combining NPN fuzz face with PNP fuzz face

[EATyourGuitar]

someone asked me on another forum if I had any info on the schumann two face.

Two germanian transistor fuzz-face style distortions linked together at the ground rail. One distortion controls the positive half of the wave. The other distortion controls the negative half of the wave

linked together at the ground rail is confusing and ambiguous. I think what he is trying to say is "half rectifier over here, half wave rectifier over there, we put the fuzz on the fuzz and bob's your uncle"

so after telling some that this is impossible I had a dream about diodes. yes thats right. a diode will give you a consistent voltage drop unlike resistors that tend to move all over the place when other things start happening around them. also, that small amount of current you need to forward bias a diode is the same for one diode as it is for 2 or 3 or 11 of the same diode.

the obvious challenge is getting the voltage drop across the diode string to match what you would like to have as a virtual ground. you are kinda working backwards from that virtual ground to have a correctly biased Q1 base with the right amount of current AND the right voltage drop while still having your Q2 collector where it should be at ~4.5v = 9v/2. in this example I assumed NPN + PNP silicon with a B E junction forward voltage 0.7 but biased at 0.6v forward. both Q1 B and Q3 B are tied together at 9v which is half way between the +18v and 0v rails. input cap is shared so impedance does not change. in a perfect world, only one transistor will be low impedance for half of the wave form at any time while the other transistor is immediately starting to get closer to B E reverse bias where the impedance becomes significantly higher. out of all the diodes, 1N4148 is probably the lower current needed to turn the diode on. there are diodes that require less but lets just say 1N4148 is better than an LED even though LED could reduce parts count with a larger voltage drop. if the diode does not turn on, we have no power supply. if we have enough current through Q1 + Q2 emitter to turn the first diode string on, we would have a virtual ground for the NPN fuzz face that sits at a voltage that will never change. same is also true for the second fuzz face. this means each fuzz face is running at a virtual supply voltage of 9.6v. they do not share virtual power supplies. all the grounds are virtual. the DC blocking caps on the input and output make this possible with referencing audio to 0v. hoping someone will share some insight to whether or not I am headed in the right direction. maybe someone has already done this?

[PRR]

> maybe someone has already done this?

I doubt it.

It may work.

Some doubt about the varying diode voltage and the supply voltage. What works with 18.000V supply and 0.6000V diodes may not work at 18.5V or with 0.65V diodes. The art of good design is NOT relying on magic-number parts.

*Personally*, I'd use two input caps. Not try to get both 1st transistor bases to come out to the same voltage.

That done, I'd go back to a 9V supply, WELL bypassed, so both "polarities" can use the well-proven circuit with only +/- changes of connections.

OTOH, a string of LEDs could do what you are doing and be eye-catching (if not eye-burning).

[Thecomedian]

The description of two transistors affecting the positive and negative voltage of the wave separately is called a push-pull amp where two class B amps deal with the opposite voltage swing.

http://www.allaboutcircuits.com/vol_6/chpt_6/10.html

[EATyourGuitar]

Some doubt about the varying diode voltage and the supply voltage. What works with 18.000V supply and 0.6000V diodes may not work at 18.5V or with 0.65V diodes. The art of good design is NOT relying on magic-number parts.

*Personally*, I'd use two input caps. Not try to get both 1st transistor bases to come out to the same voltage.

That done, I'd go back to a 9V supply, WELL bypassed, so both "polarities" can use the well-proven circuit with only +/- changes of connections.

OTOH, a string of LEDs could do what you are doing and be eye-catching (if not eye-burning).

nothing about this is going to be easy. I will take it one step at a time until I get it right. there is a reason I want to share an input cap between them and it relates to keeping the input impedance of the fuzz face as-is. if I used two input caps this project would be a lot easier definitely. I believe LED's require more forward current to switch on compared to a 1N4148. I also have BAT41 and something even better than that in SMD. if later I find I need a diode with low forward current, will do a diode shootout between them all.

The description of two transistors affecting the positive and negative voltage of the wave separately is called a push-pull amp where two class B amps deal with the opposite voltage swing.

http://www.allaboutcircuits.com/vol_6/chpt_6/10.html

I know about these circuits already. see the zvex machine, CGS wavefolder/wavemultiplier, push me pull you, triple fuzz etc... all based on the lockheart wavefolder which is basically just another application for the circuit used in push pull style power amps. the schematic I posted is intended to use each fuzz face as a half wave rectifier with the base's tied together at the zero crossing. I have done this before with opamps but I have never done it with two fuzz faces.

[PRR]

> share an input cap between them and it relates to keeping the input impedance of the fuzz face as-is.

If you tie two FF's together, already the impedance is half.

The cap(s) does not have a lot to do with that. You can always change value to keep similar bass cut-off.

> I believe LED's require more forward current to switch on compared to a 1N4148.

Experiment.

Diodes don't "switch on". They just conduct less and less. At some point they don't conduct "enough for our purpose", but that is situational.

I *will* say that back in the 1970s, even 90% of low-low-cost (reject) LEDs had substantial leakage and you had to hit them hard. But the basic theory is the *same* as ordinary diodes. The leakage was poor processing (and buying fall-out).

And you have most of a mA of current in the diodes. Modern LEDs sure run fine down there. (They light-up, which they don't do if the forward voltage is not high enough to bang the electrons/holes together).