555 Timer - Single pulse, delayed on and delayed off

Thread Starter

zcacogp

Joined Mar 10, 2015
8
Hello,

I am trying to design a 555 timer circuit that produces a delayed output pulse. The delay before the pulse needs to be around half a second, and the pulse needs to be anything from about 2 to 20 seconds long. Once the pulse has completed the output needs to remain low until the circuit is turned off and on again.

I have already built a circuit to produce a delayed output, and it looks like this;



This produces a delay of just over 0.5s and works very nicely, but once the output is high it then remains high. I need it to go low again between 2 and 20 seconds later.

I've thought about it and suspect that I could use Pin 4 (Reset) to reset the chip after this period of time has expired, making the circuit something like this;



However, I can't see any reference to any circuit like this on t'interweb and everyone who wants to create a circuit to do this seems to use other circuits. (I've had a look around on this site but don't know the technical term for this circuit so don't quite know what to look for. I think it should be a 'delayed monostable multivibrator', but this term doesn't seem to be used.)

So, my question is whether this circuit would work and, if so, what values of R1 and C1 should I use? (I have a bit of a random box of resistors and capacitors so it's be a mix-and-match affair, so maybe if would be more useful to have the formula to calculate the duration of the pulse so I can choose components to suit).

Thanks, in advance, for any help.


Oli.
 

AnalogKid

Joined Aug 1, 2013
10,987
You are trying to get two independent timers functions out of one timer chip, and that will be hard to do. There is no input to either of your circuits, so I'm guessing that the initial delay is started by applying power:

1. Power on, reed relay contacts are open.
2. No output change for 0.5 s.
3. Output change activates reed relay for 2 to 20 s. (adjustable or fixed?)
4. Output returns to initial state, relay opens, and is locked out until power cycles.

Correct?

The thing about analog timers is that many of them take as long to reset as to act. So without some care and feeding, the 20 second timer won't be ready to go again unless the power is off for 20 seconds or more between events. Still, this is not a difficult circuit once you add a second timer. This can be another 555, or you can do the whole thing with a few transistors, or a dual comparator like the LM393, or a single CMOS gate chip, or a single transistor array chip... Lotsa ways.

What is your skill set for getting components, soldering, etc?
What is this for (homework, home project, professional device, etc.)?

ak
 

AnalogKid

Joined Aug 1, 2013
10,987
I've thought about it and suspect that I could use Pin 4 (Reset) to reset the chip after this period of time has expired, making the circuit something like this. However, I can't see any reference to any circuit like this on t'interweb and everyone who wants to create a circuit to do this seems to use other circuits. So, my question is whether this circuit would work and, if so, what values of R1 and C1 should I use? (I have a bit of a random box of resistors and capacitors so it's be a mix-and-match affair, so maybe if would be more useful to have the formula to calculate the duration of the pulse so I can choose components to suit).
Looking at your 2nd schematic again, it is an interesting approach. If you have a CMOS 555 (many begin with LMC...), that's better than an original bipolar part because you can use larger resistor values, which keeps the capacitor sizes down. A simple R-C time constant for 10 seconds is RxC=10. So a 100K resistor and a 100 uF capacitor would be a good starting point. The 555 Reset input isn't a normal logic input so this will need experimenting. If the Reset input transition level is equal to 0.7xVcc (3.5 V in your case), the output pulse will be close to 10 s. If the transition level is closer to Vcc, the output pulsewidth will be short. If the transition level is closer to GND, the output pulse will be long.

ak
 

Thread Starter

zcacogp

Joined Mar 10, 2015
8
Guys,

Many thanks for your replies - AnalogueKid particularly.

SailorJoe, that's a helpful site, thanks. What I am looking at doing is what it calls a 'Bistable', but with the two states triggered automatically based on time elapsed since it was turned on (i.e. there are no external triggers other than the power on). While that site shows me a circuit, it doesn't have details of how to automate it - i.e. the two states are triggered by buttons, which I don't want.

AnalogueKid, thanks for your replies. Your synopsis is correct, thusly;

1. Power on, reed relay contacts are open.
2. No output change for 0.5 s.
3. Output change activates reed relay for 2 to 20 s. (adjustable or fixed?)
4. Output returns to initial state, relay opens, and is locked out until power cycles.

Your question in 3. is that the output needs to be stable for a period of time between 2 and 20 seconds. The time can be fixed and it doesn't matter how long it is. Closer to 2 seconds is probably preferable. And yes, it is triggered by the circuit being turned on.

Some background may be helpful; this is to turn on another device, which has a push-button switch which is meant to be pushed by hand. The reed relay is connected across this switch to automate it being turned on, and the 555 circuit is connected to the same power supply at the device, meaning that when the device is plugged in it is turned on automatically. I currently have the first circuit built and doing this job, and it effectively holds the 'on' button for the device down constantly. However I have found this to be undesirable as holding that button down for more than about 5 minutes causes the device to turn off again. Therefore I need a modification to the original circuit such that the 'on' button is held down for a period of time and is then released.

