Op-Amp Integrator (with Derivation and Solved Examples)

[@ALLABOUTELECTRONICS]

The timestamps for the different topic covered in the video is given below: 0:48 Op-Amp as an integrator (Derivation) 4:32 Output of Integrator for the different input signals 5:54 Limitations of the simple integrator circuit 8:57 Practical Op-Amp integrator 12:08 Example 1 13:10 Example 2 14:51 Example 3 17:15 Example 4 (For Practice)

[@stratupgeneralstudies2961]

Sir at 11:55 how we can go beyond 0 dB frequency? If the gain become 0 then output will be zero means completely attenuated.

[@stratupgeneralstudies2961]

In last question 17:48 how the o/p can occur if i/p at this duration 0?

[@veereshkammara9139]

how does then gain equal to xc /r at 6:34

[@rite2mohit]

@@stratupgeneralstudies2961 Gain of 0dB means Gain of 1 so if Gain drop below 0dB the signal is being attenuated by the amplifier.

[@prabhatp654]

Could you please explain, do we need our junction to be at zero potential before applying KCL there? Or you mentioned it for no purpose at all?

[@vyshnnavipiranavasothy8323]

One thing I've understand, we all are intelligent students, all we need is a good teacher to explain the concepts, and you're one among them. Thank you so much for these amazing videos.

[@anandkulkarni8313]

I am a professor of electronics. While teaching online, I make use of your video clips to make my students understand better. You have done a good job of explaining the concepts. Perhaps a few good every day applications of these circuits will help students to take more interest in basic concepts.

[@johnkyingilisi4395]

Hello professor , the concept was excellent, shame about the graph in the last Ex.

[@pafloxyq]

U really save us from wrecking up or assignments and exams....

[@Tortuga2321]

example 4 The answer to the question -0.075V for 0

[@ashishjoji1742]

Yah I got same

[@eerlasudheer4799]

I think you are wrong bro.. I should be -2v.. We have to use R value in Vout formula not RF value..

[@sivapriya1533]

I got the same...

[@kasunbuddhika3949]

@@eerlasudheer4799 why use Rin value for only this

[@AROHI-h4w]

​@@eerlasudheer4799 I got the same

[@yanndam69]

Hi, I just wanted to thank you for these amazing videos. The stuff you find on KZbin are either too complicated or designed for hobbyists, but you, sir, have found the perfect match. I study automation and we need to understand opamps in order to use them with sensors, and this has proven to be very helpful ; excellently explained with examples and exercices. I've taken a quick look at your other videos and they seem to be just as nice as this one. Here you have my thanks from Algeria !

[@aiyshafayaz6682]

My God I've notes in front of me & each step is same as u say. I think the person has made notes watching ur lectures

[@ShivamThakur-it9sg]

Can you please send those notes, It would help a lottt

[@aiyshafayaz6682]

@@ShivamThakur-it9sg send me ur email id

[@Maceta444]

Most likely both are making a summary from the same book

[@Ji-yoon]

Can u please share the notes ?

[@ujjwalkumar1689]

Please share it with me 🙏

[@alsideekbouzyan1638]

from o to -2v at the period from o to 1ms ,then it keeps the same value of -2v from 1ms to 4ms and from -2v to 0 at the period from 4ms to 5ms

[@GauravMishra-yj5yt]

Thank you sir you explained everything step by step. This made us to understand that what we are doing and why we are doing. Once again a great thanks to you sir

[@moslehuddinahmedmukit5389]

17:05 I think you’ve done a mistake here. The output should start from 0 and at 50uS, it should be at It's peak value -10V.

[@98505177229850590818]

You are right

[@arkogoswami6353]

You are absolutely right, I have been wasting my time to find my mistake till your comment came into sight

[@NadukuruPrasanthi]

Sir.. Really tomorrow is my exam.. And we are glad to have the lectures like u .. Thank u so much

[@rhp1234]

You are best lecturer. You should teach in any IIT college.

