Astable Multivibrator (using op-amp) Explained
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- Опубликовано: 31 авг 2018
- In this video, the astable multivibrator circuit is designed using the op-amp.
By watching this video, you will learn the following topics.
0:12 What is astable multivibrator?
1:10 Brief introduction of Schmitt Trigger circuit
4:19 Working of Astable Multivibrator Circuit
7:13 Design Example of Astable Multivibrator Circuit
9:17 Derivation of time period/frequency for Astable Multivibrator
Astable Multivibrator
In Astable multivibrator, both states are astable state and output used to change continuously between the two states.
When it is designed using the op-amp, the op-amp is used with the positive feedback along with resistor and capacitor.
The charging and discharging of capacitor decides the frequency of oscillation of this astable multivibrator.
The time period T of the output square wave is given by the expression
T= 2RC ln {(1+ β)/ (1- β)} ccc; for 50 % duty cycle
where β is the feedback fraction
In this video, the working of the circuit has been explained and the astable multivibrator is designed for 1 kHz frequency.
And the expression of the frequency is also derived at the end.
This video will be helpful to all the students of science and engineering in understanding the working of astable multivibrator circuit.
#AstableMultivibrator
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The timestamps for the different topics covered in the video:
0:12 What is astable multivibrator?
1:10 Brief introduction of Schmitt Trigger circuit
4:19 Working of Astable Multivibrator Circuit
7:13 Design Example of Astable Multivibrator Circuit
9:17 Derivation of time period/frequency for Astable Multivibrator
Use some Telugu words bro
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Channels like this are very helpful for last time learners😁
++ For first time learners too 😎🔥
Excellent lecture SIR .
Today i got a broad & complete view of this topic.
Please continue this ahead...
Thank you so much sir.. Because of your channel, I could understand analog electronics deeply which was my tough subject before.
Thank you so much for explaining in such a lucid manner.
Thank you so much, you have no idea how much this helped!
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Thank you sooo much for the subtitles!! Perfect explanation, but without subtitles, it would have been so hard to understand you..
Not really...
His English accent is quite easy to understand...
I definitely want to say thank-you for this
EXCELLENT as usual
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Upload such lectures .. it is very helpful for gate preparation.
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Superb video. The best I've seen. Thanks
Thank you! Great lecture!
The way you teach is very nice sir😊
Excellent video. Just did Schmitt triggers in class today and you helped me understand it better. Also I've been looking for a way to design a digital circuit clock of 25MHz. I hope this may help.
You explained it far better than my teachers.
Greatest hero of the electronics❤❤❤❤
Marvelous sir
Thank u
Very good video please make a video, using Transistor to make astable multivibrator... please...
point to be noted: capacitor charges till voltage at inverting terminal is greater than voltage at non inverting terminal.after this it gets discharges till Vc
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Same here 😀
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It's my turn to say thank you sir.
I got it
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amazing sir
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graciás hermano !
Excellent
Thank you sir
You explain better than my 2nd professor
studying a day before exam crystal cleared concept
If i am given a 20% duty cycle what would change in the derivation and the design ?
Loved it
very useful
Initially we assumed R1=R2=R and calculated R=4.5k ohm to design for 1khz. But here suddenly changed R1=R2=10k ohms. How?
What if saturation voltage for op amp is different. like VCC =6
VEE =-5 volts is there any change in the time period of charging and discharging of the capacitor ?
Muito bom explicação
There is no lecture on bistable multivibrator using opamp ? If so, can it be made?
Mantap paling OP!
5:50 thank you, my teacher told me to draw and tell how it’s works TT
tnq sir
TQ sir
Sir, what happen in capacitor charge and discharge plz explain sir!
sir can you please answer? my mind confused for real. 2:25 you said that the open loop gain of opamp can be given as Aol x (v+ - v-) ok but there is a positive feedback so isnt it closed loop? we can't apply the open loop formula for closed loop.
you are a god!
This circuit is mostly analysed to operate as a switching circuit using producing pulses as the high gain of the operational amplifier is not controlled . If the gain of the amplifier is controls to make A*K =1 then this will behave as a very good sinewave generator. All one needs to do is to control the amplifier to give the required gain hence if the feedback loop is 1/2 then the op amp gain should be about 2,
One may analyse it with vector diagram when considered to be working as a sinewave generator. This construction will even give the reason why the vectors will rotate and give the sinewave,
Take the output as being a vector,
Chose a feedback factor along this vector say K= 1/2
Draw he vectors representing the voltages across the capacitor and the charging resistor Vo = Vc + Vr = I/jwV +IR producing the normal semi circle for an c/r circuit fed by Vo.
