Why does the Foxx Tone machine still have AC present on Q3 in non-octave mode?

[mzy12]

Hey all. Quick technical question.

Just looking at the Foxx Tone machine schematic and trying to figure out why D2 doesn't rectify the entire signal to DC after Q2 when the octave is switched off. My first impression is that only the negative half of the cycle should appear as negative DC after D2.

My semi-educated guess is that the 9V through the 100k resistor is supposed to keep both D2 and D1 (when the octave is switched on) forward biased at all times.

That's all for today 

[duck_arse]

Just looking at the Foxx Tone machine schematic and trying to figure out why D2 doesn't rectify the entire signal to DC after Q2 when the octave is switched off.

did you include a schematic that we can't see?

[antonis]

Just looking at the Foxx Tone machine schematic and trying to figure out why D2 doesn't rectify the entire signal to DC after Q2 when the octave is switched off. My first impression is that only the negative half of the cycle should appear as negative DC after D2.

Νο..

https://www.diystompboxes.com/analogalchemy/sch/vulcan.html

[mzy12]

did you include a schematic that we can't see?

Sorry, I was going off common values I've seen on every schematic for it. That was dumb of me, I should have linked one regardless.
http://fuzzcentral.ssguitar.com/foxx/foxxschematic.gif

Νο..

https://www.diystompboxes.com/analogalchemy/sch/vulcan.html

I'm still not getting it I'm afraid... "The diodes in the bias network prevent hard-saturation of the transistors," doesn't explain what's going on. If I'm going to take another guess:

The base of Q1 is biased to 1.4V if you ignore the diode and aim for about 4.5V on the collector. As a negative signal develops on the cathode of the diode, it will try turn the diode on "more on", which, tbh, I'm not really sure what that does. But what about on the positive half? It tries to turn the off the diode, as a more positive voltage develops on the cathode than on the anode which causes the signal to clip, at some point.

I then did a small mock up on falstad, which kind of confirms what I'm thinking, but not really. Even if I input a square wave, the negative half of the signal on the anode of the diode is somehow spikey.

So I'm still quite lost here.

EDIT: I'm referring to the vulcan over drive schematic in this section.

[Mark Hammer]

The Foxx does not do this, but many half-wave rectifier circuits will use two diodes.  One goes to ground to effectively "dump" any AC from the unwanted half-cycle, and another to "choose" the AC from the desired half cycle.

One of the things I do with Tone Machine builds is use an on-off-on toggle for octave.  In one side position, it engages the second half-cycle for octaving.  In the middle position, it lifts that 2nd diode for non-octave fuzz.  But since that diode imposes crossover distortion on the remaining half cycle, I use the other side position on the switch to bridge the diode for the remaining half cycle.  The impact of this on the quality (and loudness) of the fuzz will depend on what diode type one used for the phase-splitter.  Schottky diodes won't sound terribly different between diode and no diode, because the signal is pretty much guaranteed to be above diode forward voltage.  Silicons will yield a more obvious difference between diode and no diode, because they will block anything below the forward voltage.

[mzy12]

I've had to re-evaluate my simulations as I wasn't quite measuring the right point.

many half-wave rectifier circuits will use two diodes

This I understand. I still cannot get my head around the single diode thing. You have a forward biased diode of some description which based on the current its passing due to the biasing through an resistor of arbitrary value, will be forward biased but not turned on fully (probably) and the cathode will be at X voltage because the diode is essentially between a resistor divider, where X depends on both the amount of current the resistors are passing and the current the diode is passing (which again depends on the first resistors value).

The A/C characteristics are as follows:

The positive half of the signal will attempt to reverse bias the diode, turning it more and more off. Depending on the V_D, which will not be the nominal Vf of the diode according to my limited testing on a simulator, I would think that this means it will stop conducting. But on falstad, it just somewhat gently clips, depending on the input level. This even occurs on square waves, softening the wave form.

Then the negative half of the signal will forward bias the diode more, turning it more on. On falstad again, it looks like it just removes some of the Vpk, around 200mV depending on the signal level and chosen diode. This makes sense to me, as it's mostly just the 'remaining' voltage drop of the diode. I think?

[antonis]

Both Diodes Anodes biased at 4.62V..