I have a device running at 12vdc that has a pair of terminals that can be opened and closed for the device to operate. I understand how to use an optocoupler to open and close the terminals, however the twist here is that I need the terminals to be normally closed (ie to be connected most of the time), and opened when the opto is activated (eg by an MCU). This is preferred to keeping the opto activated due to wear (?) and the failure state.

The below was offered here when I asked about normally closed optocouplers, but I'm having trouble understanding how to integrate it, most likely because I don't fully understand how it works.


simulate this circuit – Schematic created using CircuitLab

  1. What is the input to R4? Any available voltage? Where would I get that from? One of the terminals?
  2. I also don't have a "load" as such - I just want to complete a circuit. Is that load implied (eg behind the terminals), or should I add a resistor?
  3. Are there any other solutions to this? Although I see the value in an isolated solution, I'm interested in hearing in other ways to solve the problem. I also have regular mechanical relays which have NC terminals for instance.
  • \$\begingroup\$ Use a relay that is normally closed? \$\endgroup\$ – Huisman Aug 7 at 17:44
  • \$\begingroup\$ Do you have a datasheet for your device? \$\endgroup\$ – The Photon Aug 7 at 17:46
  • \$\begingroup\$ It's an aged but functional alarm panel. The terminals do things like set (open) and unset (closed). I don't have anything other than that, although I can measure things with a multimeter. \$\endgroup\$ – Spammy Aug 7 at 17:49
  • \$\begingroup\$ @Spammy, the trick is, you need to know which way the device will try to send current when you connect the two terminals. Your circuit expects the device to be sending current out terminal "A" and in terminal "B". It's also helpful to know what voltages will appear on these terminals. \$\endgroup\$ – The Photon Aug 7 at 18:40

Use a normally closed relay. Reasons:

  • The relay is isolated, just like the optocoupler
  • The shown circuit will only work when the (voltage) potential of "NC Terminal A" is higher than the (voltage) potential at "NC Terminal B." The relay will work in both ways.
  • The shown circuit needs to be fed as you already indicate at question 1. The relay solution doesn't need that.
  • If the circuit between "NC Terminal A" and "NC Terminal B." is unknown (what voltage?, what current?) the relay solution is probably more robust compared to the darlington that may have lower maximum ratings.
  • \$\begingroup\$ Minor point, the circuit shown is not a darlington pair. Darlington pairs have the collectors commoned and the emitter of the first transistor feeds into the base of the second. \$\endgroup\$ – Peter Jennings Aug 7 at 21:29

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