The timestamps for the different topics covered in the video: 0:29 How RC Phase Shift Oscillator Works 1:45 Simple RC Circuit (Phase Lead Circuit) 5:35 Design of RC Shift Oscillator using Op-amp 7:31 Derivation of the frequency for the RC Phase Shift Oscillator
Your derivation was so much better, my lecturer Laplace transformed the circuit and ended up getting 3x3 matrix and the derivation was worth over 5 pages
The math is kind of long but i simulated the circuit using LTspice and plotted the ratio of Vout/Vin and it came out to -1/29th as in the video. Thank you. You are so good with the math you should go back for your doctorate.
Hi, all about electronics. I just see one error on the diagram: R1 is virtually connected in parallel to R, affecting the gain calculation. Instead, R is in feedback to the input without R1.
Why is it assumed that the entire current I3 flows through R and is not split into 2 currents, one flowing through the resistor and one flowing out to the right of the circuit? When the RC circuit is connected to the amplifier circuit then there will be a current flowing in at Vin and a current flowing out at Vout, so why is only the current flowing in at Vin applied during the derivation?
Here, Phase shift per attenuator is 60. so, tan(60)=sqrt(3)=Xc/R ==> Xc=sqrt(3)*R So, Vin/Vo=1-j*sqrt(3) And we also know angular frequency , w=1/(j(Vin/Vo-1)*RC) So, in this case angular frequency , w=1/(sqrt(3)*RC) Not your mentioned : w=1/(sqrt(6)*RC) Note: I can be wrong, but I found this from my calculation. thanks at all for great video.
Ive watched all your videos for Opamp and you are just amazing , here I got a small doubt in 1:19 time of this video you said vout is in 180 phase shift but our input and feedback both are at + i.e non inverting then how come our vout is inverted ?
What I mean to say is, the feedback circuit ( RC Circuit) provides 180 degree phase shift to the incoming signal (output of the opamp) at particular frequency. Therefore at the oscillation frequency, ideally the output of the opamp ( which is input to the feedback circuit) and feedback signal ( output of the RC network) will be 180 degree phase shifted. Of course, the overall phase shift of the circuit is 360 degree, and that's what we required for oscillation. I hope it will clear your doubt.
Hi all about electronics. I have just a minor question on 6:34. I am just confused because here the op amp has a negative feedback. Isn't negative feedback used when we want our op amp to function as amplifier and positive feedback if we want to use it as oscillator? Since this is an RC Phase Shift Oscillator, shouldn't we use positive feedback in it?
At 5:41 In the figure You getting Vout From the RC feedback output I don't understand this point because generally we get the Vout from output of Op-amp.
Here this Vout is the output of the RC network. So, what I mean is, the output of the RC network is given as an input to the op-amp. Of course, as you said, the overall output of the oscillator circuit is the output of the op-amp. If I have used the different nomenclature for the two outputs (the op-amp and the RC network output) then this confusion would not have arised. But I hope, you got the point. If you still have any doubt then let me know here.
@@ALLABOUTELECTRONICS so My exam starting this month for revision I watched so many of your videos that I am used to your voice because of this I am reading your reply in your voice in my mind..
The thing is the gain provided by the op-amp is real (-Rf/R1). It does not contain any imaginary part. So, the feedback section should also have the only real part. And yes, the attenuation depends on the number of stages.
What will be the amplitude of the output sine wave??.....calculation can be made easy by using transfer functions but this is the easiest way to understand....thank you
I think this depends on the op-amp, since this is an oscillator you would have to check the slew rate for the maximum frequency and the rail voltages for the sine wave amplitude.
14:24 , can you elaborate on why the output voltage will have real part only , and can we cascade it with low pass filter and if we can , will there be any change in the result?
In oscillator, although in the input (the thermal noise) there are many frequencies, eventually we are getting the output oscillation only at the specific frequency. So, we can say that there is a sort of resonance at the specific frequency. And in resonance, the input and output are in phase. Therefore, the imaginary part will be zero.
Hi, how this circuit is in positive feed back? as per my understanding oscillator must be in positive feed back? but this circuit seems to be in negative feed back?
Try to understand in terms of the loop gain of the circuit. For oscillation, AB should be slightly more than 1 right. And around the loop, the phase shift should be either 0 or multiple of 2*pi. Here the RC feedback circuit attenuates the output signal by the factor 1/29 at the oscillation frequency. And it also introduces the 180 degrees of phase shift between the output of the op-amp and the feedback signal which is again given as input. On the other end, the op-amp provides the gain of slightly more than 29 and as it is connected in the inverting configuration, it also introduces the 180 degrees of phase shift. So, overall loop gain becomes slightly more than 1 and the phase shift around the circuit becomes zero. In short, it satisfies the conditions required for the sustained oscillations. I hope it will clear your doubt.
