Don't watch this video if your tired. I had to go through it three times especially for the non-inverting Schmitt Trigger. Has a lot of math associated with it for such a simple device. Great video, I feel I understand it much better and now can design with it.
As a professional electronics designer with decades of deep experience.... In my eyes the video misses to explain expectable problems. For example I woder why they use an OPAMP because not every OPAMP can do this. Some OPAMPs my burn (violating diff_V_IN), some OPAMPS may create crazy output (phase reversal). If you want saturated output it´s always the safe way to use a comparator (instead of OPAMP). There is a good reason why there are OPAMPS and COMPARATORS, they are not the same thing.
Thank You Sir for providing such a great content, we have our exams next week and your videos lectures really helped me in clearing all of my doubts for the subject, once again thanks a lot for posting such informative content
The timestamps for the different topics covered in the video is given below: 0:28 Limitation of the comparator circuit 2:00 What is Schmitt Trigger and how it works? 4:15 Hysteresis Curve of Inverting and Non-Inverting Schmitt Trigger 8:00 Design of Inverting Schmitt Trigger (with Derivation) 12:50 Design of Non-Inverting Schmitt Trigger (with Derivation) 18:30 Application of Schmitt Trigger
@@abaundwal V(high) can be sometimes equal to Vcc. To be more specific, V(high) is the Saturation voltage when a positive input (V+ - V-) is given to op-amp. However, many a time, Saturation voltage may be equal to Vcc or slightly less than it.
No, because it's applicable only when the opamp is used with negative feedback. In Schmitt trigger it is used in positive feedback configuration. So opamp is operating in the saturation. That's why it's not applicable.
Vut=Vref+(Vref-Vlow)*R1/R2 Vlt=Vref+(Vref-Vhigh)*R1/R2 Is the formula of upper and lower threshold voltage correct for Non-inverting scimitt trigger with reference voltage?
Here we are using positive feedback, because we want the opamp in the saturation. That means it has only two output states. + Vsat and - Vsat. And the virtual ground concept is not valid the the positive feedback. It is only valid with the negative feedback, when the opamp is operating in the linear region.
Hello, Would You explain to me, how did I get shift of transfer characteristics, for example: VH=2V, VL=-3V (not -2,5 and 2,5 as usual)? How is this possible? I've been using Hall sensor with Schmitt trigger for multipole magnetic encoder and on the output I got signal with 1/3 duty cycle, instead of 1/2 duty cycle (poles are equal, but reverse in sign). Does transfer characteristics shift depend on Vref? How did we get Vref - I mean once per measurement (on the every begining of each multipole encoder measurement it takes valuse dependent on first pole it measured and it is constant through whole measurement) or every time we cross VL or VH threshold? Does characteristics shift constant through whole measurement (if it is) or does it change every time we cross VL or Vh threshold? I would be very greatful for Your response!
AT 2:13 you have the upper thresh hold voltage sending output from low to high but that is only true for the non-inverting ST. For the inverting ST the upper thresh hold voltage signals the point where the output goes from high to low.
00:12 Schmitt Trigger is a comparator with hysteresis 02:32 Schmitt Trigger provides noise immunity over a band 04:57 Schmitt Trigger has a transfer characteristic curve known as the hysteresis curve. 07:32 Learn how to design inverting and non-inverting Schmitt Triggers using op-amp. 09:56 Inverting Schmitt Trigger has upper and lower threshold voltages 12:32 Designing Schmitt Triggers using Op-Amp 15:12 The upper and lower threshold voltages for a non-inverting Schmitt Trigger 18:01 Schmitt Trigger can be used as a level comparator, wave shaping circuit, and oscillator. Crafted by Merlin AI.
I have already started the series on it. Please check the latest videos. Or you can check this playlist. More will be uploaded in coming weeks. th-cam.com/play/PLwjK_iyK4LLArUHRm3SvPLT0XWlVhpl4h.html
00:15 Schmitt Trigger is a comparator with hysteresis. 02:38 Schmitt Trigger determines output based on input threshold voltage 05:01 Schmitt Trigger behavior explained 07:27 Inverting Schmitt Trigger's transfer characteristic curve can be shifted by providing external reference voltage. 09:56 Schmitt Trigger sets threshold voltages for input signals 12:34 Designing inverting and non-inverting Schmitt Triggers using Op-Amp 15:13 Schmitt Trigger has upper and lower threshold voltages. 18:02 Schmitt Trigger has applications in level comparison, analog to digital conversion, wave shaping, and multivibrator and oscillator design.
