I have an old wired doorbell that has an 18V DC power supply. The actually transformer has been (I think) channelled into the wall, or underfloor boards in the attic. i.e. hard to get to.
A new fancy doorbell requires connection to a 8V AC power supply.
Is there a simple way to do this? i.e. 18V DV in, 8V AC out?
18v would be unusual for a door bell. Did you measure the voltage from the power supply with no load? The sort used for this likely has very poor regulation since it doesn't much matter. Real doorbells will work over a wide range of voltage.
If it's an electronic device it will run off DC. So if no internal transformer, it will accept DC too. The DC will go through its internal rectifier OK. If so a simple regulator will give you the DC voltage you want. Exactly what that is, I'm not sure. As a nominal 8v AC supply depends on how it is rectified etc for the final DC voltage. Can this device be run off batteries? That would tell you what it needs, DC wise.
You will probably find that the new bell only requires 8V AC because it converts it internally to whatever DC voltage it requires to operate, possibly by no more than a rectifier and smoothing circuit. You could try connecting a 7.5V DC supply to it to see if it works (and change the polarity if it doesn't). If it does, and you want to be bothered, you could try one of these:
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And select the voltage output which you have checked works ok. Whatever you do, don't connect up the 18V DC supply directly to the new bell.
I'd guess 18v with no load, nominal 12v supply but 8v and transformer humming in protest when the solenoid is being powered continuously.
That is an unusual requirement. Most modern stuff uses a DC switched mode PSU (or even batteries) rather than an iron transformer and AC.
You need to know how much current the new device will draw and what its absolute maximum voltage rating is. It might be possible to put a well heatsinked 5W 10v Zener diode in series to drop the excess voltage. eg
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It only has to survive a few seconds of intermittent use.
Somewhat grubby solution and it might not work if the thing draws too much current when going ding dong or if the psu regulation is ropey. (as seems likely from the OPs description)
There is no /simple/ way to convert. Best buy a transformer - it seems that 8V AC is quite a common doorbell voltage. For something that's connected to the mains and left unattended for years, I wouldn't fiddle about.
Would you *seriously* fit that yourself rather than use a bell transformer designed for continuous use? It's not simple, not certified, certainly not as reliable and awkward to explain to the insurance company when the firemen have left.
It all depends on why it needs an AC supply. If its then rectifying the
8V AC to get a 12V DC supply you may find that the first component in the doorbell is a bridge rectifier followed by a smoothing capacitor.
If this is the arrangement you could feed in DC - and possibly the 18V you already have. Assumptions: a bridge rectifier as the first component and 18V is the no load voltage that possibly would drop under load
Do you have a photo of the innards of the doorbell showing any components that may related to a power supply?
Not if you rip out the old transformer and connect the new power supply/bell transformer to the old transformer's supply and connect the output to the existing bell wires.
A mains transformer designed specifically for continuous use is more reliable than a bunch of electronics. Why would you even argue the point?
And this might well be one. The Amazon reviews of a Friedland 8V AC doorbell suggest a 'Brrrrrrrrh!' sound.
Like this, perhaps...
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Or if it's a simple electromechanical 'ding - dong' type using a solenoid actuator then there is a difference between AC and DC solenoids, just as AC and DC relays are different.
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