[@fasahathussain8976]

Homework solution is 1V to -1Vin 0 to 1 msec and constant from 1 to 4 msec and -1V to 1V in 4 to 5msec.

[@rajeshsinghchad1109]

how? 1 to -1 ?

[@arindomphukan]

Yes. I too found the same answer. Output is 2V therefore 1V to -1V from 0 to 1 ms and -1V to 1V in 4 to 5 ms.

[@sudharsanvenkatesh966]

@@arindomphukan bro what is the value at the 1ms to 4ms

[@sudharsanvenkatesh966]

@@arindomphukan bro uf you know that ans pls reply it bro

[@inrittik]

@@sudharsanvenkatesh966 at 1ms it's -1V and then it is constant. So from 1ms to 4ms also the value is -1V

[@IbrahimKhalilUllah-yq3sv]

man you deserve more respect and more subscribers

[@jaberjeni8084]

OMG brother you are amazing , with the details you have given it all makes sense now

[@ahmadshamim3289]

Regarding the integration example at 14:59, I think to clarify, Vout(t) = -200,000t volts between 0 and 50 us (and each subsequent corresponding high pulse) and Vout is (+200,000t - 10) volts between 50 and 100us (repeating similarly). The -10 volts term is needed because the starting point of Vout is -10 for the second piecewise segment. This is consistent with what you mention at 4:20, where you correctly say that there is a second term to determine Vout, to reflect the starting point of Vout. These two piecewise functions result in a sawtooth for Vout that decreases from 0 to -10 volts and back up to 0 volts and so on. I hope this is helpful. Thanks for the videos. They are great.

[@AM-sk3lm]

I had thought the same

[@mzia1210]

That's ok But how to determine where the graph is starting

[@nathanlumbwe2015]

I got Vout = 4 V for the last example. Thank you sir for your video

[@sushmapaladi6721]

short and sweet lectures with useful ones, please upload more videos related analog electronics with examples. Exams are near by .

[@surajgarg3054]

A great explanation for what everyone is looking for. Thank you.

[@dhirajdhakal5296]

Your teaching style is 100 times better than our university teacher

[@nthumara6288]

thank youuu so much this is really helpful for my signal processing electronic class

[@medhasingh4428]

In the example of square wave ; output voltage after integration was -10v. In the graph it was drawn between +5 volts to -5 volts. But in the question for practice; output voltage comes out to be -2v(0-1ms) and you have drawn it from 0 to -2v. Both seems contradictory to me. Please clarify.

[@ALLABOUTELECTRONICS]

Yes, in the first case, it should have been from 0V to -10V. (Because at time t=0, the output will be zero, considering there is no initial output voltage). For practice example it is alright. Thank you for the correction.

[@medhasingh4428]

ALL ABOUT ELECTRONICS thank you for clarifying. your vedios are really helpful

[@solyomalbazz4925]

Thanks Medha for clafication.

[@deepakjha7189]

i had also this doubt, thanks

[@neeltej4123]

Same doubt,hehe.

[@smitakadam1760]

Very nicely explained with all minor details , many doubts cleared Thanks for sharing knowledge!!!

[@ArhamKhan05]

Thank You So Much Sir. Very Greatly and Beautifully Explained Sir. Love from Pakistan 🇵🇰 ❤.

[@raptornein2422]

Great job with the video. It helped me to finally understand concepts that I had been struggling with for awhile.

[@snehithkumar5442]

today i am not feared about my end term exams thank you soo much

[@rahulroy5206]

Made my life a bit easier now, God Bless u

[@agstechnicalsupport]

Beautiful explanation of Op-Amp integrator circuit.

[@mattg.5437]

Hi, just wanted to say thanks because this really helped me understand my homework!

[@toufiqurrahman9797]

Thank you for this amazing video May allah bless you my brother

[@sushilabhandari9268]

Bro it is an periodic function wave it should be symmetrical to the x axis

[@vatsk]

Hey man besides the obnoxious intro you make some great videos. Also love the subtitles. Thanks man

[@selfbetterment7484]

You are an absolute champion mate. Doing the lords work!!