Even with the positive feedback the output can't go beyond the supply voltage isn't it ?? And that's why the output gets saturated at the supply voltage or the Vsat of that op-amp. I hope, it will clear your doubt.
@@ALLABOUTELECTRONICS yeah but the question is that why do we see saturation voltage at the ooutput? why dont we see less output voltages? using negative feedback or positive feedback whatever using closed loops we can set gain. so what if the gain is less?
For example my +vsat =10 volts if R1=R2 then upper thershold voltage is 5 volts when voltage at inverting terminal is more than upper thershold voltage we will get -vsat then voltage at non inverting terminal is -5 then voltage at inverting terminal should be less than lower thershold voltage right if we calculate this with gain=100 then vout =A(v^+ - v^-)
At 10:07 you said to take Vout = Vh (Vh is high output when V+ >Vin)... On the next equation you asked so consider Vin > V1. TO get VLow _Doubt_ : _We have already considerd V+>Vin to obtain Vout=Vh and named it as V1 now how can you again compare Vin and V1 to get VL. And how is Vlow consider as Vut? Please sir I'm stuck badly in this confusion_
For understanding it Initial Assumption was made :- Vin < V + , then Vout = VH Now, the output will remain VH till Vin < V . This is considered as V1. Comment :- "On the next equation you asked so consider Vin > V1. To get VLow" . Yes because unless the Vin doesn't goes less than V1, the output will remain LOW only. We need to get the VH as output as well. Now when Vin > V1 , then the output becomes VL ( Low). Case 1 :- Voltage is set at VH > Vin . The output is low. (V1) When Vin < VH the Vout is becoming HIGH.
At 1.32 in the video it's shown that the output voltage becomes zero if the inverting voltage is higer than non-inverting voltage...but you are giving -Vs supply...so the output should be a negatve value(probably -Vs) at the moment...
In non inverting case if Vout is low then V^+ is less than 0 ie V^+=r2/(r1+r2) + r1/(r1+r2)*Vl not Vh.........can u explain why u wrote Vh instead of Vl
I've not understood why (physical why), the behavior of the positive feedback is so different from the negative one, looking at the negative one it is easy to see that V- follows the V+ and doing the KVL across the inverting branch and negative feedback the result will be Vout negative if Vin positive and the opposite if Vin is negative... obviously i cannot do the same with the positive feedback but i don't understand why
Please make yourself extra clear in deriving the expressions for the lower and upper threshold voltages for the two Schmitt trigger configurations, you assumed some values for V_out in the analysis and that point onwards the further analysis (in both the configurations) gets really mushy.Please redo that part.
When the op-amp is used as a Schmitt trigger then it is used in the saturation region of operation. Meaning that, the output of the op-amp is either + Vsat and +Vsat. (Vsat is typically little less than biasing voltages of the op-amp). When the op-amp is used with the negative feedback, then it operates in the linear region, meaning that the output is linear function of input. (e.g V0 = AcL x Vin). But with the positive feedback, when we are using the op-amp in the saturation region, then irrespective of the value of the input signal, the output of the op-amp gets saturated. (either to + Vsat or -Vsat). The Schmitt trigger is intended to work like that, like a comparator but with some hysteresis. I hope, it will clear your doubt.
@@ALLABOUTELECTRONICS means when we use negative feedback , then our vout will be lower than the vin, and we don't want that, we want our output to be greater than vin so that it will saturates, right?
You mean to say like inverting configuration ?? The thing is Schmitt trigger is operated with positive feedback. That means based on input, the output is either + Vsat or - Vsat.
Because the concept of virtual ground is applicable only when the op-amp is operated in the linear region. ( During the negative feedback). In Schmitt Trigger, there is a positive feedback. Means op-amp is operating in the saturation region. So, we can't apply the concept of virtual ground. I hope it will clear your doubt.
In Schmitt Trigger, the op-amp operates in the positive feedback configuration. So, when the input at one of the input is more than the other input then the output of the op-amp becomes high or low. (+Vsat or - Vsat). So, in a way, based on the input at the two terminals, there is only two outputs. (+Vsat or -Vsat) And in this way, it provides the digital output. I hope it will clear your doubt.
It has been assumed that the positive and the negative supplies of the op-amp are equal and opposite in polarity. And usually, it used to be the case for the op-amp. that's why it is symmetric about the y-axis. I hope it will clear your doubt.
