The timestamps for the different topic covered in the video is given below: 0:15 Op-Amp as a Differentiator 1:35 Derivation of Op-Amp Differentiator Circuit 3:39 Output of differentiator for the different input signals 4:22 Limitations of the simple differentiator circuit 7:56 Practical Op-Amp differentiator 11:55 Example 1 14:02 Example 2 16:48 Example 3

@aishwaryabuwa7463

5 жыл бұрын

But differentiator act as a high pass filter right? You said low pass filter..

Whoever you are, you've done a great job, I mean these lectures and even your descriptions are very much factual, you've provided a very clear and nice explication and also every topic is covered. commendable work.

I would say that this is one of the best channels for electronics on KZread.

You are great explener thanks Today I am study 1 unit of op-amp by all of your lecturer Thanku so much

You are the person who can explain compex problems in very beautiful manner.

Your voice and your English are very clear , so anyone can understand these easily.😊😊

Op-Amp Differentiator very well explained. Thank you !

the math in this video is beyond me at this point in time.. but i will get to the point were this comes as second nature to me.. so long as wonderful Humans likes yourself are willing to share their knowledge, im willing to learn from you.. thank you so much.

Outstanding, the thing I have learned from you is the better you are at math the easier an engineering education will be.

Thank you very much Sir 🙇🏿‍♂️🙇🏿‍♂️🙇🏿‍♂️. Your OP-AMP playlist saved my ass. Grasped one Night and wrote the exam the next day and made A+

Your lectures are very good . They are highly informative , a suggestion would be to teach all this concepts using Bode plots and deriving transfer function of each and every circuit for better understanding. Keep it up .Once again thanks

Great post, keep posting and your channel will grow!

You are doing great work, Blessings

Really hats off to yu sir... Bcas very very useful for examination... Keep going lik this sir 👏👏

Sir your efforts are appreciable please keep it up.

Thankyou so muchhhhh. If you werent there idk how would i pass my exm tomorrow 😭❤✨

Thanks for the support in the electronic technology to the world. keep the as per the best.

I have an exam in 2 hours.. because of you i am not scared and am understanding analog system design

@yoshikagekira7600

3 жыл бұрын

Same... Exam in 3 hrs... Bt i m afraid 😰

@AllenSA_Lymangirl

2 ай бұрын

Same, have an exam in 2 hours just like that 😨

Thank you very much! It was very helpful.

Best video sir ...concept clear ... thank uuu

it is unbelievable video of teaching,very very very very very very very very very very very very very very very thank you

Bhai tum mere professor se bhi atcha samjhate hoo ..be happy always bro and keep uploading these awesome contents

Outstanding sir...

I love sir ur way of teaching Love from haryana

Amazing Explanation

sir your videos are saviour for exams you rock!!!

Everthing is good,but the calculation have some mistake.Thanks brother for assisting us in our study.Blessing for you for surviving other vedio.

Useful material, thank you.

excellent explanation

very well explained and it contained all the details.. thanks. I just think u said both the parallel RC and the input series RC are low pass...

sir, pls tell me this thing,@9:28 u told RC pair at the input side acts as a LOW PASS FILTER,well indeed its a low pass filter when the output is taken across the capacitor,but here the current has to travel through the capacitor and at low frequencies Xc= infinity,means open ckt equivalent,then its not possible for the current to flow through it,and hence its not a low pass filter here. pls explain me as u said how this thing acts as a low pass filter. Awaiting your reply. thank you :)

@AllenSA_Lymangirl

2 ай бұрын

all pass filter Am*（s+Wz）/（s+Wp）

Sir Why do we use a feedback resistor in an integrator and a feedback capacitor in a differentiator circuit?

good job. keep it up.

best explanation

excellent explnation

mass.. superb..

Very nice video

Why cut off frequency= 1/2πfC? Why?

@user-wp7fk3sp4m

2 ай бұрын

Is education illegal in your city?

clear concept

Couldn't understand that intersection part where the gain and the frequency response meets, What happens there?

@karm00n29

Жыл бұрын

me too, i didnt understand why Zin = sqrt(r^2+xc^2) :/

good job!

Thank you so much for your help. I hope that I will be also able help someone with my knowledge that I am receiving.

