circuit needed

[Jamie]

Could someone suggest a simple circuit that would light an led for about 2-4
seconds when power is applied and then turn off and stay off until power is
cut and applied again?

Thanks in advance JJ
__________________________________________________________________

 

[Tom Biasi]

Look at some 555 circuits.
"Jamie" <Jamie4@breathemail.net> wrote in message
news:cf7i7g$sgb$1@hercules.btinternet.com...

Could someone suggest a simple circuit that would light an led for about
2-4
seconds when power is applied and then turn off and stay off until power
is
cut and applied again?

Thanks in advance JJ
__________________________________________________________________

 

[Jonathan Kirwan]

On Mon, 09 Aug 2004 10:59:58 GMT, "Tom Biasi" <tombiasi@REMOVETHISoptonline.net>
wrote:

Look at some 555 circuits.

Namely, look for "one shot" or monostable 555 circuits.

Jon

 

[John Schuch]

Jamie wrote:

Could someone suggest a simple circuit that would light an led for about 2-4
seconds when power is applied and then turn off and stay off until power is
cut and applied again?

Thanks in advance JJ

Try this:

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

John

 

[Terry Pinnell]

"Jamie" <Jamie4@breathemail.net> wrote:

Could someone suggest a simple circuit that would light an led for about 2-4
seconds when power is applied and then turn off and stay off until power is
cut and applied again?

Thanks in advance JJ
__________________________________________________________________

This should be OK:
http://www.terrypin.dial.pipex.com/Images/555MonoByPU-JJ.gif

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[Jonathan Kirwan]

On Mon, 09 Aug 2004 19:01:30 +0100, Terry Pinnell
<terrypinDELETE@THESEdial.pipex.com> wrote:

This should be OK:
http://www.terrypin.dial.pipex.com/Images/555MonoByPU-JJ.gif

Looks wonderful, Terry! And right about 3 seconds, or so.

(Personally, I prefer to _NOT_ bus the power around in a schematic. The extra
lines detract from the thrust of the idea. I suppose they are excellent for
those worrying about soldering up connections, though.)

Jon

 

[R.Spinks]

Should not the resistor (R1) and capacitor (C2) be in opposite spots to
provide a negative going pulse on powerup?

"Terry Pinnell" <terrypinDELETE@THESEdial.pipex.com> wrote in message
news:dkefh0lnm3qqmjevcer289ita1u1asbrak@4ax.com...

"Jamie" <Jamie4@breathemail.net> wrote:

Could someone suggest a simple circuit that would light an led for about
2-4
seconds when power is applied and then turn off and stay off until power
is
cut and applied again?

Thanks in advance JJ
__________________________________________________________________


This should be OK:
http://www.terrypin.dial.pipex.com/Images/555MonoByPU-JJ.gif

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[Terry Pinnell]

"R.Spinks" <rspinks1@wideopenwest.com> wrote:

Should not the resistor (R1) and capacitor (C2) be in opposite spots to
provide a negative going pulse on powerup?

No, that wouldn't work.

As shown the RC time constant is long relative to the typical ramp up
of the power supply. So pin 2 remains below required 1/3 threshold for
a short time, triggering the mono. Then it stays +ve so that it's
still high when mono ends, avoiding re-triggering.


--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[John Fields]

On Sat, 14 Aug 2004 07:30:06 +0100, Terry Pinnell
<terrypinDELETE@THESEdial.pipex.com> wrote:

"R.Spinks" <rspinks1@wideopenwest.com> wrote:

Should not the resistor (R1) and capacitor (C2) be in opposite spots to
provide a negative going pulse on powerup?

No, that wouldn't work.

As shown the RC time constant is long relative to the typical ramp up
of the power supply. So pin 2 remains below required 1/3 threshold for
a short time, triggering the mono. Then it stays +ve so that it's
still high when mono ends, avoiding re-triggering.

---
If the input pulse is longer than the output pulse the 555 won't
re-trigger, (it can't without extra external circuitry) its output
will just stay high until its input goes high, then its output will go
low. That is, it'll act like an inverter.

--
John Fields