How do I prevent the fade out beep(s) when the power is cut?

Use a transformer with a smaller filter capacitor, or connect a 12V incandescent light bulb in parallel with the beeper to drain the charge when power is removed. Another option is to switch the 12V with a relay rather than switching power to the primary of the transformer.

Thanks for the idea! I can try the 12V bulb since that should be easy to find (she says) and easy to do. And inexpensive. The transformer was just lying around so I'd rather not have to use (buy) a different one.

I will try this as soon as I can. It sounds like a resistor would also work. If so, would there be an easy way for me to calculate the resistance?

Thanks again!

Joy

Yeah a resistor would be fine too, you can calculate it using ohms law V=I*R where V is the voltage, I is the current in Amps, and R is the resistance in ohms. The I that you need depends on the size of the filter capacitor which depends generally on the capacity of the transformer, some experimentation may be needed.

Joy wrote in news:op.t8k1rkvqug52hc@user-

Try a single pole, double throw switch. Connect the common lead to the siren. Connect the battery to one of the remaining contacts on the switch and ground the other contact.

No---------------------battery common------------------siren Nc---------------------gnd

This way, when you 'cut the power' you will also ground the input lead going to the siren. If there is a capacitor inside the siren, this should discharge it and stop the siren.

Try testing first by taking the lead that connects to one end of the battery and quickly move it to the other, so that BOTH leads touch together. If that stops the siren, then the switch should work.

Joy wrote in news:op.t8k1rkvqug52hc@user-

bz wrote in news:Xns9A6C8BDDFE46DWQAHBGMXSZHVspammote@130.39.198.139:

one last try. Sorry. I missed seeing that you were using a 'DC transformer'.

They have a capacitor built in that is holding a charge.

All you need to do is disconnect the siren from the power supply. It should shut off immediately.

You should do this with a 110 vac relay.

Hook the coil of the relay in parallel with your power supply. When the power comes on, you want the relay to close and hook the siren to the power supply. When the AC goes off, you want the relay to open and disconnect the siren from the power supply.

The power supply will only use power when the beeper is activated. And you won't be wasting power with a resistor and possibly damaging your power supply.

Would a 110 vac relay be activated by a 12-15V DC transformer? And this would mean both the relay and the siren would be powered by the same power supply (the dc transformer) except that the relay would also switch the siren. Is this correct?

I WAS trying to be really simple which is why the light bulb or resistor idea sounds good.

How much damage could I do by using a resistor box and starting with a high value and slowly working down until it worked acceptably? I guess I could damage something in the process, but how likely do you think it would be?

Joy

He means connect the relay to the same power source that the transformer is connected to, and use the contacts to switch the output of the transformer. That route occurred to me but I skipped it as it seems more complex than other solutions.

You won't hurt anything with the resistor, just don't go so low in value that you draw more than the transformer is rated to produce, I would try something around 470 ohms 1/2W, that's a common value. That'll draw about

Just wanted to thank you all. A 400 ohm resistor did a pretty good job of doing what I wanted!

Joy