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)
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.
Hi, I just wanted to thank you for these amazing videos. The stuff you find on UA-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
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.
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.
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
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.
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.
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 ?
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
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
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..
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.
at 8:56 you said that adding a resistor is helpful.. and afterwards u said that the capacitor will work as an open circuit at 9:09 So if capacitor acts as an open circuit our original goal( of integrating the circuit ) won't happen as there is no capacitor. So, how does the resistor is justified?? Pleas help..
See, when we say it acts an open circuit, it is a steady-state condition. The input offset voltage remains there. It's not changing with time. So, to avoid saturation because of that, the feedback resistor is helpful. But when we use the circuit as an integrator then the input is changing after some time. If we apply a constant DC voltage then the output will be a ramp signal but in a few ms or a few seconds, it will reach saturation. So, the circuit is useful as an integrator when you want to integrate the time-varying signal or the DC signal which is on for some time. Let's say some switch is open for few ms and you want to integrate the output during that time, then its useful. But its not steady-state condition, hence the capacitor won't act as an open circuit. I hope it will clear your doubt.
@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.
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
sorry, maybe I ask it a bit late. At 3:20 , when computing the Vc, why is Vc calculated by 0-vout? Isn't voltage always flow from high potential to low? So Vc should be equal to vout-0?
If you see the current, then capacitor current Ic is going from inverting terminal to output terminal. So, based on the current direction, the inverting terminal is at a higher voltage. (Since current flows from higher potential to lower potential). It is purely based on the direction of the current. If you assume the current through the capacitor is flowing from output to inverting, then you will have (vout - 0 ) in the equation. But then, when you apply KCL at the inverting node, then you will have Iin + ic = 0, OR Iin = -Ic. So, eventually, you will get the same result. I hope, it will clear your doubt.
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 ?
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
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?
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
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 ...
18:06 that output graph which you made sir,,,,,,, the down facing trapezium......won't the right slanting line would be above the x axix as 180deg phase shift will be there??
Since the input is negative, so the slope will positive. But the slope will begin from the previous voltage. If the initial value is negative, then it will start from that value and go towards positive voltage. In this case, at 4 ms, the initial voltage is negative. So, the positive slope will start from that value. I hope, it will clear your doubt.
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.
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?
For inverting op-amp, if Z2 is the impedance in the feedback and Z1 is the impedance at inverting terminal, then Vo = -Z2*Vin/Z1 Here Z2 is Xc and Z1 is R. So, Vo= -Xc*Vin/R = -Vin/(2*pi*f*C*R) I hope it will clear your doubt. I think you got confused because you compared it with the integral equation. And still, if you have any doubt then do let me know here.
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.
If the input is a sinusoidal signal of specific frequency, then after the integration, you will get 1/2*pi*f in the denominator. And 1 / (2*pi*f*C) is Xc. So, from that you will get Xc / R. If the input is sine wave then after the integration, it will become cosine. (there will be phase shift). But if we just consider the amplitude then output amplitude is Xc / R times the input signal amplitude. Also, you can look it another way. For inverting op-amp, Vo = - Rf / R1 x Vin Now, in case of integrator, in the feedback there is a capacitor. Its reactance at the specific frequency is 1 / 2*pi*f*C (or simply Xc). Therefore, Vo = - (Xc / R) * Vin I hope, it will clear your doubt.
Why do you get an initial value different than zero while integrating the sine and square waves at 5:08? Shouldn't these start at zero and look like there was a negative offset? I mean, assuming the capacitor has initial value equal to zero.
Yes, that's true. If all initial conditions are zero and if the waveform is starting at t=0, then integration should start from zero. But the waveform which is shown at 5:08 is more of a practical case, where the signal exists for t
Ok thanks! So for signals that existed for a while there would be no "offset" in practice, but let's say i have no signal (steady 0v), discharged capacitor, ideal op amp, and then start the square wave at t=0. The thing thats bugging me is that i dont see that "offset" being a transient response that would fade away for t->infinity. Maybe I'm not getting the concept of square wave existing since beginning of time, then there would be no transient response. But I mean, in reality, don't you have to power up the circuit at some given real time then do the measure?
The first equation (with integration) is used for finding the transient output of the integrator. Like when you apply a pulse signal and you want to find the output of the integrator. (It can also be used for sinusoidal signals) The second equation is particularly used when the input is sinusoidal signal. If the input is sinusoidal signal then we can find the equivalent impedance offered by the capacitance at the specific operating frequency. (1/wc) So, considering that impedance of the capacitor, the output of the op-amp is (-Xc/R)*Vin. (Similar to -(R2/R1)*Vin) And here the reactance of the capacitor is 1/wc. So, Vo = -1*Vin/(wcR) or -1*Vin/ (2πf*R*C). If we just consider the magnitude then voltage gain |Vo/Vin| = 1/ (2πRfc). I hope, it will clear your doubt.
