David Nebenzahl wrote:
On 3/29/2009 6:43 AM Baron spake thus:
David Nebenzahl wrote:
On 3/28/2009 1:30 PM Baron spake thus:
David Nebenzahl wrote:
On 3/27/2009 3:55 PM Tim spake thus:
In article m,
kens says...
(famous last words, "simple circuit" ...)
OK, so I'm trying to come up with a simple (maybe even elegant)
solution to a simple problem. Have an idea I want to run by
y'all.
Function: person has a motion-detector light installed in their
home. They want a buzzer/bell/annunciator of some kind to go off
*momentarily* whenever the light is activated.
I think that a small sugar cube relay, diode, resistor and
capacitor will do the job !
Circuit, pleeze?
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
That is just about it ! I hadn't considered connecting the relay
contacts back to the source to feed a transformer, but yes and the
transformer provides the safety isolation.
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.)
Mmm, I tend to have trouble with decimal points. :-)
Thats a little more difficult ! You need to know the characteristics of
the relay you are going to use.
For example: If the relay requires 0.01 amp (10ma)at 90 volts to pull
in. That would give a 90/0.01 = 9K Coil resistance.
Since the capacitor will be fully discharged initially the current to
charge it up will be limited by the resistance of the relay plus R. So
in this example there may be enough time constant due to the relay
itself if the capacitor value is well chosen.
The time is basically 63% of R X C Seconds.
Lets try 100uf @ 160vwg. 0.0001F X 9000 = 0.9 Seconds /100 X 63 = 0.56.
About half a second. So for this example a 220uf @ 160vwg would be
about 1 second.
Since the current for the relay to drop out will be less than than 0.01a
the time will be less than this.
I would actually measure the relay I was going to use and adjust values
to suit. I'm sure you get the idea...
--
Best Regards:
Baron.