Connecting to the negative (inverting) terminal doesn't necessarily signifies that the OPAMP is in negative feedback. Notice that the gain of feedback loop is -1/29, which is itself negative. Hence negative feedback with negative gain (attenuation in this case) cancels each other and the overall feedback is positive.
Sir the equivalent ckt you draw the output is first taken from opam directly and in second ckt output is taken after the phase shift ckt .will it produce the same output.
For the angle of feedback circuit to be 180 degree... Imaginary part should be zero.. Then only tan^(-1)(y/x) may be equal to zero... Subject to condition that real part of beta must be negative...
Here Vo and Vin are corresponding to RC network. Although I have used the same nomenclature (Vo and Vin) for the overall oscillator circuit, but here it is the ratio of the output of the RC network to the input of the RC network. And since it is a passive RC network, it won't provide the gain in a true sense. There will be an attenuation of 1/29 at the resonant frequency.
I know it may sound dumb But is noise sinisoidal Or how the the variation in DC current is sinusoidal And it's amplified and given back So if it's not sinusoidal how we got sine wave??? Thanks
The noise does not contain single frequency. If it contains a single frequency then it is easy to identify or eliminate that from any system. But if we talk about the noise, Its combination of all the frequency. So, when it is amplified by the amplifier, all the frequencies in the noise signal gets amplified. But the feedback network acts as a frequency selective network, which passes only certain band of frequencies and attenuates remaining frequency. So, over the time only single frequency is able to get amplified and all other frequencies gets diminished. When we say a single frequency in the frequency spectra, in time domain, its a sinusoidal signal of that frequency. So, with proper gain tuning, we are able to get a sinusoidal signal of a single frequency. I hope it will clear your doubt.
sir, please clarify my doubt ... at 1:48 you said that the output voltage leads the input voltage, but generally in an RC network current leads voltage...i am not able to relate these two statements...please help me out sir...
Here, the input and output voltages have been compared. And what you are saying is true when the position of R and C is interchanged. i.e when the output is measured across C.
@@ALLABOUTELECTRONICS sir, in that case, can we reconstruct the circuit in which the feedback network consists of cascaded low pass RC network in which the resistance value is very close to infinity?..because I feel even in that case a single RC network will provide a phase shift of 90(however it will be lagging)....and thank you so much for your immediate response sir...❤❤
If you choose a very large value of R, then the AB= 1 won't get satisfied. you also need to look for the gain as well. Also, with a single RC network, you will get a 90-degree phase shift only when w is infinite.
Is there a short way or a formula for calculating the attenuation factor(beta)? It will be pretty time consuming to solve such big equations during examinations.
Gain is ratio between output and input voltage, not the value of output voltage. Let say you input is 0.1V and your output is 2.9V, so 2.9/0.1 = 29 (Gain)
If you have A + jB then the phase shift is tan-1(B/A). Similarly, for A + jB / (C+ j D), the phase shift is tan -1 (B/A) - tan -1 (D/C) In this case, the numerator does not have any imaginary part (B= 0) . Similar way, the phase shift of R/ (R - j/wc) has been found.
Great stuff! You always have clear explanations How about some videos on: In rush current, FETs, LVDS, Voltage doubler, buck, boost, buck-boost, and flybacks?
The content you provide is quite cool, but if I may ask for a favour - please use some compressor on your mic :) the clipping is quite annoying :/ And I think there is one issue said incorrectly in 14:30, i.e. you are zeroing imaginary part saying that output voltage is only the real part. I think is it incomplete, as zeroing imaginary part gives us only information that there will be zero, or 180 degree phase shift. Later on you are calculating voltage magnitude at this condition.
Actually there is a positive feedback . The op-amp provides 180 degree phase shift (since it is inverting op-amp). And the RC feedback circuit provides remaining 180 degree phase shift. So, the input which is fed to the op-amp is in phase with the initial seed signal. (noise is the seed signal). I hope it will clear your doubt. If you still have any doubt then let me know here.
The circuit diagram shown at 5:37 has Vout at terminal of Rf,while redrawn circuit diagram have Vout at terminal of R1 !.How can it be correct? Please clarify my doubt... thanks in advance.
so in this cas, depending on how much RC curcit is cascaded to each other the formula of "f" will change. and since f changed the attenation factor won't be equal to 1/29. right? so we will have to change the gain of opamp. right?
