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
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.
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.
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
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 :)
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.
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.
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.
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.
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
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
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.
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...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 😁🙏
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.
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)
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.
As long as it is within the allowed limit, it will not cause any problem. Usually, the current flowing through resistors used to be in mA. So, power dissipation across each resistor used to be within the allowed limit.(e.g example if you are using quarter watt of resistor then power dissipation across resistor should be less than 250 mW. And if it more than that then you can go for a half watt of resistor).
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.
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?
It's a very good video and a very good explanation but I think the graph that appears at 5:37 is not correct. The gain in dB is not proportional to the frequency, but it must show a logarithmic curve. For example, at f = 0 the gain is 0 but the gain in dB is 20log (0), which is indeterminate that tends to -infinite.
Since the capacitance is supposed to be independent of time, gain is just proportional to the freq. But my point is why are we taking the mod? In prev eqn there was a minus sign in gain eqn so the gradient will be negative that implies gain will fall with freq increase. I'm confused. Please explain.
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.
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.
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.
1. At 11:35 you said "Input frequency should be between f0 and whichever is lower between f1 and f2" doubt: what should be the value of f0 ? 2. At 11:45 you said, "input signal should be AT LEAST EQUAL TO fl/10” doubt: in example 2, fs=3khz and (fl/10)= 15.9khz and fs < (fl/10). Therefore input signal in example 2 can not be differentiated.!!
in the graph for gain [dB] vs f for the differentiator at 05:56, wont the gain be 0 dB when f = 0 Hz as A = -2.Pi.Rf.C.f?? i.e. shouldn't the graph start at the origin?? Thanks for the amazing lectures :)
Since the gain is in dB, the graph won't pass through origin. Because as the value of f is close to 0, then gain is also very small. In dB, it would very large negative number. (-120 or -140 dB). It would have passed through an origin, if the gain is not in dB. I hope, it will clear your doubt.
Sir for the integrator problem same as this the vout came as -10v but for that you took it from 5v to -5v but in this problem you directly substituted it from -2.4 to 2.4 and not from -1.2 to 1.2 what is the reason sir
Here we are getting -2.4 V output for the entire slope (-3V to +3V), so during that slope the output will remain constant (-2.4V). When the slope changes, accordingly output shifts to +2.4V. So, during that negative slope (from +3V to -3V), the output will remain +2.4V. So, that is why here output is either +2.4V or -2.4V. I hope, it will clear your doubt.
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?
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.
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.
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.
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.
@@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?
At 9:31, the R-C pair at the input will act as one low pass filter. I guess it is not difficult to understand how it will act as low pass filter. The one end of a capacitor will act as a virtual ground. So, it is normal RC low pass filter. Now, the other pair Rf-Cf will also act as low pass filter. At low frequencies, the reactance of the capacitor will be very high (Xc= 1/2*Pi*f*C). So, the equivalent resistance will be approximately equal to Rf. At high frequency, Xc will be small and the overall impedance will be small. So, it will attenuate the signal. And hence, it will act as a low pass filter. If you were talking about the ideal differentiator circuit where there is C in input side and R in the feedback, then yes it is high pass filter. And it is even evident from the frequency response of the ideal differentiator. I hope, it will clear your doubt.
I didn't understand how RC at input is lpf...For the RC in at the input ,at low frequencies Capacitor acts as a open ckt and at high frequency it acts like short ckt ...ie it will pass only high frequency and not low frequencies so it is a high filter.
I have the same question. There is no path between the resistor and capacitor for current to flow at low frequencies so I don't know why it is considered a low pass filter as current will not flow through the capacitor. Good Question !
@@ALLABOUTELECTRONICS for the first RC pair, the capacitor is in series with the resistor and so it should be a HPF. If the capacitor was a shunt then it would be a low pass . correct? Thanks for the great videos.
i have a small doubt the frequency response of simple differentiator should be y = mx but u drew y = mx-c why?and you also told at f = 0 A = 0 and again you are degining low level frequency as the frequency at which A = 0 iam so confused help me out please
The vertical scale is in dB. At 0 Hz or at DC, the output is not zero, but it's very small value. That's why in dB it will be negative. I hope, it will clear your doubt
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.