[@pravatkumar2575]

Sir,in this video in example -3, V output was -10V which was shown +5V to -5V in the graph. But for the last exercise question V output= -2V.How it is shown from 0(origin) to -2V rather +1V to -1V(as like your solved, example.). please tell from where V output will come, from 0V Or from +1V.

[@053_abdulhannanbhat8]

Same doubt i had output will vary from 0-(-10v)for first 50us then -10 -0v for next 50us and so on

[@sowmyaambati938]

sir u are really god for me. thank you for putting this channel. really thankful

[@ionutbuzescu4435]

FL=19.9Hz, fs=200Hz>10*FL, Vout decreases linear from 0 to -2V at 1 ms, remains -2V until 4ms and then rises again from -2V to 0V.

[@potatobits7997]

I think you explained it right.. nice tutorial.. the answer to the last quiz example is 60v? but the peak is 30v right? because an integrator config is 90 degrees out of phase and not a full 180 degrees .. that's why some are confused with your last example which is 10v but the peak is 5v.. +5v to -5v is equal to a 10v swing..

[@Saikumar-kb4lf]

You r the best youtuber for electronics.But it would be so helpful if u can really show the outputs in software like MULTISIM, so that we can understand practically too Thank you, keep going

[@ALLABOUTELECTRONICS]

Thank you, that's a good suggestion. Yes, in future I will include the results as and when required.

[@thegamerdaddycool3384]

you are the best keep it up 👏👏👏👏

[@RahulKumar-xx9or]

The opening music is as addictive as the theme of scam 1992...or let me correct myself, it is even better than that!

[@NehaKumari-bv5gd]

Thank u so much, well explained , and I got the answer of last exercise que that was given for practice 1v to -1v for o to 1 ms -1v for 1 to 4 ms Nd -1 to 1 for 4 to 5ms

[@adityamishra6954]

you got it wrong try to match with the vout graph given in the video

[@ajays6976]

Sir , I have a doubt that at last example u explained that output voltage as peak to peak but i think the vout =-10v will be for one side then total vpp=10v Can you please tell me ?

[@vishalparashar1215]

Voltage varies between -75volt to +75 volt Thanx sir

[@ronakagarwal1810]

In practice problem, during 1ms to 4ms output is constant instead of zero because you said due to the initial voltage right??? But in the case of 4ms to 5ms we didn't take care of initial voltage( -2V), we should add this voltage in the new voltage bcoz we have equation: integration of input voltage plus initial voltage.... so output will become -4V . Why we didn't add ?

[@arindamjain6892]

nah bro it adds up to become zero at 5ms

[@atifsaeed7854]

Hi your videos are very well explained I have learned a lot from them. please make some videos on op-amp configurations along with diodes. Thank you.

[@ALLABOUTELECTRONICS]

I have already covered such videos. Please check the op-amp playlist on the channel.

[@kalanadesilva9425]

Thanks a lot!! Best demonstrations ever!!!!!!!! Can you please do LTSpice simulations ? @All ABOUT ELECTRONICS

[@wkwk2o384ur]

this nigga saved my electrical life ily

[@abrahanshahzad1371]

Glass passivated sensitive gatethyristor in a plastic envelope,intended for use in general purposeswitching and phase controlapplications. This device is intendedto be interfaced directly tomicrocontrollers, logic integreatedcircuits and other low power gatetrigger circuits.

[@hashiska.5358]

at 6:50, you said that practically the op-amp gain is limited by the open-loop gain of the op-amp. What does it mean?

[@nirajprasad6468]

Explanation is simple and understandable, very helpful, Keep it up....

[@myhobbies1288]

In the homework example from 1ms to 4ms, Vo should be Zero... Because integration of 0 = 0 also it is independent of time

[@user-rf5um7wt2l]

but there's v(0+) voltage as he mentioned....so it'll be voltage for 1 ms + 0V so it'll be constant as he had shown in the curve

[@nehaiftikhar645]

Think u sir boht achha tha lecture aap ka

[@kwesigapather6964]

Thanks for the video Sir. I find the input frequency(200hz) less than the cutoff(530hz) so it works as inverter using Rf 400k

[@jagritigarg1499]

tenkww sir thanks alott....dese videos really helping me alott

[@myhobbies1288]

Vo= -(1÷(RxCf)) x Vi Here R= not feedback resistance... Am I right?

