The timestamps for the different topics covered in the video: 0:26 Introduction to Wein Bridge Oscillator 2:31 Resonant frequency and the Oscillation criteria for the Wein Bridge Oscillator 5:07 Example: Design of Wein Bridge Oscillator 6:58 Derivation of frequency for the Wein Bridge Oscillator
First of all, I hope you understood the basic working principle of the oscillator. In the oscillator, there is no input. The noise acts like as an input. And the noise has all the frequency components, few Hz to even 100s of MHz. Out of all the frequency components, only one frequency is selected by the feedback circuit. (And that's why the feedback network is also known as the frequency selective feedback network). Which eventually gets amplified and we get the oscillation. In wien bridge oscillator, the RC feedback network acts like a notch filter. So, it provides maximum output at only one particular frequency. If you look at the circuit, the output of the feedback circuit is given to the positive input terminal of the op-amp. (There is positive feedback). And because of that, at one frequency, the oscillations will build up in the circuit. Also, there is a negative feedback in the circuit through the resistor R4, which is a part of the non-inverting op-amp. So, this positive and negative feedback controls the loop gain. And by careful selection of the component we can get AB= 1 (Or for practical circuit, slightly more than 1). I hope it will help you in understanding a bit more about the wien bridge oscillator.
I like that you do the derivations of the equations. That set you apart from many other youtubers electronics videos. I have had to go back and review my math to really follow your videos. I am getting a better education now than I did in university. I think that is why you are getting so many subscribers also.
Sir At 1:05 shouldn't the positions of R3 and R4 be exchanged Because in the Wein bridge R4 is to ground and R3 is connected to output And when I analysed Wien bridge oscillator using BJT I got R3/R4 = 2 (which is correct if R3is in feedbck path and R4 is grounded) And not R4/R3=2 as in 4:46 Thank you in advance
I am passing in my Analog Ckt exams just because of your videos... Thankss a lot for ur clean and organized way of teaching... Loads of love and good wishes ❤
If you actually build this circuit, you do need some kind of active gain control. If you try to use fixed values, it either goes into saturation, creating a square wave, or, there isn't enough gain, and it simply will not continue to oscillate. I've used a very small incandescent lamp as a non-linear resistance in series with R4, and if the amplitude goes too high, the bulb filament heats up, and it's resistance goes up. It produces a beautiful sine wave, but light bulbs are microphonic, and thermally massive, sometimes causing a "Bounce" in amplitude, then it finally stabilizes. It also causes distortion at very low frequencies.(below~5 Hz) because the filament can heat and cool in real time, as the wave crests, and crosses zero. I've seen other schemes that use parallelled diodes, cathode to anode, cathode to anode, where if the amplitude exceeds the forward voltage drop of the diodes, it clamps the amplitude. This is accomplished by dividing R4 into two resistors and putting the diodes in parallel with a low value of resistance, so that at amplitude, only ~.6 volts appears across it at set amplitude. If the output amplitude goes too high, the diodes begin conducting, effectively reducing the gain of the amplifier. Unlike the light bulb scheme, it works equally well at any frequency, even down below 1 Hz. It does cause a small compression in the wave form, but you can't see it on an oscilloscope. It shows up on a harmonic distortion meter.(less than 1% THD) you have to balance distortion with stability. the less influence the diodes have, the less distortion, the more, the better the stability.
Amazing video man, thank you so much for explaining everything in detail from the roots up. Only mistake you made im pretty sure is you mixed up R3 and R4 around
Because at resonance, the circuit is purely resistive in nature and hence overall phase shift should be equal to zero. From that equation perspective, Vo/Vin will have zero phase shift when the first you mentioned is zero. I hope it will clear your doubt.
2:50 Sir How is it a notch filter Isn't notch filter a band stop filter And here this feedback RC network allows only a single frequency thus isn't this one a band pass filter rather than a notch filter?? Thank you
In starting part of video when u re-draw the circuit in that instead of R3, R4 is ground and later when u separately drawn the feedback network in that parallel combination of RC is connected to vout instead of Vin.