It is a harmonic oscillator. So, out of all the frequency components from the thermal noise, the feedback section (which is a frequency selective section) only allows a single a single frequency to grow over time. If you look at the frequency spectrum of a square wave or a triangular wave, it consists of multiple frequency components. But in this oscillator, the frequency selective feedback section does not allow the harmonics. And that's why the output is sinusoidal. I hope it will clear your doubt.
@@ALLABOUTELECTRONICSwe do not see many frequency component (from the thermal noise) at the output of the opamp. (1)Do the frequencies, which are not selected by the selective network, dies out over time / vanish ?? (2)Does the frequency components from the thermal noise die out over time??
First of all, the amplitude of the frequency components is in microvolts or even lower. So, it won't be measured by the oscilloscope. Moreover, as it contains all the frequency components, in time domain they are indistinguishable. The frequency selective network selects the particular frequency signal and that signal is boosted by the loop gain over the iteration. And yes, the remaining frequency components will die out over the period of time. For more info, please check my video on " How Oscillator Works?" It will help you. Here is the link: ua-cam.com/video/XVS8Puf4tiw/v-deo.html
@@ALLABOUTELECTRONICS Thank you very much sir. Our teachers do not interact with us. We suffering in a big problem. Again thank you from the bottom of my heart.
Isnt there a minus is absent in equation of Xc? Zc = jXc = 1/jwC = -j/wC -> Xc= -1/wC ? I am very confused because of it I could not drive the same formulas with you
I have made a video on a log and antilog amplifier using the op-amp. In that video, I have explained how the analog multiplication can be performed using the anti-log and log amplifiers. You can check that video. Here is the link: ua-cam.com/video/Nrfb-s0wl6g/v-deo.html
I have a question ? In our lab there is practical of Phase shift oscillator, In which we have three RC sections, but in the practical we vary capacitance and keep R constant, but still for different capacitance we always get sustained oscillation, According to tan^-1(xc/R) as soon as we vary xc phase shift should must vary, still we get sustained oscillations means that loop is still shifting 180° but how can this is possible if Xc is varying ???
Have you checked the frequency of oscillation ? When you are changing the capacitance then frequency of oscillation is also changing. That means now, for new frequency of oscillation, you will get 180 phase shift. For the previous oscillation frequency, of course the phase shift will not be 180.
@@ALLABOUTELECTRONICS , Yes, we have measured frequency, It means that, in product wc as c vary w will also vary such that the product remains nearly same and overall phase shift becomes 180, thank you very much 😊
Sir I had a brain storming doubt If we need each rc network to provide 60 degree phase shift then Tan^-1 (Xc/R)= 60 Thus 1/omega×C×R =root 3 So we get frequency of oscillation as 1/2pieRC(root3) and not root 6 But in derivation we get root 6 But frequency must be the root 3 case for the RC network to produce a phase shift 60 degree so that total 180 degree Thank you in advance
The thing is , when we connect three RC network togather, it won't behave like a single RC netwrok. Each stage will act as a load for the precious stage. That's why we need to consider the three stage as a single network and go through the entire derivation. I hope it will clear your doubt.
But in lecture it is told that if two 3stage are there than each one will provide equal amount of phase shift so why can't we use w=1/RCroot(3) ??? @@ALLABOUTELECTRONICS
It would be nice if you provide in the end of you each video a simulation, ex in proteus, of your circuit. So we do know what is really going on. My onpinion.
When 'R' is equal to zero, the ground resistors are not present. Therefore, the feedback circuit contains a bunch of capacitors in series with the feedback resistors. This causes the gain of the feedback loop to plummet.
In your prev video about oscillators u told that op- amp are used in relaxation oscillators(which genrate non-sinusoidal signals ) but in rc phrase shift oscillators (which genrate sinusoidal signal) we are using it ,....Why???
In relaxation oscillator, the op-amp is used in saturation. ( The output of the op-amp is either + Vsat or - Vsat). Here it is used as a linear amplifier. I hope it will clear your doubt
Without any feedback, the noise will be present in the input which contains all the frequency component. The amplifier will amplify all the frequency components by the gain factor. But as the amplitude of the noise is very small, at the output also you will observe nothing. (with out feedback) The feedback network provides the frequency selectivity.
@@ALLABOUTELECTRONICS oh sorry for my typing mistake I meant is there something we can say about the amplitude of the sine wave generated by these(RC and LC) oscillators Thank you sir
Using this circuit it is hard to control the gain. Because with slight change in the gain the output may go into the saturation. To control the amplitude, the limiter circuit is used with the oscillator ( It is implemented using the diodes and few resistors). It is difficult to explain that circuit over here. Someday I will cover that circuit in detail with simulations.
