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raja.cedt

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Wienbridge oscillator Quality Factor
Jun 24^th, 2012, 10:40am

hello,
In case of any oscillator up to my knowledge Quality factor should be grater than 1/2 (in theory and according to under damping case ). But in case of Wienbridge oscillator Q is around 1/3 (i have derived this from transfer function). So how could this oscillator is oscillating...is my first statement is correct (Q>1/2)

Thanks,
Raj.

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buddypoor Community Fellow Offline Posts: 529 Bremen, Germany	Re: Wienbridge oscillator Quality Factor Reply #1 - Jun 25^th, 2012, 6:42am Hi Raja, if a circuit in general is able to oscillate and if the oscillation criterion is met for one frequency (Barkhausen condition) the circuit will oscillate! It does not matter which value the Q factor of the feedback network has.
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aaron_do Senior Fellow Offline Posts: 1398	Re: Wienbridge oscillator Quality Factor Reply #2 - Jun 25^th, 2012, 7:44am Hi Raj, I think in order to have instability, the transfer function (from any node/branch to the output?) must have a right-hand plane pole. For oscillation, I think it must have a pair of complex right-hand plane poles. Correct me if I'm wrong here... regards, Aaron
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raja.cedt Senior Fellow Offline Posts: 1516 Germany	Re: Wienbridge oscillator Quality Factor Reply #3 - Jun 25^th, 2012, 8:05am hello aaron_do, yes, to have complex conjugate in the RHP, Q>.5 correct me if am wrong.... @buddypoor, is it possible to satisfy BH criterion when Q<1/2?
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buddypoor

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Re: Wienbridge oscillator Quality Factor
Reply #4 - Jun 25^th, 2012, 9:15am

raja.cedt wrote on Jun 25^th, 2012, 8:05am:

hello aaron_do,
yes, to have complex conjugate in the RHP, Q>.5 correct me if am wrong....

@buddypoor,
is it possible to satisfy BH criterion when Q<1/2?

I am afraid, we speak about different functions.
* Aaaron is right - the poles of the closed-loop transfer function are in the RHP (conjugate-complex).

* Raja was (and is) referring to the open-loop characteristics.
The WIEN network has two LHP poles and an open-loop pole-Q<0.5. Nevertheless, after closing the loop it will oscillate. Why not - if the Barkhausen condition is fulfilled?

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