How can we justify that by connecting non inverting terminal to output will cause +fb & with this, concept of virtual ground becomes invalid at this situation. Please justify or cover in some video lecture. Thanks 😊
At 9:41 if you see, then both I1 and I2 are outgoing currents. That's why, as per the KCL, the sum of all the outgoing currents should be zero. (There is no incoming current). I hope, it will clear your doubt.
Don't watch this video if your tired. I had to go through it three times especially for the non-inverting Schmitt Trigger. Has a lot of math associated with it for such a simple device. Great video, I feel I understand it much better and now can design with it.
bro it's trivial en tabarnak juste un diviseur de potentiel pis that's it
Yeah me too had a tough time understanding it😅
You underestimate my power to study at the last minute at 4 am in the morning. I'm very tired...
@@vaishrox66 you're extrovert and lonely, how?
Same u know, I don't get it, 5 times yesterday night, just 10min today morning, I understood very clearly🥹
I wonder why this channel has 237k subscribers only!!!
He deserves much more
Love your explanation Sir❣❣❣
Yes Anik, he deserves more subscribers....
Only 1 percent engineering student watch this channel .
As a professional electronics designer with decades of deep experience.... In my eyes the video misses to explain expectable problems.
For example I woder why they use an OPAMP because not every OPAMP can do this. Some OPAMPs my burn (violating diff_V_IN), some OPAMPS may create crazy output (phase reversal). If you want saturated output it´s always the safe way to use a comparator (instead of OPAMP). There is a good reason why there are OPAMPS and COMPARATORS, they are not the same thing.
@@klaus2t703 chup hai hume samaj mai aa rha ha dimag mt laga
@@klaus2t703 Bro, I want to have knowledge like you. Do you have any suggestions on what to do? Or any YT channel that tells stuff like this?
This Man..... Is Genious !!! Thank you sir poora youtube dekha, aapke itna best kisine nahi samjhaya... watching it 2 days before exam...
He has 6Lakhs subs now
All praises to you Sir!
you saved me for my Semester
You make me understand something that my proff can't explain it for me,
Yes bro! Better electrical engineers are made from TH-cam rather than universities! 😬
You are a good Amplifier when it comes to simplifying
Thank you for your very clear explanation sir.
Watching from Madagascar
I come back here whenever I find trouble in transfer curve. All of my doubts gets solved. Thank you ❤️
Good method and teaching. I wish you success as lecturer and best of luck in your profession.
great content! It really helps me a lot when I am learning this stuff all by myself.
me too.this is really helpfull for self study students
Thank You Sir for providing such a great content, we have our exams next week and your videos lectures really helped me in clearing all of my doubts for the subject, once again thanks a lot for posting such informative content
You should get a noble prize in teaching. thanks
Exactly....
Extremely....... Clear explanation sir......
I can't derive the expression of Schmitt trigger from book.( Coughlin)
Finally I found here.
Thanks
thank you sir for giving such a clear explanation you are superb
Too good expained.. Thank you.. 🎉🎉🎉
10:21 : you said Vin > V1 then output will be Vut
_Doubt_ : What when Vin < V1 ? How will be the output?
It's an amazing lecture. It clears all my doubts . But I have request for you can you provide written notes on opamp
bsdk sab yehi kar de tu khali laar chaat
Extremely clear explanation. Love it.
Schmitt Trigger using Op-Amp well explained !
We want more topics sir.You starts from the basic..
The timestamps for the different topics covered in the video is given below:
0:28 Limitation of the comparator circuit
2:00 What is Schmitt Trigger and how it works?
4:15 Hysteresis Curve of Inverting and Non-Inverting Schmitt Trigger
8:00 Design of Inverting Schmitt Trigger (with Derivation)
12:50 Design of Non-Inverting Schmitt Trigger (with Derivation)
18:30 Application of Schmitt Trigger
Please tell
Hi. I have a doubt. What determines V (high) and V (low)? Are they equal to +Vcc and -Vcc respectively?
@@abaundwal V(high) can be sometimes equal to Vcc. To be more specific, V(high) is the Saturation voltage when a positive input (V+ - V-) is given to op-amp. However, many a time, Saturation voltage may be equal to Vcc or slightly less than it.
Excellenttttttttt........Explanationnnnnnnnnn..
thaaaaank you, better than my uni teacher
Very helpful video, this helped me get through my finals, thanks a lot!
