Simple question about a 555 in one-shot (monostable) mode

I have a 555 timer in monostable mode. I output a 45 second pulse. No problem doing this. What I want to do though is to delay triggering this pulse while another signal is high (12V) - what control do I use to do this?

So: while input signal is 12v the output must stay high. When the input signal goes low then I want the one-shot to trigger maintaining the output (still) high for 45 seconds and then go low.

Can anyone help?

Thanks.

Reply to
AINTME
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If I understand you correctly, You want a low transition or just a low signal to start a One shot timer however, you don't want any other transitions (resets) to effect the timer until it has expired it's

45 second count? this is called a gate.

You have a couple of options here. use an OR gate where as the 555 timer out goes to one input and your Reset Line(trigger) is the other input. the OR gate output would then control the RESET line for example. with this, when the 555 output is high, it will hold the GATE's output on, this is because it's an OR gate, one or the other needs to be on for the output to be on. This also means, a low on the other input will not be seen at the RESET pin.

If you want to cut corners. place a NPN transistor in there instead.

Emitter to the 555 timer out, collector to the RESET Pin. use a pull up resistor on the RESET line. and use a bias resistor on the base. you don't need much current. the difference here is, you'll now need s + level to reset it instead of a - level because you'll be biasing the NPN. when 555 output goes high. there will be now path to pull the RESET pin to low once the timer starts.

I guess it's a matter of choice.

Btw, if you have multiple points of triggers, you can isolate them via diodes with either method. only the RESET pin directly will override.

--
"I\'m never wrong, once i thought i was, but was mistaken"
Real Programmers Do things like this.
http://webpages.charter.net/jamie_5
Reply to
Jamie

Jamie wrote in news:wV1Ei.56$ snipped-for-privacy@newsfe03.lga:

Sorry, that's not quite what I meant - I'll try to explain it better. Under normal circumstances the circuit outputs a pulse of 45 seconds as soon as it is energized (in a one shot configuration) However there is this other line that goes between o and 12 volts. Anytime this signal goes to 12 volts I want the whole process to freeze and only start the 45 second pulse after that signal goes from 12 back to

0 volts. In other words if the signal goes high (no matter if the 555 output is already high)I want the output to go low 45 seconds AFTER the 12volts go back to 0. I think there may be a pin on the 555 that will do this I just don't know which one will do it. Or maybe I need an external gate.

Thanks for your help.

Reply to
AINTME

What you want is a "retriggerable oneshot",google gives as first result

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If you insist on using the 555, then use the same signal you use for triggering, to keep the timing cap discharged. Then on last removal of the trigger, the cap starts charging, and you get 45 seconds. Without extra components you wont succeed.

Reply to
Sjouke Burry

Sjouke Burry wrote in news:46e0d2cf$0$25496$ snipped-for-privacy@text.nova.planet.nl:

I think this will put me on the right track - it is a retriggerable oneshot I'm after - I did not know what it was called - having a name for it will help me with my search and also the fact that you say that I need extra components other than the timer.

Thank you for your help!

Reply to
AINTME

Is that 12V-to-0V transition the _only_ thing that's clocking your one-shot, or is it merely an inhibit and there's a different signal triggering the one-shot?

Incidentally, it might be a convenient time for you to learn about timing diagrams. :-)

Cheers! Rich

Reply to
Rich Grise

By combining your posts, I understand it as this: You want a (+) output as long as the input is (+) and it must stay (+) for 45 seconds after the input drops. There must be no gap, no matter how small, in the (+) output when the input signal drops. You might be able to get away with oring the 555 output with the input, but the brief transition time of the 555 from non-triggered to triggered when the input drops could create a gap.

