(hopefully) simple Potentiometer question(s)

[helloesposito]

I've been searching online with no luck for the past few hours. I know that this Vol pot in the below image would be, from top to bottom, 3 coming from the end the circuit, 2 the wiper, and 1 to ground.


This one confuses me though. I'd assume from left to right it's 3 (from the 9v, after the 3.3k resistor), 2 (the wiper, but unsure how the line coming back from 1 enters into it) and 1 (going to the rest of the circuit, but again, confused about the line going back).


Would there have to be an actual lead that connects 1 back into 2?

Depending on how the wiper line shows up, does that show the orientation of the pot? Like 3-2-1 in the above image, but if the wiper was entering from the bottom, would it be 1-2-3?

Am I over thinking this and making it seem much more complicated than it is? SORRY IN ADVANCE.

Thanks

[JerS]

You had it right - just connect lug 1 to lug 2 with a conductor. In this configuration, the pot is acting like a variable resistor.

[R.G.]

It could actually be either way. The idea of a variable resistor instead of a voltage divider is to make the the resistance change the way you want it to when the shaft turns. This might be either way, depending on what the circuit that's attached to it does when the resistance changes as the shaft turns.

Well thought-out and well-drawn schematics will indicate with an arrow, "cw" or a dot which direction the wiper moves when the shaft is turned clockwise. Numbering conventions are far, far... far less reliable about this, and require a lot more interpretation to figure out what's going on. In most on-line schematics, the "well thought-out and well-drawn" stuff is ignored or not known.

[Mark Hammer]

I will echo this and suggest one interpret pot connections with respect to what produces "more" of what the control is supposed to do.  Sometimes that consists of more resistance between the wiper and another lug, and sometimes it consists of less resistance.  One needs to determine if rotating clockwise produces the desired change in resistance between wiper and that other lug.

[thelonious]

To expand on what RG and Mark said: when wiring a pot as a variable resistor (as in your second pic), you really only need to use 2 lugs. If you want resistance to go up when you turn the shaft clockwise, use lugs 1 and 2. If you want resistance to go down when you turn the shaft clockwise, use lugs 2 and 3. Which way you should do it depends on your circuit. If you wire it up and it works backward, just switch it.

These two schematics are functionally the same:



People only connect the third lug as a failsafe. With higher voltage or critical applications, it's a good idea to connect the unused lug to lug 2 (the wiper). That way, if the wiper fails, you still have the full value of the pot (from lug 1 to lug 3) in the circuit, rather than an open circuit. But we're not talking about a bias pot for a tube amp here... we're talking about 9v-18v pedals, right? If your pot dies in a pedal, it's annoying, but it won't melt your $400 matched quad of NOS Mullard power tubes.

[helloesposito]

Thank you so much, everyone! This helps a lot.