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
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
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.
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,
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).
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.
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.
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 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
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.
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.
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..
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.
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.
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 ?
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.
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.
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.
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.
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?
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.
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.
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.
Are you sure that, in the square waveform output of the astable multivibrator has a T = 50%? I thought that, by definition, T should equal 63.2%. Frank
I think that's because you're feeding Vsat to the capacitor, but comparing it to BVsat. So while it might take it one T to reach 63.2% of the voltage, it will reach BVsat sooner. But definitely don't take my word for it. I'm just a newbie trying to learn. :p
@@elendilion I was just thinking about the definition of time constant of C and L, theoretically. You know a lot more about electronics than I do. Frank
@@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.
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.
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??
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.)
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.
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.
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.
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: ua-cam.com/video/KylJ2v1_c-o/v-deo.html
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.
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.
I think you got confused because in the example I had written R1= R2 =R But what I actually mean is R1 and R2 has the same value, and it is different from the feedback resistor R. I hope it will clear your confusion.
During the discharging process when the capacitor voltage falls below the threshold voltage, the output of the op-amp switches from -Vsat to + Vsat voltage.
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
Ur content is super
Channels like this are very helpful for last time learners😁
++ For first time learners too 😎🔥
congratulation!!! you helped my more with course than the professor!!! thank god ppl like you exist
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.
You explained it far better than my teachers.
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...
The way you teach is very nice sir😊
Greatest hero of the electronics❤❤❤❤
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
Thank you so much, you have no idea how much this helped!
Upload such lectures .. it is very helpful for gate preparation.
You just made my day easy....👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻😀
Thank you so much for explaining in such a lucid manner.
u r lectures helps me a lot
You explain better than my 2nd professor
thank you so much you saved me 😭💖
🥰💜
Same here 😀
It's my turn to say thank you sir.
studying a day before exam crystal cleared concept
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.
Very good video please make a video, using Transistor to make astable multivibrator... please...
EXCELLENT as usual
Superb video. The best I've seen. Thanks
I definitely want to say thank-you for this
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,
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?
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).
@@albertfalck9481 The main reason Is the offset voltaje which creates a voltage difference between v+ and v-
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.
very well explanation.thank you sir
Where is rushal shah sir presently?
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.
I got it
Thank you so much...
maan gaye sir aapko thank you
If i am given a 20% duty cycle what would change in the derivation and the design ?
GOOD. KEEP IT UP.
Thank you! Great lecture!
Marvelous sir
Thank u
There is no lecture on bistable multivibrator using opamp ? If so, can it be made?
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
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.
Sir, what happen in capacitor charge and discharge plz explain sir!
Bro capacitor is charging and discharging interms of beta vsat right then why did we consider upto vsat???
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.
5:50 thank you, my teacher told me to draw and tell how it’s works TT
Thanku sir, very usefull,
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..
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.
Love you bro...!!!! 🔥🔥
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 : ua-cam.com/video/KylJ2v1_c-o/v-deo.html
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 ?
Excellent
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 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.
Thank you sir
Where does astable multivibrator used in ?
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
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
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
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.
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.
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.
Please make video for rrb je exam... Which contains theory and mcq.
Mantap paling OP!
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 🙏🏼
Are you sure that, in the square waveform output of the astable multivibrator has a T = 50%? I thought that, by definition, T should equal 63.2%.
Frank
I think that's because you're feeding Vsat to the capacitor, but comparing it to BVsat. So while it might take it one T to reach 63.2% of the voltage, it will reach BVsat sooner.
But definitely don't take my word for it. I'm just a newbie trying to learn. :p
@@elendilion I was just thinking about the definition of time constant of C and L, theoretically. You know a lot more about electronics than I do.
Frank
graciás hermano !
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, I guess positive feedback will be R1*vsat/(R1+R2)
The feedback voltage is the voltage across the resistor R2. So, it's R2*Vsat / (R1+ R2) (It can be found simply using the voltage divider rule)
@@ALLABOUTELECTRONICS thank you
@Allaboutelectronics..
At 9:02-9:04.. How u took R1=R2=10kohm.. after tuning with potentiometer of 5kohm???
Kindly reply me !!
how to design astable multivibrator for 35% duty cycle?
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.
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?
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
Can you draw Bistable MV using 555 timer and Op Amp.
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: ua-cam.com/video/5clfiJtRhR8/v-deo.html
very useful
Are there any informations about these electronic parts?
They are ideal
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..
How can we derive the Rx and Cx?
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
amazing sir
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 😊
Why there is not an extra resistor in series with the vin .(R1||R2)
R1 and R2 value how 10kohm
U can take any values for r1 and r2. Just keep in mind that they should be equal
include some practice questions(numericals)
It's already there. Please check the analog Electronics playlist on the second channel or Electronics quiz playlist.
You will get it.
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.
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.
thank you sir.
When he says 'beta', does this refer to the r2/(r1+r2) gain of the voltage divider?
Yes, it's feedback factor.
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:
ua-cam.com/video/KylJ2v1_c-o/v-deo.html
Thank you so much!
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
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!
9:00 but R1=R2=R !! right?
I think you got confused because in the example I had written R1= R2 =R But what I actually mean is R1 and R2 has the same value, and it is different from the feedback resistor R.
I hope it will clear your confusion.
Best.
Loved it
Nanri Anna
what is beta? how to calculate
It is the feedback fraction. It defines what fraction of output voltage is feedback to the input. In this case, it is R2* Vout/ R1 + R2.
What will happen when capacitor discharges ?
During the discharging process when the capacitor voltage falls below the threshold voltage, the output of the op-amp switches from -Vsat to + Vsat voltage.
@@ALLABOUTELECTRONICS thanks for replying 😁👍
you are a god!