I'll start by saying I have little understanding of electronic circuits. I have an old door phone to which I have connected a wireless switch in order to be able to open my door with a remote control. I power said wireless switch with an old 5 V phone charger connected to a wall socket nearby. I am tired of having an ugly cable showing all the time and I decided to design a circuit in order to attach an 18650 battery to it. Now, the problem l'm facing is that I have two cables coming from downstairs that carry 18 V AC. These cables when connected together allow the electric strike to release the door. I have wired the wireless switch to connect these two cables together.


The circuit that I'm thinking based on my little knowledge is:

1. Connect a bridge rectifier to the 18 V AC to get DC voltage
2. Connect a capacitor to smooth the DC ripple created bv the bridge rectifier
3. Connect buck converter to lower the voltage to 5 V DC
4. Connect the 18650 battery with a BMS to the 5 V output of the buck converter
5. Power the wireless switch from the BMS

Wireless switch specs:

• Input voltage: DC 4-12 V
• Working Current: 70-130 mAh (measured 110 mAh)
• Standby Current: 7-9 mAh (measured 4.5 mAh)
• eBay listing: https://www.ebay.com/itm/182812387343

I've decided to use a battery to power the switch because I only have a set of wires giving 18 V AC and when connected this voltage drops to 0 which will effectively kill the operation of the wireless switch. Maybe a capacitor could do the trick as the remote will be pressed for max 10 seconds at a time.


Can you help me design a circuit with the proper components and their values for my application?

 

I question whether you need the battery. If the wireless switch has done its job by connecting the cables does it matter if it loses power? I’d certainly try it before incorporating a battery.

You could initially try the full wave rectified and smoothed AC using a linear regulator to provide 5V which is less efficient than a buck converter but adequate to test the concept and may even be a satisfactory permanent solution

 

I question whether you need the battery. If the wireless switch has done its job by connecting the cables does it matter if it loses power? I’d certainly try it before incorporating a battery.

You could initially try the full wave rectified and smoothed AC using a linear regulator to provide 5V which is less efficient than a buck converter but adequate to test the concept and may even be a satisfactory permanent solution

If the wireless switch loses power it would prevent it from keeping the wires connected and therefore the I wouldn’t be able to open the door. Maybe a capacitor would do the trick, keeping enough power on the switch for at least 5 seconds?

 

Can you relocate the wireless switch to the location of the 18V power supply? Then it could have power while the lock is activated.

 

Can you relocate the wireless switch to the location of the 18V power supply? Then it could have power while the lock is activated.

Unfortunately no, I don’t have this capability