Simple circuit to hold relay on after input falls

Oct 27, 2003 13 Replies

I am sure this is a simple circuit but it been a while since Ive been building circuits.



I have a circuit that is controll by a PIR (not important) where the circuit is NC (12v)



When the circuit become open I want to activate a relay to switch a 240 supply. This relay needs to remain activated for a fixed period (say 10mins, either fixed in the design or adjustable - not important) and then switch off, even though the 12v input circuit will become closed after say 5 seconds - i.e the circuit holds the activation circuit.



Now for the complicated part!....... I dont want the 240v relay to hold on until the time ends and then switch off and start again (as it would cause a break in the 240 circuit), but to hold the circuit open for the time (say



10mins) after the last break in the 12v circuit.

i.e say the 12v circuit breaks, the 240v supply is activated by relay, the



12v circuit closes, five mins lates there is a break in the 12v circuit for
5 seconds, five mins later there is a second 5 second break in the 12v circuit, then no further break occurs so the 240 relay closes after a further 10 mins.

i.e after each break on the 12v circuit the 10 min timer starts, ever time the circuit "activates" the time starts again.



Can anyone suggest the simplest way to acheive this?


That isn't a good way to start a post asking for help. You are apparently not qualified to evaluate the complexity of what you are asking, and you minimize the value of what you may be about to receive.

Don

One of those small PLC's would probably do the trick. How are you at ladder logic?? Good luck on your project....Ross

circuit

10mins,
a

for

Yo,

Because of the long times envolved (minutes) and the complications in the operation, this would best be done with a small microcontroller and a solid state relay which can be driven directly with the microcontoller logic output.

The rest, of course, is software.

Luhan Monat "LuhanKnows" At 'Yahoo' dot 'Com' http://members.cox.net/berniekm

Maybe a re-triggerable one shot such as 74HC123 would work.

-Bill

There are AC time delay relays available - check local electrical supply house. I don't know if they are available with DC coils, though Wilson

circuit

10mins,
a

for

You could adapt these time delay and missing pulse detector circuits.

formatting link

formatting link

Rob.

Thanks - looks like the sort of thing I am looking for.

Right if I want to use a pulse time of about 10mins (not critical) am I right in thinking that I can use a 10M resisitor and a 50uF capacitor? (would give about 550 seconds) Is this pulse too long for the 555, and does it matter what combination is used between resistor/capacitor?

I would use a 1M resistor and a 500uF capacitor. The time will most likely run longer than calculated so a 500K pot and a 470K resistor in series would be a more practical choice.

Rob.

Yes, but I have a question. Whan you say that the PIR circuit is "NC (12v)", do you mean that it: 1. Runs on 12V and provides an isolated relay output with a normally closed contact and a common which open when the PIR detects motion? 2. Provides an output which is pulled up to +12V and is pulled low to the PIR ground when the PIR detects motion? 3. Provides an output which is pulled up to to +12V with reference to the PIR ground/common and goes open circuit when the PIR detects motion? 4. Something else? For example, the PIRs I've seen usually have a relay with SPDT contacts and a common as an output; could yours be like that? Also, if you have the instructions which came with the PIR and you could post the part about how to hook it up to external loads (or a link to a data sheet/manual) that would be a big help.

No. 1

Intended to run the 12v suplly through the relay so that there is a 12v output - but can be used just as a relay if needed.

Great idea.

Thanks for your help.

Check alt.binaries.schematics.electronic...

Join the Discussion

Have something to add? Share your thoughts — no account required.

Didn't find your answer?

Ask the community — no account required