0
\$\begingroup\$

So here is the problem:

First, I have a door opener device that gives me one wire to control it. This wire should connect to ground so the device works. We do not want to make the door opener to work all the time, obviously.

Second, I have a device for remote control, when a button is pressed on the remote, a ground signal is output on the box for a couple of seconds. This is the only output of the controller and unfortunately It can't serve as a ground to the door opener. It has simply too much impedance to make the door opener work. It works with lighter loads such as leds. Unfortunately I cannot change this control I'm treating it as a "black box".

What I need help with is figuring out a way to use the controller wire as a "trigger" to activate the door opener.

Summing up:

  • Both devices use a 12v power supply

  • Door opener wire (should connect to ground)

  • Controller wire (ground signal)

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
4
  • 1
    \$\begingroup\$ use the controller box to operate a small relay \$\endgroup\$
    – jsotola
    Commented Mar 4, 2020 at 21:40
  • 1
    \$\begingroup\$ "unfortunately It can't serve as a ground to the door opener. It has simply too much impedance to make the door opener work." - what impedance does the door opener need to work? \$\endgroup\$ Commented Mar 5, 2020 at 2:31
  • \$\begingroup\$ @jsotola I just tested with a 5 pin relay and it worked \$\endgroup\$
    – Joao
    Commented Mar 5, 2020 at 15:03
  • \$\begingroup\$ @BruceAbbott it needs a direct ground (no resistance) \$\endgroup\$
    – Joao
    Commented Mar 5, 2020 at 15:05

2 Answers 2

0
\$\begingroup\$

It seems that the control box can only power low current devices. So this is why using a relay can be helpful. To activate the relay, the pin 85 is connected to the "controller box wire" and pin 86 is connected to 12v power supply. So when the remote control is activated and in turn the control box wire is "activated" and the relay's coil is powered using a very small amount of current.

The door opener needs a direct ground connection and wer use relay pin 30 (common terminal) to connect directly to ground. Pin 87 is connected to the common terminal only when the relay's coil is powered. So we connect the relay's pin 87 (normally closed) to the door opener wire.

A diode was added across the relay's coil to prevent damage, thanks to the comment of @SamGibson

I hope this helps you.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
1
  • \$\begingroup\$ Hi, I recommend that a flyback diode should be added across the relay's coil. Otherwise, whatever component is driving that output from inside the "black box" (e.g. discrete transistor?) could be damaged over time, due to the inductive spike from the relay's coil, each time it is switched off. \$\endgroup\$
    – SamGibson
    Commented Mar 5, 2020 at 16:13
0
\$\begingroup\$

Insert a optocoupler.
Connect the LED of the optocoupler (if necessary, with a resistor) between the 12V and "output wire of control box" and the transistor of the optocoupler to the "wire to control this door opener" and ground.

My phone does not support the built in schematic editor. Below an attempt to draw the implementation:

enter image description here

\$\endgroup\$
2
  • \$\begingroup\$ Cannot use the schematic editor on this phone... \$\endgroup\$
    – Huisman
    Commented Mar 4, 2020 at 21:37
  • \$\begingroup\$ Unfortunately this won't work with most optocouplers. The device in question purportedly needs a 'direct ground (no resistance)'. A PhotoMOS relay might do it if its resistance is low enough, but standard optocouplers only have a limited output current that could even be less than the LED current (if CTR is < 1). \$\endgroup\$ Commented Mar 5, 2020 at 19:38

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.