Tellurian Drive sanity check - happy accident?

[slowpogo]

I was trying some different germanium diodes in my Tellurian, but ultimately went back to the Philips OA9, which had first been there. I found myself really loving the sound.

Then I noticed, I accidentally socketed the OA9s in such a way that the anode legs were touching! I tried separating them, and I didn't like the sound as much. It seems to me that the pedal sounds better with the germanium anodes shorted, a more lively texture in the overdrive that I really like.

That's where the sanity check comes in. Should there actually be a functional difference, with the anode of D2 and D4 connected? is there reason to think it might sound different that way?

I'm open to the possibility this is all in my head, but I could swear it sounds not only different, but really cool.

 

[swyse]

That's really interesting, the timmy has this connection in many of the schematics available for it, I believe it was done this way for switching the two sets of diodes with one pole of a switch in whatever version it was. I have been told that this makes no difference as the signal can only flow one way on each string of diodes anyways, so the connection shouldn't affect how the signal flows, but I've never tried it to compare.

 

[slowpogo]

Looking again at the schematic... normally the signal flows D1 > D2 > D3 > D4, in a loop. But the anode connection means it's also able to flow D1 > D4. In this way it kind of creates a 3rd silicon/germanium diode pair.

If you actually added a 3rd diode pair in the feedback loop, that should sound different, I would think. So it kind of makes sense to me that the difference I'm hearing could be real.

Anyway it's an easy thing to try/undo, if anyone else wants to try it. I'd be curious if anyone else thinks it sounds rad

 

[swyse]

As I understand it for the reverse parallel diodes in the clipping section, each direction is clipping the waveform on the negative or positive side depending on the direction of the diodes. In the timmy for example, when you add the bridge in the middle with the voltages we're making you will not be having either side of the sine wave go through the opposite diode backwards due to its reverse voltage threshold being higher than the voltages we can produce in the circuit. In the case of your germaniums, I wonder if there is some leakage not getting blocked by the silicon diodes. I would think the effect of that would not be clipping of any kind, but a reduced gain.

 

[spi]

Looking again at the schematic... normally the signal flows D1 > D2 > D3 > D4, in a loop. But the anode connection means it's also able to flow D1 > D4. In this way it kind of creates a 3rd silicon/germanium diode pair.

I don't think it will flow that way though... there's no potential difference between the anode of D1 and cathode of D4 (because it's the same point in the circuit). Without the voltage difference there's no current.

 

[slowpogo]

I'm admittedly out of my depth...but doesn't that statement assume the forward voltage of the diodes is identical? in reality there's probably a slight difference

 

[cdwillis]

It flows like this:

D1 > D2
D4 < D3

The signal is still only getting clamped on one half of the waveform through each diode set. If you connect the diode sets in the middle it will still only flow/clamp in the same directions.

 

[spi]

But the anode connection means it's also able to flow D1 > D4. In this way it kind of creates a 3rd silicon/germanium diode pair.

I'm admittedly out of my depth...but doesn't that statement assume the forward voltage of the diodes is identical? in reality there's probably a slight difference

I think what you were asking is if there could be signal flowing like the red line (through D1> D4).

The current only flows if there is a voltage difference between two points (i.e. potential). In this case, there would have to be a difference between point A and point A. Since it's the same point, there's no difference, thus no potential and no current.

Disclaimer: my electronics knowledge is decades rusty--so I could be wrong about this.

 

[slowpogo]

I'm confused by "the same point." The current flows through D1, and then reaches a point where it could flow to D2 or D4. They are the same part so I don't understand why there's a voltage difference toward one (D2) but not the other (D4).

 

[cdwillis]

The signal is only flowing one direction in that loop. It's trying to get from the output pin to the inverting input pin. As it passes each set of diodes on that path, each pair (D1/D2 and D3/D4) clamp opposite sides of the wave form. When the signal gets on the other side of the loop the signal is already lopped off (the diode's clipping threshold isn't met anyway) and it's not trying to travel over to the positive pin. I hope that doesn't sound more confusing.