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https://designers-guide.org/forum/YaBB.pl Design >> Analog Design >> Wienbridge oscillator Quality Factor https://designers-guide.org/forum/YaBB.pl?num=1340559645 Message started by raja.cedt on Jun 24th, 2012, 10:40am |
Title: Wienbridge oscillator Quality Factor Post by raja.cedt on Jun 24th, 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. |
Title: Re: Wienbridge oscillator Quality Factor Post by buddypoor on Jun 25th, 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. |
Title: Re: Wienbridge oscillator Quality Factor Post by aaron_do on Jun 25th, 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 |
Title: Re: Wienbridge oscillator Quality Factor Post by raja.cedt on Jun 25th, 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? |
Title: Re: Wienbridge oscillator Quality Factor Post by buddypoor on Jun 25th, 2012, 9:15am raja.cedt wrote on Jun 25th, 2012, 8:05am:
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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