Help with filters

[famous birds]

Hello all:

  I'm in the midst of designing a 4049 distortion, and I'm a bit confused as to how to calculate the cutoff frequency for the RCs in the feedback loop.  Let's take as an example the TSF:

http://www.freeinfosociety.com/electronics/schematics/audio/pictures/craigandertontubesoundfuzz.gif

  Looking at the second inverter, there is a 1M resistor and 10pf in parallel in the feedback loop.  I know these two components form a lowpass filter, but I'm at a total loss as to how to calculate the cutoff frequency from these.  Do I need to take into account other components?  I guess I'm assuming the big caps "isolate" it from the rest of the circuit, or at least form a separate highpass filter.

  It's been too long since I had to do this stuff for homework.  Maybe some EE wizard can help me out?

-Alex

[anchovie]

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

[famous birds]

Thanks, but trust me, it's bookmarked.. 

How do I use that to calculate the cutoff?  Plugging in 1M and 10pf I get a corner frequency of 15.  Not to mention this is supposed to be a lowpass filter.

Any ideas?

[PerroGrande]

The way I usually explain basic filter principles is to refer to it as a "frequency-dependent voltage divider."

For a given frequency, you simply calculate the impedance of the reactive element (cap or inductor), and then look at the circuit as a voltage divider.  Forgetting inductors for a second, the only difference between a high pass and a low pass is the location of the frequency-dependent portion of the divider.  So the corner frequency computation is the same for either low or high pass. 

Working through the f3db equation for the values you gave:  f = 1/2 * pi * R * C = 1 / (2*pi*1000000*10x10-12) = 15,923 Hz. (~16KHz)

So your corner frequency is around 16KHz with the values given.  If the circuit is an LP, then frequencies above this point will roll off at a rate of 6dB per octave, etc, etc.

As you will see in a moment, this analysis isn't quite what is needed for the Anderton circuit...

Looking at the TSF circuit, it looks like the caps are there to roll off high frequencies.  This is a fairly common approach in feedback-based design.  The resistors provide a stable feedback path at DC, while the caps serve to roll off the gain of the stage as the frequency passes a certain point.  As the frequency rises, the capacitor looks like the preferred (i.e. lower "resistance") path, and more of the inverted signal is fed back into the stage - thus reducing gain.

The first stage has the "rhythm" and "lead" switch, which really only changes the resistance portion of the feedback path.  In the lead position, the feedback loop consists of a 10pf cap and an 11Meg resistor in parallel.  The rhythm position is the same cap in parallel with a 1.1 meg resistor.  To understand the effect of this feedback, one needs to look at the point where the capacitor's impedance presents an essentially equal path as does the resistance. 

The capacitor's impedance is computed using the equation Zc = 1/2*pi*f*c.  For "quick and dirty" consideration, simply set Zc to the resistance you're considering.

For example, in the lead position:  11,000,000 = 1/(2*pi*f*10x10-12).  Solve for f and you end up with around 1,500 Hz.  In this setting, the capacitor begins to dominate the feedback loop above around 1,500Hz.  The gain falls as the frequency rises (i.e. a low pass filter).   In the rhythm position, the rolloff starts at a lower frequency, which will produce a less treble-filled sound.

The second stage is similar to the lead setting.  In this case, the cap is probably there for noise reduction as much as anything else.

[anchovie]

Thanks, but trust me, it's bookmarked.. 

How do I use that to calculate the cutoff?  Plugging in 1M and 10pf I get a corner frequency of 15.  Not to mention this is supposed to be a lowpass filter.

Any ideas?

1M = 1000000 ohms
10 pF = 0.00001 uF

Put those into the calculator and you get 15923.6Hz, same as PerroGrande.

[Mark Hammer]

Or, simply enter resistance in megohms (where 10k becomes .01M) and capacitance in microfarads (where 470pf becomes .00047uf).

BTW, 15.9khz is reasonable from a hi-fi perspective, where one wants the most bandwidth one can get.  Most guitar-amp speakers, however, will crap out above maybe 7-8khz, so I generally make a point of rolling off the highs in distortion circuits, whether invertor-based or otherwise, around 6khz.  That may seem like overkill, but: a) distortion, by definition adds more harmonic content, b) the amp, in most cases, is pushed harder by the overdrive circuit output level and generates its own harmonic contribution, c) harmonics of harmonics sound like crap after a certain point, d) the rolloff is really not all that steep (6db/oct), so even a lower rolloff at 4khz still leaves a surprising amount of treble in the output signal.

[PerroGrande]

Mark,

Agreed wholeheartedly on your frequency roll-off comments.  While it is awfully tempting to engineer for flat response from DC to Light  it isn't necessarily the "right" thing to do and doesn't necessarily sound "the best". 

[famous birds]

Thanks anchovie, for pointing out my inability to convert units.

Thanks so much for the help!  Back to business!

[Electric_Death]

Looks just like a speaker crossover.
I'm wondering if it's possible to get a 12 db boost/cut by tossing in another cap or resistor the same way it's done with speaker crossover networks. How about if we were to just add the same filter on the output of the circuit as well to get another 6 db boot/cut that way?
Should work if I'm not mistaken.