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Soldering a DC POWER JACK, MALE 2.1MM 2 POSITION

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Use an ohm meter to verify which solder terminal connects to which contact at the jack.

Be aware that there is probably a third terminal that is a switch. It is shorted to another terminal only when the mating plug is not inserted.
 
Like Chris said, the other two make a switch. The purpose is to disconnect a battery supply when an DC adapter is inserted.

So, one pin will always be connected to the center conductor, both when a plug is inserted and when it isn't.
The other pin will disconnect from the center conductor when a plug is inserted.
 
If you'll download the 305kB datasheet thats on the Jameco product page in your link, you'll see a small schematic showing that the barrel connection (pin 2) is shorted to pin 3 when the connector is unplugged. To summarize, pin 1 is + and will often have 2 wires attached (1 to the load and 1 from the battery), pin 2 is - with a wire going to the load and pin 3 is also - with a wire from the battery. On some connectors the terminals (pins) are numbered and on others you have to rely on a meter or the data sheet drawing to identify them.
 
The other pin will disconnect from the center conductor when a plug is inserted.
The datasheet actually shows that the third pin connects to the outer conductor when the plug is withdrawn, not the center one.

To summarize, pin 1 is + and will often have 2 wires attached (1 to the load and 1 from the battery), pin 2 is - with a wire going to the load and pin 3 is also - with a wire from the battery.
While a vast majority of plug packs are tip +ve, I've seen a few that are tip -ve. Check you plug pack or device just to make sure.
 
im plugging it directly into the AC outlet using a 9v dc power supply.... whatever im doing is definitly NOT right, when the power supply is cut the breadboards still has a + 9V charge for a a few min. Ive tried every combination of wiring this i can think of. The only way which seems to work is both outside terminals are connected to - 9v , the middle is +9v. I am not using a battery,,,,, and me thinks i should be using a two terminal DC power jack... not this 3 pin thing which from what everyone is sayin is intended for a battery as well???
 
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im plugging it directly into the AC outlet using a 9v dc power supply.... whatever im doing is definitly NOT right, when the power supply is cut the breadboards still has a + 9V charge for a a few min. Ive tried every combination of wiring this i can think of. The only way which seems to work is both outside terminals are connected to - 9v , the middle is +9v. I am not using a battery,,,,, and me thinks i should be using a two terminal DC power jack... not this 3 pin thing which from what everyone is sayin is intended for a battery as well???

Are you saying that you still have power on your breadboard after you unplug the output cable from your dc power supply from the jack? If so, you either have some capacitance on the board that is holding the charge, or you have found some heretofore unknown source of power. :)

As for the third pin on the jack, if you don't need it, just leave it unconnected. The datasheet for that jack shows terminal one is the center pin. Terminal two is the outside barrel. Terminal three is the switch. It is connected to terminal two only when a plug is not inserted. Check the drawing on the datasheet to know which is terminal is which.
 
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