There may be some problem in the graph of frequency response of simple differentiator .(6.24min) Open lood gain should be a straight line parallel to X-axis because there was no capacitor was added parallel across Rf. Please correct me if I am wrong.

Can someone please explain to me why at 9:34 he says both Capacitors are acting as low-pass filters here?

With those values there is overshoot and ringing on the output. To avoid it, the feedback resistance must be reduced or the fedback capacitor increased

Sir when we are dealing with practical differentiator we are connecting R1 and CF but in example why we are not taking effect of that??

@ALLABOUTELECTRONICS

6 жыл бұрын

Because the signal frequency fs is more than 10 times less than the frequency f1 and f2. So, we can use the equation of ideal differentiator. And R1 and Cf will not have much effect on output.

@bipinrathod1475

6 жыл бұрын

Thank u very much sir

@rameshkokane4611

3 жыл бұрын

hi my i you younderstand the problem

Explanation is good. It can be better if you go with little less pace considering beginners like me, it is bit difficult to follow with the current pace.

@WhyBhanshu

5 жыл бұрын

it you feel that way change the speed in settings to 0.75x or 0.5x

Sir, during calculating gain of differentiator you took impedance of capacitor 1/Xc and in limitations you said Zin = Xc .sir please calrify my doubt ..

Sir best video

14:58 Expression used to find output is for ideal differentiator, but practical diff is gven in Q

@aryandhande

Жыл бұрын

16:35

nice vidio tanks

i really like this video.

Question: why do you use ideal expression when calculating output if you have added components (resistor and another capacitor)? Dont these components have effect on the output? (Except ofcourde on the plot)

@kirtanyaarana8089

4 жыл бұрын

Aretheil for exact output we have computers , examiner isn't interested in if you can do complicated math or not but intrested in checking if you know the fundamentals of the circuit or not. Videos are exam oriented not research oriented.

I think for the differentiation of the square wave, we will get a negative spike for a positive inc of slope, since there is a minus sign in the expression

@simonlavoie2858

Жыл бұрын

was wondering the same thing

@nitinsingh8101

Жыл бұрын

Yes

Hello sir. Thank you so much for such a great explanation of both integrator and differentiator. But there's a bit of confusion with one of your examples in your video differentiator video. In example no.3 at 18.51, you had evaluated V out and the result came to out be 2.4. But when you graphed it, you had considered of square lying between -2.4v to 2.4v. Are you trying to convey, the output we get is the peak value ? If yes, then why did you consider V out in Integrator video as the swing voltage, hence the output lying between half of output values, which is in reference to your answer posed during 16.45 in Op-amp as an integrator. Thanks! Looking forward to your response asap.

@ALLABOUTELECTRONICS

6 жыл бұрын

In the third example, for the positive slope, the output will be -2.4V and for the negative slope, it is 2.4V. So, the Output voltage is swinging between -2.4 V and 2.4 V during positive and negative slopes respectively. While in case of integrator video, the output voltage represents the total voltage change in 50 microseconds, or you can say it is the slope of the output signal. And that is why peak voltage in both directions are +5V and -5V respectively. I hope it will clear your doubt.

@tjawalia17

6 жыл бұрын

Alright. That did clear it. Thank you.

can someone help me with this? i want to see the solution for the cancelation of gains provided by those practical parts. (R and Cf) he just explains it graphically.

Thank u sir🙏🙏

Sir , why are we not using circuit above upper cut off frequency?

at 4.50 when you write Vout in terms of impedances, why you removed the differentiation of Vin with respect to t ???

06:49 the gain at 0 Hz is either negative or does not exist from the graph you have drawn?

Why is it an issue if the input impedance is low when frequency is high? Does low input impedance cause some sort of problem with the output of the circuit?

Can you tell me in the last question why did we take only 125 sec? ND voltage peak to peak

God bless you!!

thank you

very nice explanation sir. nd sir can i get a link for some pdf of a book of op amps and ICs

wonderful!

when frequency is 0, how can gain be 0?? the diagram clearly shows negative gain not 0?? please tell im really confused

@ALLABOUTELECTRONICS

3 жыл бұрын

The graph is in dB. So, when frequency is zero, then gain is typically -40 dB or less. If you convert it back , then it will be very small. And for all practical purpose, it can be assume as zero. I hope it will clear your doubt.