@ALL ABOUT ELECTRONICS @ALL ABOUT ELECTRONICS ok sir thank you However can we predict or say anything about the amplitude of the sine wave we just obtained Is it any way related to any value of the components Or more like is it possible for us to create a for example Wein Bridge oscillator of +-9v output
@@abhijithanilkumar4959 Yes, the amplitude depends on the gain. As I said to you earlier, we need to fine-tune the gain to get the particular amplitude. Using the gain-adjustment circuit, it is possible to achieve the desired amplitude.
at f=1/2piRCsqr(6) => phi= tan-1(1/sqr(6)) => phi=67.79 degree . But for a Three stage RC circuit the phase angle should not be 60 degrees instead of 67.79 so that the overall feedback of RC circuit becomes 180 degree. Please Explain
The thing is, the overall phase shift should be 360. When you use, the RC circuit in the feedback then it will provide the 180-degree phase shift at one particualr frequency. so, to get a 360 phase shift, we need to use the op-amp in the inverting configuration.
@@ALLABOUTELECTRONICS okay so if we use non inverting opamp with multiple stage RC circuit ( inorder to get 360°). Does it still work the way you said?
excuse me sir, i think you has mistake at 11:30 put symbol (-)minus, it actuly (+)plus 1/omega^2C^2R^2. sorry for interrupnt and sorry for my bad english, correct me plz if i wrong
The timestamps for the different topics covered in the video:
0:29 How RC Phase Shift Oscillator Works
1:45 Simple RC Circuit (Phase Lead Circuit)
5:35 Design of RC Shift Oscillator using Op-amp
7:31 Derivation of the frequency for the RC Phase Shift Oscillator
Took me a day because there was no equation number given but it was helpful!!! Thank you
Your derivation was so much better, my lecturer Laplace transformed the circuit and ended up getting 3x3 matrix and the derivation was worth over 5 pages
Besten Dank für dieses Video
Many thanks for this video
from Switzerland
The math is kind of long but i simulated the circuit using LTspice and plotted the ratio of Vout/Vin and it came out to -1/29th as in the video. Thank you. You are so good with the math you should go back for your doctorate.
Did you simulate the rc network and it vout/vin or including the opamp?
Great that you know to use spice
Please educate me a bit about that when you have time
@@abhijithanilkumar4959 do it yourself lazy man.
What is spice
@@deitybasumatary7497 A kind of biryani
I'm here to just say "Thanks" for sharing awesome knowledge in this 17 minute video.
Good explanation and approach to the results.
Thank you so much for such video! May the Lord Jesus bless you.
Hi, all about electronics. I just see one error on the diagram: R1 is virtually connected in parallel to R, affecting the gain calculation. Instead, R is in feedback to the input without R1.
your intro is best sir,,,we feel chilled whenever we open any video.....
thanks for this type of kindness😊😊
watching before attempting my 4th try in the exam!
Me too
Why is it assumed that the entire current I3 flows through R and is not split into 2 currents, one flowing through the resistor and one flowing out to the right of the circuit? When the RC circuit is connected to the amplifier circuit then there will be a current flowing in at Vin and a current flowing out at Vout, so why is only the current flowing in at Vin applied during the derivation?
Also thought about that
@@radanh_h4691 because opamp input impedance is pretty high it will draw a very small current
The only reason to recharge my phone is to watch NESO academy and ALL ABOUT ELECTRONICS.
AOE is the real khalnayak
@@abhijithanilkumar4959 What AOE?
@@rishabhpathak1367 All About Electronics
sajan re jhooth mat bolo khuda ke paas jana hai
very cool, you nerd..
Very good video. You have made a good video which can be referred for technical exams.
sir I love the way you explain the topics. please make a video on phase locked loop.
Super sir easy to understand... thank you
Here,
Phase shift per attenuator is 60.
so, tan(60)=sqrt(3)=Xc/R
==> Xc=sqrt(3)*R
So, Vin/Vo=1-j*sqrt(3)
And we also know angular frequency , w=1/(j(Vin/Vo-1)*RC)
So, in this case angular frequency , w=1/(sqrt(3)*RC) Not your mentioned : w=1/(sqrt(6)*RC)
Note: I can be wrong, but I found this from my calculation. thanks at all for great video.
@@pallob_M10 what is j ?
Ive watched all your videos for Opamp and you are just amazing ,
here I got a small doubt in 1:19 time of this video you said vout is in 180 phase shift but our input and feedback both are at + i.e non inverting then how come our vout is inverted ?
What I mean to say is, the feedback circuit ( RC Circuit) provides 180 degree phase shift to the incoming signal (output of the opamp) at particular frequency. Therefore at the oscillation frequency, ideally the output of the opamp ( which is input to the feedback circuit) and feedback signal ( output of the RC network) will be 180 degree phase shifted.
