[sdiy] Synthex Oscillator

[rsdio at audiobanshee.com]

On Jul 9, 2017, at 4:10 AM, Gordonjcp <gordonjcp at gjcp.net> wrote:
> On Sat, Jul 08, 2017 at 09:07:35PM -0700, rsdio at audiobanshee.com wrote:
>> Ok, despite being very interested in this, I still haven't managed to finish my analysis. That said, I think it's most important to pay attention to the node connecting C93, R89, T90, and the resistor ladder "DAC" - plus some consideration of 1Nb and the analog switch at 2N.
>> 
>> Kirchhoff's current law should be able to tell us everything we need to know about C93.
>> 
>> T90 (the BC559) can quickly charge this cap to +5V when the AND gate at 1H tells it to.
> 
> *Dis*charge to +5V.  Notice that both the emitter of the tranny and the
> "free" end of the capacitor goes to +5V so the transistor shorts it out,
> just like in a normal saw-core VCO.

You're absolutely correct. The capacitor has no charge stored when both terminals are at +5V.

For some reason, I was inclined to say "charged" because I'm seeing the sawtooth voltage increase at that point in time. Because the capacitor is not grounded, its charge is the inverse of the sawtooth voltage at any point in time. Inverted logic and inverted signals are hard to talk about.


>> The analog switch at 2N is bizarre, because it either grounds the cap or connects it to +5V, depending upon the MSB of the "DAC" so I'm not quite sure what to make of that without seeing circuitry off the provided schematic.
> 
> The analogue switch 2N pin 1/2/13 switches the square wave output on and
> off.  Look to the right of the diagram and you can see on the trace
> leading to it there's a "|-|_1" symbol.

I saw the "square" wave symbol, but given the existing mysteries in this circuit, and the fact that we don't all agree with the designer's description of what it's actually doing, I don't want to make any assumptions that a symbol tells the whole story of what's connected at that point. I should just grab the full Synthex schematic...


>>> If it is a current source, I can't see how it gets a set value - it's a counter, not a latch.
>> That's the biggest mystery in my cursory glance at the circuit. I would have expected that a value would be set according to the programmed frequency, but perhaps we can figure it out.
>> 
>> Returning to my wild-ass guess, perhaps the counter adjusts the discharge resistance slowly over time to linearize what would normally be a logarithmic discharge.
> 
> The counter counts up at an audio rate, though.  We know this.

Yes. When I said "slowly" it was a poor choice of words. I was thinking in terms of an individual cycle of the waveform as the full time window, and in that perspective thinking that the adjustment would start at one value and change exactly once across the duration of the waveform cycle. So, rather than say slow or fast, I should have said that I'm imagining it in sync with the waveform cycle.

Brian