Draw the vector linking KVo and VC hence Kvo-Vc where this is the floating input to the amplifier
from this draw the output vector A*( KVo= Vc) and this will be seen leading the past Vo showing the rotation of the new output by such a vector diagram. by a full circle or a spiral out till saturation occurs ( a little)
If the amplification A or the power supply is limited then this can be shown on the vector diagram by horizontal lines cutting the circle and the output will be squaring off to produce what was shown in this video.
To use this oscillator as a sine wave generator with only one capacitor to phase shift is a great education and students will see a wider vision of oscillator rather than showing them it working as a square wave generator,
Bro capacitor is charging and discharging interms of beta vsat right then why did we consider upto vsat???
Thank u
IMPORTANT NOTE: There is no explicit input voltage and the capacitor is completely uncharged at startup. The reason why the circuit still gets a voltage is because, at startup, the op-amp sends out a current (because it is necessary for the op-amp to begin operating and for its internal circuits to stabilize).
Nanri Anna
Best.
How can we derive the Rx and Cx?
Where does astable multivibrator used in ?
Why doesn't Vout become 0 as soon as the potential across the capacitor reaches the B times Vsat?
Because here we are assuming that the op-amp works on the dual supply (it has both positive and negative voltage supply). So, whenever it works as either comparator or Schmitt trigger, the output will switch between +Vsat and -Vsat.
That's why when the voltage across the capacitor (or the input at the inverting end) just crosses the B times Vsat voltage, there will be a transition in the output from +Vsat to -Vsat.
What you are saying would have been true if the op-amp operates on the single supply.
@@ALLABOUTELECTRONICS I see, thank you very much for the explanation.
how to design astable multivibrator for 35% duty cycle?
Sir, i just started watching the video for multivibrators and i donot seem to understand coz basically its quitr complex. Can you please direct me to the very basics so i can understand better...thanks..
Can you draw Bistable MV using 555 timer and Op Amp.
I didn't understand the part where in derivation to calculate time period the Vfinal for capacitor voltage is Vsat.
While in actual capacitor charges upto BVsat. So why Vfinal=Vsat?
The capacitor is charging towards the voltage Vsat, but when the voltage reaches BVsat, the output suddenly changes to -Vsat. If that is not happening then the capacitor would have charged up to Vsat. That's why Vfinal = Vsat.
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS This is excellent from u.Thank u..
@@ALLABOUTELECTRONICS but if capacitor would reach Vsat then the reach time would be higher than the time capacitor reaches to B*Vsat. how can we apply the same formula
Please make video for rrb je exam... Which contains theory and mcq.
Thank you sir, for this video.
Can u please also explain how the Capacitor voltage is derived from transient analysis ?? Can u share link for that, please??
Thank you in anticipation.
Here is the link : ruclips.net/video/KylJ2v1_c-o/видео.html
🙏🙏
Why there is not an extra resistor in series with the vin .(R1||R2)
Sir ,i have a doubt. Here in astable multivibrator ckt When + feed back and -ve feed back are present in an op amp,then -ve feeb back is dominant and so we use virtual ground concept?? Is my statement ryt or wrong sir?
It depends, which feedback is dominant. When positive feedback is dominant then op-amp will operate in the saturation region.
Sir , why we choose of value of capacitor in microfaraday scale ..
Why we not use the capacitor of value 1faraday or 10 Faraday??
Because if you go for the larger values of the capacitor then depending on the voltage, the size of the capacitor will also increase.
Also with such large values of the capacitor, if you calculate the value of R, it would be very small. even less than ohm. And even normal copper wires which we are using for the connections have the same order of resistance.
Therefore, the capacitor value is usually chosen in uF.
How can u assume initial output Voltage as +Vsat
Output voltage opamp cannot be exceeded by biasing votlage
When opamp is open loop, even a slight positive differential voltage will cause the opamp to multiply that small positive voltage by it's very large gain, which will make the voltage required very high for the opamp.
Naturally it will saturate.
Shouldn't the voltage across capacitor be A*Vsat instead of B*Vsat(beta*Vsat)?
The capacitor will start charging towards the voltage Vsat. But the voltage at the non-inverting terminal is B*Vsat. So, as soon as the voltage reaches that threshold then there will be a transition in the op-amp output. And now it will start charging towards the -Vsat voltage.
I hope it will clear your doubt.
sir how u get voltage expression for capacitor as
vcc (t)=Vfinal-(Vinitial-Vfinal)e^(-t/RC)
It is the general equation of the capacitor voltage in first order RC circuit.
For more info please check my videos on transient analysis. Perhaps I had given a link of that video in the description.
Anyway, here is the link:
ruclips.net/video/KylJ2v1_c-o/видео.html
Sir, why do we assume that there is no transition from Vsat to -Vsat calculating Vfinal=Vsat?
