Frequency in Wien bridge Oscillator

Hi all, I have a Wien bridge Oscillator which I measured the output frequency on the oscilloscope and is about 1KHz.
Now, I am trying to calculate it theoretically, but I can't obtain the same result. I am sure that I am calculating it in the wrong way.

I know the formula is 1/(2πRC), but exactly, which resistance I have to consider in the circuit?
R3 or R4? Or the Req between them? I have the same doubt for the capacitance.

Off course I attached an image of the circuit.

Thank you for your help.
Regards,
simo_x

The circuit in you diagram is not a Wien bridge Oscillator.

Your circuit is a phase-shift oscillator .
http://upload.wikimedia.org/math/1/d/2/1d27ee136a0c12a40595a00fad82d7fd.png

Jony130 said:

The circuit in you diagram is not a Wien bridge Oscillator.

Your circuit is a phase-shift oscillator .
http://upload.wikimedia.org/math/1/d/2/1d27ee136a0c12a40595a00fad82d7fd.png

Click to expand...

I am quite surprised about it, because our professor gave me this circuit referring to the Wien bridge.. I don't understand it..
Honestly, I was not convinced about it after looking in the net other Wien Bridge schemes..
I did not know the phase-shift oscillator. Thank you Jony, now the frequency is ok with 1/(2π * 10K * 6n2 * sq(2 * 3)) = 1047 Hz

Here you have the diagram of a Wien bridge oscillator
http://commons.wikimedia.org/wiki/File:Wien_bridge_oscillator.png
And some theory about phase-shift oscillator
http://www.electronics-tutorials.ws/oscillator/rc_oscillator.html

Jony130 said:

Here you have the diagram of a Wien bridge oscillator
http://commons.wikimedia.org/wiki/File:Wien_bridge_oscillator.png
And some theory about phase-shift oscillator
http://www.electronics-tutorials.ws/oscillator/rc_oscillator.html

Click to expand...

Yes, as I edited, I was confused about the scheme, because comparing my scheme to other Wien Bridge schemes, there was no coincidence..

I really thank you for your help, You have been very helpful.
Thanks

Jony130 said:

The circuit in you diagram is not a Wien bridge Oscillator.

Your circuit is a phase-shift oscillator .
http://upload.wikimedia.org/math/1/d/2/1d27ee136a0c12a40595a00fad82d7fd.png

Click to expand...

Yep. In the phase shift oscillator you have to change R4, R5 and R6 simultaneously.

Wien bridge oscillators can produce very pure sine waves but you need an active negative feedback loop of some sort (typically using an incandescent lamp as a current regulator.

Eric

My phase-shift oscillator has low distortion because its RC filters are lowpass, not highpass and the output is taken from the 3rd lowpass filter that has the harmonics reduced.

Comparing the others circuits schemes, I noticed that no diodes are used. Why?
I think that in my circuit D1 & D2 are needed to create the sinewave.

Is it correct?
Thank you!
Regards,
Simon

Your phase-shift oscillator oscillates at only one frequency where the phase shift of each RC is 60 degrees. Then if the output of the opamp does not clip the output is a pure sine-wave.

The diodes in your phase-shift oscillator conduct a little when the amplitude of the sine-wave is about 0.6V peak at the diodes which increases the negative feedback which reduces the gain so that the output level is stabilized. If the output level becomes less then the diodes conduct less which increases the gain to increase the output level up to where it should be. The diodes add a small amount of distortion.

But diodes have a turn-on voltage that changes when the temperature changes.

Without the diodes then the amplitude at the output of the opamp will increase until it clips which produces severe distortion.

The oscillators in the articles do not have diodes but say that a method is needed to stabilize the output level. Many years ago a light bulb was used to stabilize the output level because its resistance increases when it gets hotter when the output level rises.

But diodes have a turn-on voltage that changes when the temperature changes.

Click to expand...

I know that. We are preparing a circuit based on this detail.

Thank you for the explanation, now it's all more clear.
Regards,
Simon