Of course, the overall phase shift of the circuit is 360 degree, and that's what we required for oscillation.
I hope it will clear your doubt.
Hi all about electronics. I have just a minor question on 6:34. I am just confused because here the op amp has a negative feedback. Isn't negative feedback used when we want our op amp to function as amplifier and positive feedback if we want to use it as oscillator?
Since this is an RC Phase Shift Oscillator, shouldn't we use positive feedback in it?
super ur videos with clean nd good information nd explanantion🤗🤗
Very nice explaination sir 👍👍👍👏👍👏👏
At 5:41 In the figure You getting Vout From the RC feedback output I don't understand this point because generally we get the Vout from output of Op-amp.
Here this Vout is the output of the RC network. So, what I mean is, the output of the RC network is given as an input to the op-amp. Of course, as you said, the overall output of the oscillator circuit is the output of the op-amp. If I have used the different nomenclature for the two outputs (the op-amp and the RC network output) then this confusion would not have arised. But I hope, you got the point. If you still have any doubt then let me know here.
@@ALLABOUTELECTRONICS so My exam starting this month for revision I watched so many of your videos that I am used to your voice because of this I am reading your reply in your voice in my mind..
@9:45 why we are not applying KCL at Vo? in this case it is open but what about when we connect it in the feedback path?
Thank you for detailed explanation sir
This may be a stupid question, but if you change R and C in the RC ladder do you change the gain from 1/29 to another value?
Very good explanation
1) 14:25 why the output voltage have real part only ?
2) Does the attenuation depend on the number of stages or independent ?
The thing is the gain provided by the op-amp is real (-Rf/R1). It does not contain any imaginary part.
So, the feedback section should also have the only real part.
And yes, the attenuation depends on the number of stages.
@@ALLABOUTELECTRONICS is there a formula for attenuation for the number of stages ?
What will be the amplitude of the output sine wave??.....calculation can be made easy by using transfer functions but this is the easiest way to understand....thank you
I think this depends on the op-amp, since this is an oscillator you would have to check the slew rate for the maximum frequency and the rail voltages for the sine wave amplitude.
14:24 , can you elaborate on why the output voltage will have real part only , and can we cascade it with low pass filter and if we can , will there be any change in the result?
In oscillator, although in the input (the thermal noise) there are many frequencies, eventually we are getting the output oscillation only at the specific frequency. So, we can say that there is a sort of resonance at the specific frequency. And in resonance, the input and output are in phase. Therefore, the imaginary part will be zero.
Please upload a video on RC phase oscillator using BJT
Yes please do
thanks a lot, taught me more about circuit analysis!
Hi, how this circuit is in positive feed back? as per my understanding oscillator must be in positive feed back? but this circuit seems to be in negative feed back?
Try to understand in terms of the loop gain of the circuit. For oscillation, AB should be slightly more than 1 right. And around the loop, the phase shift should be either 0 or multiple of 2*pi.
Here the RC feedback circuit attenuates the output signal by the factor 1/29 at the oscillation frequency. And it also introduces the 180 degrees of phase shift between the output of the op-amp and the feedback signal which is again given as input.
On the other end, the op-amp provides the gain of slightly more than 29 and as it is connected in the inverting configuration, it also introduces the 180 degrees of phase shift.
So, overall loop gain becomes slightly more than 1 and the phase shift around the circuit becomes zero.
In short, it satisfies the conditions required for the sustained oscillations.
I hope it will clear your doubt.
Connecting to the negative (inverting) terminal doesn't necessarily signifies that the OPAMP is in negative feedback. Notice that the gain of feedback loop is -1/29, which is itself negative. Hence negative feedback with negative gain (attenuation in this case) cancels each other and the overall feedback is positive.
Sir the equivalent ckt you draw the output is first taken from opam directly and in second ckt output is taken after the phase shift ckt .will it produce the same output.
In 14:28, why will the output voltage only have a real part?
it's a trick to find frequency
For the angle of feedback circuit to be 180 degree... Imaginary part should be zero.. Then only tan^(-1)(y/x) may be equal to zero... Subject to condition that real part of beta must be negative...
Excelente vídeo, tudo parece tão fácil. Saudações do Brasil
Man, you are my hero
Sir, at 16:07 U hv told about attenuation. But this is open loop gain(A) =Vo/Vin.
Here Vo and Vin are corresponding to RC network. Although I have used the same nomenclature (Vo and Vin) for the overall oscillator circuit, but here it is the ratio of the output of the RC network to the input of the RC network. And since it is a passive RC network, it won't provide the gain in a true sense. There will be an attenuation of 1/29 at the resonant frequency.