It's a hobby project, being done for a practical benefit. I can get hold of electronic components but would like to use what I have (zero budget is desirable!) hence the desire to simply modify the circuit I already have. I am not at home at the moment but I think that the chip I have isn't a CMOS one. Your notes about the calculation of the period of the output pulse are helpful - it looks like experimentation is the way forward.

Your point about the circuit reset time is a good one. As it is, the time between the circuit being turned off and turned on again would be quite long (a matter of minutes or even hours) so this isn't a biggie. However, if it was a problem, couldn't you connect a diode from the 0v rail to Pin 4 meaning that, should the +5v rail go low and the other side of C1 hence go down, it is shorted out and hence resets the pulse?

One question; what do you mean when you say "the 555 reset input isn't a normal logic input"?

Thanks again for your help. I think the next step is to start putting some components together and seeing what happens. I don't have a breadboard so I need to warm up the soldering iron and putting things on a board ....

Edited to Add: If the experimentation doesn't work then I'll be back to ask about circuits which will achieve what I am looking to do! AK, you mention another 555 or a few transistors or a LM393 .... it sounds like the possibilities are endless!


Oli.
 

AnalogKid

Joined Aug 1, 2013
10,987
You are correct about adding a diode for faster reset. Add a 1N914 or 1N4148 across both timing resistors, anode to GND.

A slow transition on the Reset pin might cause the output to bounce around briefly. And since the output drives the Reset, this could cause an oscillation. Or is it?

As a good general practice, add connection dots to your schematic where joining or intersecting lines are intended to touch. Also, avoid intersecting lines. Example - does pin 3 connect to R1?

I'm not a huge 555 fan, and often think of alternate approaches for timing requirements. Do you have some CMOS gate chips, or some small signal transistors like 2N4401, 3904, 2222, 7000?

ak
 

Thread Starter

zcacogp

Joined Mar 10, 2015
8
AK,

Thanks again for your help. You are right about the connection dots to make the circuit clearer - I should have done them originally.

So, a circuit with the diodes and dots would look like this, non?



(Looks like I put a spare dot in there for good measure too!)

I have a random selection of components that I was given by a friend about 20 years ago. There was a 555 timer in there as well as some resistors and capacitors and a small piece of board. And that's about all. There are some transistors and diodes I think, and the challenge (for me) is to build a circuit to meet my requirements from what I have. If that all sounds a bit odd it's probably because it is, and I am tight! :) I'm pleased to have got as far as I have but if I could perfect it to do precisely what I want then it will be even more satisfying.

I don't understand this sentence: "A slow transition on the Reset pin might cause the output to bounce around briefly. And since the output drives the Reset, this could cause an oscillation. Or is it?". I can understand that a slow transition may cause the output to bounce a little but don't understand what you mean when you say that "The output drives the Reset". Sorry to be slow.


Oli.
 

Thread Starter

zcacogp

Joined Mar 10, 2015
8
AK, thanks, that makes sense!

MikeML, thanks for your input. I don't understand the circuit diagram you posted - it's beyond me. Sorry!


Oli.
 

AnalogKid

Joined Aug 1, 2013
10,987
I don't understand the circuit diagram you posted - it's beyond me. Sorry!
It is a discrete transistor version of your 555 circuit. R2-C2 delay M2 (the output) turning on. Meanwhile, a second time constant, R1-C1, is running. When C1 is charged up high enough, M1 turns on, shorting out the drive to M2 and turning off the output. M2 is functionally equivalent to the 555 output stage, and the gate of M1 is functionally equivalent to the 555 Reset input.

ak
 

MikeML

Joined Oct 2, 2009
5,444
...
MikeML, thanks for your input. I don't understand the circuit diagram you posted - it's beyond me. Sorry!

Oli.
Your power is applied while the green line is high.

Your button is pushed (Relay L1 is pulled-in) while the yellow line is high.
 

Thread Starter

zcacogp

Joined Mar 10, 2015
8
Guys,

Thanks for your various replies. I'm updating this tread as I have now solved the problem I had, although it wasn't actually the problem that I asked about when I started the thread ....

The circuit I was working on was intended to control (turn on) another device. This other device is switched using a reed relay, and needs a small delay after power is applied before it turns on (which happens by the relay being closed). A 555 timer applied a delay then closed the relay, which was great, but the device turned itself off after 5 minutes - which I thought was a function of the fact that the relay was being held closed. I hence wanted a circuit that would delay slightly before creating a single pulse for a few seconds, thus giving a 'low - high - low' behaviour.

As it turned out, the fact that the device was turning off after 5 minutes was a feature of the device and was nothing to do with the fact that the relay was being held closed. The solution was to build a reset button into the timer circuit, so that once it had produced the high output then a press on the button would produce a brief low before going high again. This brief low is enough to turn the external device on again once it had powered off.

This 'reset' button was simply a push-to-make button that shorted out the 47MicroF capacitor in the second circuit in my first post. Yes, it's a shame that the final solution had to have manual input but it otherwise works as required and I'm very happy with it!

So, thanks for your input. It was greatly appreciated. I think that the most exciting thing for me is not that I've made this little project work but more that the world of small-scale electronics has been opened up, and I suspect I'll be dabbling in it again fairly often in the future!


Oli.
 
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