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I have a solenoid valve (12VDC, 540mA) that I would like to control with a 12VDC relay (15A SPDT). Is it possible to power the solenoid without an external power source seperate from the one powering the relay board (12V, 0.41A)? In the image, I have a logic signal in the input screws and I know that the relay is working with that signal (switching on and off with sound and light). The solenoid though does not seem to be powered. Should the wall outlet (currently going in at the DC power port) be stripped and connected to the input power screw terminal about ground? Ground power screw terminal currently is connected to ground of solenoid, and the other end of solenoid is connected to NC. Thank you for your help!

enter image description here

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  • \$\begingroup\$ the power supply for the relay is 12 VDC \$\endgroup\$
    – Meg
    Commented Jan 21, 2020 at 19:08
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    \$\begingroup\$ Welcome to EE.SE, Meg, but you haven't given us the details of the "the one powering the relay board". OK - you have now. It's 12 V. How much current can it supply? (All the details should be in the question, not in the comments.) \$\endgroup\$
    – Transistor
    Commented Jan 21, 2020 at 19:09
  • \$\begingroup\$ Thanks for your response, I will edit the post to include the 12V power supply. The plug reads Input: 100-240V, 0.2A Output: 5V, 1A \$\endgroup\$
    – Meg
    Commented Jan 21, 2020 at 19:20
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    \$\begingroup\$ If the PSU ("plug") says 5 V, 1 A, then it's not a 12 V PSU, it's a 5 V one. Please edit your question to make it very clear what you've got. Use photos, links to datasheets, etc., as well as clear text. \$\endgroup\$
    – Transistor
    Commented Jan 21, 2020 at 19:33
  • \$\begingroup\$ yes, just throw a cheap dc-dc buck converter between the 12v supply and 5v relay driver supply inlet. you could also just use a mosfet to drive the solenoid instead of adding moving parts and extra power conditioning. a logic-level fet uses no current and can be driven with just 3.3v. \$\endgroup\$
    – dandavis
    Commented Jan 21, 2020 at 21:33

2 Answers 2

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A relay is just a switch. You have to provide power to the solenoid through the relay.

You can wire the solenoid to be powered by the 12V from the relay board.

  1. Connect the red wire from the solenoid to the + pole of the terminal block beside the DC plug.
  2. Connect the black wire from the solenoid to the NO connection of the relay terminal block.
  3. Connect the COM terminal of the same relay terminal block to the - pole of the terminal block by the DC plug.

When you send a signal to the relay board, it will supply power from the DC powersupply to the solenoid through the relay.

  1. No schematic because the site's schematic software doesn't work on phones.
  2. The DC terminal block and the DC plug appear to simply be in parallel. If they are connected with a diode "OR," then this suggestion won't work. I don't see any diodes, though, so it should be fine.
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  • \$\begingroup\$ Thank you JRE for your detailed response. I understand that in step 3, I will need to wire COM to - pole of the terminal block. Is there a specific wire that is needed? WHat would you suggest to purchase? \$\endgroup\$
    – Meg
    Commented Feb 4, 2020 at 23:12
  • \$\begingroup\$ @Meg: Standard 22AWG stranded wire ought to be OK. \$\endgroup\$
    – JRE
    Commented Feb 5, 2020 at 6:36
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Think of the relay contacts as an electrically-operated switch.

The relay does not provide any power to its contacts. If you want the relay to switch the 12 volt power to the solenoid, you must provide power to the "COM" terminal of the relay.

As pointed out in the comments, you have to supply 12 volt power to the relay board - this can come from the same supply as used to drive the solenoid.

You need something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

I'm guessing that the relay board has something like Q1 to control the relay from a logic signal.

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