I'm looking to "replace" a physical keypress on a garage door remote control (RF) with a pulse (HIGH state) from a Wemos D1 mini 3.3V pin.

The remote control is this one



I measured the voltage on each button terminal (5.88V) and you can see the polarity as well.

The battery type is A23 (12v)

The electric diagram I have in mind is this:


This is the IC zoomed in:


The RF part will continue to be handled by the remote control itself, but instead of a person pressing the button, an external App will control that through a Wemos GPIO high state with the help of a transistor.

  • 1
    \$\begingroup\$ The button press sensing circuitry inside the remote could be pulled up normally and shorted to ground by the switch or pulled down and shorted to VCC by the switch. Or something else. Without opening the remote and inspecting the circuit, there's no way to tell which type of transistor/configuration is appropriate unless you want to use a relay. \$\endgroup\$
    – vir
    Sep 21, 2023 at 19:34
  • \$\begingroup\$ Are you sure the remote is operating at 12V ? Also the RF Tx input is missing in your diagram on the right. Will you keep it on high side or low side ? \$\endgroup\$
    – Prashant
    Sep 21, 2023 at 19:35
  • \$\begingroup\$ @Prashant I added some more info to the original post. I don't need to mess with the RF part. \$\endgroup\$
    – vazquezjm
    Sep 21, 2023 at 19:46
  • 1
    \$\begingroup\$ From your ‘replace this’ schematic, the switch is on the high side of the load, so your proposed circuit isn’t suitable. What you need is a high-side switch, could be a PNP or p-channel FET. \$\endgroup\$
    – Frog
    Sep 21, 2023 at 19:59
  • 1
    \$\begingroup\$ You can't really replace the button without measuring the signal first. In case it is pulsed as part of a low-current consumption scheme, there might be real-time concerns. Putting any old BJT there might not be ideal, especially if it adds Rds(on) which wasn't there in the original circuit. A more rugged solution is to use an analog switch IC. \$\endgroup\$
    – Lundin
    Sep 22, 2023 at 14:09

2 Answers 2


I tackled almost exactly the same problem except my voltages were different. I also didn't own the remote, so I wanted to be able to restore it to its previous condition.

In my case, I found carefully soldering an optocoupler across the physical switch contacts worked perfectly. It also gave me the peace of mind I couldn't possibly fry the remote PCB.

Not sure if this helps you. Maybe a transistor is simpler. Either way, good luck.

DIY garage opener using the PCB from the original remote, controlled by an esp 8266 via an optocoupler. Mounted on scrap wood.

  • \$\begingroup\$ Interesting. What optocoupler did you use? \$\endgroup\$
    – vazquezjm
    Sep 23, 2023 at 12:23
  • \$\begingroup\$ Not 100% sure, but probably a PC817. I had a few of them in a cheap assortment kit of ICs, but almost any common opto should work. The datasheet can help you calculate a good resistor value for your controler voltage. Make sure you get both polarities right or it won't work. I'll edit this answer to add a photo. \$\endgroup\$
    – aaaidan
    Sep 27, 2023 at 7:46
  • 1
    \$\begingroup\$ Authentic breadboard +1 \$\endgroup\$
    – D Duck
    Sep 27, 2023 at 7:56

It seems that from your updated photos your 4 switches are grounded at one end (center of board), this is very common.

The opposite ends of the switches look to be going to the controller chip pins. So it is likely that your schematic with the 2N2222 is close to what you need.

In your case you would remove the 1k resistor, forget the V2 side, connect the transistor's collector to the outer pin of the switch you want to trigger. The ground wires on your circuit could connect to the center part of the 4 switches (which should be ground). See the first schematic below.

Another option would be to replace the transistor with an opto-coupler. This would provide ideal isolation between the Wemos and the original board. The second schematic shows this. This option assumes the Wemos IO pin can sink about 5mA* to drive the LED side of the opto-coupler, (you may want to verify that).

*Note: The current limiting resistor on the opto-coupler could be a higher value, which would reduce the needed current sink from the IO pin. However, you would need to check the specifications on the selected opto-coupler to verify it will still allow the minimum operable input current.


simulate this circuit – Schematic created using CircuitLab


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.