The Juno 60 mixes several waveforms before the filter, and it has a clever way of controlling the sub oscillator volume. As the sub oscillator output is a square wave, the wave is considered a binary on/off signal. It is connected to the base of a transistor, which when turned on and off either sinks its collector to ground or leaves it as it is.
To control the wave volume, the collector input is connected to the panel potentiometer (through a series of buffers/mux/demuxes, but that doesn't really matter, the principle is the same). The wave's amplitude (or "on value") will then equal the level of the panel potentiometer, though negative as the potentiometer value is inverted by IC17 on panel board A. The voltage is passed through an analog switch, IC16, which in turn is controlled by the Sub oscillator on/off switch, to completely disable the sub oscillator (incidentally, this feature has been removed from the Juno 106, here the potentiometer voltage is connected directly to the transistor collector).
Interestingly, the same arrangement exists for the pulse wave, only this time the analog switch connects the transistor to -15V instead of to a variable voltage. If one would like a controllable pulse wave volume, it would likely be easy to inject the control signal here.
As for the saw wave, it also has a switch and a transistor connected to it. However, as the saw wave is not binary, it cannot be controlled the same way. Instead, the wave is connected to the collector, and the transistor base is connected to the analog switch on panel board A. The wave is then sunk to ground if the transistor is on. In other words, the saw wave has no pre-filter volume control, only an on/off switch.
The noise input has its own volume control on panel board A (all voices use the same noise generator).
Oh, and by the way - all waveforms have a range of 0 to maximum -15V (the datasheet says the saw wave is 12V p.p, I have not checked the others but they have to be about the same to be mixable - the sub oscillator level for example originates as a 0 to +5V voltage but is inverted and amplified by IC17, and the pulse wave level is always 0 to -15V). They are fed through a 10uF non-polar capacitor (C5) just before the filter, which probably centers them around 0V.
Sunday, April 16, 2017
Wednesday, April 5, 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).
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).
Monday, April 3, 2017
DCO saw core testing
I built the DCO saw core on a prototyping board Friday evening, and after some initial trouble, I got it working very well.
The core is heavily inspired by the Juno 6/60 DCO. The output of the MCU charges a capacitor, turning the output square wave into positive/negative spikes on each square wave edge. These in turn briefly switch on a transistor that discharges a capacitor. When the transistor is off, the capacitor is charged using a constant-current scheme where the cap is connected across an opamp (just like in a normal saw core VCO, see section "That old opamp trick again" of http://xonik.no/theory/vco/reference_current.html).
The Juno 60 uses a 2SA1015 PNP transistor. I tried replacing this with both a 2N3906 and a BC557, they both work very well.
I had no DAC available for the trial, so the charging current was regulated using a potentiometer as a voltage divider, buffering the output with an opamp and piping the result through a resistor to convert it to a current. After initially screwing up and connecting the center pin of the pot to ground, things started working very well.
A few challenges still persist:
1) The output wave starts at 0 and goes downwards until it is reset. It has to be centered
2) My calculations are based on an assumption that the maximum voltage should be 5V. I completely forgot that it has to be 5V on each side of the wave, making the full wave 10Vp.p.
3) For some reason, the charging stops at -8V. Not sure if this is due to the opamp or something else, but it has to be fixed. The Juno 60 service manual says they have a 12Vp-p output, so it should be possible to fix this.
4) Charging starts at the negative going edge of the MCU output wave while the MCU output wave is first high then low through the course of a cycle. This has to change, either in code or by replacing the PNP with an NPN transistor (The Juno 106 service manual says it uses an NPN but I couldn't get this working immediately. I'll look into it).
5) With a 100k charging current resistor, the charging current won't be strong enough for the highest frequencies. A 39k resistor may work (but not for a 10Vp-p wave).
6) I should check if a constant reset timer interval may work (e.g. not a 50/50 duty cycle for the MCU output wave). If so I'll save some clock cycles during frequency calculations.
The core is heavily inspired by the Juno 6/60 DCO. The output of the MCU charges a capacitor, turning the output square wave into positive/negative spikes on each square wave edge. These in turn briefly switch on a transistor that discharges a capacitor. When the transistor is off, the capacitor is charged using a constant-current scheme where the cap is connected across an opamp (just like in a normal saw core VCO, see section "That old opamp trick again" of http://xonik.no/theory/vco/reference_current.html).
The Juno 60 uses a 2SA1015 PNP transistor. I tried replacing this with both a 2N3906 and a BC557, they both work very well.
 |
| MCU output pulse train (top) vs. transistor control spikes (bottom). top is 5V/square, bottom is 1V/square and shifted down to make both visible |
I had no DAC available for the trial, so the charging current was regulated using a potentiometer as a voltage divider, buffering the output with an opamp and piping the result through a resistor to convert it to a current. After initially screwing up and connecting the center pin of the pot to ground, things started working very well.
 |
| A perfect 5v p-p saw wave generated from the pulse train on top. |
A few challenges still persist:
1) The output wave starts at 0 and goes downwards until it is reset. It has to be centered
2) My calculations are based on an assumption that the maximum voltage should be 5V. I completely forgot that it has to be 5V on each side of the wave, making the full wave 10Vp.p.
3) For some reason, the charging stops at -8V. Not sure if this is due to the opamp or something else, but it has to be fixed. The Juno 60 service manual says they have a 12Vp-p output, so it should be possible to fix this.
 |
| The charging of the cap maxes out at 8V (looks like 10V here but Y-offset was wrong). |
5) With a 100k charging current resistor, the charging current won't be strong enough for the highest frequencies. A 39k resistor may work (but not for a 10Vp-p wave).
6) I should check if a constant reset timer interval may work (e.g. not a 50/50 duty cycle for the MCU output wave). If so I'll save some clock cycles during frequency calculations.
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