The gain of differentiator cannot increase indefinetely. It is restricted by what, it is not sounding clearly. You say something "open loop..." what is that ? At 6:21 in the video this is appearing
I was talking about the gain of the differentiator. The maximum gain which is achieved by the differentiator is the intersection point of the open loop gain response and the response of the differentiator.
@@ankitpandey3774Pandey without the differentiator circuit, the gain of the op-amp would be the open loop gain. But with the differentitor circuit, the gain will change. (Due to feedback). As this differentiator circuit contains the R and C, its gain is frequency dependent and the maximum gain is the point where the two responses intersect. (open loop gain and the response of the differentiator) I hope it will clear your doubt.
@@ALLABOUTELECTRONICS yeah I get thanx bro.. a lot Currently m seeing all the Opamp videos of this channel..it's totally great & best videos among other channel for these topics. especially u see questions & give answer, so come & see this channel makes great helpful like me.
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 ???
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.
Shut up you mr. sheikh reja. Actual answer is (-1.88495559215) and which is very close to the given answer. So it would be better for you to check your strategy first.
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
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 UA-cam.
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.😊😊
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.
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
I have an exam in 2 hours.. because of you i am not scared and am understanding analog system design
Same... Exam in 3 hrs... Bt i m afraid 😰
Same, have an exam in 2 hours
just like that 😨
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+
it is unbelievable video of teaching,very very very very very very very very very very very very very very very thank you
Thankyou so muchhhhh. If you werent there idk how would i pass my exm tomorrow 😭❤✨
Everthing is good,but the calculation have some mistake.Thanks brother for assisting us in our study.Blessing for you for surviving other vedio.
Sir Why do we use a feedback resistor in an integrator and a feedback capacitor in a differentiator circuit?
Outstanding, the thing I have learned from you is the better you are at math the easier an engineering education will be.
Bhai tum mere professor se bhi atcha samjhate hoo ..be happy always bro and keep uploading these awesome contents
Thanks for the support in the electronic technology to the world. keep the as per the best.
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 :)
all pass filter Am*（s+Wz）/（s+Wp）
sir your videos are saviour for exams
you rock!!!
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.
it you feel that way change the speed in settings to 0.75x or 0.5x
Really hats off to yu sir... Bcas very very useful for examination... Keep going lik this sir 👏👏
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.
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.
Alright. That did clear it. Thank you.
Op-Amp Differentiator very well explained. Thank you !
I love sir ur way of teaching
Love from haryana
Great post, keep posting and your channel will grow!
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??
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.
Thank u very much sir
hi my i you younderstand the problem
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...
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
was wondering the same thing
Yes
Sir your efforts are appreciable please keep it up.
You are doing great work, Blessings
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
Couldn't understand that intersection part where the gain and the frequency response meets, What happens there?
me too, i didnt understand why Zin = sqrt(r^2+xc^2) :/
Why cut off frequency= 1/2πfC? Why?
Is education illegal in your city?
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.
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...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 😁🙏
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.
@@ALLABOUTELECTRONICS Yeah...I get it...Thank you, Sir.
Amazing Explanation
14:58 Expression used to find output is for ideal differentiator, but practical diff is gven in Q
16:35
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)
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.
Thank you so much for your help. I hope that I will be also able help someone with my knowledge that I am receiving.
Sir adding more resistors causes more power loss right?
Yes, it will cause more power loss.
ALL ABOUT ELECTRONICS isn't that bad for circuit?
As long as it is within the allowed limit, it will not cause any problem. Usually, the current flowing through resistors used to be in mA. So, power dissipation across each resistor used to be within the allowed limit.(e.g example if you are using quarter watt of resistor then power dissipation across resistor should be less than 250 mW. And if it more than that then you can go for a half watt of resistor).
ALL ABOUT ELECTRONICS thanks for replying
Best video sir ...concept clear ... thank uuu
Gain at 0dB frequency is 1 right...i mean f and the remaining gets cancelled and the gain would become 1
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.
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.
Capacitor will always hold the charge until any new input is introduced to it, therefore the voltage remains constant from 1ms to 4ms
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 (??)
