Making an expender that crank parameters



the idea came to me when i was playing with the Clouds, to add some variation and glitchyness, i turn the blend knob all the way cranked to just have some random grains playing.
The problem is that we can hear the run from the actual position of the knob to the full position. For live performances, the key word is handy.

So can i just use the +5v from the PSU to bring the blend knob to full, using a momentary switch?

It may seems like a ridiculous question but i really REALLY don’t want to fry my Clouds, perfect instrument but expensive ahah.

If so, is this could work?

Have a nice day! :))

Edit: The switch that i drew is a SPDT not an DPST, but it could work and so have a true bypass too…Tell me if i’m wrong, this switch thing is very large and a bit complicated ahah


That should work just fine, I would think. You’re just momentarily making or breaking a +5V connection to the CV in of “Blend” - nothing too tricky.


Alright, thank for your response!
I’ll try that soon.


Also, if you prefer using the +12V rail instead of the +5V rail (might not be available in some cases…), you just have to wire a 140k resistor between the switch and the +12V source.


I’m curious, how did you arrive at 140k instead of using something like a voltage regulator? Is that just a rough value to take the load or does that correspond to a voltage drop somehow?


Input impedance is 100k. So with a 140k series resistor, you have a voltage divider with a 0.417 ratio, which is what is needed to bring 12V to 5V.


Really need to know the Ohm law by heart now.
Damn i wish i had this epiphany in college…


@pichenettes ah ha, that makes perfect sense. Thank you for taking the time!

The full solve: we use the equation Vout = Vin (R2 / (R1 + R2)) and the input impedance R2 is on the description pages for all Mutable modules. Vin is 12v and Vout is 5! Just reading up on voltage dividers.

@YannDhou Same, totally learned this stuff but never could connect it to real world problem solving.