Sir, at 13:50 (1)fs should be atleast f1/10. (2)fs should be less than f1/10.Which is true? please tell.

super sir

How to perform XOR operation by using op amp ??

great man

At 18:52 Vout caculates to -0.24V instead of -2.4V. Is this a typo or am I missing something ? -5000 x 10^(-9) x 48000 = -0.24 (??)

@saffronminer5694

3 жыл бұрын

It is 10nF not 1F so the value is -5000*10*10^(-9)*48000.

At 6.06: Shouldn't the gain =0 when f=0 but its showing gain=-ve?

Can someone explain why differentiator gain is restricted by open loop gain gain and why open loop gain curve is like as shown ?

I'm confused. At 5:32 you say that gain is zero when frequency is zero, while the graph shows negative gain at zero frequency. And then further you say that f0 is the frequency at which gain is zero. Please clarify.

@ALLABOUTELECTRONICS

5 жыл бұрын

If you closely look at the vertical axis, it is in dB. So, when it is 0dB, the gain actually 1. I hope it will clear your doubt.

sir, at 18:45, how to get slope value 48000 ? plz...reply

@Mrlonelyuploader

Жыл бұрын

Slope = (Change in voltage) / (Change in time) Slope = 6 / 125 Micro Second Therefore , 6/125 x 10 ^ (-6) Which becomes 0.048 x 10 ^ 6 = 48000 Volt

Gain at 0dB frequency is 1 right...i mean f and the remaining gets cancelled and the gain would become 1

Sir, Is this a band stop filter?

Sir, At 4:34 Vout=-(Rf/Xc)*Vin .But Vout=-RfC dVin/dt written at top right corner. please explain

@ALLABOUTELECTRONICS

5 жыл бұрын

Both are true. The first expression represents the output response in the frequency domain and it will give you the gain at the operating frequency. The second expression represents the output in the time domain. That means with time how the output will respond to the input signal.

Great

Thank u

Why the output waveform of differentiator is starting from negative

Sir at 6:43 you have said that 0 Hz and DC Level the gain is 0. so no offset voltage. 1. What is this DC Level? 2. Why because of it there is no offset voltage? 3. Can you please explain the open loop graph that intersects with the voltage gain graph?

@ALLABOUTELECTRONICS

6 жыл бұрын

First, I was referring 0 Hz signal as DC signal. So, 0Hz frequency signal or DC signal both are same. Second, at 0Hz frequency gain is less than 0dB (Actually is it not 0dB but even less than 0dB). So, let's say at 0Hz, if the gain is -20dB, then all the DC signals will see the attention by that amount. So, if you apply any DC signal then it will get attenuated by that amount in the output. If any input offset voltage is present at the input, it will also get attenuated by that amount. (it will not be zero, but very small voltage and can be neglected, as it is getting attenuated) And third, if op-amp is ideal then the response of the differentiator should be some positive slope (Blue line in the frequency response curve in the video) But actually op-amp has finite bandwidth, and it can not amplify all the signal frequency. (Please check my video on the gain-bandwidth product for more info). So, the actual response would be the intersection of the ideal differentiator response and the frequency response of the op-amp. So, the maximum gain which can be achieved by the differentiator is limited by the frequency response of that particular op-amp. I hope it will clear your doubts. If you still have any doubt then do let me know here.

Capacitor will always hold the charge until any new input is introduced to it, therefore the voltage remains constant from 1ms to 4ms

I should've subscribed much earlier in my degree!

best ever

@6:45 at zero hertz gain is not zero (from frequency response curve). how the gain will be zero?

@ALLABOUTELECTRONICS

3 жыл бұрын

Yes, the gain is not zero. But it will be very small. Around -40 or -50 dB or even less (Y- axis is in dB), and for practical purposes, it can be considered as zero. I hope it will clear your doubt.

@bharath_rbp

3 жыл бұрын

@@ALLABOUTELECTRONICS thank you for the reply

Sir...I didn't understand the frequency response curve for simple differentiator...I mean, for frequency 0 Hz, the gain should also be 0...but in the graph, it is cutting the negative y axis...please clarify... Btw...thank you for such amazing videos...really helpful 😁🙏

@ALLABOUTELECTRONICS

4 жыл бұрын

The thing is on the y-axis the gain is in dB. And the second thing is for DC (at 0 Hz), although the output of the differentiator should be zero (Theoretically), but there would be very small voltage at the output. (few uV) And Vo / Vin would be let's say 10^-4 or something. then the gain in dB would be around -80 dB. That's why there is an intersection on the negative y-axis. I hope it will clear your doubt.