[@priyankasiwach5066]

In the example at 16:43 ,we get output voltage swing of -10V but how we know that Vout will be -5V at t=0 (it can also start from 0V and go to -10V)

[@devanshulawania9390]

Yes he has clarify our doubt in other comments. You are right , there is mistake in example but practice sample was right .👍

[@vikrammood6126]

In practice question v(output)=-2V from 0-1micro sec so the graph starts from 1 and decreases to -1 and then it will be constant till 4 micro sec .from 4 micro sec to 5 micro sec it increases to 1 V. sir, is my graph is correct? thanks you sir.

[@satyaveersinghrawat4398]

@ALLABOUTELECTRONICS I have solved practice question. My answers are: (1) for 0 to 1 ms ----- - 2V Swing in voltage (2) for 1ms to 4ms ------ 0 Swing in voltage (3) for 4 to 5 ms ----- 2V Swing in voltage. But I am unable to understand at 17:45 Min why your output waveform starting from origin? should it not be: (1) Stated from +1V and go to -1 V from 0 to 1ms (2) Then stay at -1V for 1ms to 4ms and (3) Again come back to 1 V from -1V during 4ms to 5ms. Please correct me where I am going wrong.

[@zenvir1680]

At 16:43, if we don't put limits we get Vo1= -(constant).t , which means slope of Vout vs t graph will be Vo1/t = -(constant) within 0 to 50 micro sec time. But why do we start our output curve from +5 volts? There will be 0 V at output initially, so the curve should drop from 0V to -10V and then oscillate in triangle wave between zero V and -10 V. Can you please clear my this doubt too? Thanks for clearing pervious doubt in summing video

[@myhobbies1288]

I too have same question... Moreover I get Vo = -100V not -10V

[@tpsicmin]

Amazing Explanation

[@noweare1]

Last example you did for us. I am confused because the equation gives you the slope (voltage change / time change ) not the value of the function at a certain time. For example v(t)=-1/rc * t where t1=50us t2=100us . Plugging in the values you end up with 200,000 * 50us =10 volts so in the voltage changes 10 volts in that 50us period. If you just plug in a discrete time value to the equation you end up with a number that you may think the function is at that time but isn't.

[@sauravroy5195]

Thanks this video is very helpful to understand integrator....

[@carloscardenas2815]

Thanks so much for this explanation. There is one thing that still i don't understand and is: why fs should be 10 times bigger that fl for proper integration?

[@rajatdubey8118]

9:28 gain of practical integrator is(- Zf/R) where Zf= Rf//Cf

[@omkumarsingh4194]

Ye voltage swing or Vout me kya diff hai ???

[@RohanBhardwaj-qb5ph]

in the question where first half cycle is of 2 v and second half is of -2 volt (at 15:52) the output wave form obtained, why is in the positive region also, as in starting we will consider output=0 as input is zero but when the input of 2 v is applied for 50 micro second it should decrease by 10 v, and in negative cycle it will increase by 10 volt, according to this output should not be positive i think its output should be like the last question output except the constant region in that waveform

[@akashpandey850]

Doubt For Example 3: We got Swing of 10 volt as Vout after integrating for time (0 -> 50 us) as -10V (that is our Swing), but how you determined the initial and final point as +5V and -5V ????????????????? Doubt For Example 4: Similiarly. for practice problem I got Swing as -2V, now how to decide ends ??????????

[@prabhu4323]

Actually it is not -10. It is -10t ,function of time after integration. End point will be decided by putting the value of time. I think graph is not correct.

[@moslehuddinahmedmukit5389]

@@prabhu4323 Yes. I was abt to tell this. The output should start from 0 and at 50uS, it should be at It's peak value -10V.

