Author Topic: Oddities....  (Read 1153 times)

343 Salty Beans

« on: February 19, 2006, 04:42:11 PM »
Okay, I've wired all the pots, the LED, and the board for my Rat. The last thing I need to figure out is my power jack, and then hook it all up to the 3PDT.

My power jack has 3 lug...I've seen this on a few layouts but I can't make head nor tail of the third lug. I've identified my negative (center pin) lug and my positive (outer shell) lug...I'd assume the third one is ground, but at this point with my first pedal I figure it's better to ask a stupid question rather than fry the stuff that took two weeks to ship.

I know these two things...9V+ (outer part of a male 5mm plug) should go to (duh) the 9V+ on the board, and the LED anode (or at least the middle outer pole in the 3PDT switch, if the LED anode is attached to the middle common). The center pin of the jack should go to the sleeves of the input and output jacks, the cathode of the LED, and the board ground.
So if I'm NOT including a 9V battery option, how would I wire this sucker? It's plastic, so I don't need to isolate it from the case...I just need to know WHAT that third lug is for.

Thanks in advance.

Re: Oddities....
« Reply #1 on: February 19, 2006, 04:59:07 PM »
your third lug is likely a switch contact. to find out what it does, test for continuity between the positive terminal and it, and between the negative terminal and it. when you put a plug into the jack, it will disconnect this switch.

basically, it lets you include a battery by allowing it to disconnect the battery when a jack is plugged in.

Mark Hammer

Re: Oddities....
« Reply #2 on: February 19, 2006, 05:18:44 PM »
The thing about power jacks for battery-operated devices is that they are intended to allow the battery to be providing the power UNLESS a plug is inserted and pushes the battery contact out of the way.

So-called "closed circuit" jacks  are set up so that  whatever the "active" or "hot" part of the plug  will push a movable contact away from another one.  If no plug is inserted, the contact moves back to a default position where it touches the third (non-ground) contact.

With phone plugs, it is the tip that normally does the "pushing".  In the case of barrel-type plugs, it is the outside shaft that has to do the pushing..  What you should find, if you have nothing inserted into the jack is that there is zero resistance between two of the solder lugs (with no plug inserted, the circuit is "closed").  When the plug is inserted, that connection should become open circuit.

So how can you tell which is which?

  • The jack may have 3 lugs, but the plug has two.  Identify which solder lug on the plug connects to the outside of the shaft.
  • With the cover of the plug still off, insert the plug into the (uninstalled) jack and touch one probe of your meter to the solder lug for the outside of the shaft.
  • Probe the two lugs of the jack for the one which shows continuity with the outside of the shaft when the plug is inserted.
The jack lug that shows continuity with the plug shaft is the lug that goes directly to the V+ line on the board.  The other lug is the one you attach your red battery lead to.  When the plug is inserted the battery will be disconnected from the wire connected to the board and the circuit will take its power from the plug.  Pull the plug out and the contact snaps back into place and touches the contact for the battery, connecting it to the board.

343 Salty Beans

Re: Oddities....
« Reply #3 on: February 19, 2006, 05:44:48 PM »
So in short, that lug can be left unsoldered, while I use the other two like the + and - of a battery? I'm not planning on including any battery option, I'm just going to put it on a daisy chain.