@sinchulabanerjee3087

4 жыл бұрын

@@ALLABOUTELECTRONICS Yeah...I get it...Thank you, Sir.

But why for differrntiator rc is should be less than time period of input signal

My concept is cleared

@rishabhtripathi4002

5 жыл бұрын

improve your grammar

@rishabhtripathi4002

5 жыл бұрын

I am most beautiful

For 16:02 at the Vout why it is -10^-4? -5k x 10n is -50micro right?

@ALLABOUTELECTRONICS

Жыл бұрын

Rf is 5k, C is 10nF and the amplitude of sine wave is 2V. So, 5k x 2 x 10 n = 10^-4 I hope, it will clear your doubt.

sir there are some errors, in solving 2nd examnple

sir at high frequency the capacitor reactance is zero so this act as a high pass filter. but you said this act as a low pass filter . HOW????

@chanakyaveer8257

4 жыл бұрын

:D

@H.O.79

4 жыл бұрын

I think "R" and "C" build a high pass filter, "Rf" and "Cf" build a low pass filter. Otherwise the circuit didn't make sense. Or am I wrong?

sir at 19:53 you said "at zero frequency, the gain of this differenciator will be equal to zero" but in graph at zero frequency, the gain of this differenciator is negative db...please sir tell me i am confused... and sir why Fs=F1/10 ???

@ALLABOUTELECTRONICS

6 жыл бұрын

Ideally, at zero frequency the output should be zero. But actually, you will get some voltage at the output. (Very low voltage, less than input). So, the ratio of output to the input (Gain) will be much less than 1. And in decibel, it will be negative. That is why gain is shown as negative in decibel. Now, coming to your second question, for proper differentiation, the signal frequency should be less than at least 10 times less than cut-off frequency. It is related to charging and discharging of the capacitor. If the signal is changing too fast, then capacitor will not have enough time for charging and discharging and that will affect your output. You can even try that in simulink and can see the result.

@mr.unique7689

6 жыл бұрын

its now clear...thank you sir..

3:48 output graph should cut the t-axis.

tq sir

Please make video on clipper circuit

@aniket9806

6 жыл бұрын

Hruturaj Kedar kuch b demand kar dene ki matlab... kitaab kholo aur khud b padh lo kuch....pun intended

I have a doubt in 3rd example how could u say that the time period between triangular wave form is 250us

@pijushbasak4273

4 жыл бұрын

Time period=1/frequency. =1/4kHz=250microsec

sir f1=f2; if Rf=R and Cf=C; now f0=1/2pi(Rf)C so f0 will become f1, then how will differentiation take place? please help!

@ALLABOUTELECTRONICS

5 жыл бұрын

For proper differentiation of the input signal, the frequency of the input signal should be lesser than the cut-off frequency. (At least 10 times less than the cut-off frequency for the accurate differentiation) As far as this condition is satisfied, it will work as differentiator.

@amitghosh3938

5 жыл бұрын

@@ALLABOUTELECTRONICS o got it ,i am very thankful to have the reply sir.

sir,@14:33 in the example section the circuit has a zero dB frequency(Fo) of 3.18 kHz, while solving example 2,the source has a frequency of 3 khz, but u said to use the circuit as a differentiator the source frequency shd be between Fo & F1. may be some sort of misframing the question i guess,but pls clarify sir.

@ALLABOUTELECTRONICS

5 жыл бұрын

The signal can be properly differentiated (as long as fs way below than upper cut-off frequency). But if the signal frequency is less than f0 (zero dB frequency) then its amplitude will be less than the input signal amplitude. That is what exactly we are getting in this example. Its amplitude is less than 2. I hope it will clear your doubt.

@ankitpandey3774

5 жыл бұрын

@@ALLABOUTELECTRONICS hllo sir I didn't get it answer pls elaborate again. Actually i have a doubt if Fs less than Fo then how do we get output ? Sir how we get Fo here .. any equation of it?