Timeline for How to switch separate circuit from doorbell?
Current License: CC BY-SA 3.0
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| Jun 24, 2016 at 20:42 | comment | added | Rob | Yes, it's the same here. The transformer delivers 12V@2A. Actually strange, since I measured exactly 10V when the doorbell rings. Should I have disconnected the bell before measuring the voltage? What about my questions about the relay and if it can break when the voltage is much higher as necessary to switch? | |
| Jun 24, 2016 at 20:28 | comment | added | Olin Lathrop | @Rob: Doorbell power is usually (all cases I've seen) created by a transformer connected directly to the power line. That provides the isolation and makes a lower voltage so that it is safe for people to touch the doorbell, even if something goes wrong and they are connected to one of the power leads. The diode arrangement is called a full wave bridge. I show one in a integrated package, but you can also make your own quite easily with 4 diodes. | |
| Jun 24, 2016 at 20:25 | comment | added | Rob | Thanks! Yes, the power line frequency is 50Hz at 230V. Don't you actually need the frequency of the 10VAC? Is it the same? And is "wave bridge" the name of the arrangement of the 4 diods? I understand how to calculate the capacity of the cap given the relay. I had a short look at possible relays. There are many which need 12VDC to switch. But is this small enough, since the voltage can drop down to 11.3V? The next lower step seems to be relays which switch at around 4.5-5V. Would 12V or 12.7V at maximum be too much for such a one so that it might break? | |
| Jun 24, 2016 at 12:14 | history | edited | Olin Lathrop | CC BY-SA 3.0 |
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| Jun 24, 2016 at 12:00 | history | edited | Olin Lathrop | CC BY-SA 3.0 |
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| Jun 24, 2016 at 11:29 | history | answered | Olin Lathrop | CC BY-SA 3.0 |