# How to switch separate circuit from doorbell?

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.

• You might find What are the differences between an AC and DC coil relay? useful. Jun 24, 2016 at 10:10
• This is a duplicate, if you didn't do a search on EE.SE before you posted then you should have electronics.stackexchange.com/help/how-to-ask It is required that you do prior research before posting. Thanks Jun 24, 2016 at 15:44
• Possible duplicate of Mains doorbell to Arduino Jun 24, 2016 at 15:44
• @lap: That other question is about how to get a doorbell signal into a arduino, not about how to switch another circuit, possibly requiring substantial power. A opto-isolator would be fine for the first, but not the second. Jun 24, 2016 at 17:53

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.

• 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?
– Rob
Jun 24, 2016 at 20:25
• @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 20:28
• 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?
– Rob
Jun 24, 2016 at 20:42

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.
• You should add a capacitor across the relay coil in your second circuit. It is necessary to bring the average voltage after rectification to the 12 V required by the 12 VDC relay. 10 VAC peaks are 14.1 V, dropping to about 12.7 V after the diodes. The average will be well below 12.7 V without a cap. Jun 24, 2016 at 11:48

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.

• No, I just want to connect two wires when the door push is pressed. These two wires are supposed to just be connected, without supplying the AC power to them.
– Rob
Jun 24, 2016 at 10:30
• How is that different from what everyday (SPST) Switches do? Normal switches connect two wires when switched on. Am i missing something? Jun 24, 2016 at 10:33
• Do you want two wires to be automatically connected when the bell push is pressed, without changing the working of the doorbell? Jun 24, 2016 at 10:38
• Yes, exactly! That's what I want.
– Rob
Jun 24, 2016 at 11:14