Monostable Multivibrator using 555 Timer Explained (with Working, Applications and Derivation)
ฝัง
- เผยแพร่เมื่อ 22 ก.ย. 2018
- In this video, it has been explained that how 555 timer can be configured to use as a monostable multivibrator. The working of the circuit is explained using the internal block diagram and timing diagram.
And at the end of the video, the expression for calculating the time delay is also derived.
By watching this video, you will learn the following topics.
0:36 What is Monostable Multivibrator?
1:32 Design of Monostable Multivibrator using 555 timer IC (Circuit Diagram)
2:15 Working of Monostable Multivibrator
8:44 Applications of Monostable Multivibrator
11:03 Derivation of Time-Delay expression for monostable multivibrator
What is Monostable Multivibrator?
The monostable Multivibrator has one stable state. But whenever the trigger signal is applied then momentarily the output goes into the unstable state.
The time duration for which the output goes into the unstable state can be controlled by the external resistor and capacitor.
Whenever this monostable multivibrator is designed using the 555 timer IC then it can be used in the following applications:
1) Generating Time Delays
2) Switching the Relays
3) To generate Pulse Width Modulated (PWM) Output
4) Frequency Division
This video will be helpful to all the students of science and engineering in understanding the design and working of 555 timer as a monostable multivibrator.
#555Timer
#MonostableMultivibrator
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Please Note:
The output driver the circuit is the inverting circuit. It will invert the output of the Q-bar and provides the necessary driving current.
The timestamps for the different topics covered in the video:
0:36 What is Monostable Multivibrator?
1:32 Design of Monostable Multivibrator using 555 timer IC (Circuit Diagram)
2:15 Working of Monostable Multivibrator
8:44 Applications of Monostable Multivibrator
11:03 Derivation of Time-Delay expression for Monostable Multivibrator
I would like to draw your attention towards a mistake.
At 2:57 the output is not 1,it is 0.
Actually Q-bar is 1 so output (which is Q) will be 0.
Would you please design a monostable to drive a relay which switches on a light bulb 220v 50Hz with Vcc=|Vee|=12V
I would to like to draw an attention towards a mistake you have connceted output to Q bar which is wrong the output pin is pin number 3 which is Q you have connected output to pin number 7 which is discharge pin please correct it otherwise people may get wrong concepts
Why can't we directly give q to o/p rather than having a inverter
@@ashjaahmedkhan3254 Bro if you see there is a dot before the output hence it will Invert the output Qbar to Q...
Saviour of many college students😊
Sir these videos are perfect for understanding the topics.Thanks a lot. It helps a lot
thank you to make these concepts easy dude!!!seriously its helping, keep doing the same thing really appreciated
Can't be explained better.
Great teaching. 👍
Excellent explanation, congratulation on 70K subscribers !
I wish these kinds of channels also hit 1 million subscribers
Awesome , nice lecture. Thanks for giving such a lecture.
Very nicely explained sir. Thank you!
U guys have extraordinary talent to make understand easily...thanks a lot from Bangladesh.
all dislikes are from the college teachers
Good keep it up....you have given Very good respect to your teachers... Made them really Proud...
Very helpful and interactive videos.
Nice work sir..!!
Thanks for your tutorial :D
Thank you sir very nice gide & very nice best information monostable multivibrator teaching video.👍
Very well explained
What an explanation sir great sir i got all my confussion cleared about the discharge pin and sir when the capicitor gets charged because of voltage drop will the capacitor get charged up to vcc or slightly less than vcc
Really awesome explainations tqs bro😊😊
Thanku sir it clear all my problem😊
Big fan from Nepal sir
Nice explanation 👍
u are god sir i will pass my exam because of u sir
Thanks. Sir and good explnation
It's very informative 😍
Nice work sir i salute u please upload more and more videos. there are some videos on you tube but getting more views and likes because there audience is large. This type of video must get good likes even with low audience( ece student only) everyone should like who watch.
you should have connected an inverter to q_bar instead of writing as output, it confused me a lot...
I also felt the same
Yes I agree, your comment actually cleared my doubt
Thank you
Excellent explanation compare to my college classes
Thank you 😄.. you explained very nicely....
one is me
now i am an average too oops avenger
Both control and (R4 and R5 through Vcc) connected to same point at inverting terminal of op amp.
Which one will be selected as input actually.
