I am using this Normally Closed Relay on a 24V circuit: DataSheet

The circuit is 24VDC source, which I measured the Amperage in series to be 29mA at the highest. I've breadboarded a lower voltage circuit (5V) fine with the relay. When I put the relay into the 24VDC circuit, I am able to open and close the circuit one time, but it burns out the relay: the relay can never be switched open again. I notice when switching it under the 24V load that when I close the switch again, it delays before completing the circuit.

For the switched circuit: I have 24VDC source connected to pin 1 of the Relay, and the rest of the circuit connected to pin 4 which is connected to a the magnetic door lock (and a push to exit button, which may use a capacitor to trigger the 30 second open circuit)

For the switching circuit: I have 5VDC from a Raspberry Pi connected to pin 2. On pin 3, I have it connected to the Collector side of an NPN relay, and the Emitter side is connected to a Ground pin on the Raspberry Pi. I have one of the GPIO pins connected to a 2kΩ resistor, which is then connected to the base of the relay

Reference circuit I used for my circuit is here

  • \$\begingroup\$ What kind of load are you switching? And please clarify, the relay COIL is still good, but the CONTACTS are welded shut? \$\endgroup\$
    – R Drast
    Apr 28, 2015 at 17:40
  • \$\begingroup\$ Double check if you have kept the NC terminal of the relay open it should not be grounded. Made this mistake myself. \$\endgroup\$
    – Sada93
    Apr 28, 2015 at 17:42
  • 1
    \$\begingroup\$ If your load is inductive, you should put a flyback diode across it. \$\endgroup\$
    – John D
    Apr 28, 2015 at 17:47
  • 1
    \$\begingroup\$ I'm talking about the flyback diode across the load, not the one across the relay coil (which is also necessary). What type of load is it, and what type of diode was across it? \$\endgroup\$
    – John D
    Apr 28, 2015 at 17:50
  • 2
    \$\begingroup\$ A magnetic door lock is an inductive solenoid \$\endgroup\$
    – Passerby
    Apr 28, 2015 at 18:45

1 Answer 1


If your switched circuit contains some kind of capacitor, motor or tungsten light bulb you could easily be exceeding the contact ratings (500mA maximum) momentarily.

For example, if the load has a capacitor inside, the circuit looks very much like a capacitive discharge welder, with the materials to be welded the contacts in your relay. Once they are welded together, they are damaged and you should discard the relay, though you may be able to free them up temporarily by tapping the relay.

Edit: Switching a door lock solenoid can also cause welding by arcing when the relay drops out or bounces. If the solenoid is less than 500mA (6W) you may be able to make this work by adding a catch diode such as a 1N4004 across the solenoid, but generally this kind of relay is not great for switching relatively heavy loads. You'd be better off to use a power relay or a transistor such as MOSFET (always with the catch diode).

  • 1
    \$\begingroup\$ I've been able to "revive" the relay by tapping it, so it seems like it welded the contacts together \$\endgroup\$ Apr 28, 2015 at 17:59
  • \$\begingroup\$ Would I wire the MOSFET on the ground side of the door lock, to interrupt it from grounding, and that way it would ground to that circuit, and not the Raspberry pi. I assume that voltage grounding to the Pi's ground would blow it up, right? \$\endgroup\$ Apr 28, 2015 at 22:26

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.