One qestion about "component Q" of Inductor?

[tony_taoyh]

Hi, I have one basic question about RF circuit.

The quality factor Q of inductor.

Is it right to say that the component Q of inductor goes to zero when
the frequency go to "self resonant frequency"?

If it is, how to understand it?

Thanks a lot.

Best Regards,
Tony

 

[Mr.Cool]

and inductor parasitic capacitance and series resistance inside, unavoidable. when inductor is at its resonant frequency the inductance component exactly cancels out the capacitance component and all that is left is the resistive component. a resistive component is only dampening effect, no "Q" at all.

Mr.Cool

 

[tony_taoyh]

Thank you, Mr. Cool,

That is to say, for the inductor,
if we use the serial equivilent circuit,
then:
Q = X/Rs, then X = 0, so Q = 0;

That means, the X is totally X;

When using the parallel equivilent circuit,
because X = infinity,
So:
Q = Rp/Xp = 0;

Correct?

Thank you very much.

 

[Borber]

Inductance quality factor Q is for series equivalent circuit (L in series with R) equal:
Q=ω*L/R
So the Q factor rises with frequency it is 0 at 0Hz and infinite at infinite frequency. In parallel circuit with capacitor (either parasitic or added) the series resistance of inductor determines Q factor of such parallel circuit at resonance frequency. In this case Q of inductor remains ωres*L/Rseries.

 

[sergio mariotti]

I know another definition for Q factor:
Q≡Reactive Power/Active Power
So for both serie or parallel moldels the calculated Q is 0 at resonant frequency, but may be considered as a singularity.
Practically, for resonant circuits, as Q is higher as the Extravoltage (or extracurrent for series rosonators) is higher, so Q cannot be 0 at resonant frequency.

 

[Borber]

I see problem in the reactive power calculation. Do you simply subtract PreactL and PreactC because currents through L and C in parallel connection are out of phase, or you should calculate reactive power of L and C as V²/XL and V²*YC and make addition.

 

[Debeli]

At self-resonant frequency of the inductor the parasitic capacitance annulate the inductance and equivalent reactive part of the component impedance is 0. That means that the component has just parasitic resistance on this frequency and, therefore, no reactive energy is stored in the component.
Consequence is that Q factor on self-resonant frequency of the component is equal 0.

P.S. I reconsidered what Sergio wrote, and I find him right. Singularity case must be considered and mathematical concepts should be applied only within limits where common assumptions are valid.
Thanks, Sergio

 

[sergio mariotti]

At self-resonant frequency of the inductor the parasitic capacitance annulate the inductance and equivalent reactive part of the component impedance is 0. That means that the component has just parasitic resistance on this frequency and, therefore, no reactive energy is stored in the component.
Consequence is that Q factor on self-resonant frequency of the component is equal 0.

Reactances can be arithmetically added (with sign), but power can't be subtracted, except the case of a power loose. Power is diminsionally Energy/Time.

The Q concept came from many years ago, at that time Qmeters was working exactly at RESONANT FREQUENCY, and they was measuring the Extra Voltage. As the Extra Voltage was high as the Q was high. So the Q≠0 and Q>0 at resonant frequency.
Be aware from the singularity in mathematical definitions.

 

[mixaloybas]

In the first picture a spice model for a real inductor in a 0805 package, taken from coilcraft, is illustrated. (As you can see there is a frequency dependent resistor modeling skin effect).

If we try to make an equivalent impedance having only a resistance Rs and a reactance Xs, we will end up with both a frequency dependent resistor (that shows the resistance Rs) and a frequency dependent inductor (that shows the reactance Xs). Furthermore the inductor will have negative value beyond self resonance - effectivelly it becomes a capacitor.
Zeff=Rseff + j*Xseff=Rseff(f) + j*w*Leff(f)
So if we have an inductance in series with a resistance (no matter whether they are frequency dependent), the Unloaded Q (or component Q) is given by Q=Xseff/Rseff.
Using the following equations in Genesys I took the plots shown in the second picture:
USING Linear1.Coilcraft
Qind=abs(.IM[ZIN1]/.RE[ZIN1])
Leff=1000000000*(.IM[ZIN1]/(2*PI*1000000*freq))
XLeff=.IM[ZIN1]
RSeff=.RE[ZIN1]

I used abs() in Q because after self resonance Q returned a negative value (because the inductor becomes capacitor). The freq is in MHz.

As we can see, Q is zero in self resonance. Furthermore, Leff reaches a maximum value BEFORE self resonce, then starts to decrease, it becomes zero , and then continues symmentrically..However, if the resistance in the model is decreased to zero, then the plot becomes more and more steep(is it the right word??). That is, Leff asymptotically reaches infinity (not just a max value), and after resonance suddenly (and this is a singularity or discontinuity)it becomes -infinity.


Please report me if you find any inconsistencies.

 

[Borber]

Yes if you consider Q factor is defined as X/R you are right. But it is valid only if the frequency is much lower than the resonance frequency. At resonance we have to deal with resonant circuit which has different definition of Q factor. It is defined as ratio of resonant frequency to -3dB bandwidth. It can be calculated from transfer function. At higher Q factors than 2 or 3 the Q approaches ωr*L/Rser or Rpar/ωr*L or ωr*Cpar*Rpar.