I know it may sound dumb
But is noise sinisoidal
Or how the the variation in DC current is sinusoidal
And it's amplified and given back
So if it's not sinusoidal how we got sine wave???
Thanks
The noise does not contain single frequency. If it contains a single frequency then it is easy to identify or eliminate that from any system. But if we talk about the noise, Its combination of all the frequency.
So, when it is amplified by the amplifier, all the frequencies in the noise signal gets amplified.
But the feedback network acts as a frequency selective network, which passes only certain band of frequencies and attenuates remaining frequency. So, over the time only single frequency is able to get amplified and all other frequencies gets diminished.
When we say a single frequency in the frequency spectra, in time domain, its a sinusoidal signal of that frequency.
So, with proper gain tuning, we are able to get a sinusoidal signal of a single frequency.
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS got it sir
Thanks
@@ALLABOUTELECTRONICS oh
Is it like A. Beta less than 1 for the other frequencies thus they Eventually damp and die out??!
@@abhijithanilkumar4959 Yes, exactly.
sir, please clarify my doubt ... at 1:48 you said that the output voltage leads the input voltage, but generally in an RC network current leads voltage...i am not able to relate these two statements...please help me out sir...
Here, the input and output voltages have been compared. And what you are saying is true when the position of R and C is interchanged.
i.e when the output is measured across C.
@@ALLABOUTELECTRONICS sir, in that case, can we reconstruct the circuit in which the feedback network consists of cascaded low pass RC network in which the resistance value is very close to infinity?..because I feel even in that case a single RC network will provide a phase shift of 90(however it will be lagging)....and thank you so much for your immediate response sir...❤❤
If you choose a very large value of R, then the AB= 1 won't get satisfied. you also need to look for the gain as well.
Also, with a single RC network, you will get a 90-degree phase shift only when w is infinite.
@@ALLABOUTELECTRONICS tq so much sir!!
That one south asian guy on UA-cam
Nah, it's called Indian!
@@aryanprakharsame difference
Your name is Aaryan Upadhyaya, you're a South Asian guy on UA-cam too
Thanks a lot bro.......... 01 June 2024 ko mera viva hai😅....
kudos from india
You are الأفضل يلا... استمر يلا... يا هندي...
Thank you. Very helpful.
Great! Note - you say "into" when you mean "times"
Is there a short way or a formula for calculating the attenuation factor(beta)? It will be pretty time consuming to solve such big equations during examinations.
excellent explanation sir
thank u
I don't understand op amp holds up to 18(VCC) and 18(-VCC)
We need A= 29 ' any help
Gain is ratio between output and input voltage, not the value of output voltage. Let say you input is 0.1V and your output is 2.9V, so 2.9/0.1 = 29 (Gain)
Very helpful👌
Why using 0 - tan inverse eqn for the phase shift?
Thnqq 4 the video...it's really helpful
Dear sir, why the output doesn’t have an imagery part (14:30). Thanks
Because the output of the oscillator is oscillating wave. (Sinusoidal signal), which is a real signal .
Lol what's imaginary signal.
Sir at 5:46 you have given the feedback to non inverting.Then how is it positive feedback.. can you clear it please
The input is connected to the inverting terminal, please check it once again.
How did you arrive at the formula for phase shift? 2:43
If you have A + jB then the phase shift is tan-1(B/A).
Similarly, for A + jB / (C+ j D), the phase shift is tan -1 (B/A) - tan -1 (D/C)
In this case, the numerator does not have any imaginary part (B= 0) .
Similar way, the phase shift of R/ (R - j/wc) has been found.
@@ALLABOUTELECTRONICS thanks a lot
It helped me a lot sir
Rushal Bhai..!!Will you please make a video about bootstrapping technique in emitter follower amplifier????
Yes, When I make a videos on transistors, I will cover it.
At 9:44 where did the minus sign go?...plz help
1/j = -j. Please check it once again, you will get it. If you still have any doubt, let me know here.
Great stuff! You always have clear explanations
How about some videos on:
In rush current, FETs, LVDS, Voltage doubler, buck, boost, buck-boost, and flybacks?
بارك الله فيك وهداك
Why we only change the value of capacitor to get a desired frequency ¬ change the resistor value
The content you provide is quite cool, but if I may ask for a favour - please use some compressor on your mic :) the clipping is quite annoying :/
And I think there is one issue said incorrectly in 14:30, i.e. you are zeroing imaginary part saying that output voltage is only the real part. I think is it incomplete, as zeroing imaginary part gives us only information that there will be zero, or 180 degree phase shift. Later on you are calculating voltage magnitude at this condition.
