# wein bridge oscillator

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#### qtommer

##### Member
i have done a simulation of a wein bridge oscillator (without amplitude control)
from the simulation, i noticed that the output voltage(pk-pk) of the wave is approx. 2.2V and is the sine wave is not really centered to the x axis of the graph.

My circuit and simulation results are as attached.

Considering the fact that for oscillators, any transient noise signal is able to cause oscillations,
How do i determine the value of the output amplitude signal considering that there is no input source involved? (How is 2.2Vp-p in simulation obtained?) From the simulation, it is also seen that the initial start up small oscillations start from slightly below the x axis (below 0V).Why is this so?

The power supply voltages for the op amp are +20V and -20V in symmetry. Why is the wave still not centered evenly on the x axis? Is there any theory behind this or its just a simulator related problem?

_________________________________________________________________

Besides that, it is stated in theory that if the gain on the negative feedback loop is increased higher than 3, oscillations will increase in amplitude .and if the gain is reduced to less than 3, the oscillations will diminish to nothing..

From the feedback formula Avf= Ao/(1-Ao*beta)
i calculate that
• when i increase the gain to higher than 3, AVF will be a negative number
• when i increase the gain to lower than 3, AVF will be a positive number

why does a negative gain increase amplitude while a postive gain reduce amplitude? shouldnt it be the other way around?
or is my approach on the concept wrong?

________________________________________________________________

In the Barkhausen criteria, a 0d phase shift is required.. I understand that the frequency selective feedback loop provides a phase shift of 0d. How about the gain feedback loop involving the 2 resistors? My lecturer says the opamp provides a phase shift of 180 and the feedback gain loop provides another 180 totalling a 360.. but some teaching material on the internet says there is no phase shift by the op amp...

Help is very much appreciated=)

#### ericgibbs

##### Well-Known Member
i have done a simulation of a wein bridge oscillator (without amplitude control)
from the simulation, i noticed that the output voltage(pk-pk) of the wave is approx. 2.2V and is the sine wave is not really centered to the x axis of the graph.

The power supply voltages for the op amp are +20V and -20V in symmetry. Why is the wave still not centered evenly on the x axis? Is there any theory behind this or its just a simulator related problem?
________________________________________________________________

In the Barkhausen criteria, a 0d phase shift is required.. I understand that the frequency selective feedback loop provides a phase shift of 0d. How about the gain feedback loop involving the 2 resistors? My lecturer says the opamp provides a phase shift of 180 and the feedback gain loop provides another 180 totalling a 360.. but some teaching material on the internet says there is no phase shift by the op amp...

Help is very much appreciated=)

hi,
For comparison I will run your circuit in the LTspice sim.

Ref the 180deg phase query, as you maybe aware the INV input and OPA output are inverted, so its '180deg' phase shifted.
In an ideal OPA, amplified signals should not be phase shifted.

Are these the two versions you are mixing up.?

I'll post the LTS sim later to day. #### ericgibbs

##### Well-Known Member
hi,
Using your circuit the osc dosnt start up, the gain is 2.

With a gain of ~4, it starts, but as you would expect its a square wave output.
The plot shows no dc offset in either case.

The fact that it oscillates and produces a sort of sine wave in your sim, suggests its an artefact of your simulator, which sim is it.?

To get a sinewave output you must have auto adjusting gain control, using either a small tungsten lamp or FET or diode bridge.

#### qtommer

##### Member
hey eric

thank you so much for your replies and for your time=) i really appreciate it.

The fact that it oscillates and produces a sort of sine wave in your sim, suggests its an artefact of your simulator, which sim is it.?

im using the Orcad Capture Lite Edition..
If it helps, i was using the L411 opamp but i just changed it to the uA741..using the latter, i got a square wave if R2 is exactly double R1 (Gain=3)
When i adjusted the gain higher by increasing R2, a sine wave was produced.
im actually kinda new at this stuff so im not too sure why i got a square wave if the gain was equal...
i understand that in practical-wise approach, if the loop gain =1 then there will be no oscillations unless i increase the gain slightly higher to overcome the dissipated power caused by resistors.
why is it a square wave output?

Ref the 180deg phase query, as you maybe aware the INV input and OPA output are inverted, so its '180deg' phase shifted.
In an ideal OPA, amplified signals should not be phase shifted.

so there is a 180d shift thru the resistor network but none at 0d thru the OPAMP?
if so then how does the feedback signal get in phase?

thank you thank you X infinity so much again..

#### qtommer

##### Member
hey eric, i know what the problem is about the DC offset...
i did not configure the Vcc and -Vcc correctly...apparently i was informed the wrong way on how to do it...haha...thank you so much...the graph is perfectly symmetrical

now i know why i got a sine wave because the amplitudes are clipped? thats why i need the amplitude control like diode bridge..is that correct?

im still in a blur about the phase shift ..haha=p..hopefully u cud help me understand that ..(refering to previous post)

once again my most deepest gratitude=)

#### ericgibbs

##### Well-Known Member
hey eric, i know what the problem is about the DC offset...
i did not configure the Vcc and -Vcc correctly...apparently i was informed the wrong way on how to do it...haha...thank you so much...the graph is perfectly symmetrical

now i know why i got a sine wave because the amplitudes are clipped? thats why i need the amplitude control like diode bridge..is that correct?

im still in a blur about the phase shift ..haha=p..hopefully u cud help me understand that ..(refering to previous post)

once again my most deepest gratitude=)

hi,
Must have auto gain control, a wire ended mini-lamp works well.