@@ALLABOUTELECTRONICS what's about feedback network drawn separately..in that low parallel RC combination is connected connected to vout instead of Vin.
It's just for finding the transfer function. It has nothing to do with the circuit drawn earlier. The more appropriate notation would be Vout and V+. (I mean Vout as input and V+ as output). But we are more familiar with the Vo and Vin. So, the same was used for deriving the transfer function. I hope it will clear your doubt now.
Can you please ans this AB =1 ; where A is open loop gain then how can you write A= 1+ R4/ R3 . Isn't the above equation meant for closed Loop gain Of non inverting amplifier. Correct me if I'm wrong. Besides that your video are short and to the point👍
Sir, is there any mistake in R3 and R4 in wien bridge oscillator between original circuit and the equivalent circuit that you have drawn in the screen??
in the transformed ckt you put R4 connected to -ve input and output of OpAmp,but in the basic ckt R4 has one hand grounded and another one to -ve input of OpAmp. Kindly explain this.
If you have observed the circuit, the RC circuit in the feedback is frequency selective circuit. And acts like a notch filter. So, it only allows a single frequency and rejects the remaining. If you know the Fourier transform, a single frequency in the frequency domain represents the pure sinewave in the time domain. For example, a triangular wave or square wave consists of multiple frequencies. That's why this Wien bridge or any harmonic oscillator like RC phase shift oscillator or Colpitts oscillator generates the sine wave. I hope it will clear your doubt.
Excelent videos 👏🏻👏🏻👏🏻👌🏼 Only one question, someone knows Why in colpitts oscilator we need to include the internal resistence of the output (ro) to obtain the correct equations, and in this oscilator is not necessary to the analysis? Thank you for all 🙏🏼
You call the positive feedback network a notch filter, but a notch filter attenuates a specific frequency. Isn't it rather a narrow bandpass filter? Also you did not mention the issue of amplitude stabilisation important to this circuit. With just resistors for R3 and R4 the circuit will not have stable amplitude, possible solutions include an NTC element in place of R4 or a lamp (PTC) in place of R3, but there are other solutions. Otherwise great explanation.
The magnitude of the sine wave will depend on the Loop gain and the supply voltage. Like I said earlier, for oscillations to build up, initially the gain A should be slightly more than 3. And once the oscillations build up then get gain should be reduced to 3. To achieve that a tungsten lamp can be used as resistor R3. Initially, it's value will be less than let's say R. And once the oscillation builds up then due to heating the value of R3 will slightly increase such that loop gain becomes 1. Here too, the amplitude will depend on the supply voltages of the op-amp as well the op-amp itself. By changing the supply voltages, the amplitude can be changed. The other circuit to control the amplitude of the wien bridge oscillator is using the diode. Where two diodes along with series resistor are used to control the amplitude of the output in the feedback. (In parallel with R4 resistor ). Usually, the series resistor with the diode (Let's say R5) is approx. 10 times the value of R4. And by controlling the gain of the Amplifier (either changing the value R3 or R4 slightly), the amplitude of the oscillation can be varied. I hope it will clear your doubt.
I constructed the circuit with a +-15V supply voltage to op-amp and the output that i got was a sine wave whose amplitude I can vary from 24-27Vpp with nearly sustained output. Only thing that makes me wonder is that if the gain needs to be at 3 as beta is 1/3 I shouldn't have the freedom to change gain and if I change gain the effect of AB not =1 was not that much visible in the output range of 24-27Vpp. I got your technique of how the oscillations will be built up, but theoretically can I prove the output will be close to supply voltage? Also how the two diodes along with series resistor circuit will look like.. a bit confused can you put a picture with some explanation as in how the output voltage is controlled. Thanks for your quick response. Awesome job you're doing!
