[sdiy] Synthex Oscillator

[Gordonjcp]

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.

> The op-amp at 1Nb should ideally have no current flow into or out of pin 3.

I don't see what type of opamp it is and I'm too lazy to go hunting, but
it's probably given the age of the instrument a TL0xx of some flavour.
So it'll be a FET opamp wired as a a unity-gain buffer.  The current
will be tiny.

> Assuming 2N is open and T90 is off, and also assuming that the initial state of C93 is +5V at the beginning of each cycle, then the resistor ladder made by R92 and R94-R100 will control the rate of discharge of C93. Any of the NOR gates in 1L and 2L that are outputting a "0" logic level will be combined in parallel to discharge the cap. I'm assuming that the NOR gates that are outputting "1" would initially have no effect because there would be +5V on both sides of the resistors, but as C93 discharges those resistors will start conducting current and slowing the discharge of C93. I can't quite predict the exact consequences, but see below (*).
> 

The NOR gates are driven by the counter.  If you took C93 out of the
picture you'd have a steppy 8-bit sawtooth wave.

I must admit I'm struggling a bit to work out what would go on.  At the
extreme ends of the scale the time constant would be rather short.

> Note that if there was an active stage between the resistor ladder "DAC" and the cap, such as with an op-amp follower, then it would surely be a DAC because the current flowing through the resistors would depend entirely upon the DAC and not the voltage on the cap. If C93 were acting as a smoothing capacitor, then I would expect a fixed resistance in series with the DAC output, making a simple RC low pass filter. I don't see C93 as a smoothing capacitor at all.
> 

The resistors of the DAC would probably form part of the RC circuit, but
the bit I'm struggling with is this - if you rolled over from $00 to $ff
(it counts down) on the counter, you'd be effectively driving all eight
resistors in parallel, so you'd have R = 3.5k and C = 10n for a time
constant of about 30 microseconds!  Having the reset tranny there hardly
seems worth it.

Does that then mean that toggling bit 0 up and down would give the RC
filter a time constant of about ten milliseconds, halving each time?  

> 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.

> * Here are my thoughts on this interesting digital controlled current source. The standard ramp / sawtooth oscillator is an op-amp integrator. Because an op-amp is an active circuit, an integrator has a linear charge or discharge rate, unlike the typical RC circuit which is logarithmic. In this Synthex circuit, the capacitor is not inside the feedback loop of an op-amp, so it's not really a standard integrator. My wild-ass guess is that the changing current flow of the resistor ladder ends up changing what would normally be a logarithmic discharge rate (as seen in an RC circuit) to perhaps something that is more linear. I'd have to build the whole thing in SPICE to see the curve, but since you're already planning to simulate then I can possibly get by with being lazy and waiting for your results! In light of my theory, perhaps you could start by simulating just the components around C93 and see what you get.
> 

Why model it in SPICE?  It would be easy enough to build, and then you
know it's correct.

> > 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.

> > So, what insights can you offer? Is my "discrete digital oscillator driven from divide-down technology" view of the circuit right at all? Is it really a discrete DCO design, like Mario seems to be claiming? What on earth is going on?!
> Mario does refer to C93 as the central part of the sawtooth DCO, so I'd focus on that part of the circuit first. Keep in mind that the resistor ladder "DAC" is not buffered, so it's not really outputting a Voltage. Instead, it's a bizarre, digitally-controlled discharge network for the sawtooth cap.
> 
> Now I need to focus on the LS193 parts and see what's up there. Or maybe I should move to SPICE already and stop the wild-ass guesses.
> 
> Brian Willoughby
> 

The more I think about it the less convinced I am that C93 does very
much at all.

The 193s are just wired as a pair of four-bit counters.  UP and /LOAD
are permanently strapped high and CLR is permanently strapped low.  DOWN
is clocked with the incoming "prescaled" clock, anywhere from just south
of 1kHz to 4MHz, /BORROW is wired to the clock input of the MSB counter
which has its other inputs wired the same.  Then the borrow output of
the MSB counter drives the 4H flipflop to alternate the "other" input of
the XOR gates, allowing you to approximate a triangle wave.

-- 
Gordonjcp MM0YEQ