[@aishwaryakamble5811]

Example 4 -2v how? Explain with expression plzz

[@sunildesai5531]

Such nice video. Also pls give Application example , where this integrator ckt is used

[@ALLABOUTELECTRONICS]

It is used when you want to integrate the signal. For example, if you want integrate the current and convert it into voltage then it is useful. This is particularly useful in photo diode based detector circuits for very low light detection. ( Used in bio science applications in fluorescence detection). Apart from that, it is also useful in wave shaping circuits. One such circuit I have already covered in another video. Please check triangular wave generator circuit on the channel.

[@deltonguivala4888]

Goood explanation, please recommend me books to learn about (digital, radio, analogic) electronics, and telecommunication. Please, I need.

[@prabhu4323]

In example 3 .How are you deciding end point. Output voltage is not -10. it is 10t , function of time. So end point should not be -5 to +5 .it will be decided by putting the value of time.

[@moslehuddinahmedmukit5389]

The output should start from 0 and at 50uS, it should be at It's peak value -10V. m i right?

[@lakshmis9905]

Sir please include the internal block explanation of op amp.🙏🏻

[@aindriladey2077]

It would be -1 as only half cycle is working and had it been one full cycle then it would have been -1 to +1 but I didn’t understand the previous graph properly as it starts from top although the output should have started from zero as in case of the second graph

[@98505177229850590818]

It’s incorrect It should start from 0 and goes to -10

[@warisquraishi373]

sir why u write +5v to -5v not 0 to -10v after integration of square wave? please explain .

[@shuvo4344]

should be +10 to -10

[@ahmadshamim3289]

@@shuvo4344 I think it should be from 0 to -10, -10 to 0, 0 to -10, etc.

[@mohammedumarsheriff5801]

The solution to the final problem is incorrect.

[@gireeshachari911]

As per my understanding in first example at t=0 to 50us vout should go from 0 -10V then from 50us to 100us it should be -10+10 i.e go from -10 to 0

[@anishkumargiri9490]

sir in the V(in)=sin(2*pi*5000t) its integration limit will be from 0 to t then the output will be something else because cos0=1 then 1/RCf will be multiplied with it sir

[@tarunbatchu]

in the last example we didn't used the miller integrator so in the second cycle where vi=0 before that capacitor is charged so as vi=0 the capacitor will get discharge its got path to discharge but it will right in miller integrator because it don't have resistor to discharge and and the output equation you used is for miller integrator but question is not an miller integrator

[@SasanamuPraveena]

Fabulous explanation

[@vinamrasangal8436]

And Sir, It will be great If you create one dedicated video regarding "ichimoku trading Strategy" and How can we use this strategy with combination of other strategy so as to begin our trade with multiple confirmation.

[@HimmatSingh-ju8jk]

Sir, in given example of input of +/- 2v of frequency 10kHz has squar wave. The out put of given ckt from 0 to 50us is -10 v comes. Now this output voltage hold till next input. According to given input from 50us to 100us the ckt output become +10 v + previous output that is -10 v so net out should be zero not 10 v. So out put waveform should be triangular of -10v not having any +ve value. Am I right sir? Pls clarify

[@richaphysics]

cut-off Freq in last problem come out to be approx 196Hz, and input frequency is 200Hz. Input signal is just marginally in the integration region. Right??

[@jrohit1110]

yes

[@picturization3914]

In the homework question, may I clarify why the Vout is not zero between 1ms and 4ms given that Vin is zero? The maths seems to suggest Vout=0 between 1ms and 4ms too. In the mathematics of a definite integral of a function, it represents the total area bounded by the function, x-axis and the two vertical lines representing x=1ms and x=4ms. In this homework question, the area the Vout integral is zero.

[@ALLABOUTELECTRONICS]

Capacitor will hold the charge that it has gathered during first 1 ms. So, during 1ms to 4 ms, the voltage across capacitor will remain same even if no input is applied to the capacitor. I hope, it will clear your doubt.