24V- 27V peak to peak voltage is almost near the supply rail. I mean to say if 15V is supply voltage then the output would be restricted to around 12.5- 13V. And that seems so. See, even if AB goes slightly even above 1 then also you will get the same result, as output is near saturation to the supply voltage. But if AB goes below 1 then you may not observe the oscillations. You can try that by changing the gain of A. Because in that case, oscillations will die out. So, I would say, there is not much control for changing the amplitude in this circuit. (that would be visible in the second circuit using diodes which I was talking) The only way it would be visible is by changing the supply voltages. Let's say, if you change the supply voltages to +-10 V. then the output would be limited to around 16-17Vpp. Please message me on the facebook page. (You can find the link of the page in the description) whenever I get free, I will send you the picture of the circuit which is designed using a diodes.
brother one thing is bothering me... can we interchange the terminals of feedback network??? eg you have connected series RC to output and parallel RC to non inverting terminal so can we connect series RC to non inverting input terminal of opamp and parallel RC to output of opamp?
If you have followed the video, then the feedback fraction Beta is 1/3 when the input is given at the series RC circuit and output is taken across the parallel RC circuit. If the order of these circuits is changed then we will not get the same transfer function. That's why it can't be changed. I hope it will clear your doubt.
Please do not link it with the oscillator circuit. While explaining I have considered this circuit as a separate circuit. (Just to explain the working of the circuit in general). The input and output are not the input /output of the oscillator. I just wanted to explain, if some input is given on side of this circuit then how this circuit behaves.
How is it that in the wein bridge oscillator the real part of impedance is equated to 0 but in the colpitt oscillator the imaginary part is equated to 0?
Hi, I’m newish to electronics so this may be really stupid, but what is J and w and how did you get the initial equations that you used. The rest of the maths was great and really easy to follow but I just don’t know how it started. Thanks
if I have a previous circuit that uses photocell as a voltage divider, where should I connect that voltage on the oscillator, the positive power source? or negative input?
It if is designed using the op-amp, then some op-amp requires both positive and negative power supply, while some required only positive power supply. If it is single power supply op-amp, then you can apply it it positive power supply.
Sir Bit if we are designing this with 2 BJTs in 2 stage so that phase shift is 180+180=360 How will we design the BJT biasing circuits We can design the components of the bridge by R3/R4 =2 and from frequency formula But how to get A=3 if beta of BJT is like 100 each and Vcc something like 10v Thank you
May I know exactly what you didn't understand in the derivation. In this derivation, the output of the op-amp is actually input to the feedback circuit (Vin). And the output of the feedback circuit (Vout) is going as input to the non-inverting terminal of the op-amp. So, transfer function Vout/Vin is found to get the amplitude and phase of the feedback circuit.
how parallel rc network acts as alow pass filter ???? in low pass filter if we take o/p from capacitor this is known as low pass filter but here the voltage take from both r and c r equal .
At high frequencies, the capacitor provides very low impedance. Xc = 1/(2*pi*f*c) So, the overall impedance of the parallel combination of R and C will be very low. Let's say, the impedance of the parallel combination is Z2. And series combination R1 + C1 is Z1. Then Z2*Vin/(Z1+Z2) will be very low at high frequencies. At low frequency, Xc2 will be high and can be assumed as an open circuit. So, Z2 is approximately equal to R2. And you will get finite output at low frequencies. And that's how the parallel combination of R2 and C2 acts as a low pass filter. I hope it will clear your doubt.
As long as you use ONLY solid state devices you can't get very high frequencies, no matter design or configuration, this is why a crystal is so necessary, in order to reach freqs of several hundred thousands and much higher. If you design an oscillator for, let's say, 0.5 Mhz using solid state only, you'll get it work but it will be unstable. In fact, the highr freq. the less reliable.
@@mrinmoymandal2695 No, we are not applying any AC input signal. Only DC power supply is applied to the active component. In this case, its the supply voltage to the op-amp. And the AC signal is eventually growing from the thermal noise present in the circuit. For more information about the working of any oscillator circuit, please check my video on how the oscillator works. Here is the link : th-cam.com/video/XVS8Puf4tiw/w-d-xo.html
If you watch the video 6:58 onwards, I have already derived it. If you go through the little bit of mathematics, then you will get it that how at the resonant frequency the phase shift is zero.
