Connect Relay to Google Doorbell - Page 1
Forum section: electronics / beginners

author: dr_harika
author_rank: Newbie
timestamp: May 16, 2023, 11:41:50 pm
content: Hi All,Long time viewer of the EEVblog youtube page! Love the work Dave does.I have what should be a simple question for you electrical gurus.I have an existing doorbell system (KOKOM Intercom system) which is connected to various gate / door latches and internal monitors built into the walls of my house in different locations which i use so i want to retain the existing doorbell system and add a Google Doorbell (Battery) for smart features mainly for when I am not home.The button on my existing Kocom doorbell is just a momentary switch and I have created a new facia which retains the existing camera, but removes the button, and has a nice place for the Google Doorbell to sit. The aim is to trigger my existing Kocom doorbell button from the google doorbell when pressed using the "ring existing chime" feature.I would love if someone knew a simple way for this to trigger a relay without using much standby current.I have been tinkering and have an 18V AC 500mA transformer to power the Google Doorbell, and i have drawn up a schematic using a rectifier and relay which i think will work but am seeking any advice on sizing the resistor or other circuitry to be able to trigger the relay. Thanks for any advice.I could be going down the wrong path entirely so appreciate any feedback.

author: soldar
author_rank: Super Contributor
timestamp: May 17, 2023, 10:22:58 am
content: I have used a current-sensing transistor to do something similar. I had some Passive IR sensors that worked via radio with some proprietary alarm system but they could not be used any other way. But when they detected movement their current consumption would rise and a transistor would detect it and could activate a relay or anything else. It does not add any additional power consumption when at rest.

author: dr_harika
author_rank: Newbie
timestamp: May 18, 2023, 01:15:20 am
content: Thanks - great idea to look into.Everything I've researched shows circuits for current sensing for DC circuits.I cant find anything for really low power (under 2W) AC circuits. I am not sure if an inductive sensor (around the wire) will work. My multimeter clampmeter cannot detect anything. Also, thinking about it, might not work well in my situation as there is a change in current drawn when the camera turns on, IR lights, which is not necessarily the button press.Any other ideas?In other regions google sold a "Chime Connector" with older doorbells but that is not available here in Australia.

author: Kim Christensen
author_rank: Super Contributor
timestamp: May 18, 2023, 04:21:28 am
content: In theory your circuit could work. Unfortunately, the documentation for these things is written for the home handyman and provides very little info.Does it work? ie: Does the relay click when the Google doorbell is activated? Or does it stay stuck on because the Google Doorbell draws too much standby current? (Then you'd need a lower resistance than 100 ohms)The resistor looks a bit undersized wattage wise and will dissipate 3.24W when the doorbell rings and 18VAC is across it. Not sure how long the Google doorbell will "ring the bell", when some kids are goofing around with it, so a 1/2W resistor may burn out in that situation.

author: floobydust
author_rank: Super Contributor
timestamp: May 18, 2023, 05:29:06 am
content: OP I think you are proposing to sense loop current with the google Nest doorbell battery version, to detect if it's activated.I saw another website and bonehead EE blog post a similar idea athttps://www.functionaldevices.com/blog/a-relay-for-the-nest-hello-doorbell/I saw mention of the "NEST chime driver" but it seems to be a magical unicorn module google has whitewashed off the web.google is pure torture because they've dumbed down their products to the point they're too dumb to understand. It's impressive turning hardware into spaghetti.You can't operate a 24V relay in series on a system that is powered off 16VAC. Does that make sense? The relay needs to take all, leaving nothing for the google doorbell. So neither circuit will work I think.You would need 26V drop across the 100Ω resistor to provide voltage to energize the relay coil. Would have to use a low voltage relay/opto-coupler.First I would measure the supply current the google doorbell draws to see if it draws current for a long time, or how much when charging the battery - enough to false trigger or trigger for many minutes etc. and make this idea not so feasible.

