Op-Amp Differentiator (with Derivation and Examples)
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- čas přidán 30. 06. 2024
- In this video, op-amp differentiator circuit has been discussed (with derivation) and few examples have been solved based on this op-amp differentiator circuit.
Op-Amp as Differentiator:
In Op-amp integrator circuit, if we interchange the position of resistor and capacitor then it can be used as a differentiator. The relation between the output and input has been derived in this video.
Application of differentiator circuit:
The differentiator circuit can be used to identify the rate at which the input signal is changing.
So, the differentiator circuit can be used to find the high-frequency component of the input signal and it can be used in the application of edge detection.
In early days, when digital computers were not evolved at that time for analog computation these op-amp based differentiator circuits were used.
Limitation of simple differentiator circuit:
In this simple differentiator circuit, as the input frequency increases, the gain of the differentiator will increase. So, the simple differentiator is very sensitive to the high-frequency noise. Also, in simple differentiator, the input impedance of the circuit is equal to the reactance of the capacitor. So at high frequency, the input impedance will reduce.
These problems can be overcome by using the practical differentiator circuit.
Practical Differentiator Circuit:
In practical differentiator, the series resistor is added to input capacitor. This resistor will ensure that at high frequencies, the input impedance of the circuit will be at least equal to the value of the resistor.
And because of this series resistor, the gain of the op-amp at high frequency will be restricted.
For better stability of the output signal at the high frequency and to prevent oscillations, feedback capacitor is also connected in parallel with the feedback resistor.
The condition for proper differentiation of 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)
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
The link to the related videos on the op-amp:
Introduction to Operational Amplifier:
• Introduction to Operat...
Inverting Op-Amp:
• Operational Amplifier:...
Non-Inverting Op-Amp:
• Operational Amplifier:...
Op-Amp Integrator
• Op-Amp Integrator (wit...
This video will be helpful to all students of science and engineering in understanding the working of op-amp differentiator.
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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 CZcams.
You are the person who can explain compex problems in very beautiful manner.
You are great explener thanks
Today I am study 1 unit of op-amp by all of your lecturer
Thanku 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
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.
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+
Outstanding, the thing I have learned from you is the better you are at math the easier an engineering education will be.
Op-Amp Differentiator very well explained. Thank you !
Really hats off to yu sir... Bcas very very useful for examination... Keep going lik this sir 👏👏
Great post, keep posting and your channel will grow!
You are doing great work, Blessings
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.
Thank you very much! It was very helpful.
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Best video sir ...concept clear ... thank uuu
Bhai tum mere professor se bhi atcha samjhate hoo ..be happy always bro and keep uploading these awesome contents
Everthing is good,but the calculation have some mistake.Thanks brother for assisting us in our study.Blessing for you for surviving other vedio.
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 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 😨
I love sir ur way of teaching
Love from haryana
Outstanding sir...
sir your videos are saviour for exams
you rock!!!
Amazing Explanation
Sir Why do we use a feedback resistor in an integrator and a feedback capacitor in a differentiator circuit?
Useful material, thank you.
good job. keep it up.
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
Thank you so much for your help. I hope that I will be also able help someone with my knowledge that I am receiving.
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
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.
best explanation
excellent explanation
good job!
clear concept
mass.. superb..
excellent explnation
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
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.
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 ..
i really like this video.
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）
14:58 Expression used to find output is for ideal differentiator, but practical diff is gven in Q
16:35
Very nice video
wonderful!
great man
Sir best video
God bless you!!
Thank u sir🙏🙏
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
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) :/
Can someone please explain to me why at 9:34 he says both Capacitors are acting as low-pass filters here?
nice vidio tanks
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
at 4.50 when you write Vout in terms of impedances, why you removed the differentiation of Vin with respect to t ???
Sir , why are we not using circuit above upper cut off frequency?
Thank u
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?
best ever
Why cut off frequency= 1/2πfC? Why?
Is education illegal in your city?
super sir
very nice explanation sir. nd sir can i get a link for some pdf of a book of op amps and ICs
Capacitor will always hold the charge until any new input is introduced to it, therefore the voltage remains constant from 1ms to 4ms
Great
Can you tell me in the last question why did we take only 125 sec? ND voltage peak to peak
Gain at 0dB frequency is 1 right...i mean f and the remaining gets cancelled and the gain would become 1
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.
Sir, at 13:50 (1)fs should be atleast f1/10. (2)fs should be less than f1/10.Which is true? please tell.
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.
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.
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.
tq sir
Sir, Is this a band stop filter?
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)
How to perform XOR operation by using op amp ??
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.
Can someone explain why differentiator gain is restricted by open loop gain gain and why open loop gain curve is like as shown ?
At 6.06: Shouldn't the gain =0 when f=0 but its showing gain=-ve?
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.
👌👌
I should've subscribed much earlier in my degree!
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.
3:48 output graph should cut the t-axis.
Why the output waveform of differentiator is starting from negative
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?
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.
My concept is cleared
improve your grammar
I
am most beautiful
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
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..
Feedback Rf & Cf which is how to create lpf circuit
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 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
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.
Differentiation of square wave are drawn opposite