A few basic questions about parts of this circuit

[harmless]

Last weekend I finished my first build - a Shin Ei fuzz clone (with the mid scoop mod).  It worked on the first try   Then I put it in the enclosure and managed to break some wires.  Then I fixed all that and had a short on the case...but now it's working.  One quick question here: how do you mount your veroboard inside the enclosure?  Glue spacers in or something?

Ok onto the real questions.  I've ordered an arduino with the intent of using it as an LFO for a tremolo. I found this thread ( http://www.diystompboxes.com/smfforum/index.php?topic=77471.0 ) where someone is using a PIC to do something similar and I'm looking at the circuit layout he used.  I have a basic understanding of electrical components but all the circuit design stuff I've done was DC and ages ago back in college.  I'm trying to understand why the design is done a certain way.  Either direct answers or pointers to things to read would be really appreciated.

1.  On the LFO side - why the resistors and caps when connecting the pots to the PIC?  The arduino has 5v analog pins that only provide about 40mA max.  I don't know if the PIC is different, but for those low values is there a reason why I shouldn't just wire the pot directly to the input pin?

2.  On the LFO side - I'm not sure what that transistor and 2nd LED is doing near the LDR.  I'm planning to use a vactrol, but regardless I was expecting to see the PWM output simply going through a current limiting resistor and right into the LED and then ground.  If I wanted another LED to blink for the tempo I was thinking I could just put it in parallel with the vactrol or something.  The transistor thing looks like it's inverting the LFO or something....(using the low values from the LFO to cause current through the LED)...right?  Any idea why?

3.  On the LDR side - the first op amp is just a buffer right?  The few examples I've seen just have the output looping directly back to one input with the signal coming in on the other.  This looks like it's blending the output and the signal on one input to the op amp.  Again...why?     If the circuit was wired with true bypass switching are the buffers even needed?

4.  On the LDR side - the 2nd op amp is actually doing some amplification here, right?  (Thus the gain pot).

5.  By input and output there are 1M resistors and 1u caps.  Are those blocking caps?  What are those for?

Hopefully these aren't too basic.  I'm going to buy various resistors / caps / pots / etc. and do a lot of experimenting with stuff, but I'd also like some basic idea of what's going on that I can build on.  Thanks for the help.

[PRR]

R.G. has an essay on his site with a name like "What are all these parts??". Read it.

> why the resistors and caps when connecting the pots to the PIC?

Caps are there to mellow the glitch in worn pots. Turn a pot and skip over a bit of dust, the output drops to zero. In pure analog, this may go unnoticed. The CPU may react quite fast and unfortunately.

The resistors seem non-essential. It is "good practice" to have some resistance between naked digital inputs and V+. When pot is full-up it goes to V+. Resistor limits any stray current due to supply spikes. With modern systems this may be omitted. However the board layout may go easier with a few "loose parts" to use as cross-overs.

> what that transistor and 2nd LED is doing near the LDR.
> I was expecting to see the PWM output simply going through a current limiting resistor and right into the LED and then ground.
> If I wanted another LED to blink for the tempo I was thinking I could just put it in parallel with the vactrol

This has a 10K limiting resistor out of the PIC, we may want a 1K to the LED. It appears that the designer did not want to put such a low load on the PIC. Some older ones had very low output current.

The transistor just takes the ~~4V/10K= 0.4mA from the PIC and pulls (3V/2K)*2= 3mA through the two LEDs. This demands a transistor with current-gain greater than 7.5. "Any" transistor will do that.

The second LED is surely a blinker and _IS_ "in parallel with the vactrol". You don't want to put two LEDs directly in parallel, the lowest-voltage one will hog all the current. You give each one its own resistor, to force current sharing.

The entire audio path has been taken from an old-old classic tremolo. Maybe someone knows its name; look for a description.

Buffer is needed because guitar needs to see over 100K and LDR tends to work best when loaded in less than 100K, maybe near 1K.

In/out caps are needed because there is a single polarity of supply but audio must swing "both ways".

Darn PhotoBucket is trying to infest my machine with pop-ups....

[harmless]

Thanks for the reply.
The glitch control on the pots makes sense.  I'm going to be on a breadboard for a while with this anyway, but I'll keep the resistors in mind if I ever get to the point of doing a layout.

The transistor explanation makes sense, too.  We're effectively pulling current from the power supply instead of depending on the microcontroller to provide it.  I'll look more at the Arduino's specs to see if it can provide sufficient current to drive the LEDs (I'm pretty sure it can...I seem to recall 40mA per pin).

One more followup question.  LEDs have a minimum voltage drop that has to be met before they turn on.  If my LFO is just generating a square wave then that's fine as the on is higher than what the LED requires.  What about if the LFO is generating a sine or ramp type wave?  Once the LFO voltage drops below what the LED requires it will turn off, flattening the bottom of my intended wave, right?  Do I need to do something like fix the bottom of my generated LFO wave to be just below the turn on threshold for the LED instead of 0 or am I thinking about this wrong?

I have several tabs open from R.G.'s site now that I'm working my way through.  Thanks again.