By using a high beta transistor at Q8, you can get a much more TB303-like square. This is because the original 303 used high gain 2SA733ap transistors which are no longer being produced. The x0x0b0x comes with 2SA733p transistors, which are close, but have less gain. For more information on the transistors, see HighGainTransistors
When we designed the x0xb0x, we put in a non-detent tuning pot because that was what the original 303 had. However, this may drive you a little crazy. Panasonic part number P3I9503, an equivalent 50K linear potentiometer with a center detent, is available from DigiKey. Note that you may have to retune the synth once in a while to keep the center detent 'correct'
Cut the trace between pin #5 of IC11 and where R88, R89, and R90 come together. Solder a 1MEG audio (log. taper) potentiometer between the two points, where one is the wiper (usually the middle pin) and the other is to the "lower" end. That is, when the pot is turned all the way counterclockwise, the two connections should basically be shorted and as you turn the potentiometer clockwise, the resistance should slowly increase to begin, and then increase quickly at the end.
Note: this mod will mess with your tuning a little, especially when the pot is turned all the way up. And there's no way to easily fix it.
Alternatively, I think this is a very bad way to do this. I would go about it by changing the value of c35, which is actually responsible for the creation of slide. If you want to squeeze out some hardcore math, find the cutoff frequency of the lowpass formed by c35 and whatever comes before it (DAC network). It could be 2.2K, making the cutoff ~328Hz, which makes sense I guess. I'm sure someone with more math experience could calculate how much time it would take for the voltage to change. Anyway, on to the mod:
Put capacitors in parallel with C35 to make slide time longer, place them in series to make it shorter. I'd suggest values from .1 to 10uF
Note2: normaly the frequency of the slide time Filter (C35 and unknown resistor(s)) is 7.2 to 7.5Hz (this is based on actual measurements of the output, not the schematic or circuit) thus with the 0.22uF cap the resistance is 100K (wherever it is) for a ~twice slower slide use a ~0.44uF cap
If you want to hook up your x0xb0x to a synth with CV out, you'll need a CV in jack. This is pretty straightforward, just use any panel-mount 1/8" jack (or whichever jack you'd like) and connect the sleeve to ground. Solder a wire from the jack tip to one side of a SPDT switch (so you can switch between CV in and internally-generated CV) and the middle to pin 5 of IC11. The other side should be connected to the point where R88, R89, and R90 come together.
If you have a switched jack (with an internal disconnection switch, many jacks are like this) connect the tip connection directly to IC11 and the other end of the switch to the junction point of the resistors as above. Check the datasheet or use your multimeter to verify the pins of your jack.
Wire a jack to a 100K resistor, and then wire that to the base of Q29
This mod should cause some very chaotic things to happen, as more of the frequency of the VCO is modulated by itself, it should hit the stage of chaos (or even feedback) rather quickly. The idea is to take the outcoming signal from the VCO, and feed it back into the CV in. You could either take it from the middle pin of the waveform switch, or wire up a secondary switch, with pins 1 and 3 connected to the other waveform switch, and pin 2 feeding back into the CV.
One of the cool thing is that this mod will be affected by the slide circuit. How it is affected, I am not sure.
The voltage of the waveform output is between 5V and 12V, whereas the Control Voltage goes between 2V and 5V (?). You are probably going to want to divide the voltage by 2 (at least). My guess here is that you would use a voltage divider. If this is the case, then you would wire a 50 K resistor between The VCO out (waveform switch) and the wiper of a 50K pot. You would then wire the other end of the pot to ground. This would give you a 0 to 6V signal into the CV. A 70K resistor would give you a 0 to 5V signal into the CV.
From Waveform----/\/\/\---+------ To CV 70K |___ / | 50K Pot \ <-- / | --- GND
UPDATE: Any kind of logarithmic frequency modulation with the 303 won't work unless you add circuitry. The buffer after the dac is operating in non-inverting mode and cannot be used as a summing amp (which would require a non-inverting op-amp). Simply feeding an ac signal to the non-inverting input of the op amp won't do anything. It's not a simple mod to devise a summing circuit there because of the single supply restriction.
I have no idea how this circuit is affect by current. I don't know the current that should be going into CV, or the current that is output by the VCO. You might be able to fry your VCO if you aren't careful
I dont know if "To CV" is as easy as hooking up the circuit to Pin 5 of IC 11. We need to sum the voltages basically. I don't know if this is the proper way to do it.
By taking the output from the square wave, and feeding it in to a frequency divider, you would end up with a suboscillator. This could be then fed into the VoltageControlledFilter, or depending on the level of division, used as CV for an LFO.
The easiest way to do this is to do a variation of Neil Johnsons Sub oscillator (Schematic here). You feed the output of the VCO into Pin 1 of a CMOS 4024, and then select which output you like best. You could split the output, and have one rotary switch go to CV out, and another rotary switch for Sub Osc Out/VCF in (including an off).
Ken Stone has a much more complex Sub Oscillator, but it looks very cool, and is the same basic idea.
With some (leg) work, this could be hooked into the Self Frequency Mod pretty easily.
Because the Oscillator outputs between 5 and 12v the voltage needs to be biased (like the Self Frequency Mod above) so that it is 0V and 5V respectively. It does not need to be re-biased because the filter is AC coupled, and then biased, allowing you to feed AC coupled signals into the VCF.
One way to easily do this is just divide the signal by by voltage divider by a ratio of 4, giving a signal between 1.25v and 3v. That is very iffy, and it would be better to divide by 2 and run the CD4024 off of 12v, making LOW be 3v and HIGH be 11V. This is from the datasheet and it all depends on your VCO putting out the correct voltages. I'm sure no matter what though, that it will sound crazy.
Even though this is untested, it should work.
|Schematic||Power Supply||Voltage Controlled Oscillator||Voltage Controlled Filter||Envelope Generator||Voltage Controlled Amplifier||Headphone And Mixer||Digital Sequencer||Midi, USB and Sync|
|Fabrication||Building the Ps||Building the Vco||Building the Vcf||Building the Envelope||Building the Vca||Building the Headphone And Mixer||Building the Sequencer||Building the Midi and Sync||Finishing It Off|
|Testing||Testing the Ps||Testing the Vco||Testing the Vcf||Testing the Envelope||Testing the Vca||Testing the Headphone And Mixer||Testing the Sequencer||Testing the Midi and Sync|
|Mods||Ps Mods||Vco Mods||Vcf Mods||Envelope Mods||Vca Mods||Headphone And Mixer Mods||Sequencer Mods||Midi And Sync Mods||Finishing Mods|