Sir, I think in the redrawn circuit of wein bridge oscillator the R3 and R4 should exchange their place. I think the voltages are not matching. Please clarify it.
The timestamps for the different topics covered in the video:
0:26 Introduction to Wein Bridge Oscillator
2:31 Resonant frequency and the Oscillation criteria for the Wein Bridge Oscillator
5:07 Example: Design of Wein Bridge Oscillator
6:58 Derivation of frequency for the Wein Bridge Oscillator
Please give me physical explanation of wein bridge oscillator
First of all, I hope you understood the basic working principle of the oscillator. In the oscillator, there is no input. The noise acts like as an input. And the noise has all the frequency components, few Hz to even 100s of MHz. Out of all the frequency components, only one frequency is selected by the feedback circuit. (And that's why the feedback network is also known as the frequency selective feedback network). Which eventually gets amplified and we get the oscillation.
In wien bridge oscillator, the RC feedback network acts like a notch filter. So, it provides maximum output at only one particular frequency. If you look at the circuit, the output of the feedback circuit is given to the positive input terminal of the op-amp. (There is positive feedback).
And because of that, at one frequency, the oscillations will build up in the circuit. Also, there is a negative feedback in the circuit through the resistor R4, which is a part of the non-inverting op-amp.
So, this positive and negative feedback controls the loop gain. And by careful selection of the component we can get AB= 1 (Or for practical circuit, slightly more than 1).
I hope it will help you in understanding a bit more about the wien bridge oscillator.
Why you put j=-1?
thanks man
@@ALLABOUTELECTRONICS
How are you so patient to type so much for somebody and expecting nothing in return
You are truly amazing sir
I like that you do the derivations of the equations. That set you apart from many other youtubers electronics videos. I have had to go back and review my math to really follow your videos. I am getting a better education now than I did in university. I think that is why you are getting so many subscribers also.
Keep going, you will succeed one day.
Don't give up.
Your content is very useful.
Try to add variety.
For eg, electronics projects etc.
Best of luck.
He is already succeeded for me . Trust me :)
Sir
At 1:05
shouldn't the positions of R3 and R4 be exchanged
Because in the Wein bridge R4 is to ground and R3 is connected to output
And when I analysed Wien bridge oscillator using BJT I got R3/R4 = 2 (which is correct if R3is in feedbck path and R4 is grounded)
And not R4/R3=2 as in 4:46
Thank you in advance
I am passing in my Analog Ckt exams just because of your videos... Thankss a lot for ur clean and organized way of teaching... Loads of love and good wishes ❤
You are my one click destination for any Analog electronics problem.👍
This channel is top-notch, superb work all available for free. Thank you.
Very helpful video one must watch this to clear his doubts. My all doubts are cleared regarding Wein Bridge Oscillator. Great work ! Thank you
Thank you soo sooo much...i urgently needed the explanations of these oscillators and now i come to you tube and am seeing this....♥️
thank you thank you thank you...each of your videos I learn more than entire university class
Sir i love your every video....
In your every video my concept gets crystal clear...which couldn't clear in my whole engineering .
I understand better with this video. Please come with more videos. Thank you.
I really appreciate your help with this explanations. Keep it up! May the Lord Jesus bless you.
One important thing everyone should notice he uses Vout/Vin in the smaller ckt which should be named Vf/Vout according to main ckt to avoid confusion.
