Electronic – How does the fuzz face guitar pedal actually output any gain to the input signal

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I'm referring this this circuit right here:

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I do not understand why the circuit would be designed like this and why the output wouldn't be drawn from the collector of Q2. I have 2 questions, which are interrelated.

  1. What is the purpose of the 330 ohm resistor? I've never come across something like this in audio stages
  2. If the output is being drawn from the potential divider circuit formed by the 330ohm resistor and 8K2 resistor, wouldn't the resulting signal output voltage range be very small (with a large DC voltage) since it is quite close (in terms of impedance) to the power rail?

Best Answer

These fuzz circuits (Fuzz Face, Tone Bender, whatever) actually sound quite good by themselves (that is if you can find transistors with the right leakage, hfe, noise and various other mysterious properties). So unlike some distortion pedals, the fuzz is not supposed to overdrive the input of your amp. The 330 / 8K2 forms a voltage divider that taps into only a few hundred mV of the output. In fact, this is quite a bit less than the peak output of a typical guitar which can generate a couple of volts if you bang on it just right. Meaning the fuzz actually limits the output compared to the input. When I plug a fuzz into the clean channel of my tube amp, I have to turn the level control way down. Otherwise it's just completely saturated a la Hendrix. It's when you back off the Attack, Level and Volume on the guitar that the fuzz really generates interesting tones.