Why not Vfinal = B*Vsat?
Because after transition, the capacitor starts charging towards vsat or - vsat voltage. That means if the capacitor charges fully then the voltage across the capay would have been Vsat or - Vsat. But due to the comparator, when the voltage across the capacitor reaches either +B*Vsat or -B*Vsat, there is a transition in the output. Therefore, Vfinal = Vsat .
I hope it will clear your doubt.
What changes I have to do to the circuit if I want duty cycle other than 50%?
You can use Zener diodes of different Zener voltages and connect them back to back at the output. With that, you will get different feedback voltages for both positive and negative saturation. And hence, you can alter the duty cycle.
@@ALLABOUTELECTRONICS thank you sir ... I got you👍.
how to make asymmetrical(t1 not equal to t2) astable multivibrator ??
I have the same dought , if u know plz share with us
@@malathit7468 bruh, it's been more than 2 years even if I knew at that time I don't remember now
@@srijansrivastava6834 if u have answer for that question plz inform me
I need that
When he says 'beta', does this refer to the r2/(r1+r2) gain of the voltage divider?
Yes, it's feedback factor.
How can I vary the output amplitude of the square wave?
You can use the back to back Zener diodes at the output to change the output voltage to particular voltage level.
Can i adjust the R value through a pot so that i can get a adjustable frequency?
Yes, you can adjust the frequency with the POT.
@@ALLABOUTELECTRONICS thanku very much
Why during +Vsat the voltage at the inverting terminal become greater than the non inverting terminal? The capacitor charges till +vsat and the positive n inverting terminal of opamp should have same voltage??
And also why is the V final not equal to +beta Vsat in the derivation.. final capacitor voltage is +beta Vsat not +Vsat?
Vfinal is the voltage towards which the capacitor is charging for the given applied voltage. But as soon as it reaches the beta Vsat voltage, the output of the Schmitt trigger will change, which will prevent it from charging towards the +Vsat voltage. (And same is true for the -Vsat voltage)
That's why it is +vsat and not beta Vsat. (Due to the abrupt change in the input voltage to the capacitor at Beta * Vsat, it is not able to charge till Vsat.)
Ok i understand now... Thanks
Are there any informations about these electronic parts?
They are ideal
sir, is it like.. the input voltage varies accordingly to the output voltage?
There is a feedback from output to input. It's closed loop system.
@@ALLABOUTELECTRONICS yes sir. so its the varying voltage of capacitor. so does it mean that its the graph of input voltage?
what are are the main different between astable and monostable multivator?
I have already made a seperate video on that.
Please go through it.
Here is the link: ruclips.net/video/5clfiJtRhR8/видео.html
Sir, how can we amend the circuit to initiate the loop? What if the initial voltage of the output and the capacitance are zero? Thank you.
I mean how to trigger the circuit
@@chantony9318 Typically, the op-amp output can be positive or negative (+Vsat or -Vsat). So, even if capacitor is uncharged, then also at the positive terminal of the op-amp, B*Vsat or -B*Vsat voltage will be there. And with that, the voltage transition at output will start.
Now, there are some op-amps which works with single supply. In that case, the initial output might be zero ideally. And the capacitor can be totally uncharged. But actually some offset voltage at the output will be there. And based on that, few mV signal will be there at the positive terminal of the op-amp. So, that can also start the multivibrator action. Because, in that case, even if capacitor is uncharged, but due to some positive voltage at the + terminal, there will be transition in the output. And then the multivibrator action will continue.
I hope, it will clear your doubt.
Sir can you please tell how beta is 0.5 when both R1 =R2=R
Thank you Sir...
beta is R2/(R1 +R2).
So, when R1=R2=R, beta= 0.5
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS Thanks you Sir
Sir here Why the Vfinal= Vsat. The capacitor charges from -Vsat to +Vsat
Vfinal is the voltage where capacitor supposed to reach in the steady state condition. When the capacitor starts charging from -BVsat voltage, it supposed to charge up to Vsat voltage. That's why Vfinal is equal to Vsat.
I hope it will clear your doubt.
Hello sir..could you please tell when is virtual short not applicable in opamp??
Please tell all the cases where it's not applicable.
The concept of virtual ground is only applicable when op-amp is used in the linear region.
Here, like you can see, the op-amp is operating in the saturation region (Between plus-minus Vsat). So, the concept of virtual ground is not applicable even if it seems to appear, there is a negative feedback. (along with the positive feedback)
@@ALLABOUTELECTRONICS ok but if there is no negative feedback but some positive feedback through a combination of resistors is the virtual short still valid??