Thanks! Really helpful😮😊
Thanks for the video. I have one doubt. Will the concept of virtual ground not applicable in Schmitt trigger? Kindly reply.
No, because it's applicable only when the opamp is used with negative feedback. In Schmitt trigger it is used in positive feedback configuration. So opamp is operating in the saturation. That's why it's not applicable.
@@ALLABOUTELECTRONICS thanks for your prompt reply. Sir, what is the reason that virtual ground is only applicable in negative feedback?
Amazing as usual
Your explanation was awesome 🤗. Thank you
Schmitt trigger is a ___ voltage comparator.(+ve,-ve,both,none) sir what is the answer of this
Vut=Vref+(Vref-Vlow)*R1/R2
Vlt=Vref+(Vref-Vhigh)*R1/R2
Is the formula of upper and lower threshold voltage correct for Non-inverting scimitt trigger with reference voltage?
For the inverting Schmitt trigger is VL and VH the voltage supplied the op-amp as in VCC
yes, both for inverting and non inverting
THANKS FOR YOUR EFFORT SIR
You are great. Clear explanation.
you are the best
Sir I have some doubts
Why we use positive feedback?
Does virtual ground concept act here too?
How can we fix a specific voltage as UTV and LTV?
Here we are using positive feedback, because we want the opamp in the saturation. That means it has only two output states. + Vsat and - Vsat.
And the virtual ground concept is not valid the the positive feedback. It is only valid with the negative feedback, when the opamp is operating in the linear region.
@@ALLABOUTELECTRONICS thank you very much for your valuable reply sir 🙏
Very good presentation. Thankyou very much!
Half an hour before exam💀
Many thanks for the video :)
really very good and simple explanation
Is it possible to detect Reverse and Forward flow of IR using this schmitt trigger ?
Which software is used to make circuit diagram and graphs 📊
Sir can u pls make a dedicated video on hysterisis ?
excellent appriciated,wellexplained.liked vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv much
and watch out when applying reference voltage because it also has a factor attached to it
thanx again
Hello,
Would You explain to me, how did I get shift of transfer characteristics, for example: VH=2V, VL=-3V (not -2,5 and 2,5 as usual)? How is this possible?
I've been using Hall sensor with Schmitt trigger for multipole magnetic encoder and on the output I got signal with 1/3 duty cycle, instead of 1/2 duty cycle (poles are equal, but reverse in sign).
Does transfer characteristics shift depend on Vref? How did we get Vref - I mean once per measurement (on the every begining of each multipole encoder measurement it takes valuse dependent on first pole it measured and it is constant through whole measurement) or every time we cross VL or VH threshold? Does characteristics shift constant through whole measurement (if it is) or does it change every time we cross VL or Vh threshold? I would be very greatful for Your response!
Explanation is very nice..thanq sir
But the voicenotes are disturbing the video during explaination 😊
Easy to learn 👍
Thanks a lot
great video bro
Sir in non inverting st you say ki v1>0 then output is high similarly when v2
Dear Ashish, something is missing ....
Hi.I like your video very much. It's really great. I'll keep an eye on your channel. I am your fan and I will support you.
Yes, friend you said it right....
Super video
Very useful
Can you upload "555 timer as Schmitt trigger" topic please?
Awaiting
Hello friend, in my channel, these videos are available...
AT 2:13 you have the upper thresh hold voltage sending output from low to high but that is only true for the non-inverting ST. For the inverting ST the upper thresh hold voltage signals the point where the output goes from high to low.
indeed !
Thanku so much 4 ur guidance
00:12 Schmitt Trigger is a comparator with hysteresis
02:32 Schmitt Trigger provides noise immunity over a band
04:57 Schmitt Trigger has a transfer characteristic curve known as the hysteresis curve.
07:32 Learn how to design inverting and non-inverting Schmitt Triggers using op-amp.
09:56 Inverting Schmitt Trigger has upper and lower threshold voltages
12:32 Designing Schmitt Triggers using Op-Amp
15:12 The upper and lower threshold voltages for a non-inverting Schmitt Trigger
18:01 Schmitt Trigger can be used as a level comparator, wave shaping circuit, and oscillator.
Crafted by Merlin AI.
Thanks its so clear
Very good video
Yes Deepty, really good video....