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 ??????????
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.
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.
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.
U really save us from wrecking up or assignments and exams....
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
Hi, I just wanted to thank you for these amazing videos. The stuff you find on UA-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 !
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...
@@eerlasudheer4799 why use Rin value for only this
@@eerlasudheer4799 I got the same
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 🙏
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
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
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
man you deserve more respect and more subscribers
OMG brother you are amazing , with the details you have given it all makes sense now
Sir.. Really tomorrow is my exam.. And we are glad to have the lectures like u .. Thank u so much
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
I got Vout = 4 V for the last example.
Thank you sir for your video
thank youuu so much this is really helpful for my signal processing electronic class
A great explanation for what everyone is looking for.
Thank you.
You are best lecturer. You should teach in any IIT college.
Great job with the video. It helped me to finally understand concepts that I had been struggling with for awhile.
Very nicely explained with all minor details , many doubts cleared Thanks for sharing knowledge!!!
Your teaching style is 100 times better than our university teacher
Thank You So Much Sir. Very Greatly and Beautifully Explained Sir. Love from Pakistan 🇵🇰 ❤.
Hi, just wanted to say thanks because this really helped me understand my homework!
short and sweet lectures with useful ones, please upload more videos related analog electronics with examples. Exams are near by .
Hey man besides the obnoxious intro you make some great videos. Also love the subtitles. Thanks man
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
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.
You are an absolute champion mate. Doing the lords work!!
Made my life a bit easier now, God Bless u
Beautiful explanation of Op-Amp integrator circuit.
you are the best
keep it up
👏👏👏👏
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
you got it wrong
try to match with the vout graph given in the video
sir u are really god for me. thank you for putting this channel. really thankful
today i am not feared about my end term exams thank you soo much
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?
Explanation is simple and understandable, very helpful, Keep it up....
tenkww sir thanks alott....dese videos really helping me alott
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.
The opening music is as addictive as the theme of scam 1992...or let me correct myself, it is even better than that!
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)
Yes he has clarify our doubt in other comments.
You are right , there is mistake in example but practice sample was right .👍
16:50 How do U know the Vout starts at 5V @ t=0 ?
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.
Think u sir boht achha tha lecture aap ka
Brother, at 17:06, Does your graph should not start from 0 and move to 10 with a positive slope???
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.
Amazing Explanation
Thanks a lot!!
Best demonstrations ever!!!!!!!!
Can you please do LTSpice simulations ? @All ABOUT ELECTRONICS
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 ?
In the homework example from 1ms to 4ms, Vo should be Zero... Because integration of 0 = 0 also it is independent of time
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
Thank you for examples 🤗😊
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.
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
I too have same question... Moreover I get Vo = -100V not -10V
Voltage varies between -75volt to +75 volt
Thanx sir
9:28 gain of practical integrator is(- Zf/R) where Zf= Rf//Cf
Thanks this video is very helpful to understand integrator....
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..
this nigga saved my electrical life ily
Great video.Keep up the good work
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.
Same doubt i had output will vary from 0-(-10v)for first 50us then -10 -0v for next 50us and so on
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.
Nice lecture
Great explanation thanks a lot
For practical Integrator how does the integration equation holds, I couldn't get.. @13:55
Fabulous explanation
Thank you ! Hope I see more video from you
Thanks for the video Sir. I find the input frequency(200hz) less than the cutoff(530hz) so it works as inverter using Rf 400k
sir why u write +5v to -5v not 0 to -10v after integration of square wave? please explain .
should be +10 to -10
@@shuvo4344 I think it should be from 0 to -10, -10 to 0, 0 to -10, etc.
Truly amazing 👍👍
Well presented video.
at 8:56 you said that adding a resistor is helpful.. and afterwards u said that the capacitor will work as an open circuit at 9:09
So if capacitor acts as an open circuit our original goal( of integrating the circuit ) won't happen as there is no capacitor. So, how does the resistor is justified??
Pleas help..
See, when we say it acts an open circuit, it is a steady-state condition. The input offset voltage remains there. It's not changing with time. So, to avoid saturation because of that, the feedback resistor is helpful. But when we use the circuit as an integrator then the input is changing after some time. If we apply a constant DC voltage then the output will be a ramp signal but in a few ms or a few seconds, it will reach saturation.
So, the circuit is useful as an integrator when you want to integrate the time-varying signal or the DC signal which is on for some time.
Let's say some switch is open for few ms and you want to integrate the output during that time, then its useful. But its not steady-state condition, hence the capacitor won't act as an open circuit.
I hope it will clear your doubt.
@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.
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
Best video of integrator
sorry, maybe I ask it a bit late. At 3:20 , when computing the Vc, why is Vc calculated by 0-vout? Isn't voltage always flow from high potential to low? So Vc should be equal to vout-0?
