The timestamps for the different topics covered in the video: 0:22 What is Monostable Multivibrator ? 1:28 Working of the monostable multivibrator 4:48 Design of Monostable Multivibrator Circuit 6:34 Derivation of the equation for the monostable multivibrator 9:38 Use of RC Differentiator circuit for the generation of the trigger signal
Sir, when the diode is in forward biased condition than how we know that the forward biased volatge always less than beta vsat voltage at non inverting node?
@@sumitkumarpal2306 Because forward bias = 0.7 volt. So Vsat*(R2/(R2+R1)) must be larger than 0.7 volts. R2 would have to be very small to have less than 0.7 volts across it. Good Question.
Obisly i understand.... Dude... Help me a lot.... Just watching your videos & putting the equipment on bread board & see the results & calculate my own one & modify... Awesome
just excellent. from ur video, I made 5 pages of detailed explanation of monostable multivibrator. very interesting! Could you simulate a design using NI MultiSIM?
@@sreeharibalachandran Trigger pulse should be less than the unstable time. It should be sufficient enough so that the output can switch from the positive to negative saturation voltage. (and that will depend on the slew rate of the op-amp). For fast op-amps, the switching will occur in less than 1us. For op-amp like 741it may take 2o us to go from 10 to -10V. So, roughly it will be 20-25 us. So, trigger pulse should remain low for that duration. For fast op-amps, even 1us pulse width is enough for the transition. I hope it will help you.
Can the astable pulse be a time period that is less than the trigger pulse? Example: i press the button for 20 milliseconds. The astable state switches on for
Why does the diode become forward biased when the Vout=Vsat and reverse biased when Vout=-Vsat? Cause if we use Kirchhoff on the external mesh, we don’t consider the voltage across R?
The op amp circuit and trigger circuit are both good. Add them together like this and the monostable can't work, as the voltage divider on the output is compromised. This combination seems to be ubiquitous on the internet, however it's useless.
Sir, I always watch ur series on electronics and find quite easy to understand but this explanation was a messed up! And quite fast.Still ur videos makes easy to work out!..
Because here we have assumed that, the output of the op-amp is +vaar initially, and when transition occurs then it goes to - Vaar. That means capacitor starts charging towards the -Vsat Voltage. I hope it will clear your doubt.
The diode will ensure that the voltage at the inverting node is less the input voltage. Because normally the Vin is high voltage (Vcc) and during the triggering action only it will go low. So, until the trigger voltage goes low, the voltage at the non-inverting node will be more than the inverting node. Hence, the output of the op-amp will be high during that time. I hope it will clear your doubt.
Why are we getting (- BVsat) for a time period of t by using momentary negative trigger? By using - ve trigger, the Vid will be negative momentarily, but how are we getting the unstable state for a long time? does capacitor have any role? please explain.
Yes, the capacitor does play a role. And if you see the time duration T, it depends on R, C, and R1, R2. As T= RC ln (1+R2/R1) Please check the derivation at 6:34. You will get an idea about the working of the circuit.
It is possible to do that using an astable multivibrator. But if you are designing it using the op-amp, then you need to check the slew rate and the gain-bandwidth product of the op-amp. I mean it should be able to support that rapid transitions.
The thing is, the different multivibrators are used for the different applications. The monostable multivibrator can be used to generate the pulse width of the finite duration based on the triggering instance. For example, let's say you want to design the alarm system, which rings or shows red light when particular even occurs. The monostable multivibrator can be used in such project. This is just simple example, but there are many other use of this multivibrator. I hope it will clear your doubt.
Because the capacitor is actually charging towards the -Vsat voltage. But due to the comparator, you will see a transition at -Beta times Vsat voltage. If that transition is not happening at -B Vsat then the capacitor would have charged up to -Vsat voltage. I hope it will clear your doubt.
Because we want to change the output of the op-amp from +Vsat to -Vsat. That will happen when the voltage at the inverting node is more than non-inverting node. And to make that happen, we need to momentarily reduce the voltage at the non-inverting terminal. When the negative spike is applied then only, momentarily the voltage at the non-inverting terminal will go below inverting terminal. And due that, we will see a transition in the output voltage. I hope, it will clear your doubt.
want the unstable output to be a minimum of 10 sec; Vcc = +3V; using cmos op-amp; using C=10exp-6 F, R = 14.4M ohm If Vcc = 3V, +Vsat = 3V, & -Vsat = 0V Is R = 14.4M ohm too large?
Instead of going for large value of R, you can change the value of C. I mean just use 100 uF instead of 10 uF. And then you can use 2M ohm potentiometer.
@@ALLABOUTELECTRONICS yes sir because in our banaras hindu university it is based on using transistor and in practical I unable to understand if you provide lecture as recent as you can it will so helpful for us thanku 🙏💕sir
The timestamps for the different topics covered in the video:
0:22 What is Monostable Multivibrator ?
