"David" wrote in message
...

"David Nebenzahl" wrote in
message
s.com...
On 3/29/2009 6:43 AM Baron spake thus:

David Nebenzahl wrote:
Ok but excuse the ascii art.

live ----diode---relay---resistor---capacitor---neutral

You're excused. That illustrates the circuit perfectly.

Size the resistor to give the time constant with a
particular value
capacitor. The relay contacts are isolated and can be
connected to
your low voltage circuit.

The resistor should have sufficient wattage rating with
respect to the
current and the capacitor should be rated for at least
the maximum
voltage applied. ie 120v X 1.414. The relay can be
almost anything
with a suitable coil voltage. ie 100 - 120.

The circuit works by utilising the charging current into
the capacitor
to energise the relay. As the capacitor voltage
increases the relay
will drop out. The time difference between energising
and dropping out
is how long the annunciator will sound.

So I take it you'd like this circuit:
http://www.geocities.com/bonezphoto/...-shotBell2.gif

And now can you give us the R and C values to give, say,
a 1- or 2-second on time? I don't know how to calculate
such things. (Understand how they work, just never
learned the actual math involved.)

I thought you learned that there should be a discharge
path for the capacitor unless you want to wait hours or
days before it will work a second time.

David

Following up on my own post, the circuit is poor for other
reasons. Think about what happens on the negative half
cycles of the AC input. You need at least a 'catch' diode in
there.

David