x0x CV / Gate out to Arp Odyssey

[E:pp:ik]

Hey all - quick question ~

If I trigger CV Gate out from the x0x to an Odyssey the notes from the Oddy are really high. If I use the x0x to trigger an MC-202 for example they're basically spot on but it's way off with the Arp. Is the Arp expecting a different voltage? I thought it was 1V / Oct. ?

[Brassteacher]

Hey all - quick question ~

If I trigger CV Gate out from the x0x to an Odyssey the notes from the Oddy are really high. If I use the x0x to trigger an MC-202 for example they're basically spot on but it's way off with the Arp. Is the Arp expecting a different voltage? I thought it was 1V / Oct. ?

Yes, the Odyssey is 1V/Octave. However, the Odyssey has dual supplies, therefore low "C" can be (and is) zero Volts. Check the position of the transpose switch on the Odyssey. The CV's move to the following ranges using the transpose switch:

Picture 33.jpg (51.99 KiB) Viewed 3009 times

To get the x0x to play down at the lowest end of the Odyssey's range, some way will have to be found to shift the x0x's CV out to accurately go down to 0V. This most likely will require some type of external circuit be built.

[E:pp:ik]

Thanks Teach! I'll have to dig around and find a way to convert this. I think the Octave Cat is a similar situation.

[Brassteacher]

Thanks Teach! I'll have to dig around and find a way to convert this. I think the Octave Cat is a similar situation.

E:pp:ik, I think this is the circuit you want, it will certainly do the job:

CV input mixer.jpg (70.31 KiB) Viewed 2962 times

It's a common CV mixer circuit that appears just about anywhere there are multiple CV inputs into a circuit. Build the portion inside the red box, with a couple changes.

It MUST be powered from the ARP, or with an external power supply, using the +15V and -15V rails (±12V may work fine as well) to power the opamp. Obviously, using the ARP power supply will be cheaper. The KOV/ECV switch is not necessary, just use whichever input jack you favor. At any rate, that input is the one to use coming from the x0x. The bottom of the pot marked P8 (Frequency) in the schematic MUST be connected to the -15V rail, not ground. The circuit will not be able to take the x0x CV out to 0V otherwise. P8 can be a normal pot, or a trimpot, your choice. This pot will now function as your "zero adjust" pot. Have the x0x output its CV for its lowest note, and adjust P8 until the output of A3 is 0V.

The V/Oct trimpot (P3) and R17 are optional, and probably redundant. You can just take the output off the opamp. You should be able to use just about any dual-supply opamp you like in this circuit, as TL074 would be a waste of 3 other opamps.

All this could be built in a small box. I would still recommend running wires out of the Odyssey to power it, ±15V is all you'll need, the signal cables will take care of ground.

If you don't mind modifying the Odyssey, you should be able to build this circuit on a tiny "daughter board", and connect the CV input directly to the Odyssey's CV input jack. The the wire formerly connected to the "tip" contact of the Odyssey's CV input jack and connect it to the output of the opamp. If you choose to use a trimpot for P8, you will not need to drill any holes in the Odyssey's case. WARNING: Doing this will make it difficult to use any other analog synth's CV out to control the Odyssey, as it will need to be re-tuned for the output of the opamp to equal 0V. You may be able to get by with using the tuning knob on the Odyssey, or you can add a normal pot to the back of the synth for P8. You could also add an additional CV input jack for this circuit and, via a switch, connect the output to the "tip" connection of the factory CV input jack. The switch will allow you to turn the "x0x CV in" input off, so that no knob twiddling is necessary when not using the x0x.

I hope you find this useful!

[E:pp:ik]

Sweet! Yes. this is really useful! I'll see how I fare and I may also see if I can get this to play nice with the Cat. Great find. Where's this from?

[guest]

that looks like an inverting amp configuration
which i think will make the scale play backwards

also
how many octaves off is it
with the arp octave switch
in its lowest position

[E:pp:ik]

I have to wire it up again to be sure but it was way off. 3 octaves perhaps? I recall the Cat being a similar situation but I'll have to confirm it again.

[Brassteacher]

that looks like an inverting amp configuration
which i think will make the scale play backwards

also
how many octaves off is it
with the arp octave switch
in its lowest position

In this case, it HAS to be an inverting amp, or it will not work as a mixer.

If the x0x is playing its lowest C, and sustained, and you adjust the Frequency pot until the summed voltage =0V, it should play the Odyssey's lowest C when the octave switch is in the lowest position (unless the ARP goes into a negative control voltage, I'll have to check later when I get my Axxe back). If the bottom of the Frequency pot isn't connected to the negative power rail, it won't work. Or, rather, it won't be able to go to 0V.