My suggestion is that you add 3 transistors and support components to the 45 second 555, like this:

PNP +12 ---------------------+----------------- ----+ | e\\ /c | | --- Q2 | | | | [10K] [1K] | | | | +-------------------+ +---> Out | | | /c |c | +----+-------+---| NPN / | | | | \\e Q1 +------| NPN | [.1uF] [100K] [D1] | | \\e Q3 | | | |a | [3.3K] | [1K] | | | | | | | | | Input-+---- | --[555]--+ | | | | | | | Gnd --+----+-------------+-------------------+------+

(+) on the input will turn Q1 on through D1. That will turn Q2 on, so you get a (+) out. When the input goes (-), the .1 uF in parallel with the 100K will hold the base of Q1 (+) briefly, keeping Q1 (and therefore Q2) on while the 555 triggers. That way, there will be no dropout of the (+) output during transition. D1 prevents the (+) on the base of Q1 from interfereing with the (-) trigger on the 555, so the 555 will trigger when the input goes (-). The 555 output goes (+), which turns Q3 on. Q3 turns Q2 on, so the output stays (+). When the 555 times out 45 seconds later, Q3 (and therefore Q2) turns off, and the output goes (-).

Ed

Reply to
ehsjr

Rich Grise wrote in news: snipped-for-privacy@example.net:

I did not include a timing diagram because the ascii drawings look all jumbled in my reader (Xnews) the only way I can see them is by emailing the posting to myself - I can't however, reverse that process (write the email and then post, it doesn't work). Thank you for your input.

Reply to
AINTME

ehsjr wrote in news:EukEi.16$qV3.10@trndny02:

Yes, thank you, this seems like it will work. Like another poster suggested... what I need if a retriggerable monostable cct. Your cct here seems pretty close to what I wanna do. Thank you.

Reply to
AINTME

Rich Grise wrote in news: snipped-for-privacy@example.net:

The 45sec one shot triggers as soon as power is applied to the chip.The output goes high right away. But it will stay on as long as the other signal is 12volts after this signal goes to 0 then the output goes low 45 seconds later. Everytime the control signal goes hi it keeps the output hi until 45 seconds after the signal goes back to 0v.

Reply to
AINTME

Jamie wrote in news:iRhEi.21$ snipped-for-privacy@newsfe02.lga:

I'm getting a lot of good ideas from you guys - thank you very much.

Reply to
AINTME

"and" the 2 inputs, you'll have to get levels compatible if using ttl

Reply to
Dleer

---

View in Courier:

+12V>----------------+------+---------+--------+ |R1 | | |R4 [10M] | | [3K3] | | | | Q1 | | +---|--[1M]--+ 2N3904 | |R2 | | R3 | C----+ [10K]--[10K]---B | | | >------+-->VOUT E +------|--------|-/ U1A | |+ | | LM393 | [4.7µF] | | | |C1 | | GND>------------+----+------+---------+----------->GND

With no +12V, C1 will be discharged and 0V will appear on U1A-.

Then, when power comes up, U1A+ will go more positive than U1A-, causing VOUT to rise to +12V and stay there until C1 charges to a voltage higher than that on U1A+. At that time VOUT will fall to about 0.4V worst case.

Any time VIN goes high it will turn on Q1, which will discharge C1 quickly, making the voltage on U1A- more negative than the voltage on U1A+, forcing VOUT high until VIN goes low and allows C1 to charge up again.

R2 should be set somewhere in the vicinity of 8V for a 45 second timeout and it would be a good idea to use a polyester (Mylar) cap for C1 in order to get low leakage and a reasonable tempco. If you want to/have to use an aluminum electrolytic, observe polarity.

Also, connect all of the unused pins of the second comparator to ground.

-- JF

Reply to
John Fields

--
Slow day...

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Reply to
John Fields
[posted and mailed]

John Fields wrote in news: snipped-for-privacy@4ax.com:

Thank you John.

Reply to
AINTME

Just the inputs. Leave the output floating.

Cheers! Rich

Reply to
Rich Grise

--- Interesting...

Their 1989 General Purpose Linear Devices databook, 5-65 states: "All pins of any unused comparators should be grounded", while page

8 of the datasheet at:

formatting link

states: "All input pins of any unused comparators should be tied to the negative supply."

-- JF

Reply to
John Fields

I will grant you that grounding an open-collector output would do no harm, but I wonder if it's really beneficial, i.e., would it make any actual difference in the operation overall?

Thanks, Rich

Reply to
Rich Grise

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