How to switch separate circuit from doorbell?

Question

My doorbell rings as long as somebody pushes the button. I measured 10V AC at the bell when it is ringing. Now I would like to use this AC voltage to switch another circuit. How can I achieve this?

Here is what I want to happen when the doorbell button is pressed:

1. Let the doorbell ring. That is currently the case.

2. Close another circuit, which is independent from the first one.

Edit: As Olin says in a command, the proposed duplicate is about how to get the doorbell signal into a arduino and not about closing another circuit.

Answer 1

What you want is called a "relay". Ideally you want one that runs from 10 VAC. That's not a common coil voltage, but 12 V and 6 V are. You could use a 6 VAC relay with a resistor in series, or full wave rectify the 10 VAC to make close enough to 12 VDC on the capacitor after the diodes.

Either way, the relay is basically a electrically operated mechanical switch. At minimum, you will probably find a SPDT (single pole double throw) relay. That gives you the option of using it as a switch that is normally open except when the button is pressed, or normally closed except when the button is pressed.

Here is a schematic of what I'm talking about:

Note the capacitor after the full wave bridge. This is necessary to keep the average rectified voltage high enough to reliably activate the relay. The particular relay I'm showing in this example draws 27 mA at 12 V. The peaks of the 10 VAC waveform will be 14.1 V. The full wave bridge subtracts two diode drops for about 1.4 V, leaving 12.7 V peaks. That will be what the cap gets charged up to twice per power line cycle. For a 12 V average, the voltage on the cap can drop to 11.3 V before the next peak, for a total drop of 1.4 V. You didn't say what your power line frequency is, so I assumed 50 Hz. That means the cap will get recharged every 10 ms. This gives us enough to compute the target cap value:

   C = (27 mA)(10 ms)/(1.4 V) = 193 µF

So the common value of 200 µF will work fine.

Answer 2

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1 (a) original circuit and (b) with monitoring relay.

How it works

• The 10 V AC is applied to the bell.
• Diodes 1 to 4 (or a small bridge rectifier) rectifies the AC voltage to give DC.
• C1 smooths out the DC and holds the voltage up during AC voltage drops.
• The 12 V relay will energise when the bell-push is pressed.
• The contacts should be rated for greater than or equal to the voltage and current you are switching.

Answer 3

If I understand your question correctly, you want to connect two wires when the switch is pressed and supply both of them with the AC power. To do this, replace your switch with a SPDT Switch.