It is 10nF not 1F so the value is -5000*10*10^(-9)*48000.
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?
It's a very good video and a very good explanation but I think the graph that appears at 5:37 is not correct. The gain in dB is not proportional to the frequency, but it must show a logarithmic curve. For example, at f = 0 the gain is 0 but the gain in dB is 20log (0), which is indeterminate that tends to -infinite.
Since the capacitance is supposed to be independent of time, gain is just proportional to the freq. But my point is why are we taking the mod? In prev eqn there was a minus sign in gain eqn so the gradient will be negative that implies gain will fall with freq increase. I'm confused. Please explain.
@inu Nope, it's 20log(1)
when frequency is 0, how can gain be 0?? the diagram clearly shows negative gain not 0?? please tell im really confused
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.
06:49 the gain at 0 Hz is either negative or does not exist from the graph you have drawn?
Sir mujhe Duuhhhh Bhai ne recommend ki video Duhhh bhai jindabaad🥰😍🤤💅💅💅🫂
Can someone please explain to me why at 9:34 he says both Capacitors are acting as low-pass filters here?
excellent explanation
Sir, At 4:34 Vout=-(Rf/Xc)*Vin .But Vout=-RfC dVin/dt written at top right corner. please explain
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.
It was a great video but the frequency response for the differentiator must have a more in depth video explanation
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!
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.
@@ALLABOUTELECTRONICS o got it ,i am very thankful to have the reply sir.
Outstanding sir...
Very nice video
1. At 11:35 you said "Input frequency should be between f0 and whichever is lower between f1 and f2" doubt: what should be the value of f0 ?
2. At 11:45 you said, "input signal should be AT LEAST EQUAL TO fl/10” doubt: in example 2, fs=3khz and (fl/10)= 15.9khz and fs < (fl/10). Therefore input signal in example 2 can not be differentiated.!!
at 4.50 when you write Vout in terms of impedances, why you removed the differentiation of Vin with respect to t ???
in the graph for gain [dB] vs f for the differentiator at 05:56, wont the gain be 0 dB when f = 0 Hz as A = -2.Pi.Rf.C.f?? i.e. shouldn't the graph start at the origin?? Thanks for the amazing lectures :)
Since the gain is in dB, the graph won't pass through origin. Because as the value of f is close to 0, then gain is also very small. In dB, it would very large negative number. (-120 or -140 dB). It would have passed through an origin, if the gain is not in dB. I hope, it will clear your doubt.
Can you tell me in the last question why did we take only 125 sec? ND voltage peak to peak
Sir for the integrator problem same as this the vout came as -10v but for that you took it from 5v to -5v but in this problem you directly substituted it from -2.4 to 2.4 and not from -1.2 to 1.2 what is the reason sir
Here we are getting -2.4 V output for the entire slope (-3V to +3V), so during that slope the output will remain constant (-2.4V). When the slope changes, accordingly output shifts to +2.4V. So, during that negative slope (from +3V to -3V), the output will remain +2.4V. So, that is why here output is either +2.4V or -2.4V. I hope, it will clear your doubt.
I think that there is one error in your vedio. F1(RS CS series cut off frequency) is the cut off frequency of high pass filter.
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?
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.
at 4:01 sir isn't Vo = -(Rf)(C) [Vpeak sin(wt)](w)
thankyou 😊
both are correct...
@6:45 at zero hertz gain is not zero (from frequency response curve). how the gain will be zero?
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.
@@ALLABOUTELECTRONICS thank you for the reply
Thank you very much! It was very helpful.
Sir, at 13:50 (1)fs should be atleast f1/10. (2)fs should be less than f1/10.Which is true? please tell.
Sir , why are we not using circuit above upper cut off frequency?
Why the output waveform of differentiator is starting from negative
Useful material, thank you.
very nice explanation sir. nd sir can i get a link for some pdf of a book of op amps and ICs
Sir, Is this a band stop filter?
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.
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.
@@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?
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????
:D
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?
good job. keep it up.
At 6.06: Shouldn't the gain =0 when f=0 but its showing gain=-ve?