No, it is not valid. When there is a positive feedback in the circuit then op-amp will operate in the saturation region and virtual short is not valid in that case.
@@ALLABOUTELECTRONICS ok thanks! You should start teaching microprocessors as well. Your videos are great!
At 5:40 why we assumed that input at inverting terminal is higher than non inverting terminal?
Because the capacitor starts charging towards the +Vsat voltage. The output of the op-amp is +Vsat. Therefore, due to the feedback, the voltage at the positive terminal is B*Vsat. And as soon as the capacitor reaches the B*Vsat voltage, there will be a transition in the output.
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS thank you...now it is cleared
10:36 Vfinal should have been BetaVsat no? Because the capacitor is charging from -betaVsat and +betaVsat
The capacitor is charging towards the Vsat. But because of the comparator, it is not able to charge up to Vsat. Otherwise, it would have charged up to Vsat. That's why Vfinal is Vsat. I hope, it will clear your doubt.
@@ALLABOUTELECTRONICS Thank you sir 🙏🏼
Sir,you told that Vfinal= +Vsat and you told that it is supposed to go to that voltage.But by voltage division rule some of voltage get dropped across feedback resistor.Then how it can be +Vsat.I think
Vfinal = (Xc/(Xc+Rf))(+Vsat).
You can apply that formula, when the input to the RC circuit is AC. Here, we are considering it transient situation, that means when there is a sudden change in the circuit condition. And we are dealing with the DC voltages for that transient condition. (e.g Vsat)
So, consider it as normal RC charging circuit.
Thank you sir
sir you've said that "The capacitor is charging towards the voltage Vsat." but this is not true. because if we look at curcit there is an Resistor above the opamp (potentiometer). so that there will be a voltage drop. it should be Vsat - I*Resistor. so the capacitor will never reach Vsat even if we ignore that cap will charge till B*Vsat.
How vfinal =vsat capacitor won't charge till vsat it charges till beta vsat
The capacitor starts charging towards the Vsat. But before that due to the transition at B vsat , it won't be able to charge till Vsat.
If there is no transition at Bvsat, it would have charged up to Vsat.
That's why Vfinal = Vsat
I hope it will clear your doubt.
I didn't get how the switching happens btw +Vsat & -Vsat in output .Sir ,can u plz give explanation?
Since there is a positive feedback in the configuration, the opamp acts as a comparator. In this configuration, when the voltage at the positive terminal of opamp is more than negative terminal, then output will switch from + Vsat to - Vsat. And vice versa, from - Vsat to + Vsat when V- is greater than V+.
I hope, it will clear your doubt.
@@ALLABOUTELECTRONICS Got it. Thank you sir 😊
R1 and R2 value how 10kohm
U can take any values for r1 and r2. Just keep in mind that they should be equal
Shouldnt the Vfinial be B*Vsat for t1 not Vsat ?
When the capacitor is charging, it is charging towards the Vsat voltage. But because of the comparator, as soon as the capacitor voltage reaches B*Vsat, there is a transition in the output. If there is no comparator in the circuit, then the capacitor would have charged upto Vsat voltage. That is why Vfinal = Vsat. I hope, it will clear your doubt.
Initially we assume R=R1=R2.
From where we get value of R1=R2=10 k ohm???
For 50 % duty cycle, R1 and R2 should be the same. So, as far as R1 and R2 are of the same value, any of R1 and R2 can be selected, (for 50 % duty cycle). With R1 = R2 = 5 k ohm also, feedback fraction β would remain same.
But here 10 k ohm is selected during the design.
while deriving the expression for time period,why have you taken final voltage as Vsat and not as BVsat while charging from -BVsat?
The thing is, the capacitor charges towards Vsat voltage, but as soon as it reaches BVsat, there is a transition, in the op-amp output. If that transition do not occur than the capacitor would have charged upto Vsat. Thats why final voltage is Vsat.
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS thanks sir..
can we generate wave with different duty cycle
Yes, but for that positive and negative reference voltage ( + B*Vsat and - B*Vsat ) needs to be different. You can use diode in the R1 R2 path to achieve that. I mean instead of just one set of R1 and R2, you may have two set of R1 and R2, at the output. In both branch you can have diode in series with those resistor, but with opposite polarity in both branches)
During positive output voltage only one diode will conduct and feedback will be through one set of R1 and R2. During negative output the feedback will be through another set of R1 and R2.
I hope you got the point, what I mean to say.
If have any doubt then let me know here.
i got it..thank u sir
What's the difference between Vsat and betaVsat? Please reply sir
Beta is the feedback factor. That means the fraction of the Vsat which is being sent as feedback. Vast is the saturation voltage of the op-amp.
@@ALLABOUTELECTRONICS thank you sir