Right 👍🏻
Thank you sir very nice gide & very nice best explain wien bridge oscillator teaching video.👍
If you actually build this circuit, you do need some kind of active gain control. If you try to use fixed values, it either goes into saturation, creating a square wave, or, there isn't enough gain, and it simply will not continue to oscillate. I've used a very small incandescent lamp as a non-linear resistance in series with R4, and if the amplitude goes too high, the bulb filament heats up, and it's resistance goes up. It produces a beautiful sine wave, but light bulbs are microphonic, and thermally massive, sometimes causing a "Bounce" in amplitude, then it finally stabilizes. It also causes distortion at very low frequencies.(below~5 Hz) because the filament can heat and cool in real time, as the wave crests, and crosses zero. I've seen other schemes that use parallelled diodes, cathode to anode, cathode to anode, where if the amplitude exceeds the forward voltage drop of the diodes, it clamps the amplitude. This is accomplished by dividing R4 into two resistors and putting the diodes in parallel with a low value of resistance, so that at amplitude, only ~.6 volts appears across it at set amplitude. If the output amplitude goes too high, the diodes begin conducting, effectively reducing the gain of the amplifier. Unlike the light bulb scheme, it works equally well at any frequency, even down below 1 Hz. It does cause a small compression in the wave form, but you can't see it on an oscilloscope. It shows up on a harmonic distortion meter.(less than 1% THD) you have to balance distortion with stability. the less influence the diodes have, the less distortion, the more, the better the stability.
such a clear explanation in so easy manner. thanks!
Thanks a lot. 0:57 R3 and R4 must be interchanged so the diagram be the same as in 0:59
Amazing video man, thank you so much for explaining everything in detail from the roots up. Only mistake you made im pretty sure is you mixed up R3 and R4 around
In redrawing from bride model to op amp model R3 and R4 -> R3 must be feed back resistor in inverting op amp config
Yeah. They must interchange. R3 and R4
Excellent explanation 👏👏👏
Very good video! Enriched with proper mathematical explanation.
Keep it up🤠
Beautifully explained. Thank you so much.
At 9:50 you had said that at resonant frequency the term 1-w^2R1R2C1C2=0 : but why?
Because at resonance, the circuit is purely resistive in nature and hence overall phase shift should be equal to zero. From that equation perspective, Vo/Vin will have zero phase shift when the first you mentioned is zero. I hope it will clear your doubt.
man,you nailed it,very good explanation.Thank you for such a clear explanation.
Sir, A=1+(R3/R4)
BUT AWESOME ❤️❤️ VIDEO SO HELPFUL
2:50
Sir
How is it a notch filter
Isn't notch filter a band stop filter
And here this feedback RC network allows only a single frequency thus isn't this one a band pass filter rather than a notch filter??
Thank you
No word to admire u r explanation 👏👏👏👏
It's very helpful. Please make Hartley oscillator explanation video also
Thank you for every think. it is very beneficial for me. İ wish you will very succesful.
Thanks for giving theory below and everything 👍
Thanks sir for such a great lecture
谢谢！
Thank you. I really appreciate your support.
Thanks for the information. It is helpful.
awesom truely fantastic explaination
Thank you very much . The flow is very simple to catch
i love u love u so much ^^
Your explanation has greatly helped me to understand my major.
Nice explaination
Thanku for such a helpful content thanku so much 🙏
Tq so much.....u nailed it!!!!
The only mistake is just reverse R3,R4 positions in the first circuit diagram of we in bridge oscillator
And upload videos frequently 😊
In starting part of video when u re-draw the circuit in that instead of R3, R4 is ground and later when u separately drawn the feedback network in that parallel combination of RC is connected to vout instead of Vin.
Yes, instead of R3 it should have been R4. (The positions of R3 and R4 got interchanged). Rest is fine. Thanks for pointing it out.
@@ALLABOUTELECTRONICS what's about feedback network drawn separately..in that low parallel RC combination is connected connected to vout instead of Vin.
It's just for finding the transfer function. It has nothing to do with the circuit drawn earlier. The more appropriate notation would be Vout and V+. (I mean Vout as input and V+ as output).
But we are more familiar with the Vo and Vin. So, the same was used for deriving the transfer function.
I hope it will clear your doubt now.
You are great man thank you.
Can you please ans this
AB =1 ; where A is open loop gain then how can you write
A= 1+ R4/ R3 .
Isn't the above equation meant for closed Loop gain Of non inverting amplifier.
Correct me if I'm wrong.
