Op-Amp Integrator (with Derivation and Solved Examples)
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- เผยแพร่เมื่อ 16 ก.ค. 2024
- In this video, op-amp integrator circuit has been discussed (with derivation) and few examples have been solved based on this op-amp integrator circuit.
Op-Amp as Integrator:
In inverting op-amp configuration, by replacing the feedback resistor with a capacitor, it can be used as integrator circuit. The relation between the output and input has been derived in this video.
Limitation of simple integrator circuit:
In this simple integrator circuit, for DC input or for very low-frequency signal the capacitor will act as an open circuit and the input signal will see a very high gain (Open loop gain of the op-amp). So, even if very small DC signal is present at the input, it can lead the output into the saturation.
So, even if your signal does not contain any DC signal, but because of the input offset voltage, the output of the op-amp may get either saturated or distorted.
Practical Integrator Circuit:
The problem of the simple integrator circuit can be overcome by connecting feedback resistor in parallel with the feedback capacitor.
So, because of the feedback resistor, the gain of the circuit for DC signal will get restricted and saturation of the output voltage can be avoided.And the circuit will behave as a low-pass filter.
The condition for proper integration of input signal:
For proper integration of input signal, the frequency of the input signal should be higher than the cut-off frequency. (At least 10 times the cut-off frequency)
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)
The link to the related videos on the op-amp:
Introduction to Operational Amplifier:
• Introduction to Operat...
Inverting Op-Amp:
• Operational Amplifier:...
Non-Inverting Op-Amp:
• Operational Amplifier:...
This video will be helpful to all students of science and engineering in understanding the working of op-amp integrator.
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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)
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.
In last question 17:48 how the o/p can occur if i/p at this duration 0?
how does then gain equal to xc /r at 6:34
@@stratupgeneralstudies2961 Gain of 0dB means Gain of 1 so if Gain drop below 0dB the signal is being attenuated by the amplifier.
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?
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.
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.
Hello professor , the concept was excellent, shame about the graph in the last Ex.
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
U really save us from wrecking up or assignments and exams....
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
Can you please send those notes, It would help a lottt
@@ShivamThakur-it9sg send me ur email id
Most likely both are making a summary from the same book
Can u please share the notes ?
Please share it with me 🙏
Hi, I just wanted to thank you for these amazing videos. The stuff you find on TH-cam 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 !
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
Great job with the video. It helped me to finally understand concepts that I had been struggling with for awhile.
A great explanation for what everyone is looking for.
Thank you.
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.
You are right
You are absolutely right, I have been wasting my time to find my mistake till your comment came into sight
example 4 The answer to the question -0.075V for 0
Yah I got same
I think you are wrong bro..
I should be -2v..
We have to use R value in Vout formula not RF value..
I got the same...
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.
how? 1 to -1 ?
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.
@@arindomphukan bro what is the value at the 1ms to 4ms
@@arindomphukan bro uf you know that ans pls reply it bro
@@sudharsanvenkatesh966 at 1ms it's -1V and then it is constant. So from 1ms to 4ms also the value is -1V
Very nicely explained with all minor details , many doubts cleared Thanks for sharing knowledge!!!
You are an absolute champion mate. Doing the lords work!!
OMG brother you are amazing , with the details you have given it all makes sense now
Beautiful explanation of Op-Amp integrator circuit.
man you deserve more respect and more subscribers
short and sweet lectures with useful ones, please upload more videos related analog electronics with examples. Exams are near by .
I got Vout = 4 V for the last example.
Thank you sir for your video
sir u are really god for me. thank you for putting this channel. really thankful
Explanation is simple and understandable, very helpful, Keep it up....
Sir.. Really tomorrow is my exam.. And we are glad to have the lectures like u .. Thank u so much
Hi, just wanted to say thanks because this really helped me understand my homework!
thank youuu so much this is really helpful for my signal processing electronic class
You are best lecturer. You should teach in any IIT college.
Thank you ! Hope I see more video from you
Hey man besides the obnoxious intro you make some great videos. Also love the subtitles. Thanks man
Great video.Keep up the good work
Thanks this video is very helpful to understand integrator....
Good work here, thanks again.
Made my life a bit easier now, God Bless u
tenkww sir thanks alott....dese videos really helping me alott
today i am not feared about my end term exams thank you soo much
Great explanation thanks a lot
Thank you for examples 🤗😊
Amazing videos... Thank you
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.
I had thought the same
That's ok
But how to determine where the graph is starting
Awesome video! Thank you!
thank you very much! good job
Thanks a lot sir for teach us opamp
Amazing Explanation
this nigga saved my electrical life ily
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?
Best video of integrator
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 ?
Truly amazing 👍👍
Thanks a lot!!
Best demonstrations ever!!!!!!!!
Can you please do LTSpice simulations ? @All ABOUT ELECTRONICS
Zooper class sir.thank you very much
Think u sir boht achha tha lecture aap ka
you are the best
keep it up
👏👏👏👏
Your teaching style is 100 times better than our university teacher
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.
helped a lot. thanks
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 ...
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..
Well presented video.
Well explained sir
Simply great
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?
Nice lecture
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.
I have already covered such videos. Please check the op-amp playlist on the channel.
Fabulous explanation
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.
so good ! thanks
Thanks for the video Sir. I find the input frequency(200hz) less than the cutoff(530hz) so it works as inverter using Rf 400k
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.
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.
ALL ABOUT ELECTRONICS thank you for clarifying. your vedios are really helpful
Thanks Medha for clafication.
i had also this doubt, thanks
Same doubt,hehe.
Thank you so much.
Thank you sir
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.
thank you for the...... video.
Thank you for this amazing video
May allah bless you my brother
Bro it is an periodic function wave it should be symmetrical to the x axis
Voltage varies between -75volt to +75 volt
Thanx sir
Thanks a lot i understand it :)
Very good
what is the relationship between input and output waveform frequencies? Are those frequencies same?
How did you obtain the relation FL=1/(2pi*Rf*Cf)??
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
Thanks a lot sor
Ye voltage swing or Vout me kya diff hai ???
The opening music is as addictive as the theme of scam 1992...or let me correct myself, it is even better than that!
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
Thank you, that's a good suggestion. Yes, in future I will include the results as and when required.
In the practice question, when input is zero, shouldn't the output be also zero?
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
Can you tell me what is op-amp advantage for SC integrator? why we use opamp?
I'm very confuse. Thank you.
Why is V out the same for both the simple RC circuit and practical opamp integration of example 2 solved?
thaaaank you very much
Make video on what is use of integration and derivative in engineering various field ( with detailed)
Awesome vid!!! Is it possible if you could provide the solution of the homework problem? Really need it. :)
Vo= -(1÷(RxCf)) x Vi
Here R= not feedback resistance... Am I right?
Goood explanation, please recommend me books to learn about (digital, radio, analogic) electronics, and telecommunication. Please, I need.
Bless your soul
sir why we didnt use 0,6V as Vinput in the last example
Sir please include the internal block explanation of op amp.🙏🏻
You are a 💎 gem 💎
bro please upload more videos ur explanation ulti
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..
Is the graph between dB and f is straight line or parabolic?
so nice
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