UrielOhringer wrote:1) The Elektra-variation had a resistor connecting the Collector to Base. When I look up Muff schematics I see something like that happening there too. But Face schematics don't have that. Which is all fine and dandy, but when I try to breadboard it, I can't get output from the first transistor. As soon as I do bridge C and B with a resistor, it works. So... what does that connection do?
I hope you're ready for a somewhat long answer.
The base of a transistor does not start leveraging current through the transistor until it reaches a certain voltage (I can explain why, but it'll make this post overly convoluted). So, when you plug your signal directly into a base it will ignore EVERYTHING below that threshold voltage and only turn the transistor "on" during peaks. The threshold voltage is established between your base and emitter, and is roughly 0.6 volts for silicon (Si) and 0.3 for germanium (Ge).
The resistor between the collector and the base of the transistor in some schematics is a path by which to charge the base of the transistor so it is always warmed up and ready to go. This trickle voltage is referred to as bias.
Now, why was a bias resistor omitted from the Face?
In all likelihood, the Face schematic is based on Ge transistors. Ge tends to be somewhat "leaky", either through the metallurgy of the PN junction (this is the transition point between collector and base inside of the transistor) or through primitive fabrication methods (process contamination, etc.). The point is that a lot of Ge transistors would essentially self bias and that external path of a bias voltage was not needed.
By adding the large value resistor from collector to base, you provided a bias path and turned your transistor on so it'll work. Another method would be to hit it with larger signal, but that will incur higher levels of distortion (gated fuzz), as your signal will allow the transistor to turn off when it dips below the threshold for conduction.
UrielOhringer wrote:2) Why the double resistors between power and the Collector of Transistor 2? I understand that the output of the pedal is the differential between Q2 and Power, and I've read that the 330Ohm resistor is there to reduce output level to prevent overloading the amplifier. But why? If I want less output to prevent overloading my amp/other pedals, can't I just turn down the pedal's volume? I've built it with just one resistor from power and the output being the differential between that and Q2 and it works. (Don't know how it compares, though, I don't actually have a FF.)
Signal collected from a collector of a transistor (when in a "common emitter" configuration) is generally the result of the load resistor in that circuit. If you visualize for a moment a transistor:
The collector is connected to the power rail through a 10K resistor and the emitter is connected to ground. For this I'm calling the power rail a +9 volt battery
While the transistor is "off" the effective resistance between the collector and emitter is very high, let's call it 1M.
So, you have a voltage divider consisting of a point between 10K and 1M across the power supply. The "output signal" will be resting at very close to 9 volts, but - this is a DC voltage and the capacitor you have coupling this output node to the rest of world does not pass DC so you have nothing.
Now, let's feed in a bias voltage and turn the transistor "on". Quiescent current through a biased transistor will reduced the effective path resistance between collector and emitter. Let's call it 100K now, without signal.
So, we've got our "output" quietly sitting at a DC point around 8 volts.
Let's hit the base with some signal.
As the signal applied to the base of the transistor goes more positive the transistor conducts more readily. This is to say, the effective resistance across the transistor (from collector to emitter) decreases allowing greater conduction of signal across the power rails.
So let's hit the transistor hard enough to reduce its path resistance to 10K at the peak.
If you are familiar with ohms law, you'll see that a point (the output) between a 10K load resistor and 10K worth of transistor will reflect a voltage exactly one half of the power supply, which in this case is 4.5 volts. Meaning your output signal is swinging between the aforementioned 8 volt quiescent and 4.5 volt peak, a signal that is now 3.5 volts in magnitude.
One way to limit that output signal size (if you are so inclined) is to build in another voltage divider in the load resistor, which is where you've wound up looking at two resistors in series on the collector instead of one. Your resulting signal size will be a ratio of the output swing, bringing your signal back into the expected design range of everything else in your chain.
The reason "why" is an engineered decision by the manufacturer to eliminate the possibility of the end user setting things outside nominal levels and laying waste to stuff down the line. As technology has progressed, we see less instance of actual failure as the parts are more robust, but in 1970 you could overstep the boundary of SOA and the hand crafted PN junction of your prize fuzz pedal would deteriorate, and no one wants that.
The example I painted is a gross generalization/simplification, but I hope that sheds some light on it anyway.