In PWM how control voltage changes the output pulse width as output pulse time period is independent of voltage and as capacitor always charges to 2/3 Vcc.
Please upload lecture of a/d converters soon...
Could you please give brief explanation on applications of monostable multivibrator using 555 timer
Excellent 👌👌👌
Was helpfull
Thank you Sir for the clear explanation. How do you generate the trigger? Especially when using the pulse width modulation. Do you use a switch? Is the there to make it automatically trigger itself?
There are many ways. You can use a switch or external signal generator for the trigger. Or even you can generate using 555 itself. (By using 555 timer as astable multivibrator )
@@ALLABOUTELECTRONICS Thank you so much! I am currently using this for a design project. I am going to use two 555 timers like you said :-)
Awesome!
as you said that initially pin 2 is at vcc and when we apply trigger then what wil be potential at pin 2, will it be (trigger + Vcc)
Nice video sir
I think base of the transistor should connected to 'Q' then only at unstable state it will act like open switch and capacitor start charging from Vcc
Please help me out!
I just made a water motor automatic turn on/off controller with 555 timer,i was very happy as it was working fine initially but then i noticed the triggering pin is too damn sensitive and getting self triggered with even a small shake in the circuit..
What m i suppose to do? Please give answer with specific value of component
Circuit :-
Pin 1 - Negative
Pin 2 - is connected to the ground/negative with 1 mohm and through 22 k ohm resistor this lead is put in the bottom level of the tank ( around 25 % level,as if the water goes below this level motor should turn on )
Pin 3 - connected to base of bc547 with 1 k resistor which is driving the relay and also connected to the ground with 220 k ohm resistor
Pin 4 & 8 - Connected to the positive 6 volts
Pin 5 - connected to the ground through 104 ceramic capacitor
Pin 6 - connected to the ground with 1mohm resistor and to the top level of the tank through 22k resistor ( to turn off the motor )
Pin 7 - Left it just like that
A common/positive is given at the bottom most level of the tank
How is pin2 getting Vcc supply greater than Vcc/3 by default(to keep S=0 ). I mean i dont see any wire connecting pin 2 to Vcc. please reply
Hi Man, I have a chip that powers my Electronic power steering IT has Timer 555, ic 151 SMD, M7 diode and some led that I am unable to identify. The circuit stops working after 8-9 hours of continuous driving. The output should be between 4-6 v with 12 v vcc from battrey. Can you suggest a better and robust circuit that can better support my Electronic power steering
I loved it
You made a mistake in last minute. In derivation you should take R.H.S. part in denominator so that '-' sign in power of e would eliminate. And ln(1/3) is not equal to 1.1. ln(3) equals to 1.1 . Besides this video has lots of great information. Thank you for uploading.
bro take the negative sign to LHS. -ln(1\3)=ln3
sir can u please explain the working of metal detector circuit using an IC 555 timer
Best👍👍👍👍
Please upload ad/da coverters and memories
Sir,I want to know how you make videos..please help me in making videos for JEE students
Can we make a 50% duty cycle frequency divider
there are many wires intersecting each other but i am unable to find out how to distinguish that where is the connection joint or where not
Also upload FSK using IC 555
Please suggest authentic book for reading about IC 555 and comparator!!!!
With due respect sir, @3:28/14:02 in the IC 555 timer,the output should be connected to Q not Q'.
Yes, it seems so. But if you closely observe, there is a bubble at the output driver circuit. Basically, its inverting output buffer, which will invert the output of the Q-bar.
I should have mentioned it in the video. Anyway, I will add that in the pinned comment.
Sir, I think Q is the o/p for the timer IC and Q-bar must be fed into the transistor.
The output box next to Q-bar creates confusion. I assume it must be placed next to Q.
Please correct me if I'm wrong...I don't mean any offense to the enormous knowledge you are providing to all of us.
After the Q bar, the output driver circuit is inverting circuit. So, actually, the output would be Q.
There is a bubble for inverting logic (just before the output driver circuit). But I think should have mentioned clearly. Anyway, I hope it will clear your doubt.
I think Vedant Kashyap is right.
But Q should be feed to transistor.
And Q' as O/P
the discharging should be upto one-third of the vcc
Sir , i have seen a different diagram of monostable using 555 timer where transistor is connected to Q of flip flop and then connected to Vthreshold .Then, sir both the circuit of it are correct ?