I may have missed something here: If you added the V/Oct trimpot, marked P3 and R17 in the schematic, that should be connected to the negative rail also, instead of ground. My mistake, sorry. If you don't have a 250 Ohm trimpot, use a larger one, but keep the proportion with R17 the same. If you wired the output of the opamp directly into the CV input and it didn't work, or can't be adjusted to the proper range with the frequency pot, try adding the V/Oct trimpot.

As for where this came from, it is from just one of the gazillions of schematics online for the "Moog" filter. There seem to be as many variations of the circuit as there are schematics floating around.

Let me know how it goes, and I'll try it from this end when I get a chance (Read: Whenever I can find my ^$%@%^ breadboard).

[guest]

but since its an inverting amp
wont that mean that the output
will do the opposite of the input
so 1v -> 2v in becomes 2v -> 1v out
basically higher notes played in
will make lower notes played out

you can make a noninverting mixer
by using a differential configuration
basically make r15 and r16 100k
and connect your input to one end of another 100k resistor
and the other end of this resistor to the + pin
and then p8 and r11 stay as they are
as you turn p8 closer to +15v the output voltage will drop

[Brassteacher]

but since its an inverting amp
wont that mean that the output
will do the opposite of the input
so 1v -> 2v in becomes 2v -> 1v out
basically higher notes played in
will make lower notes played out

True, but a lot of voltage controlled inputs actually respond to negative voltages anyway. Besides, most of what we think of as voltage controlled is actually current controlled. I'm looking at the schematic for the filter inside the Minimoog right now, the CV input mixer is a single PNP transistor, with the output taken off the collector. The expo converter is a single NPN transistor, with the collector wired to the bottom of the ladder via a 680 Ohm resistor, and the emitter tied directly to ground. In other words the CV actually opens or closes the amount of current flowing through the ladder. I know you know this, just saying for the benefit of others who may be reading.

@E:pp:ik - I may need to go take a look at this again, I had forgotten about the CV scheme in the Odyssey for VCO 2 when operating it in duophonic mode, i.e., with the sync switch off. The reason you can play two different pitches at the same time is that VCO 2 senses the voltage drop across the keyboard when you press two keys at once, and mixes that into its CV input, while VCO 1 takes its CV input only from the voltage divider (keyboard resistor chain) itself. So, try turning the sync switch on if it is not already and see what happens. Also, I don't remember you saying if the notes being fed in from the x0x go the opposite direction on the Odyssey, rather that they were just too high, is this correct? If they go the opposite direction, build the circuit as guest described. If they go the same direction, I'll need to find the breadboard and tweak things a bit.

[E:pp:ik]

This is really helpful and interesting just the same (and thanks for elaborating too). The Cat also has this arrangement. Mono / Off / or Duophonic with VCO1.

Guest - wasn't that the idea though? A specific voltage drop? 2V --> 1V, 1V --> 0V, etc.?

[guest]

sort of
the equation for the circuit shown is
Vo = -1*(r15/r10)*CV + -1*(r15/r11)p8
where p8 is the voltage on the wiper of the pot

the circuit has r15 = 2.37k and r10 = r11 = 100k
so the gain will be 2.37k/100k = .0237
so the x0x will have to play up 40 octaves to get one octave on the arp
so r15 should be replaced with a 100k resistor
which simplifies the equation above to
Vo = -1*CV + -1*p8

so p8 can be used to shift everything down
by applying a positve voltage at p8
so 1v in equals 0v out and so forth

but it will also invert the polarity of the CV signal as well
so 1v CV in will be 0v out
but 2v CV in will be -1v out
so as you play higher notes on the x0x
the ARP will play lower and lower notes
at least if higher voltages equal higher notes on the arp

you should build it up and give it a try
and if it ends up making the notes go backwards
you can just put another inverter before it

[Brassteacher]

the circuit has r15 = 2.37k and r10 = r11 = 100k
so the gain will be 2.37k/100k = .0237
so the x0x will have to play up 40 octaves to get one octave on the arp
so r15 should be replaced with a 100k resistor
which simplifies the equation above to
Vo = -1*CV + -1*p8

Dammit, THAT would be the part I missed. This particular schematic came from the mixer feeding the expo converter of a Moog filter and I wasn't paying enough attention, I just assumed it would be unity gain.

I need to get my insomnia issues fixed.