Hello, and thank you in advance.
I'm making a privacy friendly "ring" cam/doorbell following this guide: https://tristam.ie/2023/758/ which has been great, but requires running a micro-usb cable down to the doorbell for power. I'm hoping to improve on this by using the existing doorbell power instead.
The problem is that I'm a DIY electronics noob and I can't create a mental model for how it should all work. The picture I attached is my existing doorbell wiring scheme, which is as simple as it comes. I totally get how this works. Pressing the doorbell completes the circuit and makes the bingbongs. But this will have to change so the new door cam gets power full time. Ideally without the chime bingbonging full time.
In addition to the ESP-32CAM, button, ring lights, etc., I also bought these: https://www.amazon.com/dp/B079FJSYGY which I thought might be needed to complete the circuit?
I measured the voltage after the transformer and it was around 18 volts, but maybe this is AC and I want DC?
Generally I don't know where in "the loop" to put things. Also, all the existing components are very far apart from each other, so I would love a solution that doesn't involve running any new wires through the walls.
Any help is appreciated. Thank you!!
xoJimbabwe
The easy way: take your existing wiring. Put an 18VAC to 5VDC power supply in parallel with the pushbutton. Then, the output of that power supply goes to the 5V USB input of your camera setup.
The downside is that it will reboot the camera every time someone rings the doorbell, because you are shorting across the camera's power supply.
You can put a resistor in series with the button to fix this. You will need to find a resistance that's low enough to still cause the chime to ring, but high enough not to disrupt the camera's power supply. Maybe start around 20 ohms. If you can't find a working resistor value, you can change the transformer to a 24V or 36V transformer, but make sure to keep that resistor high enough not to burn out the chime, and make sure your 5VDC power supply can handle the increase in input voltage.
Hi, thanks for your reply. I didn't see it because I was drawing up this awful diagram that I responded with to @ch00f with. Is that more or less what you're describing, with the 18VAC -> 5VDC power supply being in that top ? box?