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555 timer circuit

--- A 555 (even a CMOS 555) is kind of iffy at 5 minutes, so you might want to try this instead:

+---------+-----+------+ | Vcc | | | +V [510K] | +-->[250k] [0.1µF] | | [100K] | | HC4060 +-------+ | | +-------+------+------+ |K | O----->C +-------|Rs Rtc Ctc | [1N4148] [COIL]- - -| | | | | | O--> | +--|MR Q7 Q13 | +-------+ | | +------------+----+---+ | +---------->NO | | | C +---[1N4148>]---+ +--[510]--B 2N4401 | | E +---[1N4148>]--------+ | GND

Set the pot to 160K and the output pulse will be about 5 seconds long with a time between pulses of about 5 minutes and 20 seconds.

-- John Fields Professional Circuit Designer

Reply to
John Fields
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I'm trying to build a 555 or 556 timer circuit that creates a 5 second pulse to activate a small reed relay every 5 or 6 minutes. Can anyone help me with that. Thank You. John

Reply to
John Winstead

In message , John Winstead writes

You'll have trouble getting a 555 to run that slowly. You'd have to use a very low charging current for the capacitor, and it may be less than the capacitor's leakage current.

I'd use a 4060 (

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) to produce a pulse every 5 or 6 minutes then use that to trigger a 5 second monostable. I seem to recollect that the 555 can be used as a monostable and 5 seconds would be do-able.

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Graham
Reply to
Graham Orme

[...]

If you can live with the nominal 5 seconds VS 5 minutes ratio being fixed you don't even need the 555:

Q(N) ---/\\/\\/-----+------------- Reset ! V --- ! Q(N+A) -------------+-------------- Pulse

Q(N) needs to be such that it has a period of 10 seconds so that it will be low for 5 seconds after Q(N+A) goes high.

From the reset, it will be 10 seconds times 2^A until Q(N+A) goes high. It will then remian high for 5 seconds. If you make A 5, the full cycle will take 325 seconds.

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

So you want it to work in astable mode with Tl = 5min and Th = 5s

It's not that hard, read the datasheets.

Reply to
OBones

See my reply in sci.electronics.basics, pointing you to this solution:

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--
Terry Pinnell
Hobbyist, West Sussex, UK
Reply to
Terry Pinnell
[...] A slight change:

You don't need two diodes.

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith 

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Clever. :-)
Reply to
John Fields

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