Thank you sir, great video
Pls upload videos on Communication systems
I have already started the series on it. Please check the latest videos. Or you can check this playlist. More will be uploaded in coming weeks.
th-cam.com/play/PLwjK_iyK4LLArUHRm3SvPLT0XWlVhpl4h.html
Sir may you please explain what is noise immunity here
00:15 Schmitt Trigger is a comparator with hysteresis.
02:38 Schmitt Trigger determines output based on input threshold voltage
05:01 Schmitt Trigger behavior explained
07:27 Inverting Schmitt Trigger's transfer characteristic curve can be shifted by providing external reference voltage.
09:56 Schmitt Trigger sets threshold voltages for input signals
12:34 Designing inverting and non-inverting Schmitt Triggers using Op-Amp
15:13 Schmitt Trigger has upper and lower threshold voltages.
18:02 Schmitt Trigger has applications in level comparison, analog to digital conversion, wave shaping, and multivibrator and oscillator design.
if it is positive feedback how the output doesnt get infinite with time?
Even with the positive feedback the output can't go beyond the supply voltage isn't it ??
And that's why the output gets saturated at the supply voltage or the Vsat of that op-amp.
I hope, it will clear your doubt.
@@ALLABOUTELECTRONICS yeah but the question is that why do we see saturation voltage at the ooutput? why dont we see less output voltages? using negative feedback or positive feedback whatever using closed loops we can set gain. so what if the gain is less?
@@Siamg-b4b when the positive feedback is more than negative feedback then eventually, the output will get saturated.
@@ALLABOUTELECTRONICS got it. thanks
For example my +vsat =10 volts if R1=R2 then upper thershold voltage is 5 volts when voltage at inverting terminal is more than upper thershold voltage we will get -vsat then voltage at non inverting terminal is -5 then voltage at inverting terminal should be less than lower thershold voltage right if we calculate this with gain=100 then vout =A(v^+ - v^-)
Thank you so much sir got it please make vedio on examples numaricals of s.trigger
Yes, very soon I will make it.
why is V+ assumed to be greater than Vout???
why i take Vo= Vth why we dont assuming Vo =0v
Bro, Vo cannot be zero as the output of Schmitt trigger will be either at +Vsat or at -Vsat.....
At 10:07 you said to take Vout = Vh (Vh is high output when V+ >Vin)... On the next equation you asked so consider Vin > V1. TO get VLow
_Doubt_ : _We have already considerd V+>Vin to obtain Vout=Vh and named it as V1 now how can you again compare Vin and V1 to get VL. And how is Vlow consider as Vut? Please sir I'm stuck badly in this confusion_
Did u get the answer?
For understanding it Initial Assumption was made :- Vin < V + , then Vout = VH
Now, the output will remain VH till Vin < V . This is considered as V1.
Comment :- "On the next equation you asked so consider Vin > V1. To get VLow" . Yes because unless the Vin doesn't goes less than V1, the output will remain LOW only. We need to get the VH as output as well.
Now when Vin > V1 , then the output becomes VL ( Low).
Case 1 :- Voltage is set at VH > Vin . The output is low. (V1)
When Vin < VH the Vout is becoming HIGH.
Plz send the note in pdf or give the link for notes in description
in the beginning when the input crosses lower threshold voltage, why does the output not change accordingly?...just in the beginning?
Great video! , Can you also make a video of Schmitt trigger with BJTs or MOSFETs?
At 1.32 in the video it's shown that the output voltage becomes zero if the inverting voltage is higer than non-inverting voltage...but you are giving -Vs supply...so the output should be a negatve value(probably -Vs) at the moment...
He doesn't say it becomes Zero, he says it becomes low voltage, (what ever the negative supply is)
one word...UYIR
Make an video for schmitt trigger using IC555
In non inverting case if Vout is low then V^+ is less than 0 ie V^+=r2/(r1+r2) + r1/(r1+r2)*Vl not Vh.........can u explain why u wrote Vh instead of Vl
Need to have practical example ,GL
Why you don't use inverting sc hmit trigger circuit? Is it not happen with referrence voltage
I've not understood why (physical why), the behavior of the positive feedback is so different from the negative one, looking at the negative one it is easy to see that V- follows the V+ and doing the KVL across the inverting branch and negative feedback the result will be Vout negative if Vin positive and the opposite if Vin is negative... obviously i cannot do the same with the positive feedback but i don't understand why
the video was great and intresting
Thanks a lot ☺️
@all about electronics schmit trigger is a circuit positive feedback applied to the comparator.is it right?
Hello, Schmitt Trigger is basically a comparator, with positive feedback...