If you see the current, then capacitor current Ic is going from inverting terminal to output terminal. So, based on the current direction, the inverting terminal is at a higher voltage. (Since current flows from higher potential to lower potential).
It is purely based on the direction of the current. If you assume the current through the capacitor is flowing from output to inverting, then you will have (vout - 0 ) in the equation. But then, when you apply KCL at the inverting node, then you will have Iin + ic = 0, OR Iin = -Ic. So, eventually, you will get the same result. I hope, it will clear your doubt.
Zooper class sir.thank you very much
Good work here, thanks again.
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 ?
nah bro it adds up to become zero at 5ms
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
It’s incorrect
It should start from 0 and goes to -10
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?
Thanks a lot sir for teach us opamp
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
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 ...
Ye voltage swing or Vout me kya diff hai ???
18:06 that output graph which you made sir,,,,,,, the down facing trapezium......won't the right slanting line would be above the x axix as 180deg phase shift will be there??
Since the input is negative, so the slope will positive. But the slope will begin from the previous voltage. If the initial value is negative, then it will start from that value and go towards positive voltage. In this case, at 4 ms, the initial voltage is negative. So, the positive slope will start from that value. I hope, it will clear your doubt.
In 6:50 why we don't use the minus sign for gain? We have already written A=-Xc/R. Please give me explanation sir🙏
I forgot to write a minus sign. Otherwise there will be a minus sign. I mentioned it at 6:37.
at 17:53 are you sure that o/p for 4 tp 5 microsecond will remain in negative direction for Vout ?
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Such nice video. Also pls give Application example , where this integrator ckt is used
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.
Where can we find the solution for this problem at 17:15
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?
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
I think fs is not greater than 10fl...10fl is 1.5khz and fs is 1/4ms
Amazing videos... Thank you
At 6:29 shouldn't it be Vout = -(omega*X)/R? As 1/C = omega*X
For inverting op-amp, if Z2 is the impedance in the feedback and Z1 is the impedance at inverting terminal, then Vo = -Z2*Vin/Z1
Here Z2 is Xc and Z1 is R.
So, Vo= -Xc*Vin/R = -Vin/(2*pi*f*C*R)
I hope it will clear your doubt.
I think you got confused because you compared it with the integral equation.
And still, if you have any doubt then do let me know here.
How did you obtain the relation FL=1/(2pi*Rf*Cf)??
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.
Well explained sir
06:40 how did the equation became Vout = -Xc/R * Vin from the enclosed equation at 05:36???
If the input is a sinusoidal signal of specific frequency, then after the integration, you will get 1/2*pi*f in the denominator. And 1 / (2*pi*f*C) is Xc. So, from that you will get Xc / R. If the input is sine wave then after the integration, it will become cosine. (there will be phase shift). But if we just consider the amplitude then output amplitude is Xc / R times the input signal amplitude.
Also, you can look it another way. For inverting op-amp, Vo = - Rf / R1 x Vin
Now, in case of integrator, in the feedback there is a capacitor. Its reactance at the specific frequency is 1 / 2*pi*f*C (or simply Xc).
Therefore, Vo = - (Xc / R) * Vin
I hope, it will clear your doubt.
Sir please include the internal block explanation of op amp.🙏🏻
Why do you get an initial value different than zero while integrating the sine and square waves at 5:08? Shouldn't these start at zero and look like there was a negative offset? I mean, assuming the capacitor has initial value equal to zero.
Yes, that's true. If all initial conditions are zero and if the waveform is starting at t=0, then integration should start from zero. But the waveform which is shown at 5:08 is more of a practical case, where the signal exists for t
Ok thanks! So for signals that existed for a while there would be no "offset" in practice, but let's say i have no signal (steady 0v), discharged capacitor, ideal op amp, and then start the square wave at t=0. The thing thats bugging me is that i dont see that "offset" being a transient response that would fade away for t->infinity. Maybe I'm not getting the concept of square wave existing since beginning of time, then there would be no transient response. But I mean, in reality, don't you have to power up the circuit at some given real time then do the measure?
6:21 how did you get the expression for Vout I’m confused
same
The first equation (with integration) is used for finding the transient output of the integrator. Like when you apply a pulse signal and you want to find the output of the integrator. (It can also be used for sinusoidal signals)
The second equation is particularly used when the input is sinusoidal signal. If the input is sinusoidal signal then we can find the equivalent impedance offered by the capacitance at the specific operating frequency. (1/wc)
So, considering that impedance of the capacitor, the output of the op-amp is (-Xc/R)*Vin. (Similar to -(R2/R1)*Vin)
And here the reactance of the capacitor is 1/wc.
So, Vo = -1*Vin/(wcR) or -1*Vin/ (2πf*R*C).
If we just consider the magnitude then voltage gain |Vo/Vin| = 1/ (2πRfc).
I hope, it will clear your doubt.
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 ??????????
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.
@@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.
Example 4
-2v how? Explain with expression plzz
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