# How do Sinusoidal output comes out in the Wein-Bridge Oscillator

## Main Question or Discussion Point

This question was asked to me in a VIVA.

How do Sinusoidal output comes out in the Wein-Bridge Oscillator.

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I tried to solve the problem using the control system. That is, by deriving the transfer function of the Wein-Bridge Oscillator, but first i got confused on where to take the input as there is no input and oscillators work due to noise input(this is what written in the books). Then i assumed the input to be the negative Op-Amp terminal, this time the confusion was on what function the input Voltage should be?

I have studied two Sinusoidal Oscillators, another one is phase-shift oscillators, and I don't know the 'How' of Phase-Shift oscillators also. The book i follow is Sedra-Smith and doesn't say anything about the 'How' behind the generation of Sinosoidal output. I have one more book on Op-Amp Circuits and nothing is written about it this in that book also.

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NascentOxygen
Staff Emeritus
The input to the passive network is the output of the op-amp. As far as the transfer function goes, you need the combination to give 2.Pi radians of phase shift at a gain of just over unity, at some frequency, for oscillations to be self-sustaining.

I suggest that you include a scan of the particular circuit you have in mind, if you have further questions.

I suggest that you include a scan of the particular circuit you have in mind, if you have further questions.
This is the Circuit, the wein-bridge oscillator.

The input to the passive network is the output of the op-amp. As far as the transfer function goes, you need the combination to give 2.Pi radians of phase shift at a gain of just over unity, at some frequency, for oscillations to be self-sustaining.
I know the two conditions that you just told. This is the only two things that we have studied about oscillators. Which comes from Av*β = 1
1. β is real.
2. ∠ Av*β = 2n∏

I derived the Transfer function taking non-inverting input as the input, but the question is how does the Sine wave comes out. Without knowing input i can't predict the output.

NascentOxygen
Staff Emeritus
The potential divider at the - input sets the gain of the amplifier stage (the OP-AMP), at a value that just makes up for the attenuation of the passive Wein bridge, restoring loop gain to 1. Input to the bridge comes from the output of the OP-AMP. Output of the bridge is seen at the - input of the OP-AMP.

So long as the loop gain is not much over 1 where phase shift is 2 Pi, and this condition is not met at another frequency, the system noise will develop into a self-sustaining sinusoidal oscillation within a short period of time. The more loop gain you allow, the wider the band over which feedback can give sustained oscillation, and the more frequency components you introduce into your signal, meaning it starts to depart from a pure sinusoid.

1 person
The potential divider at the - input sets the gain of the amplifier stage (the OP-AMP), at a value that just makes up for the attenuation of the passive Wein bridge, restoring loop gain to 1. Input to the bridge comes from the output of the OP-AMP. Output of the bridge is seen at the - input of the OP-AMP.
That means the principal of operation is same as that of a square-wave generator and monostable multivibrator. I understood this part.

So long as the loop gain is not much over 1 where phase shift is 2 Pi, and this condition is not met at another frequency, the system noise will develop into a self-sustaining sinusoidal oscillation within a short period of time.
Why sinusoid?. I mean when we study the square wave generator(whose working is similar to this), we derive how output comes out to be the square wave.

The more loop gain you allow, the wider the band over which feedback can give sustained oscillation, and the more frequency components you introduce into your signal, meaning it starts to depart from a pure sinusoid.
Again same question, why. Isn't there a General output Voltage as function of time and Circuit components as in the case of square wave generator?

NascentOxygen
Staff Emeritus
It's a linear amplifier/network, with a loop gain great enough at only one frequency to stop oscillations dying out. A squarewave can't leap out of nowhere in a linear system. Squarewaves originate from abrupt non-linearities, there are none here.

It's a linear amplifier/network, with a loop gain great enough at only one frequency to stop oscillations dying out. A squarewave can't leap out of nowhere in a linear system. Squarewaves originate from abrupt non-linearities, there are none here.
So you mean there is no proof for the Sinosiodal output. Is is just an observed fact?

Output signal is a sinewave because only for this one frequency wein-bridge and the amplifier meets the conditions necessary to start oscillations.

Output signal is a sinewave because only for this one frequency wein-bridge and the amplifier meets the conditions necessary to start oscillations.
We assume that the output is Sine wave that is wave we put s=jω. Putting s = jω is only true for sinusoidal transfer functions. My question is why do we assume the output to be sinusoid?