[@picturization3914]

​@@ALLABOUTELECTRONICS Thanks you very much for the response. I think this is a very good question that prompts learners to think in details. And I have three more questions based on your response. 1. Is the measurement of the Vout between Vout point and ground? And is it the same as the voltage across the capacitor which is between Vout and the negative input terminal (virtual ground)? 2. When there is no input from 1ms to 4ms, is the capacitor going to discharge through the 400KΩ resistor connected in parallel? 3. For a periodic on and off input signal, e.g. -1V (0 to 5ms), 0V (5ms to 10ms), -1V (10ms to 15ms), 0V (15ms to 20ms), and so on, is the Vout going to be in triangular wave form? Or is the line going to be flat at where the last signal ended in periods that have no input signal?

[@apoorvsharma6999]

sir this is really an amazing video... you have really helped us a lot ...just one question in the example for square wave input show in the video..if someone asks for the minimum slew rate for the amplifier what would be its value..plz let me know the answer and approach ...

[@shubhamdesai7749]

Thank you for examples 🤗😊

[@caleb_wole]

Output voltage is -2V. Hope I got it?

[@charmonnicolaih649]

Great video.Keep up the good work

[@neshu4044]

In the practice question, when input is zero, shouldn't the output be also zero?

[@manojkumarsahu7929]

Nice lecture

[@SimonYells]

Thank you ! Hope I see more video from you

[@benjaminanane7029]

@ALL ABOUT ELECTRONICS .....You should solve the problem and post the link for us to verify our answer... How do you expect us to know whether if what we have solved is correct or not?

[@ateekkhan1249]

PLS PROVIDE US PPT OF THIS LECTURE THIS WOULD BECOME VERY HELPFUL FOR US AND THANKS FOR PROVIDING SUCH A GREAT LECTURE WITH ALL BASICS..

[@georgethomask9818]

For the practice problem, do we assume that the time period of the input signal is 5 ms? How do we know that the signal frequency lies between 10*Fl and Fo?

[@sain8827]

We can find 1 = 1/(2*pi*f*R*C) and we know R and C. This would be the 0 db frequency. That is Fo

[@chitrasingh9200]

I think fs is not greater than 10fl...10fl is 1.5khz and fs is 1/4ms

[@sreeragsethukumar5104]

Great explanation thanks a lot

[@md.anwerhossain5479]

I have a question : If input signal frequency is greater than cut off frequency but not greater than 10 times the cut off frequency ( Fcut

[@RajeshYadav-mo6zd]

Thanks a lot sir for teach us opamp

[@keukenrol]

Together with the playlist of analog filters, this is very helpful for me! I suppose the instrumentation amplifier and VI convertors (floating and grounded -> howland current pump) will be handled as well? :)

[@ALLABOUTELECTRONICS]

Yes. definitely, it will be covered.

[@leeplatt4038]

Well presented video.

[@prabhakardas4261]

nice one...when will be the next video on opamp?

[@ALLABOUTELECTRONICS]

Very soon. In 2-3 days.

[@gautamkumarsingh2574]

Brother, at 17:06, Does your graph should not start from 0 and move to 10 with a positive slope???

[@moslehuddinahmedmukit5389]

Yes. I was abt to tell this. But the slope should be negative, as Vo=-10. The output should start from 0 and at 50uS, it should be at It's peak value -10V.

[@shubhamkumargupta3478]

just messed up by something in last question i took T=5ms,f=200Hz,fL=39.78Hz,fs=159Hz for integrable it should be between both, Even after that, i took then for 0-1ms:Vout=-2V and for 1-4ms:Vout=0(as no Vin) and for 4-5ms:2V so is it that form -1 to 1 or how to figure out Graph,Help?

[@geraldcanaansohn8235]

@Kumar Gupta .. your answers are similar to mine. To graph, you should use the integrated function on each given intervals and you will see the form of the graph the author hinted at the end of the video.. thanks bro.

[@senthilmurugan8298]

Zooper class sir.thank you very much