A probably pretty simple pedal electronics question. for all you pedal builders

[almix12]

I've been working on an "expression box" for my moog for awhile now. Not an expression pedal, but let me explain:
On moogerfoogers there are a bunch of jacks on the back for expression pedals, in the manual, it says the pedals have to be 100k pots.
I want a knob/pot i can set, and switch on and off with a stomp switch. (presumably to switch between where the moog knob is set and where the knob in this box is set.
I got a layout and everything that should work fine.

the problem is, with my 100k linear pot, it only has a very limited range. It is working, but just not enough, it only "turns" the moog knob a tiny bit rather than being a representation of the entire knob.

what do I ned to do to make it act as a complete second pot/knob?

thanks a lot,
Alex

[niftydog]

How is it wired?

[almix12]

like this

[dannybuoy]

Perhaps you need to use a TRS jack hooked up to all 3 legs of the pot?

[almix12]

with no offense at all intended, are you just speculating or do you think that's what it might be?
before i go out and buy a TRS jack

edit: pretty much all pots have 3 connections coming off of them

[niftydog]

From the horses mouth:

Originally Posted by Moog

An expression pedal for use with moogerfoogers should contain a 50K or 100K potentiometer which is connected from the sleeve to the ring terminals of the plug. The potentiometer wiper is connected to the tip terminal. The pedal cable should be shielded, with the shield connected to the sleeve.

So the diagram above is wrong and you will need a TRS plug.

[almix12]

hmmmmmmm
thanks very much, i'll give that a try when i get my TRS cable back
would i then need a 3pdt switch then?

[niftydog]

Maybe not, but it's hard to say without more details about the pedal itself.

However, I have a feeling this isn't going to work how you want it to. I'm willing to be proved wrong, but usually with this type of setup the TRS socket has built-in switches that disconnect the internal control as soon as a plug is inserted. So, with your idea, merely disconnecting the external pot from the TRS jack might not be enough to return control to the built-in pot. The only thing that will do that is removing the plug from the socket.

Easy to test, take a TRS plug with nothing connected to it and plug it in - if you loose control with the built-in pots you're idea is in trouble!

[almix12]

it seems no matter what i do (including short circuiting a TS cable that's plugged into it) I can still change it with the moog knob itself. however i suppose it's still possible that with a TRS cable it might be different, i'll see. At the moment it's working almost how i want it too, just doesn't have very much range at all. But when i disconnect the circuit, it goes back to the moog knob position.

Oh, and another thing that may on may not be important, the tiny change that I do get from turning the knob is always around where the moog knob is. So if it's really low, i'll be able to go plus minus a tiny bit in the low range
if it's really high, i'll be able to go plus minus a tiny bit in the high range

[niftydog]

Cool, so it may actually be of some benefit. Like perhaps you'll be able to set a maximum or minimum adjustment with the moog knob.

Please keep us posted, I'm interested to know how you go.

[almix12]

cool, i'll let you know when i have a proper TRS cable to test it with
the next step after that will be figuring out how to attach a stomp switch to it... and what kind of switch

[niftydog]

Oh yeah... I reckon you'll find the sleeve of the connection is probably ground, and there'd likely be no harm leaving that connected permanently. So just switch the tip and the ring connections with that DPDT switch and you should be sweet.

[SamanthaCay]

could it be that it needs a audio taper pot not linear?

[almix12]

i believe in the moog manual it says linear. But something along those lines had crossed my mind as well.
What's an audio taper pot? How does it differ from linear?

[rratajski]

I was just thinking about this...It's interesting that Moog tells you that the expr pedals need 100K pots b/c inside the Moog expression pedal, there is a 50K pot...hmmmmmm...weird!

For more info on audio and linear pots, go here: http://www.beavisaudio.com/techpages/Pots/

[Toasted]

Save yourself some hastle and get the moog controller, eh?

[niftydog]

Originally Posted by rratajski

...It's interesting that Moog tells you that the expr pedals need 100K pots b/c inside the Moog expression pedal, there is a 50K pot...hmmmmmm...weird!

What am I, chopped liver?

That info comes DIRECT from Moog.

Also, having a look at another manual I discovered that the sleeve connection is definitely ground, so you can leave that connected and it shouldn't cause a problem. The tip is +5V and the ring is the control voltage input. The pot simply works as a voltage divider, and the supplied voltage is current limited, so the value of the pot almost doesn't matter. My suggestion would be to use a linear pot - a log pot will have virtually 0 effect for 80% of it's rotation.

[almix12]

maybe 50k works better?
i might try a 50k and see

[almix12]

and those moog things are WAY more expensive. I'll be spending $20 tops on this, including everything

[Toasted]

... If it works.