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We need to use the audio output of a wireless doorbell to trigger a stable DC voltage which is going to keep switched a 3.3V DC relay as long as the audio output is present.

The relay is supposed to power/open an electronic lock until the doorbell audio finishes.

The doorbell I'm using is also a 3.3V device.

When applying the audio output directly to the relay, the relay was playing the audio just like a loudspeaker. This happened even when I tried to even out the audio using capacitors.

Here's an image of the Doorbell used

The Wireless doorbell I used

And this is the wireless switch for the same doorbell

enter image description here

These are the capacitors I tried for smoothing away the audio

enter image description here

The relay was still playing the audio like a loudspeaker (instead of being switched) even after I used these capacitors. do you think capacitors with different values could help or are there better solutions?

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  • \$\begingroup\$ The audio will be AC. The relay requires DC. Did you rectify the signal? \$\endgroup\$
    – Transistor
    Commented Jan 19, 2020 at 9:07
  • \$\begingroup\$ @Transistor with a diode rectifier? no, I didn't use any diodes since it seemed like only the positive wire is modulated for the audio. can a four diodes rectifier solve my problem? \$\endgroup\$
    – S. Goody
    Commented Jan 19, 2020 at 9:19
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    \$\begingroup\$ I don't know. You haven't given enough details on the doorbell or on the relay. A bridge rectifier will drop 1.2 to 1.4 V which you probably can't afford on a 3.3 V circuit. \$\endgroup\$
    – Transistor
    Commented Jan 19, 2020 at 9:23
  • \$\begingroup\$ @Transistor I also have a 5-volt power source; I could use the 3.3V source to drive the 5V source and power the relay using that. \$\endgroup\$
    – S. Goody
    Commented Jan 19, 2020 at 9:28
  • \$\begingroup\$ @Transistor I can change the relay with whatever relay you think can do the job. about the doorbell, the loudspeaker of the doorbell was a 5 ohm one. let me know if there's a specific parameter you need in your calculations. thanks \$\endgroup\$
    – S. Goody
    Commented Jan 19, 2020 at 9:30

1 Answer 1

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If the relay isn't switching, you will likely need to buffer the audio output with a transistor. Just because the audio output is 3.3V doesn't mean that it can supply enough current to switch the relay. If you feed the audio signal into a transistor that switches the relay, and then put your capacitor from the base of the transistor to ground, you should be able to make the relay switch.

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