I have always had problems with my wireless doorbell solution. Since my driveway is around 50m (160ft) the signal is never quite strong enough to trigger the bell all the times.

So I was thinking about creating my own wired solution, but my question is how I would make this work.

Would a let’s say 5v DC signal be able to travel this distance and still be picked up by a microcontroller? Or is there maybe a better way to make this happen (using AC and converting it to DC at the gate)?

  • \$\begingroup\$ There are hundreds of what they claim to be "long-range" doorbells. Just search \$\endgroup\$
    – Eugene Sh.
    Jul 11, 2018 at 19:35
  • \$\begingroup\$ Yeah I have already tried many, but none seem to get through the concrete/steel house structure. \$\endgroup\$
    – Caske2000
    Jul 11, 2018 at 19:37
  • \$\begingroup\$ Which kind of power is already available where you want to place the doorbell button (mains AC, low voltage AC, DC, etc., no power)? \$\endgroup\$ Jul 11, 2018 at 19:42
  • \$\begingroup\$ Moreover, how is the terrain between the button location and the building? Is it concrete, some kind of pavement, simple terrain? \$\endgroup\$ Jul 11, 2018 at 19:44
  • \$\begingroup\$ Add a longer wire antenna at both ends so there are parallel to each other yet farther from ground effects \$\endgroup\$ Jul 11, 2018 at 19:55

2 Answers 2


5V signal is sufficient to make travel till the end of 50 m but you cannot connect that directly to your micro controller.

Best thing is to isolate the noisy weak digital signal from the MCU port. One way is to use optocoupler as mentioned by Michael.

You can use say from example below circuit. enter image description here

  1. Replace 3.3 V (only the one connected to R1) with your 5 V supply and a series switch at the entrance.
  2. Choose a optocoupler with good CTR (current transfer ratio, higher is better)
  3. The optocoupler will turn on(the internal LED) when you close the switch
  4. It will in turn, turn on the transistor.
  5. Microcontroller will sense it because now the output is shorted to ground via transistor in the optocoupler.
  6. Microcontroller can run with independent supply. (3.3 V) for example.
  7. You have isolation between signal and MCU.. Hence there will never be pseudo triggering of the signals due to voltage coupling because of a generator close by or some power spikes..
  8. When switch is closed a finite current will flow through the series resistor and the LED in the opto coupler. Choose R according to optocoupler data sheet.

You can also use higher voltage (instead of 5 V) there by reducing ohmic losses in the cables.

This idea is feasible with just a battery for 5V side. Because the current will flow only when the user activates the switch.


A workable way to do this is to use a pair of wires to the remote switch operating as a current loop. At the near end bias the current loop with a resistor to your 5V supply. The other side of the loop pair would go to the anode of the LED in an optocoupler. Cathode of the coupler LED to GND. Output side of the optocoupler can go to trigger your device. Select the resistor to allow about 10mA to flow in the current loop when the switch is closed.

  • 1
    \$\begingroup\$ I’m sorry, but my electronics knowledge isn’t quite up to par with yours, could you please further explain what you’re talking about? \$\endgroup\$
    – Caske2000
    Jul 11, 2018 at 20:09
  • 2
    \$\begingroup\$ @Caske2000 - This site is not going to be a detailed design hand out site. I gave a quite detailed description of what you could try. The ball is really back in your court to start doing some research and learning about what you want to achieve. If you want a solution that is ready made then you should probably buy it outright or hire a consultant to design and install it for you. \$\endgroup\$ Jul 12, 2018 at 0:20

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