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Replace a pot/knob with a CV input? (For dummies version)
MUFF WIGGLER Forum Index -> Music Tech DIY  
Author Replace a pot/knob with a CV input? (For dummies version)
Hi all,

Context: I can solder well, can sort of read schematics, and I've managed to fix more things than I break when I rip them apart. That's the baseline for my electronics aptitude. smile

I'm trying to figure out how hard it is to swap out a piece of kit's pot knob with a 1/4" plug so I can replce the knob twiddling with electrons dancing.

I've googled it, and everything I find immediately seems to progress to an Electronic 400 level class. :(

For example, on a Vermona DRM MkIII it seems like a cool thing to do would be to replace the pitch knob with a CV input...

It "seems" like it should be fairly straight forward conceptually, but that just may be my naivete. The fact that I haven't seen more examples of people doing this makes me think perhaps there's much more to it that I thought?

does anyone have a really "For Dummies" high level version of how this would be done, or a link? I've search sites like Instructables, etc. but so far not finding anything. Maybe I'm just using the wrong search terms...?
I dont really know but I think you can use vactrols
Link-> VAC-PAK CV to Variable Resistance – A Circuit Bender’s Dream
You can also use a JFET to replace a pot being used as a variable resistor as it will essentially function as a voltage-controlled resistor, but there is some nuance to choosing and setting up your JFET in this situation that I can't explain in brief. If you're curious, just start out by learning about JFETs in general... this tutorial seems pretty good:
I would say this is a cleaner way of doing it than using a vactrol as the JFET will have a far more linear response than the vactrol.

However, replacing a pot with CV control is often not as simple as just putting in a JFET or vactrol. You can't just slap a single JFET in there if your pot was being used as a voltage divider, and you can't slap a single vactrol in there in that case either.

If you're not sure what I mean by this: generally, a pot with only two of its pins connected or with two of them connected to each other is being used as a variable resistor, and a pot with all three leads connected to different things (often including your +supply and/or -supply and/or 0V) is being used as a voltage divider.

If your pot is a voltage divider, things are more complicated, and you'd actually have to create a whole circuit to replace that pot... what you need is a voltage controlled voltage divider.
This may be used.. there are examples for both situations explained by belltones com/images/VtoR.jpg
It really depends on how the pot is used in the circuit.

It is very common for a panel pot to be wired as a voltage divider from one of the supply voltages to ground, so turning the knob just supplies a variable voltage to some part of the circuit. In that case it is pretty straightforward to turn this into a CV input: just replace the pot with a jack. You may want an op amp buffer after that to set a high input impedance and perhaps to scale the incoming CV so that the circuit receives the same range as it did from the pot.

On the other hand if the pot is really being used as a variable resistor, things are more complicated and the previous posts apply.
also makes sense! maybe look at the Mi stuff for inspiration. 0.pdf
berenie's got you with some actual example circuits for how to do it!

cygmu's suggestion that you can just drop the CV in in place of a pot wired as a voltage divider is true only under specific conditions. A little more on what the qualifiers/concerns would be:

1. the voltage being created by the original voltage divider must be in the same range as your incoming cv would be. if you have a voltage divider pot being used to give you a range of voltages between -10V and -15V, you don't want to just put in a CV patch point and feed it 10V p/p oscillating about 0V for instance.
In a situation like this, feeding in a voltage so different from what the circuit is designed to have at that point could actually destroy a transistor or IC!

(If you're using one of the circuits berenie linked to, you also have to be aware of what incoming voltages would not work in your original circuit, and set up the circuit you're making to be sure to avoid them, and also keep the output in the range you need.)

2. the voltage being created by the original voltage divider can't be subject to pretty much any current draw. The whole concept of control voltage hinges on the CV having to be brought into a circuit and right away sent to high impedance devices (normally op amps) so that the CV merely controls the circuit without powering it. Otherwise your CV will drop... the module sending the CV could even stop working if the current draw is too great.

If your pot voltage is being subjected to current draw, that's when/why you'd have to use an op amp buffer to protect your CV source from having its current all drawn off. An op amp that is not configured with unity gain can also help you get your incoming voltage into the correct range.

3. Is it OK for there to be NOTHING at that point? If your CV input is the only thing sending the voltage to that part of the circuit and there is nothing plugged into that jack, will it still work?
It might be necessary to sum your pot's original voltage with your incoming CV, or reduce your pot's original voltage and then sum it with your CV, or sum your pot's original voltage with an attenuated version of your incoming CV, etc etc

Basically there's no easy answer and it's just going to be very application-specific, which methods will and won't work, and what you'll have to do to make any of them work better.
belltones is 100% right.

The Branches schematic linked above is a great example both of how to have a pot in straightforward voltage divider configuration and a CV, and of the fact that things can be more subtle than you expect.

[Ignore the capacitor for now...]

Here you can see a pot acting as a voltage divider between GND and -5V. So turning the knob sends a voltage of 0 to -5 into the op amp, which inverts it, so you get 0 to 5V out, which goes to the rest of the circuit.

Alongside, there is a straightforward jack CV input. It goes through the same op amp, so again the CV will be inverted, and added to the CV from the pot. But I bet the CV you apply is not in a 0 to -5V range. So the designer (Olivier Gillet) has obviously decided that the typical use of the module calls for a sum of whatever your typical CV range is with 0 to -5V coming from the pot. Why it's like that, I don't know.
oh yeah, a very good example from cygmu!
All of the above proves the difficulty one can encounter in electronics. But you could start simple:
Trying to mod some Doepfer modules can be a great start in eventually bending your Vermona smile
I wouldn't replace a pot of a pitch Control with just a CV in jack... i would leave the pot were it is and add the CV in as further Control... The day will come, when you will need the pot too... having both seems to me to be more desired as i can't have enough control Options...

The vactrol method imho is the easiest and simplest solution to add further control possibilities.
I have various old external effects that I'd like to integrate with my modular, so I've wondered about this. You could do this digitally:The advantages of this approach would be:
    Speed of response.
    The CV and knob response could be customised in the PIC code.
Disadvantages include cost (I guess), relative complexity and a limited range of digital pot resistance values (i.e. nothing greater than 250kΩ, unless you ran several in series).
this seems like an interesting opportunity for a simple retrofit board with jumpers and or trimmers to define the range.

i am heavily in favor of adding cv or gate/trigger control over all pots and switches wink
synchromesh wrote:
[list]Connect the existing pot and a CV input to e.g. a PIC16F684 (as used by Tom "Electric Druid" Wiltshire, costs less than $2).
The PIC drives e.g. a Maxim MAX5483EUD+ 10-bit digital potentiometer (available in 10kΩ or 50kΩ for about $2.50) or the Analog Devices AD5235BRUZ250 (10-bit, dual-channel, 250kΩ for about $6.50).

I just sponged this piece of text into the deepest part of my mind. I needed that! we're not worthy
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