Hi. I just finished my Nurse Quaky. I did a pcb layout based on Runoffgroove's perf.
I was reading through the posts here, and I think I have the so-called undersignalitis... like described here:
http://diystompboxes.com/sboxforum/viewtopic.php?t=15825&highlight=quacky
BUT, what I get from the circuit is not a muted signal (although the level is indeed low). I get a reverse sweep...i.e. If I play softly, the signal is full of highs, and then if I hit the strings harder, the sweep goes down.
I'm at work right now, so I don't have the board with me, but I used almost the same values that B.Tremblay did, except for the two 0.0047 caps. I think I used a couple of 0.0056, but that shouldn't make such a big difference...should it?
Now, let's see...if I take the tranny out, the leds keep working, but the sound comes out clean. I reversed the tranny (I know for a fact that the pinout is correct) and I got basicaly the same sound. Then I places the 2n3904 back and took the IC out (tryed either 4558 and 5532) , and got clean sound, but no led blinking.
Also, the caps I used differs a lot in size (not value). Is that a problem at all?
Any idea how to make it work?
First of all, make sure you used the inverting input of the opamp of the EF. Then increase the resistor in the feedback loop to increase the gain, this will probably solve the problem. the 0.05uF caps set the sweep range not direction. If that doesn't work, try a JFET input buffer.
The transistor is the control element, so taking it out would make the sound clean. Maybe trying another transistor would work.
A Dr. Quack / Nurse Quacky hybrid is the next project in line for me :D
I,ve been coming to the conclusion that reverse sweep is qualitatively different in some ways than upward sweep.
The analogy I like to use is this. Go up to any cat or kitten and quickly move your arm up. The cat tracks it instantaneously, as evidenced by the head movements. Now move your hand downward over the same distance at the same speed, and the cat fails to follow adequately, and certainly not with the same speed or precision. Clearly movement in each direction evokes a different sort of response because movement in each direction has different biological/evolutionary meaning to the animal.
Likewise, I think it is the case that the information we need to hear, and perhaps the rate it needs to change, is different for upward and downward sweep of a filter. Especially a bandpass filter as in the DQ/NQ.
Consider that any musical instrument signal is going to consist of fundamentals and harmonics. The harmonic content is usually going to vary as a function of emotional intent/intensity, the same way that the human voice gets raspy (harmonically rich) when stressed and more muted (less harmonic content) when more at ease.
More importantly, the harmonic content and sources of signal amplitude are going to shift over time. There will be rich harmonics at the initial attack, which quickly evolve into primarily fundamental as the note decays. The typical envelope controlled filter, when sweeping upwards, provides degrees of focus on those initial harmonics, but after the initial high peak of sweep, comes back down, de-accentuating the harmonics and restoring the fundamental. Not unlike a tensing and relaxing of the throat, corresponding to emotion-related variations in speaking.
Now consider a downward sweep of a bandpass filter. First off, it *starts off* up in the upper midrange where the fundamental doesn't live. Then it sweeps down to the fundamental for a brief instant, and settles back to where the fundamental doesn't live. Because the fundamental is essentially the meat and bone of the note and the harminics are the fur and feathers, the note seems to go from "not quite there" to "there" and then "gone" as the filter sweeps down then up again. The same degree of filter sweep, in terms of number of octaves covered, is actually perceived as moving *faster*, largely because of the relative importance of the spectral content at the ends of the sweep range, and because of how long they last as well. Remember, those harmonically rich transients don't last long. Downward sweeps effectively "run away" from harmonics, and then come back after the harmonics have gotten on the bus and left town.
The bottom line is that downward sweep cannot be, and should not be, the exact mirror image of upward sweep.
So what *should* it be? My hunch is that, first off, it can't simply start at the opposite end of electronically feasible sweep. In other words, if the control element is capable of being varied over a range of 0-5vdc control voltage, you don't automatically invert sweep by feeding it with a steady 5vdc and then subtract the envelope voltage from that for downward sweep. That's fine for some things but not for envelope controlled filters applied to acoustically-generated (but electronically sensed) string sounds. It should probably be adjusted so that it starts somewhere down from the complete opposite end of the sweep range. Maybe even as much as halfway down.
Second, because downward sweep is perceived as moving faster, and because perceived rate of change depends on how far the sweep goes in a given period of time, not only the start point, but the amount of sweep needs to be decreased for pleasing downward movement.
It is common for bass-oriented EFs to blend in some straight signal so that things don't thin out too much as the filter sweeps past the range of fundamentals. That is probably true of downward sweep as well.
So how do you implement this on a DQ/NQ? If it is fixed at downard sweep only, then I suggest the easiest way to adjust the "start" point is by the very cap-value change you suggest. Something between 6n8 and 10n is probably about right for guitar.
The second thing you may want to do is adjust how much sweep occurs per pluck. With LDRs it is a simple matter of adjusting the response of the control element by tweaking parallel and series resistors to adjust the effective amount of resistance change. I'm not sure what you'd do here but certainly the circuit should accommodate parallel resistors. A parallel resistor will also, of course, determine the start point since it will set the maximum equivalent resistance to ground.
Hope this helps.