DaveC said:
A new twist to the challenge:
The output needs to be a one-shot, momentary-close event. In other words, 5
minutes after power fails, the contacts need to close and open again. This
alerts the alarm to the power out-condition, and clears the alarm sense line
for another event to be reported.
Ideas?
Two relays, two capacitors, one resistor.
Refer to the circuit mentioned in my post above -
a 5.5 super cap and RLY-639 or RLY-405. When power
drops, the relay stays energized for ~4 minutes, then
drops.
Use one of the closed points on the first relay to
complete the circuit to a second relay from a charged
cap.
One point of the second relay is the output. Another
point of the second relay to switch a low value resistor
across the cap that energizes the second relay.
Here's the circuit: (the output point is not shown)
A +-----0 \---relay2coil---+
| R1P(nc) |
+---/\/\/\---o \--+ |
| R1 R2P | |
Cap2 | |
| | |
B +-----------------+------+
+ 5 volts is supplied to point A, ground is at point
B. R1P is the normally closed point of the relay
mentioned in my earlier post and shown in the diagram
below. R2P is one of the normally open points of
relay 2. The other normally open point of relay 2 is
the output you need.
Here's the ~4 minute delay circuit:
A -----+---relaycoil1----+
| |
SuperCap |
| |
B -----+-----------------+
The super cap is a 1 farad 5.5 volt cap
and the relay is RLY-639, both specified
in the earlier post, from Allelectronics.
Use a regulated 5 volt supply with series
diode protection, or a regulated 6 volt supply
with two 1N4001 diodes in series to produce
the supply to points A and B. The diode
protection prevents discharge of the capacitors
back into the supply. Cap2 does not need to
be a super cap. On an untested guess, I'd
say a 470 uf cap with a second RLY-639 would
work. A 100 ohm 1 watt resistor will work for
R1 with that cap. Let me know if you want me to
test it - I have the parts.
Ed