In tank circuit, what does changing ratio between L and C for a given frequency do?

I know the formula for tank resonance, but am wondering how having big L and small C (or vice versa) differs from medium L and medium C if both give the same frequency.





Does the ratio affect the Q, the efficiency, or something else?





How do I calculate what L and C I need in order to ensure maximum energy sloshing between the two?

In tank circuit, what does changing ratio between L and C for a given frequency do?
The ratio of inductance to capacitance sets the Q





For a series resonant circuit Q can be given by:





Q = (ω0*L)/R = (1/R)*sqrt (L/C)





For a parallel resonant circuit Q can be given by:





Q = R/(ω0*L) = R/sqrt(L/C)





Where ω0 = 1/sqrt(L*C) and is the resonant frequency.





One should note that the Q is affected differently by L %26amp; C depending on whether you are designing a series or parallel resonant circuit. It is clear by looking at the series circuit equation, that the Q increases as L increases and C decreases. For the parallel circuit, the Q increases as L decreases and C increases.





Finally, Q is a measure of how much energy a tank circuit can store.





Q = ω * (energy stored / power lost)





If you want the maximum energy "sloshing around" as you say, design your circuit for the maximum Q.





Regards,


Dave
Reply:In theory, a wide range of L and C values will resonate at the same frequency as long as the ratio remains the same. In practice, there is never any such thing as a pure L or C component. Capacitors have inductance and resistance in their leads, and Inductors (coils) have parasitic capacitance between their ends as well as resistance in their conductive parts. So in practice, your choice of appropriate L and C values will be limited by these practical considerations.





The ratio of L to C will also affect the selectivity of the circuit.


A parallel LC circuit will have greater selectivity (narrower band pass of frequencies around the resonant frequency) with higher capacitance values and lower inductance values.





Maximum energy storage in a tank circuit is attained by having high "Q" coils (inductors) - this means low resistance conductors, and if a ferrite core is used, choosing one with the lowest hysteresis losses. High Q also gives the narrowest passband of nearby frequencies - not necessarily a good thing if you intend to pass modulation sidebands that bear information.