Look thru this OPA guide for the phase etc, useful pdf. #### Attachments

• opamps.pdf
273.2 KB · Views: 498

#### BrownOut

##### Banned
From the feedback formula Avf= Ao/(1-Ao*beta)

why does a negative gain increase amplitude while a postive gain reduce amplitude? shouldnt it be the other way around?
or is my approach on the concept wrong?

Your concept is correct. When the the gain is sufficient to make the denomanator of your equation zero, then you have gain that increases without bound, which is what is requried for oscillations. An oscillator is basically an infinite gain element, as the equation clearly shows.

#### audioguru

##### Well-Known Member
The frequency determining RC networks are a lowpass filter with lagging phase and a highpass filter with leading phase. The two phases cancel and produce no phase shift at the frequency where they have the same amount of loss and produce a peak with a loss of 3 times.
They produce positive feedback for the opamp.

The opamp is a non-inverting amplifier with no phase shift and a gain of slightly more than 3 times. Its amplitude will keep increasing until the output of the opamp clips the positive and negative peaks.

#### qtommer

##### Member
to eric:
thanks alot for the useful pdf and also thanks once again for helping me simulate the circuit.=)

to brownout:

Your concept is correct. When the the gain is sufficient to make the denomanator of your equation zero, then you have gain that increases without bound, which is what is requried for oscillations. An oscillator is basically an infinite gain element, as the equation clearly shows.

yeah i understand that if the denominator is 0 then only will oscillations occur...due to infinite gain..
but if the gain is increased lets say to a gain of 4, the total gain would then become a -ve number..
Avf=4/{1-4(1/3)}
=-12

at a negative gain, why does the opamp still oscillate??

to audioguru:

The opamp is a non-inverting amplifier with no phase shift and a gain of slightly more than 3 times. Its amplitude will keep increasing until the output of the opamp clips the positive and negative peaks.

so is it safe to say that the opamp as well as the voltage divider resistor network produces no phase shifts?? cuz my lecturer says op amp produces 180 and the resistor network produces another 180 totalling 360 resulting in no phase shift..

#### BrownOut

##### Banned
yeah i understand that if the denominator is 0 then only will oscillations occur...due to infinite gain..
but if the gain is increased lets say to a gain of 4, the total gain would then become a -ve number..
Avf=4/{1-4(1/3)}
=-12

You're only considering the absolute value of the loop gain. In fact, the loop gain is a complex number. I won't go through all the math, but essentially, a negative loop gains indicates positive feedback, which then means we have infinite system gain, as we previously discussed.

#### qtommer

##### Member
You're only considering the absolute value of the loop gain. In fact, the loop gain is a complex number. I won't go through all the math, but essentially, a negative loop gains indicates positive feedback, which then means we have infinite system gain, as we previously discussed.
is it because the imaginary part of the loop gain equal to 0 resulting in no phase shifts thus causing positive feedback?

#### BrownOut

##### Banned
The imaginary part is not zero. The poles of the oscillator are complex conjugates. The point at which the loop equation goes negative is the point at which the feedback becomes positive. If you plotted the poles on an s-plane, you would see this.

Don't feel bad. I had to dig out my old text book and try to remember how this works. #### audioguru

##### Well-Known Member
It does not matter what your lecturer says because he is wrong.
The opamp, the two cascaded RC networks and the voltage divider resistor network produce no phase shift in this circuit.

Don't you have an oscilloscope? It will show that the phase at the output of the opamp is exactly the same as the phase at the (+) input of the opamp.

#### BrownOut

##### Banned
Did you the math for the feedback circuit? You should get, for 1 + AB:

1 + A*jw/RC/((jw)^2 + jw*3/RC + 1/(RC)^2)

Now, in your feedback equation, you have in the denominator:

(jw)^2 + jw*(3 - A)/RC + 1/(RC)^2

Now, the dominator dissapears for W = 1/RC, and A = 3. The poles are on the imaginary axis at this value of A, and the system is marginal unstable (which means it will almost oscillate). For largers values of A, that terms changes sign. Also, the poles of the expression are now in the right half-plane. Now, you have increasing positive feedback, and your system will oscillate even more.

A bode plot would show positive phase at frquencies lower than W, and negative phase at frequencies higher than W. Also zero phase at W = 1/RC.

Does that make sense? Sorry I can't plot the poles. Look up pole plots of 2nd order feedback networks.

