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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] 8 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] 1 points 6 months ago (2 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] 10 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.

[–] [email protected] 4 points 6 months ago (1 children)

Remember that voltage is measured across a pair of wires, so you can't power the chime with only a single wire. That's part of what makes this difficult -- these doorbell systems only have the bare minimum of wiring in them. Powering the camera and the chime in series with each other is quite difficult. I think a lot of these things just accept the short circuit, and use a battery to power the camera while the button is being held.

Here is what I was proposing, and I think what ch00f was also proposing. Replace "5VDC power supply" in this diagram with "a full bridge rectifier and a bunch of caps" in their description, and also note that your camera probably requires a well regulated 5VDC supply.

[–] [email protected] 3 points 6 months ago (1 children)

Awesome, thank you both. This is more complicated than I thought, so on the plus side I don't feel as dumb as before.

Maybe i'll just ditch the dinky chime I have and setup some alert automations or something. In that case, I could just connect the two chime wires to create a closed loop, then replace the transformer with a 120vac outlet plug, plug in a generic 120va -> 5vdc power supply, chop the two output wires from the power supply and connect them to the existing doorbell wires. And violá, 5vdc delivered right to the front door? Is this plan insane?

[–] [email protected] 3 points 6 months ago (1 children)

Actually, I think that's a really good idea. Just be aware of possible inductance in the long wires from the transformer area going to the camera. Putting a large electrolytic capacitor next to the camera should fix that.

As for your bell, there are dead simple systems that use a battery-powered stick-on button with a transmitter, and a chime/receiver that plugs into a wall. If you want to be fancy, you could find a way to power the transmitter with that 5VDC, in parallel with the camera... or you could just keep it simple and use the battery that comes with the button. They last near forever.

[–] [email protected] 1 points 6 months ago

Awesome. You and @[email protected] are true OGs. Thanks (again)!

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

So when I installed my amcrest doorbell it came with a small component that jumps across the doorbell chime leads so that the push button and camera unit get continuous power. I didn't open it up but I assume it's just a small resistor. Here's the link to the manual. It's got some info on wiring too that might help. https://drive.google.com/open?id=1JmduLOXZetxixoFjSTGhf5IbDlLUHB7M

[–] [email protected] 1 points 6 months ago

Oh, dang, maybe I can just buy their "relay module" thing and it will solve all my problems.

[–] [email protected] 3 points 6 months ago

Smart switches/cameras tap into the power directly and actuate the chime circuit via a separate control (or use batteries). The ring pro does this and there’s a little power box that gets wired in at the chime unit for that purpose.

I’m not sure how it actually does the controls, however. A short term capacitor to run the door end temporarily and cutting the circuit to release a relay, maybe?

[–] [email protected] 2 points 6 months ago (1 children)

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.

[–] [email protected] 1 points 6 months ago

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?