I can't understand how rc ckt work as a low pass filter? Acc to me it is high pass filter
At 9:31, the R-C pair at the input will act as one low pass filter. I guess it is not difficult to understand how it will act as low pass filter. The one end of a capacitor will act as a virtual ground. So, it is normal RC low pass filter.
Now, the other pair Rf-Cf will also act as low pass filter. At low frequencies, the reactance of the capacitor will be very high (Xc= 1/2*Pi*f*C). So, the equivalent resistance will be approximately equal to Rf. At high frequency, Xc will be small and the overall impedance will be small. So, it will attenuate the signal. And hence, it will act as a low pass filter. If you were talking about the ideal differentiator circuit where there is C in input side and R in the feedback, then yes it is high pass filter. And it is even evident from the frequency response of the ideal differentiator.
I hope, it will clear your doubt.
I didn't understand how RC at input is lpf...For the RC in at the input ,at low frequencies Capacitor acts as a open ckt and at high frequency it acts like short ckt ...ie it will pass only high frequency and not low frequencies so it is a high filter.
I have the same question. There is no path between the resistor and capacitor for current to flow at low frequencies so I don't know why it is considered a low pass filter as current will not flow through the capacitor. Good Question !
@@ALLABOUTELECTRONICS got it bro;)
@@ALLABOUTELECTRONICS for the first RC pair, the capacitor is in series with the resistor and so it should be a HPF. If the capacitor was a shunt then it would be a low pass . correct? Thanks for the great videos.
i have a small doubt the frequency response of simple differentiator should be y = mx but u drew y = mx-c why?and you also told at f = 0 A = 0 and again you are degining low level frequency as the frequency at which A = 0 iam so confused help me out please
The vertical scale is in dB. At 0 Hz or at DC, the output is not zero, but it's very small value. That's why in dB it will be negative. I hope, it will clear your doubt
best explanation
But why for differrntiator rc is should be less than time period of input signal
i really like this video.
mass.. superb..
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.
For 16:02 at the Vout why it is -10^-4? -5k x 10n is -50micro right?
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.
good job!
The gain of differentiator cannot increase indefinetely. It is restricted by what, it is not sounding clearly. You say something "open loop..." what is that ? At 6:21 in the video this is appearing
I was talking about the gain of the differentiator. The maximum gain which is achieved by the differentiator is the intersection point of the open loop gain response and the response of the differentiator.
@@ALLABOUTELECTRONICS bhai how then open loop gain comes here ... why we comparing this ! I didn't get bro..
@@ankitpandey3774Pandey without the differentiator circuit, the gain of the op-amp would be the open loop gain. But with the differentitor circuit, the gain will change. (Due to feedback). As this differentiator circuit contains the R and C, its gain is frequency dependent and the maximum gain is the point where the two responses intersect. (open loop gain and the response of the differentiator)
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS yeah I get thanx bro.. a lot
Currently m seeing all the Opamp videos of this channel..it's totally great & best videos among other channel for these topics. especially u see questions & give answer, so come & see this channel makes great helpful like me.
Differentiation of square wave are drawn opposite
4:01 sir we know the derivative of sinx is cosx . can u explain why did u write d(sinwt) = -coswt ?
The negative sign is coming because the op-amp is operated in the inverting configuration. (As I have shown in the equation at 3:38)
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 ???
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.
its now clear...thank you sir..
can I have the objective type questions based on OPAMP in pdf form?
You can check this playlist for op-amp.
It might be helpful to you.
ua-cam.com/play/PLH9R5x7JVXCFFScqREEGiFCSyKVM3RTM8.html
excellent explnation
clear concept
Can someone explain why differentiator gain is restricted by open loop gain gain and why open loop gain curve is like as shown ?
How many wonder 3:45 spikes should be reversed?
What will we get when we give input as square wave ??
The spikes at the rising and falling edges.
I should've subscribed much earlier in my degree!
16:00 CALCULATION IS WRONG!
WARNING!
Shut up you mr. sheikh reja.
Actual answer is (-1.88495559215) and which is very close to the given answer.
So it would be better for you to check your strategy first.
@@gamerzzz1809 thank you shutting that guy up!! although i dont think he's a hater just wanted to point out how clever he or she is..
sir, at 18:45, how to get slope value 48000 ?
plz...reply
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