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I am working on parallel LC tank.

It is Rp // Lp // C .

Rp = Rs( 1 + Q^2 )

Since inductor dominates capacitor under high frequency, Rs is parasitic resistance of inductor.

It seems to me Q factor is determined by the inductor.

Also, Q factor is determined by ratio of L and C components.

If I want to have 100 Q factor, should I consider Q factor of certain inductors from the datasheet or Q = Rp * root( C / L ) equation.

Rp value is given.

noah k
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  • There are lots of references/articles on the interweb (e.g. http://www.radio-electronics.com/info/formulae/q-quality-factor/inductor-q.php) - perhaps a little research would be in order. Basically you are trying to reduce the resistance of the inductor OR you actively try to reduce the effective resistance by (positive) feedback (to replace the lost energy per cycle ) OR you try to add a dynamic negative resistance to cancel out the resistance of the coil. – JIm Dearden Jun 03 '16 at 15:40
  • i would add a dynamic negative resistance to cancel it out – noah k Jun 03 '16 at 15:58
  • Poor choice (IMHO) , much easier to produce a low resistance coil, but here's where you start https://en.wikipedia.org/wiki/Negative_resistance – JIm Dearden Jun 03 '16 at 16:17
  • If you have parasitic design issues its probably better to find or build a component with less parasitics – Voltage Spike Jun 03 '16 at 16:21
  • Welcome to EE.SE Could you please edit your question ask a question? http://electronics.stackexchange.com/help/how-to-ask It's good to ask specific question (and in the guidelines of the site). You should also poke around the help center for other guidelines relating to EE.SE. – Voltage Spike Jun 03 '16 at 16:25

1 Answers1

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The inductor is nearly always the weakest link when trying to get decent Q so concentrate on high Q inductors. When you do the math on the Q factor it comes out to be equivalent to the Q of the inductor but, as others have commented the Q of the capacitor cannot be ignored but in nearly all cases the Q of the inductor is nowhere near as good as that of the capacitor.

Q = \$\dfrac{1}{R_S}\sqrt{\dfrac{L}{C}} = R_S\sqrt{\dfrac{L^2}{LC}} = \dfrac{\omega_0L}{R_S}\$ noting that \$\omega_0 = \dfrac{1}{\sqrt{LC}}\$

Andy aka
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  • I would also add that the total Q is the "parallel" combination of the Q of all the components, so he may need a higher Q inductor than 100 since the capacitor Q won't be infinite. – Captainj2001 Jun 03 '16 at 15:53
  • I have to consider two different Q factor. One for inductor itself and the another for parallel combination – noah k Jun 03 '16 at 15:54
  • How high Q inductor would be good? A higher Q inductor than 3000 is available ? – noah k Jun 03 '16 at 16:02
  • @noahk Yes, since you're building a parallel tank circuit you need to consider the Q of all the components you place in there. The equivalent Q of the tank will be the combination of the Q of the inductor (with its series resistance) and the capacitor with its equivalent parallel resistance. – Captainj2001 Jun 03 '16 at 16:30