Series to Parallel inductance conversion

Reviewing some material I thought I understood, but it turns out there's some subtlety I'm not getting. The usual formula for calculating the Q of a non-ideal inductor is $\frac{Xs}{Rs}$, where Xs is the series reactance and Rs is the series resistance. For a parallel circuit, it's the inverse, $\frac{Rp}{Xp}$. The two are related by the equation $Rp = (Q^2 +1)Rs$, where Q is the unloaded Q of the element. Consider a non ideal parallel LC circuit. You have an R and L in series, and this series combination is itself in parallel with a capacitor. Using the above transformations, one can convert the series resistance into a value of parallel resistance, and then using the value of parallel resistance solve for a value of parallel inductance using the second formula above.

But then, what is the resonant frequency of the tuned circuit? Is it given by the original value of series inductance, or the new value of parallel inductance? The texts I've read make the assumption that the reactance of the parallel inductance is the same as the series inductance, and working out the equations with the Laplace transform of the original circuit shows that the resonant frequency is solely determined by the value of the original capacitance and non-lossy inductance. Indeed, it would hardly make sense to try to calculate a new resonant frequency based on this equivalent parallel inductance value, because the transformation is only valid at one frequency! So what's the significance of this new value of parallel inductance, then? For low Qs it can be significantly larger than the original series value. Does any of this make sense?

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You are confusing the Q of the inductance with the Q of the circuit. In a series LCR circuit, the Q of the inductor is the ratio of the voltage across the inductor to the voltage across the circuit. As you say, this is also the ratio of the inductive reactance to the resistance. In a circuit with series R & L in parallel with C, this relation also holds.

In a parallel LCR circuit, the resistance is no longer a property of the inductor itself which is why you get a different definition for Q. The Q in this case is more a property of the circuit than the inductor. You cannot convert one circuit to the other although you could find two circuits with identical Q.

The Laplace transforms of the impedance for (r + L) // C is ...

$$\frac{\frac{r}{LC}+\mathrm s\frac{1}{C}}{\mathrm s^2+\mathrm s\frac{r}{L}+\frac{1}{LC}}$$

for which Q=$\frac{1}{r}\sqrt{\frac{L}{C}}$ (at resonance)

The Laplace transform of the impedance for R // L // C is ...

$$\frac{\mathrm s\frac{1}{C}}{\mathrm s^2+\mathrm s\frac{1}{RC}+\frac{1}{LC}}$$

for which Q=$R\sqrt{\frac{C}{L}}$ (at resonance)

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I think I may not have worded the question well. In the parallel RLC circuit I am describing, the Q of the circuit will depend only upon the inductor (assuming the capacitor is lossless), because it was created by forming an equivalent circuit out of the series inductor and its losses modeled by a series resistor. You obtain the equivalent parallel RLC circuit by applying the equations in my post. What I'm not understanding is why when the value for the inductor calculated in this equivalent circuit is different, the resonant frequency of the tank stays the same. –  Bitrex May 19 '11 at 12:53
Let's say I have a 50 nH coil in series with 10 ohms, representing its internal losses. In parallel with that I have a capacitor, value doesn't matter. Applying the transformations above, its equivalent parallel representation is a 108 ohm resistor in parallel with a 55 nH inductor. The Q of the series combination or parallel combination at a frequency is the same, since they represent the same inductor. But is the resonant frequency of the tank calculated using the original inductance, or the new inductance? The above analysis indicates the "old" inductance. So what's the 55nH about? –  Bitrex May 19 '11 at 13:00
Those two R/L combinations will have the same impedance at about 100MHz, but then so would 50nH in parallel with a 98Ω resistor and that combination would have the same Q with a resonant capacitor of about 51pF. That is, it's only consistent if you have $R_p=Q^2R_s$ which you get by multiplying $Q=\frac{X_L}{R_s} by Q=\frac{R_p}{X_L}$ –  MikeJ-UK May 19 '11 at 13:43
Right, it's only consistent if the series reactance is equal to the parallel reactance after the transformation. But that's just an approximation: for low values of inductor Q the parallel reactance is not equal to the series reactance. So what happens to the resonant frequency of the tank? The resonant frequency is always going to be fixed by the lossless inductor and capacitor values. But at low Q values it seems the equivalent parallel value of inductance increases.I know that with low damping the peak gain frequency of a tuned circuit will shift, perhaps that's how it's showing up here? –  Bitrex May 19 '11 at 14:02
I think I understand what's going on now. The "undamped" resonant frequency is in the equations, but at low values of Q the peak of the response is going to shift, which is why the equivalent value of parallel reactance will change at low Q - the frequency where the impedance of the circuit is at maximum has changed. –  Bitrex May 19 '11 at 14:14
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