Thank u so much for this.
शानदार explanation... Hartley oscillator भी krao
Can anyone tell me how the formula for phase shift came at 2:36
sir pls write the point which u explain so that it take less time to make notes during watching the lecture
at the ending, you called Vo/Vin the attenuation, I though it was the gain and attenuation was the inverse?
@allaboutelectronics As you mentioned the oscillators have positive feedback then why positive feedback is not employed in this circuit?
Actually there is a positive feedback . The op-amp provides 180 degree phase shift (since it is inverting op-amp). And the RC feedback circuit provides remaining 180 degree phase shift. So, the input which is fed to the op-amp is in phase with the initial seed signal. (noise is the seed signal). I hope it will clear your doubt. If you still have any doubt then let me know here.
The circuit diagram shown at 5:37 has Vout at terminal of Rf,while redrawn circuit diagram have Vout at terminal of R1 !.How can it be correct? Please clarify my doubt...
thanks in advance.
Yes. It's by mistake. It has to be on the left side.
Anyway thanks for pointing it out.
Sir, why can’t we use cascaded low pass filters instead of high pass filters?
13:16 what's the formula for voltage drop across capacitor?
Sir ,there is a resistance in parallel with capacitance in high pass filter!!
Sir actually l don't like the electronics... After seeing your video I understood, about the electronics now I began to love it
so in this cas, depending on how much RC curcit is cascaded to each other the formula of "f" will change. and since f changed the attenation factor won't be equal to 1/29. right? so we will have to change the gain of opamp. right?
Not only the gain, but you also need the look at the phase. You also need phase of 180 degree, so that the overall phase is 360 degree.
WHY THE OUTPUT WAVEFORM IS SINUSOIDAL , not a square or triangular waveform???
It is a harmonic oscillator. So, out of all the frequency components from the thermal noise, the feedback section (which is a frequency selective section) only allows a single a single frequency to grow over time.
If you look at the frequency spectrum of a square wave or a triangular wave, it consists of multiple frequency components.
But in this oscillator, the frequency selective feedback section does not allow the harmonics. And that's why the output is sinusoidal.
I hope it will clear your doubt.
@@ALLABOUTELECTRONICSwe do not see many frequency component (from the thermal noise) at the output of the opamp.
(1)Do the frequencies, which are not selected by the selective network, dies out over time / vanish ??
(2)Does the frequency components from the thermal noise die out over time??
First of all, the amplitude of the frequency components is in microvolts or even lower. So, it won't be measured by the oscilloscope. Moreover, as it contains all the frequency components, in time domain they are indistinguishable.
The frequency selective network selects the particular frequency signal and that signal is boosted by the loop gain over the iteration.
And yes, the remaining frequency components will die out over the period of time.
For more info, please check my video on " How Oscillator Works?"
It will help you.
Here is the link:
ua-cam.com/video/XVS8Puf4tiw/v-deo.html
@@ALLABOUTELECTRONICS Thank you very much sir.
Our teachers do not interact with us. We suffering in a big problem.
Again thank you from the bottom of my heart.
Isnt there a minus is absent in equation of Xc? Zc = jXc = 1/jwC = -j/wC -> Xc= -1/wC ? I am very confused because of it I could not drive the same formulas with you
Would you make a video about op- amps used to perform analog multiplication??
I have made a video on a log and antilog amplifier using the op-amp. In that video, I have explained how the analog multiplication can be performed using the anti-log and log amplifiers.
You can check that video.
Here is the link:
ua-cam.com/video/Nrfb-s0wl6g/v-deo.html
I have a question ?
In our lab there is practical of Phase shift oscillator,
In which we have three RC sections, but in the practical we vary capacitance and keep R constant, but still for different capacitance we always get sustained oscillation,
According to tan^-1(xc/R) as soon as we vary xc phase shift should must vary, still we get sustained oscillations means that loop is still shifting 180° but how can this is possible if Xc is varying ???
Have you checked the frequency of oscillation ? When you are changing the capacitance then frequency of oscillation is also changing. That means now, for new frequency of oscillation, you will get 180 phase shift. For the previous oscillation frequency, of course the phase shift will not be 180.
@@ALLABOUTELECTRONICS ,
Yes, we have measured frequency, It means that, in product wc as c vary w will also vary such that the product remains nearly same and overall phase shift becomes 180, thank you very much 😊
Sir
I had a brain storming doubt
If we need each rc network to provide 60 degree phase shift then
Tan^-1 (Xc/R)= 60
Thus 1/omega×C×R =root 3
So we get frequency of oscillation as 1/2pieRC(root3) and not root 6
But in derivation we get root 6
But frequency must be the root 3 case for the RC network to produce a phase shift 60 degree so that total 180 degree
Thank you in advance
The thing is , when we connect three RC network togather, it won't behave like a single RC netwrok. Each stage will act as a load for the precious stage. That's why we need to consider the three stage as a single network and go through the entire derivation.
