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This is the schematic for a fuzz circuit. Even knowing that the diode is probably what's providing the signal clipping, I still can't figure out how I'm supposed to analyze the circuit.
When V(collector - base) < 0.7volts, the diode acts as an open circuit, and there's no feedback path, and as long as the base voltage is forward-biased, the transistor acts like an amplifier (linear region).
When V(collector - base) > 0.7volts, the diode is... shorting the base and the collector?
Simulating the circuit gives this:
simulate this circuit – Schematic created using CircuitLab
So it is replacing the positive excursions of Vin with the downward excursion of a square wave. Negative excursions of Vin are basically unaffected.
The downward square wave excursion is largely independent of the amplitude of Vin. Basically when Vin is positive it saturates Q1 causing the output to be pinned to -800 mV in the graph.
This answer goes to explaining the behavior found in the simulation.
The basic idea is that either the diode conducts or the transistor conducts but neither of them conduct at the same time.
The DC operating point of the base is a standard NPN Vbe -- about 600mV.
The DC operating point of the collector is about 800 mV. There is only microamps of current going through the diode.
Because the base is right at 600 mV, any increase in the base voltage will quickly turn on the transistor. So positive excursions of Vin will quickly saturate the transistor and pull the collector voltage down.
Likewise, a negative excursion of Vin will quickly cutoff the transistor. We can therefore take it out of the picture and are left with this situation:
Vc
+9V ---/\/\/---o----|>---(Vin)
100K
Therefore, the voltage at the collector, Vc, will track Vin, which is what we see in the simulation.