Besides that your video are short and to the point👍
U r really superb sirr🤗🤗🤗
Sir, is there any mistake in R3 and R4 in wien bridge oscillator between original circuit and the equivalent circuit that you have drawn in the screen??
yes i am also having the same doubt
in the transformed ckt you put R4 connected to -ve input and output of OpAmp,but in the basic ckt R4 has one hand grounded and another one to -ve input of OpAmp. Kindly explain this.
Sir why this Wien bridge oscillator generates only the sine waves. ? Could u please explain this Sir
If you have observed the circuit, the RC circuit in the feedback is frequency selective circuit. And acts like a notch filter.
So, it only allows a single frequency and rejects the remaining.
If you know the Fourier transform, a single frequency in the frequency domain represents the pure sinewave in the time domain.
For example, a triangular wave or square wave consists of multiple frequencies.
That's why this Wien bridge or any harmonic oscillator like RC phase shift oscillator or Colpitts oscillator generates the sine wave.
I hope it will clear your doubt.
Great work sir 😊
There is mistake in low pass filter
no mistake see it again please
@@ephrembedhaso7741 yes there is small mistake
Thank you so much!!
R3 & R4 are misplaced in bridge circuit
yes
Yes
Right
Great presentation. I wounder if you could change R with L in the feedback and do the calculation. We should get better phase noise.
Sir, your video s are excellent and very clear. May I know what is the software you are using to draw the circuits?
yeah this is awsome...Can you please make a video on Hartley Oscillator?
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Now, as we know this op-amp is configured in a non-inverting configuration, so the gain
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Very nice sir
Excelent videos 👏🏻👏🏻👏🏻👌🏼
Only one question, someone knows Why in colpitts oscilator we need to include the internal resistence of the output (ro) to obtain the correct equations, and in this oscilator is not necessary to the analysis?
Thank you for all 🙏🏼
Do make videos on feedback amplifiers also
You call the positive feedback network a notch filter, but a notch filter attenuates a specific frequency. Isn't it rather a narrow bandpass filter? Also you did not mention the issue of amplitude stabilisation important to this circuit. With just resistors for R3 and R4 the circuit will not have stable amplitude, possible solutions include an NTC element in place of R4 or a lamp (PTC) in place of R3, but there are other solutions. Otherwise great explanation.
Long time ago but you are quite correct....still !
first of all thank u so much it really helped me, but I have a question
why the ratio between the output and input voltages are 1/3?
From 7:00 onwards, I have explained the derivation for the same. You will get to know that, why it is 1/3.
it will be useful if you use current directions (while applying kirchoff curent law ) and define numbers to all equations.....
Yes, will keep that in mind in the future videos
Thank you sir
can anyone explain to me why the RC section has a gain of 1/3? Many Thanks!
He explains at 10:52
Nicely explained. I have a doubt though.. what's the magnitude of sinusoidal output voltage we get of win bridge how can we evaluate that?
The magnitude of the sine wave will depend on the Loop gain and the supply voltage.
Like I said earlier, for oscillations to build up, initially the gain A should be slightly more than 3. And once the oscillations build up then get gain should be reduced to 3.
To achieve that a tungsten lamp can be used as resistor R3.
Initially, it's value will be less than let's say R. And once the oscillation builds up then due to heating the value of R3 will slightly increase such that loop gain becomes 1.
Here too, the amplitude will depend on the supply voltages of the op-amp as well the op-amp itself.
By changing the supply voltages, the amplitude can be changed.
The other circuit to control the amplitude of the wien bridge oscillator is using the diode.
Where two diodes along with series resistor are used to control the amplitude of the output in the feedback. (In parallel with R4 resistor ).
Usually, the series resistor with the diode (Let's say R5) is approx. 10 times the value of R4.
And by controlling the gain of the Amplifier (either changing the value R3 or R4 slightly), the amplitude of the oscillation can be varied.
I hope it will clear your doubt.