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS oh ok
Got it
Thank you so much sir
But in lecture it is told that if two 3stage are there than each one will provide equal amount of phase shift so why can't we use w=1/RCroot(3) ???
@@ALLABOUTELECTRONICS
It would be nice if you provide in the end of you each video a simulation, ex in proteus, of your circuit. So we do know what is really going on. My onpinion.
Can you explain why you said that gain will be 0 in two cascaded RC circuit when R=0? I am finding the gain to be 1.
When 'R' is equal to zero, the ground resistors are not present. Therefore, the feedback circuit contains a bunch of capacitors in series with the feedback resistors. This causes the gain of the feedback loop to plummet.
In your prev video about oscillators u told that op- amp are used in relaxation oscillators(which genrate non-sinusoidal signals ) but in rc phrase shift oscillators (which genrate sinusoidal signal) we are using it ,....Why???
In relaxation oscillator, the op-amp is used in saturation. ( The output of the op-amp is either + Vsat or - Vsat). Here it is used as a linear amplifier.
I hope it will clear your doubt
What software do you use for this presentation? thanks
Hello, can we use voltage divider method?
Sir what factor plays the role of input in starting when we not connected any feedback network ???
Without any feedback, the noise will be present in the input which contains all the frequency component. The amplifier will amplify all the frequency components by the gain factor. But as the amplitude of the noise is very small, at the output also you will observe nothing. (with out feedback)
The feedback network provides the frequency selectivity.
your videos were useful . Can you give the equations and pictures as a note or text in the description ? thank you
Really nice video :)
Loved it👍
Thank you so much
Thanks a lot.great explanation
Sir is there something we can say about these(RC and LC) oscillators}???
Thank you
I didn't get. Would you please elaborate it.
@@ALLABOUTELECTRONICS oh sorry for my typing mistake
I meant is there something we can say about the amplitude of the sine wave generated by these(RC and LC) oscillators
Thank you sir
Using this circuit it is hard to control the gain. Because with slight change in the gain the output may go into the saturation. To control the amplitude, the limiter circuit is used with the oscillator ( It is implemented using the diodes and few resistors).
It is difficult to explain that circuit over here. Someday I will cover that circuit in detail with simulations.
@ALL ABOUT ELECTRONICS @ALL ABOUT ELECTRONICS ok sir thank you
However can we predict or say anything about the amplitude of the sine wave we just obtained
Is it any way related to any value of the components
Or more like is it possible for us to create a for example Wein Bridge oscillator of +-9v output
@@abhijithanilkumar4959 Yes, the amplitude depends on the gain. As I said to you earlier, we need to fine-tune the gain to get the particular amplitude. Using the gain-adjustment circuit, it is possible to achieve the desired amplitude.
Why there is voltage gain from v2 to v0?it should be drop in voltage
There is no voltage gain. We are just writing the nodal equations to find the transfer function Vo/Vin.
How it can acts as a positive feedback?
what will be the final output voltage level (Vout) ?
Great sir
Sir by using transistor also we can do the same there sir
at f=1/2piRCsqr(6) => phi= tan-1(1/sqr(6)) => phi=67.79 degree .
But for a Three stage RC circuit the phase angle should not be 60 degrees instead of 67.79 so that the overall feedback of RC circuit becomes 180 degree. Please Explain
Sir why 60 deg each nw equally ,is possible to three diff cimbination to achieve 180?
if the opamp provide 180 phase shift , then why dont we use non-inverting opamp??
The thing is, the overall phase shift should be 360. When you use, the RC circuit in the feedback then it will provide the 180-degree phase shift at one particualr frequency. so, to get a 360 phase shift, we need to use the op-amp in the inverting configuration.
@@ALLABOUTELECTRONICS okay so if we use non inverting opamp with multiple stage RC circuit ( inorder to get 360°). Does it still work the way you said?
excuse me sir, i think you has mistake at 11:30 put symbol (-)minus, it actuly (+)plus 1/omega^2C^2R^2. sorry for interrupnt and sorry for my bad english, correct me plz if i wrong
@@umarqalbibinmohdluai3604 not it would be (-) sign as there is j^2 which equal -1
If I substitute w resonance into the phase formula, the phase is not equal to 60
Hello , have you samthing about open and closed loop gain?
How charging and discharging of capacitors occur through the resistor. can u please explain