12VDC transformer / relay / alarm panel -- need a capacitor and resistor!

Question

I'm holding a relay coil shut with a 12VDC / 500ma transformer (wall wort). The output of the relay is connected to a normally closed zone in my alarm system. I use this to tell me if the power has gone off while I'm not at home. However, with the transformer and relay only, I get no "power off" delay before the zone goes into alarm. Ideally, I'd like to get about a 30-60 second delay so that it won't go into alarm for short power outages. I don't have any way to do this through the alarm panel, so I need an electronic solution. I'm aware of a device which already does this (http://www.smarthome.com/7154.html) but I think I can do it cheaper with a capacitor (2, 3 .. ?) and a bleed resistor. Does anyone know the math and/or can you provide an answer to what size caps I need?

Answer 1

capacitor...900ma...just for start up....will reduce to 350-500 ma

Answer 2

Here is the most simple electronic to delay the relay switching on when 115 volts AC goes out. Use a comparator,one side input sets to referent voltage,other side of input connect to a capacitor and a shunt resistor. Output of comparator control a power transistor connecting as a voltage fellower,this follwer controls the relay coil. This circuit needs 12 volts  rechargable battery source as a backup power,at the same time it is power by the 115 volts power source (convert it into 12 volts DC). An independent circuit to convert AC 115 v  into DC volt (say +5 volt). Use this 5 volts to charge the input capacitor. Select the reference voltage into +5 v. As AC power source cut off,no more 5 v charges the capacitor. Capacitor starts discharge by the shunt resistor. Comparator becomes active as capacitor drops to a certain level.Then turn on the relay.

Here I cannot give you the exact value of capacitor and shunt resistor that can last 60 seconds delay time. Because they depend on the reference voltage,monitor voltage,comparator that is used. It is much better to select them by experiment. Capacitor can be chose 10uf to start with, use a 100k ohms potential meter as the shunt resistor to adjust the timing.

Answer 3

Trying to answer but YA taking breather. Or is it???

Edit:

No it just doesn't want to display my answer. Try again...

I may be reading your question wrong, but it looks like you want to charge the capacitor from the power supply, in such a way that if the power goes off the voltage on the capacitor will hold the relay closed for 30-60 seconds, until the capacitor discharges.

That would work, but it is clunky, and you will need a big capacitor.

Assuming that the relay will continue to hold closed until the voltage drops to about 60% of normal,

C = 2 * T / R where R = the coil resistance of the relay

Let's say you have a very small, low power relay with coil resistance 1200 ohms. For 30 seconds:

C = 60 / 1200 = 0.05F = 50,000uF

You would probably go for an aluminium electrolytic cap 47,000 uF, 16V.

If your relay has a lower resistance coil, which it probably does, you will need a bigger capacitor, or several in parallel.

You may need to charge the cap through a rectifier diode like 1N4001, so that when the power fails it doesn't discharge back through the power supply.

The way I would do it is to use a darlington transistor, FET or solid-state relay to switch the line. Then you can do the same thing but with a much smaller capacitor, and you can run it from a battery or power direct from mains through a capacitor / rectifier circuit.

Don't try that latter idea unless you know what I'm talking about, it's very easy to kill yourself...