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[SOLVED] Quality Factor Calculation

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Zayzoon

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Hello all,


I want to know if Quality Factor can be measured using the following:

by finding admittance: Y=G+jB

does that mean Q=|B/G| ?


PLease respond, Thanks.
 

Zayzoon

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I need a solid answer. reason: ive seen a form completely opposite in a published paper but i suspect it is wrong. the math just doesnt make sense....

thanks for the attempt....fyi im looking at narrowband amplifier circuits however it really shouldnt change the answer..
 

saiaditya

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Hi,

I may be wrong, but i dont think that Q=|B/G|....the general definition of Q is that it equals w*( energy stored/avg power dissipated)
Consider the example of a parallel RLC circuit. Here Y = G + j/wL (w^2*L*C - 1) and Q = w0 * (1/2*C*((I*R)^2))/(1/2 * I^2 * R) = w0*R*C.

Hopefully it clarifies or else let me know if something is wrong
 

FvM

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Y=G+jB refers to a parallel equivalent circuit. Consider an ideal capacitor or inductor with a parallel loss conductance. Obviously a high quality factor requires the conductance G to be low related to the susceptance B. So Q=|B/G| is right. But it only applies to simple lossy L or C elements, not to RLC circuits.
 

Zayzoon

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Thanks for the help FvM.
 

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