Simple circuit to hold relay on after input falls
 

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?

Simple circuit to hold relay on after input falls
 

John wrote:

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?






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

Simple circuit to hold relay on after input falls
 

"John" wrote in message ...
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?

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

-Bill

Simple circuit to hold relay on after input falls
 

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


"John" wrote in message
...
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?

Simple circuit to hold relay on after input falls
 

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

http://home.cogeco.ca/~rpaisley4/LM555.html#20

http://home.cogeco.ca/~rpaisley4/LM555.html#22

Rob.


"John" wrote in message ...
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?

Simple circuit to hold relay on after input falls
 

"Rob Paisley" wrote in message
om...
You could adapt these time delay and missing pulse detector circuits.

http://home.cogeco.ca/~rpaisley4/LM555.html#20

http://home.cogeco.ca/~rpaisley4/LM555.html#22


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

Simple circuit to hold relay on after input falls
 

"Rob Paisley" wrote in message
om...
You could adapt these time delay and missing pulse detector circuits.

http://home.cogeco.ca/~rpaisley4/LM555.html#20

http://home.cogeco.ca/~rpaisley4/LM555.html#22

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?

Simple circuit to hold relay on after input falls
 

"Pete" wrote in message ...
"Rob Paisley" wrote in message
om...
You could adapt these time delay and missing pulse detector circuits.

http://home.cogeco.ca/~rpaisley4/LM555.html#20

http://home.cogeco.ca/~rpaisley4/LM555.html#22

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.

Simple circuit to hold relay on after input falls
 

On Mon, 27 Oct 2003 14:04:51 GMT, "John" wrote:

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?
---
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.

--
John Fields

Simple circuit to hold relay on after input falls
 

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?

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.

Simple circuit to hold relay on after input falls
 

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.

Great idea.

Thanks for your help.

Simple circuit to hold relay on after input falls
 

On Fri, 31 Oct 2003 15:06:08 GMT, "Pete" wrote:

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?

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.

---
Check alt.binaries.schematics.electronic...

--
John Fields