"Wein network" work as a highly selective bandpass filter. So even if you put square wave ta the input the output voltage will be sinusoidal. So the op amp input will see a sine wave at his non-inverting input. And next the amp will amplifier this sine wave 3x times. So the output also will be sinusoidal. Because just for one single frequency F = 1/(2 * pi RC) circuit meets conditions necessary to start oscillations.

On the other hand in multivibrator we have a all pass filter network so for all frequancy circuit meets the conditions necessary to start oscillations. And this is why we have a square wave output.

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sophiecentaur
Gold Member
We assume that the output is Sine wave that is wave we put s=jω. Putting s = jω is only true for sinusoidal transfer functions. My question is why do we assume the output to be sinusoid?
For some reason you keep ignoring the explanation you are being given. The feedback is only 'right' at one frequency so no other frequencies will be sustained - that makes the output a single frequency with no harmonics (i.e. a sine wave).
PS In general, it's pretty fruitless to question why the (well established) analysis of a common system happens to be done in a particular way. Science and Maths is full of apparently arbitrary approaches to solving problems - which just happen to work. Very annoying until you know the particular crafty dodge. Why do you divide both sides of an equation by the coefficient of x (etc. etc.)? Because you know it will work. Why do you try a sinusoidal solution to this sort of problem? Because you know it works.

NascentOxygen
Staff Emeritus
We assume that the output is Sine wave that is wave we put s=jω. Putting s = jω is only true for sinusoidal transfer functions. My question is why do we assume the output to be sinusoid?
It's a second order linear system, and the natural response of every second order system is a sinusoid, just as you see in a child's swing. Give it a jolt and it adopts SHM; give this filter/amplifier system a voltage jolt at switch-on, and it responds with SHM. The sinusoid doesn't decay away because of accurately setting the loop gain to just sustain the response.

NascentOxygen
Staff Emeritus
That means the principal of operation is same as that of a square-wave generator and monostable multivibrator. I understood this part.
The principal of operation of a square-wave generator, and of a sinusoidal oscillator, are totally different. Granted, superficially their schematics may sometimes bear a resemblance, but one is highly non-linear, and the other very linear. The two are horses of a different colour!

It's a second order linear system, and the natural response of every second order system is a sinusoid, just as you see in a child's swing. Give it a jolt and it adopts SHM; give this filter/amplifier system a voltage jolt at switch-on, and it responds with SHM. The sinusoid doesn't decay away because of accurately setting the loop gain to just sustain the response.
A second order linear system. If you could tell me where did you get this from.

For some reason you keep ignoring the explanation you are being given. The feedback is only 'right' at one frequency so no other frequencies will be sustained - that makes the output a single frequency with no harmonics (i.e. a sine wave).
I am not ignoring that explanation. Okay the feedback is right at only one frequency-agree. Then you say feedback is right at only one function(sine wave). How do match these two things-frequency and function. I mean the output could have been a different function with the same frequency!

sophiecentaur
Gold Member
I am not ignoring that explanation. Okay the feedback is right at only one frequency-agree. Then you say feedback is right at only one function(sine wave). How do match these two things-frequency and function. I mean the output could have been a different function with the same frequency!
Yes, but the phase and amplitude characteristic of the feedback in this oscillator is such that only one component of your hypothetical waveform would satisfy the requirement for oscillation. There is only one function that has one frequency component and that is a sinewave. The amplifying device in this sinewave oscillator needs to be linear so you can use superposition and treat signals of all possible frequencies independently. When you have a non-linear amplifying device, other waveforms may result - but the same thing could happen if you followed the nicely sinusoidal Wien Bridge with a non linear amplifier.
In practice, of course, there will be a certain amount of harmonic content, even in the Wien Bridge output.

NascentOxygen
Staff Emeritus
A second order linear system. If you could tell me where did you get this from.
It's from mathematics. Differential equations, etc., it's all a part of physics. The current through a capacitor is determined by the derivative of the voltage across it, and so on.

It's from mathematics. Differential equations, etc., it's all a part of physics. The current through a capacitor is determined by the derivative of the voltage across it, and so on.
I am pretty okay at maths and physics. The EE curriculum at my university is quite mathematical. For me understanding circuits with mathematical equation is the easiest method. I mean this is way we are taught circuits at our University. So if you could tell me those Differential equations describing wein-bridge oscillator, i would be very keen to solve those myself.

sophiecentaur
Gold Member
I googled "wein bridge oscillator analysis". Did you try that? You get loads of hits with treatments at various levels of difficulty. I suggest you look at them. It's a far more efficient way of learning than just asking questions and getting off - the - cuff answers, many of which will come straight off the net as people refresh their memories of stuff they knew quite well in the past.