I constructed the circuit with a +-15V supply voltage to op-amp and the output that i got was a sine wave whose amplitude I can vary from 24-27Vpp with nearly sustained output. Only thing that makes me wonder is that if the gain needs to be at 3 as beta is 1/3 I shouldn't have the freedom to change gain and if I change gain the effect of AB not =1 was not that much visible in the output range of 24-27Vpp.
I got your technique of how the oscillations will be built up, but theoretically can I prove the output will be close to supply voltage?
Also how the two diodes along with series resistor circuit will look like.. a bit confused can you put a picture with some explanation as in how the output voltage is controlled.
Thanks for your quick response. Awesome job you're doing!
24V- 27V peak to peak voltage is almost near the supply rail. I mean to say if 15V is supply voltage then the output would be restricted to around 12.5- 13V. And that seems so. See, even if AB goes slightly even above 1 then also you will get the same result, as output is near saturation to the supply voltage. But if AB goes below 1 then you may not observe the oscillations. You can try that by changing the gain of A.
Because in that case, oscillations will die out.
So, I would say, there is not much control for changing the amplitude in this circuit. (that would be visible in the second circuit using diodes which I was talking)
The only way it would be visible is by changing the supply voltages. Let's say, if you change the supply voltages to +-10 V. then the output would be limited to around 16-17Vpp.
Please message me on the facebook page. (You can find the link of the page in the description)
whenever I get free, I will send you the picture of the circuit which is designed using a diodes.
Bro video on " concept of nodal analysis"
brother one thing is bothering me... can we interchange the terminals of feedback network??? eg you have connected series RC to output and parallel RC to non inverting terminal so can we connect series RC to non inverting input terminal of opamp and parallel RC to output of opamp?
If you have followed the video, then the feedback fraction Beta is 1/3 when the input is given at the series RC circuit and output is taken across the parallel RC circuit. If the order of these circuits is changed then we will not get the same transfer function. That's why it can't be changed.
I hope it will clear your doubt.
ALL ABOUT ELECTRONICS thanks man .. I really appreciate your efforts
1:40 how come vout is connected to parallel circuit of RC
Please do not link it with the oscillator circuit. While explaining I have considered this circuit as a separate circuit. (Just to explain the working of the circuit in general). The input and output are not the input /output of the oscillator. I just wanted to explain, if some input is given on side of this circuit then how this circuit behaves.
What if the capacitor is replaced by inductor. Can you confirm the oscillation frequency is f =R/(2*pi*L)
In rc phase shift oscillator, why we will consider -jixc...can you please explain
Are you referring to Vo/Vin expression, Vo/Vin = R/(R - jXc) at 2:34 in RC phase shift oscillator video?
How is it that in the wein bridge oscillator the real part of impedance is equated to 0 but in the colpitt oscillator the imaginary part is equated to 0?
Hi, I’m newish to electronics so this may be really stupid, but what is J and w and how did you get the initial equations that you used. The rest of the maths was great and really easy to follow but I just don’t know how it started. Thanks
It's the reactance of the capacitor. It can be represented as 1/jwc.
where w is the frequency in radian. w= 2πf
Wow! No calculus! Beautiful!
Frank
Frank Reiser Video/Audio Service
Let Newton bless you to like calculus.....
if I have a previous circuit that uses photocell as a voltage divider, where should I connect that voltage on the oscillator, the positive power source? or negative input?
It if is designed using the op-amp, then some op-amp requires both positive and negative power supply, while some required only positive power supply. If it is single power supply op-amp, then you can apply it it positive power supply.
Sir if you can say what are the advantages and disadvantages then it should be more efficient
Sir
Bit if we are designing this with 2 BJTs in 2 stage so that phase shift is 180+180=360
How will we design the BJT biasing circuits
We can design the components of the bridge by R3/R4 =2 and from frequency formula
But how to get A=3 if beta of BJT is like 100 each and Vcc something like 10v
Thank you
can you kindly explain that in order to show the phase shift zero, why you had equated the real term in the denominator to zero ?
i too have the same doubt......
resonant frequency:w^2*R1*R2*C1*C2 = 1, may i ask why the equation works?
