Q-factor in resonant RCL circuit

  • Thread starter temujin
  • Start date
  • #1
temujin
47
1
Dear Forum,

I am not sure how I can calculate the Q-factor in the circuit attached.
(The circuitresonates at 13.56MHz and tuned to 50 ohm input impedance, thus the two capacitors)

Is the Q-factor of the circuit simply [tex]Q = \frac{\omega L}{R_{a}+R_{coil}}[/tex] ?

I went through some examples and it looks like this is the right way to do it, however, I´m not sure why the capacitorscan be disregarded, since they also store reactive energy.

best regards
eirik
 

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Answers and Replies

  • #2
mezarashi
Homework Helper
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The left most capacitor is probably used for coupling purposes. Don't forget that the omega term is the resonance frequency, which is indeed a function of the capacitance. It's quite a mess, but I think if you derive the equations based on first principles (3db amplitude drop - bandwidth), you will see why this is so.
 
  • #3
temujin
47
1
Hi again,
One additional thought...I think that the total reactive energy in the circuit is constant. At resonance the total energy swings back and forth between the capacitors and the inductor. Can it be that at one instant when the stored energy is at maximum in the inductor, the stored energy is zero in the capacitors? So that I just need to consider the inductor at that instant??

e.
 
  • #4
temujin said:
Hi again,
One additional thought...I think that the total reactive energy in the circuit is constant. At resonance the total energy swings back and forth between the capacitors and the inductor. Can it be that at one instant when the stored energy is at maximum in the inductor, the stored energy is zero in the capacitors? So that I just need to consider the inductor at that instant??

e.
As mezarashi said, [tex]\omega = \frac{1}{\sqrt{LC}}[/tex], so [tex]\omega L = \sqrt \frac{L}{C}[/tex]. There you have both L and C.
 

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