, the parasitic values around the stage & things like frequency response  - none of that really mattered as long as what you were getting was in the ballpark (probably even in the same county  was good enough ).   It wasn't until I switched to the diploma course that we started talking about the effect of (say) dynamic resistance & miller capacitance etc.  The servicing level courses assumed that the designer had already considered the more esoteric things & the role of the serviceman was to just restore correct operation to a failed stage/block. In the diploma course you WERE the designer, you could quite happily design using 'ideal' components initially, but then you had to understand what was causing that stage that was supposed to be amplifying to be oscillating!  All the same, the servicing models were basically adequate - one should have an idea about what to expect from a working circuit stage in general terms before the DMM is even turned on!!

david

[clarkguitars]

What i don't understand is, WHAT IS PI i really need to know this as i am designing my own distortion pedal :/

[DavenPaget]

A mind-blowing math symbol .
Mostly math is useless for real-life simulations .

[clarkguitars]

Hahahahahahahaha oh man I feel SO stupid right now 

[amptramp]

Let someone else do those messy calculations for you:

http://www.muzique.com/schem/filter.htm

[DavenPaget]

Or ... Using Electronics Assistant 2000 ... 

[PRR]

pi is equal to "3".

This is legal when selling trees for fence-posts in Indiana.

This is close-enough for all single-capacitor musical-instrument figuring.

If you must obsess, use "3.14".

If you feel Babylonian, use 22/7. This is more than good enough for building temples and such; only the best surveyers on extra pay will prove a difference between 22/7 and the true value.

[Rob Strand]

There is a bigger pie,

      pi * e = pie = 8.53973422