Yes, but the phase and amplitude characteristic of the feedback in this oscillator is such that only one component of your hypothetical waveform would satisfy the requirement for oscillation.
Why are the phase and amplitude characteristics satisfying only sine function? If you could go into some mathematical proof, that would be easier for me to understand.

There is only one function that has one frequency component and that is a sinewave. The amplifying device in this sinewave oscillator needs to be linear so you can use superposition and treat signals of all possible frequencies independently.
And finally we assume it to be linear. I don't understand why this are assumptions is made. Not many Op-Amp circuits are linear, including the square generator, comparator, monostable multivibrator. I will definitely agree that this circuit is linear just because you and others saying this. I just want to know how do we came to conclusion that this is linear.

I googled "wein bridge oscillator analysis". Did you try that? You get loads of hits with treatments at various levels of difficulty. I suggest you look at them. It's a far more efficient way of learning than just asking questions and getting off - the - cuff answers, many of which will come straight off the net as people refresh their memories of stuff they knew quite well in the past.
I did googled "wein bridge oscillator" before posting this thread. Didn't find much. Didn't googled "wein bridge oscillator analysis". I am going to do that now!

NascentOxygen
Staff Emeritus
And finally we assume it to be linear. I don't understand why this are assumptions is made.
We don't assume anything. We design the circuit to be linear throughout. Resistors are linear, capacitors are linear, and the OP-AMP-based amplifier is given a linear feedback network so it presents as a linear amplifier. The only non-linearity possible occurs if the OP-AMP output hits a supply rail, and in a practical circuit steps are taken at the design stage to ensure this doesn't happen (not shown in the simplified circuit you attached).

sophiecentaur
Gold Member
Why are the phase and amplitude characteristics satisfying only sine function? If you could go into some mathematical proof, that would be easier for me to understand.

And finally we assume it to be linear. I don't understand why this are assumptions is made. Not many Op-Amp circuits are linear, including the square generator, comparator, monostable multivibrator. I will definitely agree that this circuit is linear just because you and others saying this. I just want to know how do we came to conclusion that this is linear.
If you read the articles you will find, you will notice they deal with the problem of linearity and specify that the amplifier must be operated well below its maximum output power. You will find this sort of analysis throughout electronics. It's a general Engineering approach to build in as many ideal components as possible to the model (and the realisation of it). It's the only way, really.

f95toli
Gold Member
Why are the phase and amplitude characteristics satisfying only sine function? If you could go into some mathematical proof, that would be easier for me to understand.
Maybe you already know this, and I missunderstood this comment; but the point is that if you have a single frequency you have -per definition- a sinewave.
Every other waveform will -again by definition- contain multiple frequencies. This is no such thing as a "single frequency square wave".
You can of course have many other periodic functions, but if you look at their frequency distribution you will find that they always contain other frequency components in addition to the fundamental.
.
This is just a consequence of the math (Fourier analysis).

1 person
If you read the articles you will find, you will notice they deal with the problem of linearity and specify that the amplifier must be operated well below its maximum output power. You will find this sort of analysis throughout electronics. It's a general Engineering approach to build in as many ideal components as possible to the model (and the realisation of it). It's the only way, really.
We don't assume anything. We design the circuit to be linear throughout. Resistors are linear, capacitors are linear, and the OP-AMP-based amplifier is given a linear feedback network so it presents as a linear amplifier. The only non-linearity possible occurs if the OP-AMP output hits a supply rail, and in a practical circuit steps are taken at the design stage to ensure this doesn't happen (not shown in the simplified circuit you attached).
I am very much familierr with the notion of taking resistors, capacitors to be linear, we do that in analysis of almost all the circuits. and i know these assumptions are necessary in circuit analysis. I asked for the non=linearity because we don't take op-amps as linear element in all circuits, specially comparators and wave-generators are mostly non-linear ones(among those i have studied). However most of the op-amp circuits i studied are linear.

Maybe you already know this, and I missunderstood this comment; but the point is that if you have a single frequency you have -per definition- a sinewave.
Every other waveform will -again by definition- contain multiple frequencies. This is no such thing as a "single frequency square wave".
You can of course have many other periodic functions, but if you look at their frequency distribution you will find that they always contain other frequency components in addition to the fundamental.
.
This is just a consequence of the math (Fourier analysis).
Thanks. Did know this but couldn't connect Fourier series to this circuit till now. Things are becoming clear now for me.