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#### qtommer

##### Member
By BrownOut
Did you the math for the feedback circuit? You should get, for 1 + AB:

1 + A*jw/RC/((jw)^2 + jw*3/RC + 1/(RC)^2)

Now, in your feedback equation, you have in the denominator:

(jw)^2 + jw*(3 - A)/RC + 1/(RC)^2

Now, the dominator dissapears for W = 1/RC, and A = 3. The poles are on the imaginary axis at this value of A, and the system is marginal unstable (which means it will almost oscillate). For largers values of A, that terms changes sign. Also, the poles of the expression are now in the right half-plane. Now, you have increasing positive feedback, and your system will oscillate even more.

A bode plot would show positive phase at frquencies lower than W, and negative phase at frequencies higher than W. Also zero phase at W = 1/RC.

Does that make sense? Sorry I can't plot the poles. Look up pole plots of 2nd order feedback networks.

thanks again..I am so sorry but my mathematical understanding of transfer functions is not solid yet because where i come from. they teach you how to derive but they do not explain its concept and its application in the circuit itself.its just plain number manipulation without understanding..
i might need some time to review it and digest...

however, i know how to derive the feedback transfer function to get 1/3 when w=1/RC..
and i understand that the poles are on the imaginary axis...

the transfer function ive worked out is

jwRC/[ 3jwRC +1 -(wRC)^2 ]

however from the formula you've shown me, i dont understand how the gain, A is connected into the transfer function equation as well..
really sorry..hope ull be patient with me...

By audioguru:
It does not matter what your lecturer says because he is wrong.
The opamp, the two cascaded RC networks and the voltage divider resistor network produce no phase shift in this circuit.

Don't you have an oscilloscope? It will show that the phase at the output of the opamp is exactly the same as the phase at the (+) input of the opamp

thank you for your reply...I understand that the final phase shift is 0...but for instance lets say the RC lead-lag network...there is a lag and lead in phase thus resulting in a final 0 phase shift(so there is phase shifting "along the way")..im interested to know the "along the way" process wheter there is any phase shifting..so is it safe to say that the opamp and voltage divider resistor networks produce no phase shifts "along the journey"?

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++

in other news, ive just done a simulation of the wien brige now with amplitude stabilization using a diode bridge. from my simulation results i notice that the oscillatory start up from a small wave signals to a large constant wave is much faster compared to when there is no amplitude stabilization...

ive also noticed that the sine wave is still clipped at the top and bottom edges even with the diode bridge.

Does this mean that the purpose of amplitude stabilization is NOT to ensure that the sine wave edges are distorted (clipped) but it IS to ensure that a sine wave with constant amplitude is generated right from the start without waiting for it to have to grow?

Thank you all so much

#### BrownOut

##### Banned
i dont understand how the gain, A is connected into the transfer function equation as well..

You've already written the equation: Av(closed loop) = A/1 + A*B. I just showed the math for the denominator of the equation, because that determines the poles, and I was hoping that would give you some insight into why the oscillator works. Sorry if I confused you.

#### qtommer

##### Member
You've already written the equation: Av(closed loop) = A/1 + A*B. I just showed the math for the denominator of the equation, because that determines the poles, and I was hoping that would give you some insight into why the oscillator works. Sorry if I confused you.

ahh i see i see..thank you very much.
so the conditions for oscillation are...A=3 and w=1/RC..

however, when u said the poles are on the imaginary axis, all in understand is that when w=1/RC the Real part gets cancelled out leaving the imaginary part behind..

however, i do not understand from this part onwards..
Also, the poles of the expression are now in the right half-plane. Now, you have increasing positive feedback, and your system will oscillate even more.

do you mean the right half of the cartesian graph? and if it is on the right side..why does positive feedback increase?

thank you =)

#### BrownOut

##### Banned
I'm guessing you haven't studied s-plane graphs? Basically, a feedback system is stable if the poles are in the left half plane, and unstable in the right-half plane. Without that background, it suffices to say that the loop gain of the oscillator must be >= 1. And I think you already know how to determine that.

If you've studied 2nd order differential equations, you can apply that to see that the solution is a growing exponential function when the real parts of the poles are positive. That would be identical to what I've been saying about right half plane.

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#### qtommer

##### Member
yes that guess would be correct...they do not teach the s plane at all..to be honest with you ive never heard of it *sigh*...im really tired of not being able to understand a concept to the fullest of its essence...been trying my best to self-learn though ..altou it takes a longer time...but its definitely catalyzed by people like you=) thank you so much for your help=)

however i have just one last little question...if the loop back gain is >1 that means it is on the right side of the s plane deeming it unstable..

therefore is it correct to say that if the loop gain is less than 1, it would be on the left side of the s plane and becomes a stable op amp thus resulting in attenuations in the circuit?

#### BrownOut

##### Banned
therefore is it correct to say that if the loop gain is less than 1, it would be on the left side of the s plane and becomes a stable op amp thus resulting in attenuations in the circuit?

The position of the roots on the left side affect the response of the amplifier. You can have a damped system or undamped, depending on exactly where the poles lie. However, you're right that the system would be stable.

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