Please make yourself extra clear in deriving the expressions for the lower and upper threshold voltages for the two Schmitt trigger configurations, you assumed some values for V_out in the analysis and that point onwards the further analysis (in both the configurations) gets really mushy.Please redo that part.
Can you write your lecture in a blog so that we can collect it as a note .
Yes, you will find some of the notes on allaboutelectronics.org
thank you very much
17:20
Why negative vh?
very helpful videos sir...
In non inverting u interchange vL and Vh.. While calculating vi
Ur english is like a computer talking , u have to talk with ur hearts
Thank you sir,
But can you provide us with notes
Sir why we have only taken positive feedback i.e non inverting feedback in Schmitt trigger?
When the op-amp is used as a Schmitt trigger then it is used in the saturation region of operation. Meaning that, the output of the op-amp is either + Vsat and +Vsat. (Vsat is typically little less than biasing voltages of the op-amp).
When the op-amp is used with the negative feedback, then it operates in the linear region, meaning that the output is linear function of input. (e.g V0 = AcL x Vin).
But with the positive feedback, when we are using the op-amp in the saturation region, then irrespective of the value of the input signal, the output of the op-amp gets saturated. (either to + Vsat or -Vsat).
The Schmitt trigger is intended to work like that, like a comparator but with some hysteresis. I hope, it will clear your doubt.
@@ALLABOUTELECTRONICS means when we use negative feedback , then our vout will be lower than the vin, and we don't want that, we want our output to be greater than vin so that it will saturates, right?
Can you make a comparator with Vh and Vl are NOT equal?? Perhaps a diode in series with a resistor?
THANKS FOR ALL OF THE GREAT VIDEOS!!!
Yes, definitely. I think I have already covered such example. Please check, opamp solved examples playlist. You will get it.
In case of non inver schmit trgr why we can not put vin at + and feed back resistor on - with ground?
You mean to say like inverting configuration ??
The thing is Schmitt trigger is operated with positive feedback. That means based on input, the output is either + Vsat or - Vsat.
If there is negative feedback, then it's not a Schmitt trigger circuit. I hope it will clear your doubt.
thanks and keep up the good work.
waiting for lectures on transistor models amplifiers and oscillator
Nice
Sir why doesn't we consider the virtual ground thing in the case of non inverting Schmitt trigger
Because the concept of virtual ground is applicable only when the op-amp is operated in the linear region. ( During the negative feedback).
In Schmitt Trigger, there is a positive feedback. Means op-amp is operating in the saturation region.
So, we can't apply the concept of virtual ground.
I hope it will clear your doubt.
ALL ABOUT ELECTRONICS thankyou sir ❤️
How does a Schmitt trigger convert a analog input signal to a digital output signal ? Which part of this video is the answer?
In Schmitt Trigger, the op-amp operates in the positive feedback configuration.
So, when the input at one of the input is more than the other input then the output of the op-amp becomes high or low. (+Vsat or - Vsat).
So, in a way, based on the input at the two terminals, there is only two outputs. (+Vsat or -Vsat)
And in this way, it provides the digital output.
I hope it will clear your doubt.
@@ALLABOUTELECTRONICSNow it is clear to me ! Thank u ALL ABOUT ELECTRONICS for this easy explanation !
How in hysteresis curve,if Vh and Vl equal and opposite in polarity makes hysteresis curve symmetric to y-axis.
It has been assumed that the positive and the negative supplies of the op-amp are equal and opposite in polarity.
And usually, it used to be the case for the op-amp. that's why it is symmetric about the y-axis.
I hope it will clear your doubt.
How did u decide current direction?
Direction of current depends on polarity of output....
mst padhaya ekdm
How can we justify that by connecting non inverting terminal to output will cause +fb & with this, concept of virtual ground becomes invalid at this situation. Please justify or cover in some video lecture. Thanks 😊
because of positive feedback
KCL states sum of the current entering the point equal to sum of current leaving the point , then how that I1+I2=0? at v+
At 9:41 if you see, then both I1 and I2 are outgoing currents. That's why, as per the KCL, the sum of all the outgoing currents should be zero. (There is no incoming current). I hope, it will clear your doubt.
At 13:50 I1 is entering and I2 is leaving. Am I right?
You applying for a point v+.right?
Yes, applying at V+. And here, both the currents are leaving the node. ((V+ - Vin)/ R1, and (V+ - Vo) / R2))