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I'm working on a circuit to activate a relay with an Arduino, adapted from this schematic:

enter image description here

The relay is a RTE24005, which has a 5V coil. When I use the 5V and ground from the Arduino it works perfectly.

When I use a wall wart power supply instead (5V, 550mA), it fails to switch.
The only real difference I can tell is that the wall wart isn't the same ground as the the input from the Arduino.
Does this make a difference?

Is there something else that I'm missing?

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You do need a common ground connection between the Arduino and the added power supply, otherwise the transistor doesn't have a reference for the Arduino output.

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  • \$\begingroup\$ Where would I learn more about this? I can just use the Arduino for both the 5V and ground for this project, but I'm planning on using this circuit layout with a larger relay later. I'll need 12V coil for that one, so the 5V from the Arduino isn't going to cut it. Would I just splice the ground from the Arduino and the negative from the power supply together? Thanks! \$\endgroup\$ – AndyD273 Feb 8 '17 at 3:11
  • \$\begingroup\$ Ok, just tested out hooking the ground from the Arduino to the ground for the power supply and it worked just fine! Thanks for the info! \$\endgroup\$ – AndyD273 Feb 8 '17 at 3:35
  • \$\begingroup\$ Oh, side question because the person that helped me with the schematic didn't explain it... What's the purpose of the backward diode parallel to the relay? \$\endgroup\$ – AndyD273 Feb 8 '17 at 3:39
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    \$\begingroup\$ The diode across the relay coil (called a "catch diode" or "Freewheel diode") is used to limit the high voltage spike that would otherwise be produced by the relay coil inductance when the relay is switched off. \$\endgroup\$ – Peter Bennett Feb 8 '17 at 6:16
  • \$\begingroup\$ Ok cool! I noticed it worked without it when I was setting it up originally, and so it made me wonder if I really needed it. I'll do some reading on freewheel diodes. \$\endgroup\$ – AndyD273 Feb 8 '17 at 15:19

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