1:28 Working of the monostable multivibrator
4:48 Design of Monostable Multivibrator Circuit
6:34 Derivation of the equation for the monostable multivibrator
9:38 Use of RC Differentiator circuit for the generation of the trigger signal
Waiting for 2nd order active low pass ,high pass filter derivation using opamp...
Sir, when the diode is in forward biased condition than how we know that the forward biased volatge always less than beta vsat voltage at non inverting node?
@@sumitkumarpal2306 Because forward bias = 0.7 volt. So Vsat*(R2/(R2+R1)) must be larger than 0.7 volts. R2 would have to be very small to have less than 0.7 volts across it. Good Question.
Which softwear ur using for making such writings? Diagrams?
🙏🙏🙏 thank you so much sir🙏🙏🙏
Exactly this is the video what I need. A voice of guy from tamil nadu
Ayin nammal enth venam
Best channel on UA-cam about electronics
best channel for electronics engineer
And electrical
Has barely covered all the topics tho
Thank you very much for this lecture series on opamp. They are very clear and much useful.
you are a very clever guy, beautifully explained, thankyou, Paul Clements, electronics designer
Obisly i understand.... Dude... Help me a lot.... Just watching your videos & putting the equipment on bread board & see the results & calculate my own one & modify... Awesome
Nicely explained.
You are a great teacher
Best videos available on youtube for electronics
so clear and great explanation
Good explanation of monostable multivibrators. Thank you for sharing !
Proper vedio to understand the process.... Thanks sirji
Very good explained.. thank you..!
Sir U are best in electronics
Very Insightful lecture... Sir
thanks sir greeting from nepal
sir upload more lecture yours lecture are awesome
Excellent presentation. Thank you very much.
Muchas gracias, es justamente la explicación que estaba buscando.
Thanks you so much for the effort. Really helpful 👍
Just Great 👍
Suggestion : Make a video of Bistable Multivibreter using Op-Amp.
just excellent. from ur video, I made 5 pages of detailed explanation of monostable multivibrator. very interesting!
Could you simulate a design using NI MultiSIM?
Will you please do a video on bistable multivibrator... It will be very very helpful for all the students.
Great thank you sir.i don't know how to express it
you would have the best electronics channel if you make a practical examples anyway thanks for sharing and consider making a real examples
Very good presentation sir, Please tell me how to modify this circuit so that it can accept a positive trigger signal instead of negative
Just use inverter circuit. you can even use transistor or op-amp to invert the triggering signal.
@@ALLABOUTELECTRONICS Thankyou Sir, Is there any relation between the unstable state time and the trigger pulse time?
@@sreeharibalachandran Trigger pulse should be less than the unstable time. It should be sufficient enough so that the output can switch from the positive to negative saturation voltage. (and that will depend on the slew rate of the op-amp). For fast op-amps, the switching will occur in less than 1us. For op-amp like 741it may take 2o us to go from 10 to -10V. So, roughly it will be 20-25 us. So, trigger pulse should remain low for that duration.
For fast op-amps, even 1us pulse width is enough for the transition.
I hope it will help you.
Thank you sir
Can the astable pulse be a time period that is less than the trigger pulse? Example: i press the button for 20 milliseconds. The astable state switches on for
10:52 *Why we should eliminate the positive spike* ? I didn't get it...
Please anyone response
Here, the monostable is getting triggered by the falling edge.
Why does the diode become forward biased when the Vout=Vsat and reverse biased when Vout=-Vsat? Cause if we use Kirchhoff on the external mesh, we don’t consider the voltage across R?
The op amp circuit and trigger circuit are both good.
Add them together like this and the monostable can't work, as the voltage divider on the output is compromised.
This combination seems to be ubiquitous on the internet, however it's useless.
best video...thanks Sir
Great video. But there is no video from you on bistable multivibrator using op amps. Any plans?
Dude u r d best 😉
Sir, I always watch ur series on electronics and find quite easy to understand but this explanation was a messed up! And quite fast.Still ur videos makes easy to work out!..
sir could you pls explain how those spikes were coming while using the differentiator circuit
becouse when we differentiaate a constent we get a 0 voltage and the terminals give a valu becouse is a unit step function
Thanku very much 🙏
Bro make a video on "concept of nodal analysis"
Thanks for grate Explanation! how to do this job with single supply op amp ????? Any suggestion?
Very good work
please make a class of transistors and mosfet
I have already started the new series on it. Very soon you will get the videos on JFET and MOSFET.
Please make video on Transistor biasing and amplification both BJT and MOSFET.
i have a doubt regarding ...why we used a negative trigger why not positive trigger?...if yes wt are the problems we may face..
Thankyou sir
7:54. .. why cant we take Vfinal = +Vsat ... please reply...!???
Because here we have assumed that, the output of the op-amp is +vaar initially, and when transition occurs then it goes to - Vaar. That means capacitor starts charging towards the -Vsat Voltage.