THANK YOU SIR
If all the variables we keep constant and we change only the value of R3 how can the frequency will be affected,
What happens if the gain
𝐵𝐴>1 instead of 𝐵𝐴=1
What kind of signal will I get at the output vout?
Is it still called a Wien Bridge?
Square wave
plyzz upload a video on LC oscillator
I have already uploaded the video on LC oscillator. (Colpitts Oscillator)
Here is the link:
th-cam.com/video/1fgw-ONlAcc/w-d-xo.html
Could u plz explain the math derivation after sub of z1&z2
May I know exactly what you didn't understand in the derivation.
In this derivation, the output of the op-amp is actually input to the feedback circuit (Vin). And the output of the feedback circuit (Vout) is going as input to the non-inverting terminal of the op-amp.
So, transfer function Vout/Vin is found to get the amplitude and phase of the feedback circuit.
@@ALLABOUTELECTRONICS @8.37 after u substitute the values of z2/z1+z2 plz explain the math derivation
Using LCM the equation has been simplified.
very good explanation but your video is not include bistable multivibrirator using op-amp
platform you are using to create this videos?????
how parallel rc network acts as alow pass filter ????
in low pass filter if we take o/p from capacitor this is known as low pass filter but here the voltage take from both r and c r equal .
At high frequencies, the capacitor provides very low impedance. Xc = 1/(2*pi*f*c)
So, the overall impedance of the parallel combination of R and C will be very low. Let's say, the impedance of the parallel combination is Z2.
And series combination R1 + C1 is Z1. Then Z2*Vin/(Z1+Z2) will be very low at high frequencies.
At low frequency, Xc2 will be high and can be assumed as an open circuit. So, Z2 is approximately equal to R2. And you will get finite output at low frequencies.
And that's how the parallel combination of R2 and C2 acts as a low pass filter.
I hope it will clear your doubt.
thanks sir
Nice explanation....
But can you please tell me why there is no phase shift at only (( f=1/2πRC)) in non mathematical way?
This is in ideal form
what is the highest frequency you can design the wien bridge for and what type of wave does it produce?
As long as you use ONLY solid state devices you can't get very high frequencies, no matter design or configuration, this is why a crystal is so necessary, in order to reach freqs of several hundred thousands and much higher. If you design an oscillator for, let's say, 0.5 Mhz using solid state only, you'll get it work but it will be unstable. In fact, the highr freq. the less reliable.
Sir can you plzz make video on designing of rc phase shift oscillator
I have already made video on RC phase shift oscillator. Please check on the oscillator playlist.
Sir can you tell me why does the wein bridge oscillator Will not work when we don't supply any input voltage
Do you mean to say why it will not work without a dc power supply?
Sir do we applying any input sigal here for which getting a sinusoidal out / it's generat due to leakage current???
@@mrinmoymandal2695 No, we are not applying any AC input signal. Only DC power supply is applied to the active component. In this case, its the supply voltage to the op-amp. And the AC signal is eventually growing from the thermal noise present in the circuit.
For more information about the working of any oscillator circuit, please check my video on how the oscillator works.
Here is the link :
th-cam.com/video/XVS8Puf4tiw/w-d-xo.html
how did u convert diagram from bridge to non bridged
Nice vedio
Kdk bhawa
Sir my doubt is how phase shift is zero at resonant freq can you explain plz
If you watch the video 6:58 onwards, I have already derived it. If you go through the little bit of mathematics, then you will get it that how at the resonant frequency the phase shift is zero.
Thanks sir
Sir, I think in the redrawn circuit of wein bridge oscillator the R3 and R4 should exchange their place. I think the voltages are not matching. Please clarify it.
Feedback factor is vout /vin.i think so
This is minor, but this circuit is often mislabeled. It is really spelled Wien bridge oscillator.
Thank you for pointing that out.
how do we get the exp
A=1+R4/R3
The op-amp is configured in non-inverting amplifier configuration. That's why its gain would be 1 + Rf/R1.
In this case, it would be 1 + R4/R3.