Q factor of an AM radio RLC circuit

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SUMMARY

The discussion focuses on calculating the minimum quality factor (Q) for a series RLC circuit used to tune into an AM radio station broadcasting at 1.20 MHz while minimizing interference from a nearby station at 1.10 MHz. The participants derived Q values using various formulas, including Q = wL/R and Q = 100/(0.16), resulting in estimates ranging from 582 to 625. The circuit's resistance is fixed at R = 0.1 ohm, and the challenge includes determining the appropriate inductance (L) and capacitance (C) values to achieve the desired resonance.

PREREQUISITES
  • Understanding of RLC circuit theory
  • Familiarity with quality factor (Q) calculations
  • Knowledge of resonance in electrical circuits
  • Basic proficiency in logarithmic and exponential functions
NEXT STEPS
  • Calculate the inductance (L) and capacitance (C) values for resonance at 1.20 MHz
  • Explore the implications of varying resistance on Q factor in RLC circuits
  • Study the effects of bandwidth on signal tuning in AM radio applications
  • Review the Radio Engineers Handbook for advanced circuit design techniques
USEFUL FOR

Electrical engineers, radio frequency (RF) engineers, and students studying circuit design who are interested in optimizing RLC circuits for radio frequency applications.

RyanP
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Homework Statement


Suppose you want to use a series RLC circuit to tune in your favorite AM radio station which broadcasts at a frequency of 1.20 MHz. You would like to avoid the obnoxious easy listening station which broadcasts at 1.10 MHz, right next to the one you like. In order to achieve this, for a given EMF from your antenna, you need the current flowing in your circuit to be 10-2 times less at 1.10 MHz than at 1.20 MHz. What is the minimum Q for this circuit? Now, note that you cannot avoid having a resistance of R = 0.1 ohm, and practical considerations also dictate that you use the minimum L possible. What values of L and C must you use?

Homework Equations


Q = wL/R = w/(delta w) = w/2a where a = R/2L
f = w/2pi

The Attempt at a Solution



.01 = e^(-a(x-w))
a = ln100 / (2pi * (1.1*10^6 - 1.2*10^6))I don't know if this is right or where to go from here.
 
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Terman* gives universal resonance curves for engineering purposes, so it is obviously a complicated calculation to do exactly.
I have slightly adapted his formula for a case such as ours where we are more than 3Fo/Q (MHz) from resonance:-
Vfo/Vf = Q((f/fo)^2 - 1)
100 = Q (1.1/1.2)^2 - 1)
100 = Q (0.16)
Q = 100/0.16 = 625

*Radio Engineers handbook page 138.
 
I also found this formula:

I = Io / [1 + Q^2 (w^2 - wo^2) / (w * wo)] where wo is the resonant frequency. With this formula I got Q = 23.8. I'm very confused now...
 
RyanP said:
I also found this formula:

I = Io / [1 + Q^2 (w^2 - wo^2) / (w * wo)] where wo is the resonant frequency. With this formula I got Q = 23.8. I'm very confused now...
Using your formula I obtain Q = 582. I think you have a mathematical error somewhere.
 
It may be expected that you solve by analysing the 3-element series circuit, rather than quoting a formula from somewhere.

The LC product is set so as to give resonance at 1.2 MHz
At resonance, the series impedance is 0.1 Ω

Now set the impedance at 1.2 MHz to be 10 Ω, and solve for L and/or C

My answer for Q is just under 600
 

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