The circuit part is relatively simple. Domestic doorbells run on low voltage AC. An optocoupler, 2 resistors, and one capacitor will give you an isolated two-state signal indicating a button press. That can drive any kind of interface circuit needed.What's the likelyhood of working out what signal would be required; and building a circuit to send it?
The problem is the cordless phone handset. Based on ***zero*** information, it probably has a microcontroller inside that runs all signalling and function selection. Mine have an LCD screen with three text lines. Getting it into Intercom mode takes a total of 6 button presses on two buttons to get through the menu selections to the Intercom mode. That is a difficult thing for an external circuit to activate.
A basic ring video doorbell is about $60. It will alert your mobile when anyone is present.
I assume your budget is less than $60?
nd at the other extreme, some DECT phones have a "locate" button on the base unit that makes all handsets ring until cancelled
The problem is the cordless phone handset. Based on ***zero*** information, it probably has a microcontroller inside that runs all signalling and function selection.
nearly any of these https://bellman.com/en/product/ transmitters will work. They have an external trigger.
They use the Holtek 12e and 12E chips. **broken link removed**
It should be easy to create a receiver. This
Indestructables: https://www.instructables.com/Make-a-RF-Transmitter-and-Receiver-With-T12E-HT12/
Transmitter/Receiver modules: https://www.componentsinfo.com/fs10...-xy-mk-5v-receiver-module-explanation-pinout/
FS1000A 433MHZ RF transmitter & XY-MK-5V RF receiver module pair
Transmitter Voltage 3-5V. Receiver: 5V
design should work. It;s missing a few bits.
the Bellman transmitters are battery only,
Watch the number of handsets your system supports.
Most handsets we've had use 3x 1.2v batteries.What's the likelyhood a circuit that uses 2x1.2V batteries would be happy with 3.2V?
This one definitely only has 2Most handsets we've had use 3x 1.2v batteries.
Add a diode in series to drop it by around 0.6V, to be on the safe side.What's the likelyhood a circuit that uses 2x1.2V batteries would be happy with 3.2V?
Add a diode in series to drop it by around 0.6V, to be on the safe side.
Good idea. How do I calculate the required power rating for the diode?
Determine the peak current through the diode, and use one rated for at least twice that.How do I calculate the required power rating for the diode?
They are typically rated in amps, not watts.
Determine the peak current through the diode, and use one rated for at least twice that.
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