Oscillator drone box pitch bend?

So I’m gonna make a drone box using the reverse avalanche oscillators but I want to modify it to add a universal pitch bend. If I’ve got it right I just build the oscillators separately and combine their output wires to the tone pot and wire my jack output there to get the mixed oscillator outputs.

My question is what else do I need to add besides another pot for to make a master pitch bend? I found this PINK DRONE OSCILLATOR | SOUND BENDER 36 based off the super simple oscillator design but don’t quite understand it. image

What’s the purpose of the led and switch?

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hi & welcome here @TheArcticKitten !

it’s an add (the switch open/close the power and the led indicated it) but you can take off the switch and the led, and keep only the power starve potentiometer :slightly_smiling_face:

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Ah so a pitch bend is as simple as adding a resistance to the output of an audio signal? like i can just take the output from the oscillators and throw a potmeter on it to pitch bend? sorry would test this out if i could but i messed up breadboarding it so still working on fixing that

No. That pot is not on the audio signal output.

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Funny, I just asked this about another circuit here: A litany of dumbassery - #355 by BlackDeath and analog broke it down here: A litany of dumbassery - #359 by analogoutput

I’m pretty sure that add a variable resistor (pot) to the power decreases the speed of oscillation, adding it to the audio line will only decrease the volume.

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That’s exactly what’s done in the reverse avalanche oscillator as shown at the above link:


The pot is on the supply voltage, and it affects the pitch. By limiting the current it increases the time needed to charge the pot, so the frequency decreases.

In the picture posted above one pot is added to the supply voltage for multiple VCOs, changing all their pitches at the same time.

Be careful not to put too much current through the pot or it’ll burn out.

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Ah thank you so much! so much clearer now I wasn’t sure what they had meant by VCO+ but now I understand.

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That’s a penis. I’ll get my coat.

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Hey so I finally got the circuit properly working and i’ve run into a bit of problem with choosing the proper pot for detuning. I read your reply to blackdeath and i’ve tried a 1k, 100k, a 4.7k 2W pot, and all of them will detune properly but if i turn the knob all the way they end up smoking. I understand this is because i am putting through much current in them (the wall wart im using to supply is 12vdc 250ma but when I read the voltage with my multimeter it reads 20v).

How do I go about implementing this without having to just not turn the pot all the way?? Would a voltage divider type circuit work? the circuit needs 12-18V. Thanks again!

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You can see above the individual oscillators have a fixed resistor in series with the pot, to limit the current and avoid burning things out. But presumably if you have multiple oscillators you’re trying to control with a single pot, their total current is too much for the pot when it’s fully open. You could increase those series resistors, or add another series resistor (maybe need a higher power than the usual 1/4 to 1/2 watt), to limit the current even more, but that will also limit the pitch range more.

Basically trying to control current rather than voltage with a pot is a dicey proposition and you’re running up against the limits. You could presumably use a high power rheostat instead, but that’ll set you back $60 or so.

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