author: Kim Christensen
author_rank: Super Contributor
timestamp: May 18, 2023, 05:03:27 pm
content: Quote from: floobydust on May 18, 2023, 05:29:06 amNot correct. The doorbell system is 18 VAC... A 24V relay will still pull in at a lower voltage, but if it needs a bit more, the OP can put a capacitor across the relay coil which would bring the DC voltage up to 24V. The Google doorbell has a battery in it that powers it when it's ringing the bell, so the voltage across the Google device will be near zero (Shorted out by Google DB itself) and the remaining 18VAC will go to the bridge rectifier.Below is a typical old fashioned doorbell circuit with an illuminated button. The bulb is lit as long as the button isn't pressed because it draws a small amount of power through the chime solenoid without ringing it. The Google doorbell would present a similar load as the lamp and is designed to replace that lamp/button combo. In the OP's diagram, the chime solenoid has been replaced with the 100ohm resistor + bridge rectifier load and the lamp/button combo by the Google DB. He may have to use a resistor with a lower resistance than 100 ohms and a higher wattage rating.

author: floobydust
author_rank: Super Contributor
timestamp: May 19, 2023, 02:10:43 am
content: I couldn't find any basic tech docs or wiring for google Nest Doorbell products. Just a website railroad into "buy the product". Got a URL?It's loop powered for battery charging and I still think OP's circuit will not work.My doorbell solenoids measure about 10Ω each, and in series with the google doorbell, apparently even this can sometimes be a problem. The solenoid voltage drop or impedance too much for the current it requires. This is relevant to OP's 100Ω resistor choice, and preventing the yellow ring of "insufficient power" alert.Nest fix appears to be the (now designed out) "chime connector"has electronics in it.Inside is aLCB710mosfet SSR 60V 1A normally CLOSED plus a bunch of other hardware. It seems to normally shunt the solenoid to allow full power charging and then briefly open circuit when full voltage/high current happens. Apparently you don't need one if the transformer voltage is high enough to overcome the solenoid losses.For no chime, you can also apparently short out the chime solenoid and run the transformer directly to it. So the doorbell's output switch doesn't appear to get damaged (chime turned off in the app) but who knows.So I think OP wants to spoof a chime solenoid well enough for the google product to have sufficient power, and activate a relay as a chime solenoid.Inside I assume the Nest/google has a rectifier, filter cap and buck-converter but we have no idea how much current it pulls and where are the teardown pics?The relay coil sees less than 15VRMS with no filter cap, and I would worry about inrush if there was one.Hard to come up with a solution given google's hardware support- a well known hot, steaming turd.

author: Kim Christensen
author_rank: Super Contributor
timestamp: May 19, 2023, 03:48:59 am
content: Quote from: floobydust on May 19, 2023, 02:10:43 amI guess a solution could be to use a 12V relay with a series resistor to drop the extra 3V. Or even a zener in series with a 9V or 6V relay so that low level current draw doesn't produce any relay coil current at all.*After reading your Reddit link, it sounds like this thing was designed by people ignorant of how the classic doorbell system works. They then bodged in the "chime connector" to fix a poorly designed product.

author: floobydust
author_rank: Super Contributor
timestamp: May 19, 2023, 05:08:34 am
content: I have never seen such a mess in how google is handling these home IoT devices, it's terrible. The ultimate dog's breakfast. 1,000's of posts asking for help that have no conclusion or ending or answer, and then a zillion low content youtube videos that are stupid installation pics never mind the many different configurations. Idiocracy is real, it's here. Tech docs are kindergarten-grade they assume people are that dumb. google kaiboshed the Nest Secure system, many people not happy about it, $500 toss it in the garbage now. Terrible product management across the board.I should know better than to get involved in these Nest threads...Anyway, I think we'd have to know how much current the doorbell requires and what running voltage is adequate for it, sense high loop current with a relay or transistor/opto and live with the constant few volt drop.Spec is "8-24VAC 10VA" but people sometimes have troubles with 16VAC so I think the buck-converter is a little cheap. I expect a few watts drain for the camera etc. This is not when it's trying to trigger the door chime solenoid.Bell transformers have poor regulation, 16VAC rated can give 20-23VAC at no load.A lower voltage relay coil would be OK, it only gets activated for a few seconds so overload is fine.But sizing the sense resistor I'm guessing 33Ω 10W as a start and a 9-12VDC relay coil w/no cap.OP's approach may not be not feasible if the operating current is near enough the solenoid current, and some people have buzzing humming solenoids all the time, so current must be high hence the "chime connector" module was created.If we say it's 0.1A and 33Ω for a 5.3V loss (incl. bridge) for ~11VAC available to the google doorbell, is it enough?OP you'd have to do an experiment and see if the google doorbell will run with a 33Ω 10W resistor in series, and no relay or anything else yet.

