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bemis

join:2008-07-18
Reading, MA
Reviews:
·Comcast

using a lighted doorbell button in/output from uC

I'm looking to use a lighted doorbell as an input and output from a microcontroller.

The idea is this--The light being on represents a "ready" condition. Pressing the button causes a function to "execute" and the light to turn off and remain off until the system is back in the "ready" state.

The way that lighted doorbells work, as far as I understand, is that the button is a N.O. switch and the light is wired in parallel with it. The resistance of the bulb is insufficient to trigger the coils at the bell, for that it takes a dead short caused by the switch being pressed.

What I'm envisioning is a couple of relays, let's call them A and B. "A" is used to switch power to/from the button circuit (i.e. indicate the ready condition). "B" is used to detect a "button pushed" condition.

My thought here is that "B" could behave similar to a traditional doorbell in that the bu (in fact, if needed, I would consider a MacGuyver hack of mounting a switch to the sounding plate in an actual doorbell so that when the button is pushed the bell strike will physically wack the switch).



cowboyro
Premium
join:2000-10-11
Shelton, CT

Maybe a nice LED-lighted switch would be a better option than a doorbell switch...
Furthermore you'd have separate light and switch contacts, but your options will remain open with a low voltage environment.


bemis

join:2008-07-18
Reading, MA
Reviews:
·Comcast

said by cowboyro:

Maybe a nice LED-lighted switch would be a better option than a doorbell switch...
Furthermore you'd have separate light and switch contacts, but your options will remain open with a low voltage environment.

I should have mentioned in my first post, this is being on an actual house. So I'd like to avoid changing the exterior or re-wiring. I'd like it to re-use the existing button/light. Though I do have a 3 or 4 conductor cable going to the button now, so if this can't be made to work, your option is certainly available.


cowboyro
Premium
join:2000-10-11
Shelton, CT
reply to cowboyro

On another thought the straight 24VAC option to a lighted bell switch could work too. Just hook your system via an optocoupler and do appropriate debouncing in software to get rid of the 120Hz noise you'll have. Use a higher-power bulb as a ballast...



tschmidt
Premium,MVM
join:2000-11-12
Milford, NH
kudos:9
Reviews:
·G4 Communications
·Fairpoint Commun..
·Hollis Hosting
reply to bemis

If you really want to use the existing lighted button your two relay solution will work.

Wire the coil pf one in series with the switch and doorbell. Pressing the button will cause the relay to close and the doorbell to ring. May need to add a resistor in parallel with the relay coil to prevent current through the light from triggering it. If you want to use a DC relay instead use the appropriate power supply and use a dual pole relay one set of contacts to ring the bell and another to provide notification to the controller.

Use a second relay with N.C. contacts in series with the push button and driven by the controller to turn off power to the button. Hopefully the off state will not be too long - or the person at the door will be unable to ring the bell a second time - or perhaps that is the idea.

/tom


bemis

join:2008-07-18
Reading, MA
Reviews:
·Comcast

said by tschmidt:

Hopefully the off state will not be too long - or the person at the door will be unable to ring the bell a second time - or perhaps that is the idea.

The delay is intentional to prevent overuse.

The idea is an automated candy dispenser--since I cannot be home on halloween night.

I have the gable end of a small attic about 1-2' above my front door with a vent opening. The candy is on a "conveyor belt" built within an 8' long 3" PVC pipe that sticks out of the vent slightly. Total capacity is about 90 pieces of candy. It will dump 2 pieces at a time into a flexible hose / shoot which terminates at one of those pumpkin buckets at about "kid height" in front of my door...

Instructions are simple and will posted on the door "I'm not home tonight, if the doorbell light is on, please press the button to get a piece of candy. It will take 10 seconds for the machine to fill up and light to come back on. If the light is out longer than 10 seconds, that means I've run out of candy, sorry".


pferrie3

join:2005-01-27
Boston, MA
reply to bemis



i think that sums it up nicely


Zach1
Premium
join:2006-11-26
NW Minnesota
reply to bemis

I think I would start the project by connecting a full-wave bridge with the AC input side in series with the existing doorbell circuit. This would allow current monitoring and control using off-the-shelf DC components and techniques with very minimal disruption to the existing doorbell circuit. For the current monitoring portion, I'm thinking two standard diodes in forward bias between the '+' and '-' of the bridge output. This configuration would give you 1.2~1.4V across the diode array when there is a load on the circuit. A resistor (guessing about 5~10 Ohms) in parallel with the diode array should allow the button light to stay lit while keeping the voltage across the diode below forward bias. An OP amp would work as input buffering for the PIC. A transistor would work to cut power once candy runs out and the tricks begin.

What a cool idea though!
--
Zach



cowboyro
Premium
join:2000-10-11
Shelton, CT

said by Zach1:

For the current monitoring portion, I'm thinking two standard diodes in forward bias between the '+' and '-' of the bridge output. This configuration would give you 1.2~1.4V across the diode array when there is a load on the circuit.

That's a really bad idea when dealing with separate circuits. There should be galvanic separation between the uC circuit and the doorbell circuit. An optocoupler or relay should be used in such an instance. Imagine the transformer fails and you end up with high voltage on the 24V line... Not to mention you are starting with a floating circuit (24V) so you don't have a defined potential in respect to the ground or to the uC circuit.