[SOLVED] 555 Timer IC monostable

[Eshal]

Hello everyone!

I am improving my past paper at university so I reviewed my past papers from MSc (While I am doing BS) so I found some questions which I know that you people can surely help me. :-D I am by hope..

Q#1) In monostable, if required frequency is 5KHz and duty cycle is 50% then what is the applied trigger and timing circuit.
I know its conduction time is given as 1.1RC. But what more? How to reach actual answer. :-(

Q#2) In monostable, how long does the triggering voltage have to go in order to initiate the output pulse and why duty cycle must be greater than 70%?
I don't know how long triggering is required. I read from book but not proper answer is defined there. And about duty cycle.... I think this question is wrong because it is not necessary for monostable that it must have duty cycle greater than 70%. I know why duty cycle of astable multiviberator is greater than 50% always but don't know about monostable.

PS: I know the circuit operation please don't ask me to post about circuit operation. If answer is in the circuit operation then take me to the point so that I could grab the answer. Thank you.

Regards,
Cute Princess.

 

[Syncopator]

The trigger pulse must be shorter than the desired duration of the output pulse. Apart from its amplitude that is the only criteria.

Wherever did you get the figure of 70%, and 70% of what?

It is usual to pass the trigger pulse/s through a differentiator, as shown below.

 

[Eshal]

70% of the total duty cycle.

Thanks for reply

 

[Syncopator]

A monostable doesn't have a duty cycle.

 

[Eshal]

What do you mean doesn't have duty cycle? How did you say that?

 

[Syncopator]

I can understand why you ask Eshal :smile: It's not very easy to explain well.

A monostable produces one pulse - "mono", from the Greek monos, means one.

It gives one pulse each time it is triggered, and it is triggered by the fast rising or falling edge of an applied voltage.

The duration of the applied voltage is irrelevant - with the caveat that it must revert to its original state, high or low, before the monostable may be re-triggered.

Duty cycle is the ratio of the two parts of a regularly repeating series of events, often, but not always, from an oscillatory source.

If a monostable is repetitively and regularly re-triggered, the output pulses will certainly have a duty cycle, but that is a combined result or function of the trigger source plus the monostable, not the monostable itself.

I hope that makes sense.

 

[Eshal]

So you are saying my electronics teacher guided us wrong and she gave us wrong question in exam? Obviously you are saying this. Huh?

 

[Syncopator]

So you are saying .....

I am saying what I wrote in my post.

I have no idea what your teacher said, but it is obvious that someone, or something, probably because of a lack of rigour, has led you to harbour misconceptions. Believe me, it's much more common than you might wish to believe.

 

[goldsmith]

So you are saying my electronics teacher guided us wrong and she gave us wrong question in exam? Obviously you are saying this. Huh?

Hi Eshal
I agree with Syncopator .
Sorry because i have to say it too ! your teacher is wrong !
Let me try to conduct you to a good path !
See below , please :
**broken link removed**
or perhaps this one :
http://www.technologystudent.com/elec1/5553.htm
You could have a bit search during the google ! to see gang of guidelines !

Recall : as i can remember , me and other members here , have suggested you to , don't believe your teachers ! just try to believe correct things ! ( you shouldn't expect that your teacher be able to teach you ! you should learn yourself )
But i don't know why you don't want understand that your way is completely wrong ! ( just relying on university isn't ok ! )
( when i'm in a class , and when i see my professor's description isn't correct , i won't accept and i will try to tell him / her , the problem ! )

Best Wishes
Goldsmith

 

[Eshal]

OK.
I respect you people, what you say.
Can you kindly tell me why monostable doesn't have duty cycle?
And why astable and bistable have duty cycles?

 

[goldsmith]

OK.
I respect you people, what you say.
Can you kindly tell me why monostable doesn't have duty cycle?
And why astable and bistable have duty cycles?

Hi Eshal
A question :
Do you know meaning behind the word Duty Cycle ? if yes tell me what you know . ( i want improve your consideration then you'll understand your answer ! yourself ! )
And a suggestion :
Take a look at those two links that i've attached in my former post . and read it with enough patience !
Good luck
Goldsmith ;-)

 

[Eshal]

Duty cycle is the time spend by devices in active mode. Is this?
And yes bro I will surely check them, first link is in depth I will check.

 

[goldsmith]

Duty cycle is the time spend by devices in active mode. Is this?

Consider please that we have a square wave .
(it will have an on time and off time ? ok ? ) ( do you know what on time and of time are meaning ? ) and a T ( period time )
D.C ( duty cycle ) is Ton/ T ok ?

Good luck
Goldsmith

 

[Eshal]

Yes, on time is the time when we get output. and off time when we don't get any output.
I know formula bro.

 

[Syncopator]

The duty cycle of a rectangular wave is the ratio of the two time periods, the one during which the output is high, and the one during which it is low.

An astable gives an output which is a continuous and regular series of pulses (rectangular waves), which have a known duty cycle, which is a function solely of the astable's operation and totally independent on any applied stimulii. It is therefore said that the astable has a duty cycle.

A monostable does not, on its own, produce a continuous series of pulses. It produces one only pulse when it is triggered. When considering the operation of a monostable we consider just one trigger. We are not concerned with when another trigger might be applied, so for that reason a monostable doesn't (can't) have a duty cycle.

As I said in a previous post "If a monostable is repetitively and regularly re-triggered, the output pulses will certainly have a duty cycle, but that is a combined result or function of the trigger source plus the monostable, not the monostable itself."

Similarly, a bistable changes state when either a set or reset is applied. That's all we know about a bistable. We have no idea from its operation when either of them will be applied, so the bistable doesn't (can't) itself have a duty cycle.

Again, if the bistable is set and reset on a continuous and regular basis, the ouput waveform/s will have a know duty cycle, but that is a function of the time relationship between the set and rest pulses; nothing to do with the bistable itself.

 

[Eshal]

I am still confuse. You said about duty cycle which is the ratio of two times, one is the time when the output is high to the total time.
Monostable also has a high time when pin2 in triggered with -ve pulse. Then why monostable has not duty cycle. This is a simple question.

 

[pebe]

The total time is the time On plus the time Off. The time Off can be as long as you want to make it, ie. a couple of seconds, an hour, a day, or a week. It depends when you decide to trigger the monostable again.

Unless you define the Off time, and it is repetitive, you cannot define the duty cycle.

 

[Eshal]

Sorry experts I am still confused. You guys using difficult or deep english, English is not my native language that's why I am having problem. :-(
But listen what I understand uptil now, if I go wrong then correct me please,

You people are trying to say if we could control off time then we could define that monostable has a duty cycle, but we just trigger it and after a defined period it goes back to its stable state (off time). Is it right?

 

[pebe]

Let's give a 'for instance'.

For instance, if the On period of the monostable was 1minute and you triggered it every 10minutes then its duty cycle would be 1/10 or 10%. If you triggered it every 20minutes it would have a duty cycle of 1/20 or 5%.

The time between triggers is the trigger period, and so you cannot define duty cycle without first defining the triggering period. As soon as you do that you have an astable oscillator - not a monostable one.

I hope that makes it clearer.

 

[Eshal]

So you mean triggering period should be stable. In monostable, how long we trigger is the trigger time. If I trigger it for 10 minutes then during this time I get output high. right? similarly if I trigger it for 20 minutes then during this time is output high. and after this defined period I get to stable state of it.
BTW: I got what are you trying to say or make me to understand you. If there is another clear example you can provide then please pebe give me. So that remaining confusion which I have in my mind and you see what should more clear to me then kindly help me. I will tell this to my madam who taught us wrong.

Regards,
Cute Princess.