author: dr_harika
author_rank: Newbie
timestamp: May 23, 2023, 04:26:26 am
content: Thanks for all the help -I've pieced together some parts, and did a bit of trial and error using the original schematic I pictorially showed in the first post.I have an 240V AC to 18V AC transformer designed for doorbells (10m lead), it actually produces 21.53V AC without load.I changed the resistor to a 82W 5W resistor, and added a 220uF 25v capacitor in parallel with the coil of the relay and removed the diode at the relay.The relay is a 24V DC coil, but it triggers from as low as 13V from my tests.Results:In normal mode (stable after 10min or so):-Across the 82 ohm resistor I get 1.15V AC at 13.48mA. If I am not mistaken, that is dissipating 0.0155W (and for a year that would consume 136W (24hrs*365days)-Across the Google Doorbell (20.68V AC),When the doorbell is pressed (instantaneous):-Across the 82 ohm resistor I get OL on my meter (but assume it gets to 21.53V AC and that the auto ranging on my meter is not catching the voltage in time) at 232mA. If I am not mistaken, that is dissipating 4.99W (P=IV)-Across the Google Doorbell is shorted as expected.The only addition was a 220uF capacitor in parallel with the coil of the relay, because I think I was getting some odd ripple or something that I noticed when putting a red LED in place with a 1k ohm resistor instead of the relay for testing purposes – it was flickering without the capacitor. I didn’t size up, I just had one lying around so tried it, hopefully I don’t have any inrush issues.Summary of my tests:With 100ohm resistor, across the resistor I get 1.60V AC at 15.8mA (222W for a year) and 21.53V AC at 168mA when pressed (3.62W).With 82ohm resistor, across the resistor I get 1.15V AC at 13.48mA (136W for a year) and 21.53V AC at 232mA when pressed (4.99W)  My chosen resistorWith 50ohm resistor, across the resistor I get 0.80V AC at 17.0mA (119W for a year) and 21.53V AC at 376mA when pressed (8.10W).  I’d need to buy 10W resistors which were not in stock locally so I didn’t botherAll options sufficiently power the google doorbell battery, and triggers the relay on door press, so I think unless there is anything fundamentally wrong or concerning with my solution (i don't want anything to burn down my house), I can close this out.On the google doorbell app settings, it has options for ringing an internal chime. With this on, it shorts the contacts momentarily which is the feature I am trying to utilise. It also has an option for ringing an internal electric chime where you can select how long (1-10seconds) it rings for. With this enabled, it just shorts the contacts for the time specified.

author: Kim Christensen
author_rank: Super Contributor
timestamp: May 23, 2023, 05:09:47 am
content: Quote from: dr_harika on May 23, 2023, 04:26:26 amI wouldn't worry too much about lowering the resistance to "save power" because most of the power used will be consumed by the Google doorbell.ie: While you may lose 136Wh over a year in the resistor, you are "losing" 2442Wh per year in the Google doorbell itself. (20.68V * 13.48mA * 24hrs * 365days) So reducing the power dissipated in the resistor by going to 50 ohms isn't going to save you much.The only thing I can see going wrong here would be if the power went out for a long enough time for the Google doorbell's battery to get drained enough that when the power comes back on the Google doorbell draws much more current (when it tries to recharge the battery) and activates the relay when it shouldn't.

author: dr_harika
author_rank: Newbie
timestamp: May 24, 2023, 12:55:33 am
content: Thanks for bringing it in perspective. I have been testing with a 15% battery on the google, so it is charging as i have been testing, but I will check it again when fully charged so the relay works as expected.I will be running the cabling in the wall this weekend for my purpose to trigger the existing doorbell.

author: floobydust
author_rank: Super Contributor
timestamp: May 25, 2023, 12:53:54 am
content: I wasn't sure what (power hungry) features you have enabled. The older google doorbell apparently did video streaming, which used up a lot.