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Building DIY a smart doorbell but would appreciate some help with the power and wiring (lemmy.world)

submitted 6 months ago by [email protected] to c/[email protected]

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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

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[–] [email protected] 0 points 6 months ago (1 children)

So a doorbell transformer steps 120VAC down to a lower AC voltage. That’s what you’re measuring (surprised you can’t tell if it’s AC. Your multimeter should have an AC/DC setting).

The circuit puts the transformer, the button, and the bell (classically an electronagnet that accelerates the hammer) in series. When you close the switch (push the button) the hammer strikes the bell.

An electromagnet is just a coil of wire, and it is possible to pass a small amount of current through it without activating the hammer. If you’ve ever seen a doorbell with a light up button, the light bulb is placed in parallel with the button. That way it’s always on drawing a small amount of current. It also explains why the light switches off when you press the button. You’re shorting across the bulb, so it’s voltage is zero.

If you want to use this power for an electronics project, you’ll need to find a way to draw just a small amount of power from the transformer. It has to be very small or you’ll activate the hammer.

Drop a full bridge rectifier in parallel with the button, then a bunch of caps to give you a steady-ish voltage and then some kind of voltage regulator to make it useable.

Just remember that you can’t draw too much power, and you’ll lose that power whenever the button is pressed.

[–] [email protected] 0 points 6 months ago (1 children)

I couldn't remember if it was AC or DC, but you're almost certainly right that it's stepped-down AC. I just went and checked and there's no light on the button, so unfortunately no bonus parallel always-on power source to tap into. This is my dumb guess about how this all could work, but there are some question marks. Does this seem kinda correct though?

[–] [email protected] 1 points 6 months ago* (last edited 6 months ago)

Your drawing is a little confusing. Here's what you need:

There's nothing special about a setup with a light in the button. It's literally just an incandescent bulb across the button. If you connect your circuit across the button terminals, it'll work fine.

The 4 diodes in a loop is a "full bridge rectifier" that gives you DC from AC. The 5V regulator could be something like an LM7805.

The "large capacitor" is to keep power applied while the button is pressed, though you might be better off in that case with some small onboard battery. You just have to make sure that your battery charging circuit doesn't draw too much power.