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Monitor voltage on a circuit when powered off?

lingyueqing

Plastic
Joined
Jan 16, 2018
Here's the situation: I have a baby. Soon, I'll have two. So I don't really like spending half an hour trying to get him to sleep and then somebody rings the doorbell, waking him up.

I've already installed a switch on the doorbell circuit which disables it. I'd still like to know if somebody is trying to ring it though. For simplicity sake, lets say it's a simple light bulb that I'm always looking at, so it doesn't need to stay on. (Just a momentary flash will do).

The diagram on the right is the simple doorbell circuit. On the left, I have my indicator circuit with the 12v relay (EP2-3N1S : EP2-3N1S - KEMET NEC-Tokin - Automotive Relays - Kynix Semiconductor)I assume I'm going to be needing. I believe the relay coil needs to go in parallel with my switch, but I'm not sure if this will work. Do I need a resister between the two?

16.1.png

What am I missing to complete this solution?

Maybe I did myself a disservice by oversimplifying. The indicator isn't a lamp, it's actually my alarm panel which can monitor for either an open or closed circuit (but not a good idea to put the actual bell on that same circuit). This will allow me to be notified of a doorbell signal by email/text message, or whatever method I choose, even if I'm away from home.

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Updated diagram

Additionally, I have a dog that barks regardless of whether the doorbell is disabled or not. This is probably outside the scope of the question, however.
 
Can you supply the 12 VDC to the alarm panel? If so, you can change your manual switch to a SPDT (single pole, double throw) switch, which would work as a selector switch between alarm and bell.

I don't understand the intent of your relay. It does not appear to be attached to anything.
 
Instead of the single pole switch and the relay:

Use a single pole double throw switch. When the switch is in position #1 (ring) the connection is like what you have on the right. When the switch is in position #2 (lamp) there is a connection to a RC discharge circuit with a LED indicator. When the door bell switch is pushed the capacitor is charged and lights the LED. When the door bell switch is released the capacitor will discharge through the resistor and LED for a short or long time, depending on the RC time constant.
 
I don't understand the intent of your relay. It does not appear to be attached to anything.
It's attached to the website for the company he is promoting :D

But it's a very nice job so heck, when a guy puts in that much work, he deserves the link :D
 
Yes, it's a clear case of the guy just looking for a clever way to advertise. I won't give him credit for the nice job, because it's an elementary-school electrical 'problem'.

If he really 'had a baby' he'd do what the rest of us did: Put out a dish of anti-freeze for the dog, and put a sign on the door that says "baby sleeping- go away".
 








 
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