I think yes because in most of the books Q is connected to transistor Qd
What is the bias of the transister here?
when Q=0 then output should be '1' how it could be '0' as output is connected to Q^ (Q bar)
The output driver circuit is inverting circuit. There is bubble at the starting of that box, which indicates that the signal gets inverted. Although it would have been better if I had mentioned it in the video itself. But I hope it will clear your doubt.
in this circuit one correction 'the Q of SR Flipflop ONLY connected to output of 555 timer not the Q bar.'
Can u explain y time(T)=1.1RC ?
Instead of rsff you have to take srff because when s=0,r=1 then q=1, qbar=0 but u have said reverse
I have seen, many have this misconception about RS and SR flip flop. Both are actually same. In programming, just to avoid indeterminate state ( when both S and R = 1), they have assigned priority. E.g in SR Flip flop , the set input has higher priority when both inputs are 1. And similarly, for RS flip flop, the reset input has higher priority when both S and R =1.
I hope it will clear your doubt. If you still have any doubt then let me know here.
3:38 at
How the trigger pin voltage without the triggering signal is Vcc?
The thing is, in this case, the trigger signal is high to low transition. So, untill that transition occurs, the voltage at the trigger pin is VCC. ( In the figure, it is now shown )
And when the trigger signal is applied, there will be a high to low transition ( VCC to 0 V) at that pin.
I hope, it will clear your doubt.
Why do we short 6th and 7th pin in 555 monostable
I didn't understand this lecture. Can you please explain how the capacitor and transistor work together? It's confusing me a lot.
Here transistor just acts as a switch. Whenever transistor is in saturation then it will act as a closed switch and capacitor gets the path for discharging. When the transistor is in cut-off, it acts as an open switch and capacitor remain isolated from the discharging path.
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS Oh ok! thanks a lot. I was confused about how the SR circuit o/p affects the charging and discharging. Now, it's pretty clear.
Sir I have a doubt ,u have told that when ever the output is 1 the input to the first comparator is logic zero..how ??
And how did Trigger Pin get Vcc? From the Block diagram it is clear that there is no connection from voltage divider ckt to the trigger pin. And even though Pin 6 and 7 is connected to VCC, there is no connection from Pin 2 to VCC in the Block diagram here.
You need to externally connect the trigger pin to Vcc with provision to connect it to the ground (via a switch) momentarily. (For monostable operation)
@@ALLABOUTELECTRONICS Did you forget to add that to the circuit. Or ?
And don't be offended Ok. I find your channel very helpful to prepare in the days before exam. So I'm just trying to see everything clearly.
No, it has been assumed that the connection to the trigger pin is readily available. (So, that the circuit looks a bit cleaner)
Do you have any videos on counters?
No, digital electronics will be covered soon.
At (6.55)as R=1,Q=1but how Q become "0".and Qbar become 1??
Nicely explained sir. I had confusion.
Why time duration output pulse should be greater than triggering pulse at 10:12
maybe it gives the same frequency output
Lets restart from the first case.When o/p is high Qbar = low. So Base of transistor gets low and as a result turns the Transistor to an Open Ckt/OFF. Which means Capacitor is NOT AT GROUND POTENTIAL. Which means what you stated in Duration 2:58 is Wrong. If you're right, explain how?
Yes, By mistake I said the output of the 555 timer is high. Because initially at 2:35 it has been assumed that the output of 555 timer is low. And assuming that the working was explained.
@@ALLABOUTELECTRONICS I'm surprised nobody noticed it.
is the triggering voltage is equal to Vcc voltage?😅
Sir according to the video unstable state means 0 or 1?
Unstable state is when output is 1. Because momentarily output goes to high.
@@ALLABOUTELECTRONICS thank u sir
Sir I have an doubt
If it is an multivibrator we use 2 transistor . But in our channel we use one transistor y?
It is designed using two transistors.
Please check this link for the video:
th-cam.com/video/nTm-fB1OrqE/w-d-xo.html
At 3:20 ,the pin 6 is grounded then how come the it's voltage is greater than the non inverting voltage that is 2/3Vcc
If you closely observe, the pin 6 is not grounded. It is connected to the capacitor C1. So, the voltage at pin 6 is the voltage across the capacitor C1. I hope, it will clear your doubt.
can you explain the frequency division in brief and also relate how the pin diagram is to draw from 555 timer relating monostable multivibrator sir.
