Low voltage smoothing (doorbell) SORTED.

Nod

Nod

Tootles
Suspended / Banned
Messages
49,339
Name
Nod (UK)
Edit My Images
Yes
HELP, please!!!

A recently fitted doorbell (ding/dong type) has a small transformer powering it (since the ding is activated by a door opening and the dong by it closing, it was eating batteries at a rate of knots!) but the ringer now buzzes while the door is open. I know why (cheap transformer isn't completely smoothed) but can't remember how to do the final smoothing. From memory, it's a simple circuit using a diode or 2 and a capacitor but I'm not sure of the components' values or the exact circuit I need to build. I'm confident that I can build the smoother, just not design it! The transformer supplies 8V (almost) DC, not sure what current the ringer solenoid draws but I doubt it's very much.
 
How many batteries does the doorbell use, multiply by 1.5 to determine the total DC voltage required. If you are only using a transformer its output will be AC, not DC, not suitable. If rectified DC, it is unlikely it will be well smoothed or regulated based upon the post above, and does its voltage match that from the batteries? You can easily use an LM317 variable regulator if you need stable DC supply. The LM317 needs a min of 3V overhead, if not suitable select a low forward drop regulator. So many variables and not enough info.
 
steveb said:
How many batteries does the doorbell use, multiply by 1.5 to determine the total DC voltage required. If you are only using a transformer its output will be AC, not DC, not suitable. If rectified DC, it is unlikely it will be well smoothed or regulated based upon the post above, and does its voltage match that from the batteries? You can easily use an LM317 variable regulator if you need stable DC supply. The LM317 needs a min of 3V overhead, if not suitable select a low forward drop regulator. So many variables and not enough info.
Click to expand...


1 battery - a PP3. The transformer was sold as an 8V DC supply, specifically for doorbells. As you say, the problem is that the supply isn't smooth enough to eliminate the buzz (more of a hum) as the striker oscillates while the circuit is closed (i.e. the button is pressed [in practice, a microswitch is closed]). Looking at post #3 (are you an Alan or a Richard, Mr Clogwyn/Jones?), I think I could get away with a capacitor across the supply to the ringer, I just need to know what value I need to do the job.

FWIW, the sounder operates at 6V but takes a 9V battery. The way I have it wired is so that the ding sounds when the door is opened and would dong when the door's closed but I've stuck a felt pad on the dong blade so there's just a ding. I could add another microswitch to the door end of the circuit so the supply is disconnected slightly after the ding goes off but smoothing the power supply to eliminate the buzz/hum would look tidier.

Thanks for the help so far - I'm an old bear with a very little brain!
 
Unfortunately not an option.

The bell is for Mrs Nod's Yoga studio and is basically to gently alert the teacher (or someone!) that the door's been opened. There's a magnetically locked door downstairs but one of the other businesses in the building keeps switching it to unlocked, meaning that anyone can wander upstairs. There's a lock on the top door but students often arrive late so the top door's often left unlocked. The dong has been disabled (!) so there's just a gentle ding as the door opens - currently followed by a gentle buzz until the door shuts.
 
Solved the problem by butchering the cable and using an old 'phone charger. Enough power to ring the ding and smooth enough not to buzz.
 
You must log in or register to reply here.
Back
Top