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Background

I have an Fridland AC doorbell that I am about to make "smart". My approach is to connect a device named a shelly uni inside the doorbell that draws power from the doorbells own transformer. The shelly device has inputs to read the button press and I can then decide if I want to sound the doorbell (activating an output) besides being able to get notifications on my phone when someone is at the door. The doorbell transformer outputs 12V AC.

The problem

I have connected the Shelly to the transformer and power is good. The issue here is the input from the doorbell as the cables pick up noise. I am getting everything between 0V and 17.5V on the input making it impossible to tell when the button has been pressed. If this was a DC circuit it could easily be solved with a PullDown resistor and I'm wondering if this can be done here too? The frequency of the transformer should be the same as on the mains which I think is 50Hz.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Place the pull-down between the input and the Sensor_GND pin. \$\endgroup\$
    – Klas-Kenny
    Commented Sep 6, 2022 at 12:27
  • \$\begingroup\$ Thanks, thats what I was hoping for but not sure if it was the same approach for an AC circuit. I'm thinking a resistance around 10k would be suitable? \$\endgroup\$
    – Philip Johansson
    Commented Sep 6, 2022 at 12:37

1 Answer 1

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Put a resistor around 10k+ between the input and power ground pins of the shelly. You can also add a diode like a 1N4148 to the input. enter image description here

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  • \$\begingroup\$ Thanks for you comment and time to draw the diagram. May I ask what the advantage of the diod would be? The Shelly's inputs are rated to work with both AC and DC as long as it's powered by the same. \$\endgroup\$
    – Philip Johansson
    Commented Sep 7, 2022 at 20:36

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