For frequency division, you need to select the RC time constant which is more than the time period of the trigger signal. In that case, monostable will not get triggered until the RC time period. And in this way, one can divide the frequency. I have already shown the 555 timer pin diagram in the video.
For more information, do check this simulation. You will get it. In case, if you still have any doubt then let me know here.
www.multisim.com/content/oUMurxgpGky9a6w45AneA7/555-timer-monostable_frequency-division/open/
The explanation is very good. But in all your diagrams, Q should be connected to the discharge transistor, not Q dash
The output of the Q-bar is inverted by the output driver circuit. If you notice there is a small bubble at the input side of the output stage.
But as it is very small, it appears as if the Q-bar is given to the output.
But I hope it will clear your doubt.
@@ALLABOUTELECTRONICS Ohh thanks. Out of curiosity, why dont you just conenct Q to the transistor?
Because we want that as soon as the trigger signal is applied, the transistor should start charging. Before that it should remain uncharged. As, soon as we apply the trigger signal, the output of the 555 timer becomes 1. That means Q = 1 and Q-bar is 0.
So, now the transistor is in cut-off. And now it is no more attached to capacitor. That means now, the capacitor starts charging through R1.
I hope it will clear your doubts.
After applying trigger the o/p of rs flip flop at logic 0 and q bar at logic 1 then again transistor act as closed switch in that case capacitor will not charge ..
Before the trigger signal is applied, Q-bar is 1, and Q= 0. So, the transistor will be ON and the capacitor is connected to ground. Once the triggering action occurs, the output of the second comparator becomes high. (i.e S= 1 and R=0). So, Q-bar = 0 and Q= 1.
And as soon as Q-bar = 0, the transistor will be turned off and the capacitor starts charging towards the supply voltage. (untill it reaches 2/3 Vcc)
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS for rs flip flop the when r=0 and s=1 the q =O and q bar =1
As per rs flip flop truth table
@@ramanapolireddy5040 I would recommend you to go through any standard text book. S stands for SET and R stands for RESET.
That means when S= 1 and R=0, it will set the flip-flop or in other words, Q= 1 and Q-bar = 0.
Design a monostable multivibrator using IC556 with a pulse width of 5 milliseconds.... please explain
The design procedure wil remain the same as 555 timer. In fact, you can use the same equation which is used in the video. The only thing is 556 is dual 555 timer. So, you just need to see the pin diagram of it. Depending on the model number, the control or discharge pin number might change. So, please check that. Otherwise, for calculation wise and for selecting the R and C, it will be same as 555 timer.
Sir,,,why do we use 0.01microfarad capacitance with the control pin to the ground??
Plz give me answer
Can it work on 3v Vcc sir ..
Operating supply range of 555 timers is 4.5V. to 16V. So, it won't work properly below that.
Thank you for your response
I didn't get the point why capacitor is grounded when logic is 1 and y it starts charging when logic zero
please explain sir
when the logic is 1, then transistor will turn ON and it will act as a short circuit (provides low resistance path). When the output is logic 0, then transistor will act as an open circuit. So, capacitor will charge through resistor. I hope, it will clear your doubt.
Sir I have a doubt,u have told that whenever the output of the 555 timer is 1 the capacitor will remain at ground potential so when output of q bar is 1 the input to the first comparator is logic 0 ...how sir?
because as Q bar is 1, so at the base of transistor is high which activates the transistor so the VCC passes to the ground through transistor instead of going to pin 6. So, voltage at pin 6 is 0. which is input to comparator 1. that's why input to first comparator is 0 when Q bar is 1
i think he wanted to say output 0 there
@@kushagraagrawal509 thanku so much
What is the role of capacitor C at pin 5 in multivibrator, plz explain in detail
Through pin number 5, it is possible to provide the control voltage to the timer. (other than 2/3 Vcc). When it is not in use, it can be left open. But in that case, it can pick up external noise and may override over reference voltage. To prevent that, capacitor C1 is connected. Which prevents any abrupt changes in the voltage or pick up due to external factors. I hope it will clear your doubt.
Thanks much sir even I had the same doubt...
At 4:47 if output of 2nd comparator is logic '1' , how can Q bar be logic '1'. ??
When the output of the second comparator will become high i.e when s=1 and R=0, Q= 1and Q bar = 0. And the output of the 555 timer will be 1. Because the output driver stage is inverting stage. Meaning that if the input is low, the output will be high.
I hope it will clear your doubt.