I hope it will clear your doubt.
wooooooooooooo ........sir ji nice
amazing man
what is the use of diode in parallel with capacitor? pls........ rply to my question it is my doubt
The diode will ensure that the voltage at the inverting node is less the input voltage. Because normally the Vin is high voltage (Vcc) and during the triggering action only it will go low. So, until the trigger voltage goes low, the voltage at the non-inverting node will be more than the inverting node. Hence, the output of the op-amp will be high during that time.
I hope it will clear your doubt.
thanks for clarification........
Why are we getting (- BVsat) for a time period of t by using momentary negative trigger?
By using - ve trigger, the Vid will be negative momentarily, but how are we getting the unstable state for a long time? does capacitor have any role? please explain.
Yes, the capacitor does play a role. And if you see the time duration T, it depends on R, C, and R1, R2.
As T= RC ln (1+R2/R1)
Please check the derivation at 6:34. You will get an idea about the working of the circuit.
11
Koppana veera GANESH
😀😍😍😍😎😎
Ok
Sir, why do we actually need that trigger circuit as the Vin itself is a short span signal?
Yes, here Vin itself is a trigger signal.
What is max frequency i can reach whith this system? For example, can i make a LEDONn for 1 microsecond and OFF 1 microsecond ?
It is possible to do that using an astable multivibrator.
But if you are designing it using the op-amp, then you need to check the slew rate and the gain-bandwidth product of the op-amp.
I mean it should be able to support that rapid transitions.
Awesome
Amazing
❤
Sir,
How to make -vsat as the stable state instead of +vsat. Does the diode reversal suffice?
Yes it will be enough. If you are using diode in differentiator circuit for taking trigger signal input, you have to reverse it as well.
thank youuuuuuuuuuuuuuuuuuuuuuuuuuuuu
Not discussed about recovery time
which software do you use for making tutorials ?? Please tell me..
The op amp man
Can I use a single Vcc = +3V & 0V - confused by +V & -V supply to op-amp - so that Vsat = Vcc & -Vsat = 0V
yes, it can be used.
why we move towards monostable multivibrator from a stable multivibrator??
is there any use of its using???
The thing is, the different multivibrators are used for the different applications. The monostable multivibrator can be used to generate the pulse width of the finite duration based on the triggering instance.
For example, let's say you want to design the alarm system, which rings or shows red light when particular even occurs.
The monostable multivibrator can be used in such project.
This is just simple example, but there are many other use of this multivibrator.
I hope it will clear your doubt.
Why Vfinal is -Vsat and not Beta times-Vsat.Please explain Sir.
Because the capacitor is actually charging towards the -Vsat voltage. But due to the comparator, you will see a transition at -Beta times Vsat voltage. If that transition is not happening at -B Vsat then the capacitor would have charged up to -Vsat voltage.
I hope it will clear your doubt.
TQ sir
Why only negative spikes are used at non inverting?
Because we want to change the output of the op-amp from +Vsat to -Vsat. That will happen when the voltage at the inverting node is more than non-inverting node. And to make that happen, we need to momentarily reduce the voltage at the non-inverting terminal. When the negative spike is applied then only, momentarily the voltage at the non-inverting terminal will go below inverting terminal. And due that, we will see a transition in the output voltage. I hope, it will clear your doubt.
want the unstable output to be a minimum of 10 sec; Vcc = +3V; using cmos op-amp; using C=10exp-6 F, R = 14.4M ohm
If Vcc = 3V, +Vsat = 3V, & -Vsat = 0V
Is R = 14.4M ohm too large?
Instead of going for large value of R, you can change the value of C. I mean just use 100 uF instead of 10 uF. And then you can use 2M ohm potentiometer.
Why the period of trigger pulse is made small? Pls explain
It has to be smaller than the pulse it creates at the output or the output will be continually low, I believe.
Ap consi class me parte ho?
How to choose the value of Rd and Cd
Hey, make a video on multivibrator.
Thankoo sir
I am an advanced member of this channel, where to access the notes for op amps and vibrators??
Check the membership tab. I have already replied to your comment.
Waiting for 555
V nice
Sir can you explain then multivibrator using transistors
Yes, recently I got so many requests for that. Very soon I will make the video on it.
@@ALLABOUTELECTRONICS yes sir because in our banaras hindu university it is based on using transistor and in practical I unable to understand if you provide lecture as recent as you can it will so helpful for us thanku 🙏💕sir
Please explain triangular wave generator Tomorrow I have exam
Integrator cascade with astable multivibrator... Btw how was the ExM..?
@@bhuvaneshs.k638 I did well
Me too bro
La simulazione del flusso eletrico perché non la fate vedere sul simbolo circuitale
Pls give me formula to calculate pulse width in monostable multivibrator
It’s already explained in the video. Please check at 6:16. The timestamps are also given in the description.
@@ALLABOUTELECTRONICS Actually sorry for this que . I got confused between time and pulse width. After wards, my doubt get cleared.
nice
bro bistable multivibrator using opamp
Cc: t series
i dont understand sir
What part in the video you didn’t understand?
First like
explation is not in detail
Waiting for 555