connecting pot lugs

[notnews]

Hey all,

I'm having an argument with a friend and we need the forum to settle it.

When a layout calls for a single hole to connect to two lugs of the same pot (i.e. when you see "Delay 1 & 2" for a single hole on a vero), is it ALWAYS okay to wire that hole to just one of the pot lugs and bridge the pot lugs with a small piece of wire, like a resistor leg? OR are there some instances where the wires coming off lug 1 and lug 2 need to be independent, and you can't just bridge the lugs?

Thanks buddies.
Alex

[Ciaran Haslett]

The connection is the same regardless.  In fact, if the pot is wired as a variable resistor you could happily JUST connect to lug 2.  Tying two pot lugs together like you mentioned is just for redundancy (and OCD!!!)

[notnews]

EXACTLY, redundant. No matter what purpose the pot serves in the circuit it is OK to just bridge the lugs together and wire one lug to the board. Thanks, C!

My friend's argument was this:
let's say you have one spot on a board that's supposed to connect to lugs 2 and 3 of a pot. If you wire the board to lug 3 and then bridge 2 and 3, when the circuit is trying to output from lug 2, the output has to go across the bridge to lug 3, and then goes inside the pot via lug 3 AND down the wire to the pcb... his argument was that the signal travelling from lug 3 back into the pot caused issues, whatever they are.. be it impedance, feedback, or otherwise via "signal competition" as he calls it.

False, right? In my mind the things he was saying were starting to make sense and I got confused. thanks guys.

[Ciaran Haslett]

Electrons will follow the path of least resistance...so the "inside" of the pot between lugs 2 and 3 won't even be seen by the circuit.

[induction]

Your friend is confused. The only difference between the two scenarios is the length of the wire bridging the wiper to the unused lug.

Remember a signal is just a variation in voltage over time. On an ideal level (ie. assuming the wire bridging lugs 2 and 3 is a perfect conductor) lugs 2 and 3 are at exactly the same voltage at all times. Thus they carry the same signal. Since there is no voltage difference between the wiper and lug 3, no current will flow between them inside the pot. That section of the pot may as well not exist.

On a less ideal and slightly more realistic level, we can consider the wire to have some small resistance. In this case, there may be some slight voltage difference between lugs 2 and 3. How much difference depends on the length of the wire connecting them, but the voltage difference will be tiny unless the wire is very, very long. Connecting the two lugs to the board with separate wires would make the voltage difference between the lugs larger, and would cause a tiny amount of current to run inside the pot, which could not possibly affect the operation of the circuit unless it was already unstable. It would be insane to design around the specific length of wire connecting an unused pot lug to the wiper because the circuit doesn't care about the voltage on each pot lug, it can't even see that. The circuit only cares about the voltage seen by the surrounding components. If you want the circuit to care about the difference in voltage on all three pot lugs, you'd use a voltage divider instead of a variable resistor.

Bottom line: It's not important whether you bridge the lugs directly on the pot or on the board. But if it did matter, it would be better to bridge them directly on the pot.

[Frank_NH]

The electrons will not know the difference between the two approaches.  (That is, either way is OK).  

Having said that, the fewer wires dangling around in your 1590B the better, so I'd go with method 1 given the choice.   

Edit:  I see Ciaran and induction are quicker on the draw than I.  

[BuGG]

It really doesn't even make a functional difference if you don't bridge the two lugs together....  If you just connect the wire to the middle lug it'll work just the same.

There is some logic to connecting the lugs together, wipers are noisy and may (very) briefly lose contact with the resistive track while rotating.     During this one instantaneous moment (or if the pot fails) if the lugs are not connected together you have a totally open circuit.   If the two lugs are connected together the resistance doesn't go higher than the maximum value of the pot.     This is a hypothetical situation that I've never actually encountered myself.....

[tabbycat]

am on the same page as bugg with this. i remember reading (i think it was the great rg keen, so a reliable source) that while just attaching to the wiper (lug 2) gets the job done and is all that is needed technically (perfect world), pots are imperfect things so it is worth attaching a failsafe wire from lug 2 to 3 so if the wiper dies there will still be a connection there, albeit the pot wide open. better that than a completely non-functioning pot.

also worth screwing the nuts on your pots down really tightly when you box up to stop any electrons escaping out through the top of your enclosure. better safe than sorry.

[ξεναγος νεκροπολης]

tabbycat wrote

also worth screwing the nuts on your pots down really tightly when you box up to stop any electrons escaping out through the top of your enclosure. better safe than sorry.

haha!
i always use electrician's tape to secure  the jacks holes when i don't have a cable plugged in!