I have a solenoid door lock that requires 3 different voltages to operate - Unlocked Idle: Requires low voltage (6V) that'll hold the solenoid seated - Locked: No power (0V) - Unlock: 50-100ms pulse at full current (12V) to move the mechanism

I've been attempting to do this and failing. I've seen this work using x2 relays and x2 buck-converters and a Pi Zero. I feel like that's over-complicating it and I'm looking for a simple solution.

Here's what I drew up (I'm very new to this so bear with me)

enter image description here

I came to this after someone told me to use PWM and a single FET. But this did not work at all.

Edit: I also added a flyback diode to this after I drew it up, still nothing; the lock doesn't budge.


1 Answer 1


If your diagram is correct, the power supply +(Positive) is connect to the lock -(Negative). This will not work.

Also, I'm not sure how this is going switch from 6v to 12v. There is no 2nd relay/transistor in you diagram. You will have to either use a buck converter or a voltage divider circuit And a second transistor/relay.

EDIT: You could use a single relay with a 'Normally open" pole and a "Normally closed" pole. If relay is activated 12v... If relay not activated 6v... If transistor off, 0v

  • So if I had a relay with NC, NO, and COM - the 6V would go into the NC connection, and the 12V would go into the COM?
    – M.Traynor
    Oct 22, 2018 at 17:32
  • I have edited your circuit above to kinda show what I'm talking about. i.imgur.com/PoPPNNf.jpg . Sorry for the bad picture. Oct 23, 2018 at 0:13
  • Also forgot to mention, in the picture I posted, 12v is on when relay in off. 6V is on when relay is on. Circuit is open when the transistor is off. But 12v and 6v could be switch if you would rather 6v be on when relay is off. It really doesn't matter. Oct 23, 2018 at 2:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.