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I have recently bought a dcp917S and hoped I finally found a wireless chime that I can tie into my apartment buildings bell system. Since I live for rent there I can't do any changes that are not reversible, nor anything that is destructive to the whole thing, nor things that afflicts other residents.

Details: There are 5 wires coming in. Wire one is ground. Wire 2 is AC power 25V. Wire 3 is going out to the door buzzer. Wire 4 is the bell and provides 20V AC when the button at the building is pressed and something else (pulse to short for my measurement equipment) when the button at my door is pressed. Wire 5 is the intercom wire.

Since mains is 230V+-23V @ 50Hz I assume the AC frequency also to be 50Hz.

Also nothing is default or normed since judging from the circuit board in my apartment it (the intercom system) was constructed in the 70s (hand drawn and cut circuit board with solder reinforced copper tracks)

The chime has 6 lines in 1,2 is the buzzer button 1 circuit (with 3V). 3,4 button 2. 5, 6 is power in to up to 12AC.

Line 5 can alternatively also be used as reference to line 1 and 3 to detect button presses, though none can be higher than 12V ac and have to be high enough to the transformer voltage to be detected ... so that's out of question, I think.

(Official wiring)


My two problems are:

  1. The new chime only accepts up 12V ac so I need to bring down the 25V provided to 12V ... since I only need a few mA I thought I can put two 1M ohm resistors between the 25V AC line (line 2) and ground (line 1) and leach of one of it in parallel to run the wireless bell converter.

  2. When one of the buttons is pressed I get between some volts and 20V (AC) on line 4 I need that to close a circuit on the chime. I thought of a capacitive dropper circuit (since it is low power and won't be unplugged I probably save the discharge resistor) with 10 µF or something to switch a transistor which closes the chime circuit.


  • Do I have any mistakes in my thoughts?
  • Any easier ways? (tend to do things "right through the knee into the eye" way)

Oh! And I need to fit it all into the (relative small) box of the buzzer so anything than some soldered components is out of question. (e.g. Circuit boards micro controllers, larger circuits, transformers, etc)

Thought I do it like this:

schematic

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  • \$\begingroup\$ The physical logic diagram does not look logical with a shorted bridge and open loops on Vac to "gnd". So it does not make sense without concise specs and makes it hard to read. A logic diagram = Schematic should be a logical symbolic connection of inputs and outputs in a block diagram then realization with components can be done easily. this physical diagram will make some stand on their head. j/k \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 5 '18 at 2:50
  • \$\begingroup\$ Within the box all lines use ground as neutral. I measured the lines 2 against ground plane and against line 2. \$\endgroup\$ – Bizz Keryear Sep 5 '18 at 3:03
  • \$\begingroup\$ All I know about the system I know through experiments. Pressed buttons while monitoring power (with the cheapest multi-meter you can imagine). Removed cables and see what doesn't work any more. And I have no clue about electronics (though I have some 30 years ago as kid some electronic classes ... made a primitive radio) nor schematics. That is also the reason why I ask. Please don't let me die as an idiot. \$\endgroup\$ – Bizz Keryear Sep 5 '18 at 3:11
  • \$\begingroup\$ Why is bridge shorted inside the box? That would DC clamp the 20Vac rms inside the box to from 0 to 58Vpp. Sorry I can't help \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 5 '18 at 3:15
  • \$\begingroup\$ I only need 0.7V (max) to trigger the transistor (which closes the chime circuit where a button should be) \$\endgroup\$ – Bizz Keryear Sep 5 '18 at 3:32
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If you only need "a few mA" (I'll assume <10 mA), then a simple shunt regulator will do what you want:

schematic

simulate this circuit – Schematic created using CircuitLab

The RMS current through the resistor and diodes is about 10 mA, so the resistor dissipates about 120 mW and the diodes dissipate at most about 60 mW each. The load can draw about 10 mA before the voltage starts to droop significantly.

The waveform at the load is more square than sine, but that shouldn't matter in this application.

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