onsdag 5. april 2017

Juno 106 vs 6/60 DCOs

There are some differences between the DCOs in the Juno 106 and the Juno 6/60 - except for the fact that the DCOs on the 106 are integrated into a single chip.

First of all, according to the datasheet, the 106 uses an NPN instead of a PNP transistor to reset the integrator (saw wave converter).

More interestingly, the way they sum the saw wave and the pulse waves are very different.

The Juno 60 has a separate control line for the pulse wave, connected to TR2. It looks just like the one for the saw wave (TR3) so one can assume that it turns on and off the pulse output. Also, the output of the pulse is sent through D2 which will block any positive halves of the pulse (? which sort of makes sense as the saw wave is also negative only).

The 106 on the other hand, has no such control line, which means that the pulse output is always on. The pulse output has a diode to ground which I assume means that it will never be negative (?).

So how can the 106 output a saw wave and no square wave? One theory may be that if the user selects saw wave only, the pwm is set to 50%. Summing the sqare and saw waves will chop up the saw wave just as it reaches its half period, and move the remainder upwards. If the amplitude of the saw and square waves are equal, the saw wave will magically realign with its phase shifted half a period, this time centered around 0V.

DISCLAIMER: This is only my initial theory after studying the datasheets, no measurements have been made. 

But how about when both the saw and pulse waves are on? Wouldn't this mess up how the wave turns out? Since the saw wave is positive and the pulse is negative, couldn't the total amplitude end up being double?

Well, the high part of the pulse will always come at the second half of the period, thus it will never "lift" the saw wave from higher that minus half the total voltage p-p. Also, The parts that are lifted will only be lifted by 1 x the voltage p-p.

In practice, this means that if we could set the duty cycle to 0%, the output would be a saw wave that starts at 0 and drops to minus the total voltage p-p. If the duty cycle is 50%, we get a saw wave with the same amplitude centered around 0V.

This is interesting and may reveal a flaw in my assumptions. As the wave is centered around 0V by a capacitor later, summing a pulse and saw wave this way will only shift the saw wave phase! I need to check if the polarity and phase assumptions for the square wave actually holds. What happens if instead the polarity is reversed or the phase shifted by half a period?

TODO: Write about SUB OSC and level control (done by using the output from the DAC) and how the passing through a diode affects the signal (reversed in one)- one uses NPN and the other uses PNP, 60 has a resistor to GND at the transistor base. The 106 has no on-off switch for the SUB OSC, only a volume control.

Both the 106 and the 60 DCOs have their outputs mixed with noise (with adjustable volume) right before a 10uF non-polarized cap and injected into the IR3109/800170 filter chips. The mixing point has no opamp connected to it so I assume this job is done by the chips. I guess the cap is there to filter out any DC component and center the waves around 0V. I have not looked closely at this - what happens here when the frequency changes for example, will still stay centered during the transition because the amplitude is still the same?

BTW: NP in the juno datasheet means Non-polarized (capacitor). MF means metal film (resistor).  G (capacitor) may mean 2% (needed for accuracy in the integrator).

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