Same question sir...
The o/p q of flipflop is the 555 timer o/p
Please kindly use a big red pointer for pointing something...!!
That small blue pointer is not visible
At 5.49, how voltage at pin 2 becomes equal to supply voltage
@J. Vanitha the output will be stable only when both s=0,r=0.to make this happen 2 is connected to Vcc,when trigger is applied,pin 2 momentarily goes below 1/3Vcc
How triggering voltage get to vcc
Just for understanding purposes, let's say, you applied a triggering signal through a switch. The connection is made such that when you press the switch, the triggering pin will get connected to the ground. And soon as you release the switch, it will go back to the VCC voltage. So, that is how, momentarily, the voltage will go below the reference voltage and then it will come back to Vcc. I hope, it will clear your doubt.
Even college teachers watch here to learn 😂😂
It is confusing that you reference two outputs and call them output during the video 1) the output of the RS flip flop and the output at pin 3. They are next to each other which adds to the confusion.
At 10:20 how frequency of output is half of the frequency of trigger please explain🥺
Since the pulse width is more than the trigger signal time period, when the second trigger arrives, the output is already is high and it will not get triggered again. In this case, monostable gets triggered at alternate trigger pulses and in this way, it can be used for frequency division.
Sir at 11.34 you have written an alternative expression for charging of capacitor. I couldn't get it from your previous video of transient analysis. Will you please write the intermediate step?
Please watch this video.
Here is the link: th-cam.com/video/KylJ2v1_c-o/w-d-xo.html
I have discussed it at around 22:00 in that video.
If you go through that portion, you will get it.
And after that still if you have any doubt, then do let me know here.
Sir, I have watched it. In that video at 22.00 you have discussed shortcut method. I need to know that how could we reach to the general equation.
V =Vfinal+(initial-final) e^-t/T
What does capacitor C do
The capacitor bypasses the noise or ripple from the supply
in the diagram, trigger value is greater than one-third of Vcc which is wrong i think .
The trigger signal is usually high. When it goes below 1/3 Vcc, then the trigger action happens.
anyone who noticed the peculiar buzzing sound of cricket(brown cicada) in the background ? :)
In frequency division circuit, why t should be greater than T?
please explain
To avoid the re triggering. If t is less than T, then once again at the negative edge of the trigger signal, once again we will get the pulse at the output with per-defined pulse width. And eventually the frequency of the output pulse would be the same as the input trigger pulse.
I hope it will clear your doubt.
At 3:45 how the pin2 at Vcc ??
Because if pin 2 is at Vcc, then the output of the second comparator will be low. So, S and R both will be zero and output of the timer will remain in the previous condition. The output of the timer will change when this pin 2 goes low momentarily. Or in other words whenever the trigger signal is applied at pin no2.
I hope it will clear your doubt.
I have the same question.. pin 2 has not connected to Vcc
How Vpeak is Vcc why not less than Vcc?
I think you are reffering to the charging of the capacitor right ?
If it is so then if you see, the capacitor is connected to the Vcc via resistor. So, if it charges fully then it will charge upto Vcc.
When I'm using pcb pot then its working fine as per calculation, but when I use potentiometer with two legs as resistance then it doesn't work as per calculation ! Kindly suggest me the solution !
Are you using 3 leg POT with two legs shorted ?
No, I'm using two legs only (one end & middle) , its third leg is remaining disconnected,
Just short the middle leg with any of the two legs of POT.
@@ALLABOUTELECTRONICS I did but multimeter is not showing either exact or approx value of resistance so thus it become hard to mark up the time indication after connecting pot in circuit and also the timer is being higher than calculated. 1000uF with 433K ohm resistance which means it should run 7min 56s but its running approx 9min 20s !
Is there any possible way to calculate pot resistance after installing and any formula for calculating more accurate time?
@@aloklakhera866 For very large value, you might get an error. You also need to consider the tolerance of the capacitor. Use low tolerance components.
At 5:13 how flip flop logic 1 when r=0 and s=1 then i think flip flop logic 0
For the RS flipflop, when S= 1 and R=0, the flip-flop will get SET. That means the output Q =1 and Q bar = 0.
Please refer the truth table of RS flip-flop.
Sir pls reply I have exam tmrw
Keep your arrow little bit large,t o see where ur explaining in the circuit
make pointer bigger, it becomes hard to follow
Sir,please provide notes