Q factor of an AM radio RLC circuit

AI Thread Summary
The discussion revolves around calculating the minimum quality factor (Q) for a series RLC circuit tuned to an AM radio station at 1.20 MHz while minimizing interference from a nearby station at 1.10 MHz. Participants explore various formulas to derive Q, with results ranging from 23.8 to 625, indicating confusion over calculations. The circuit's resistance is fixed at 0.1 ohm, and the goal is to achieve a current that is significantly lower at the unwanted frequency. Additional considerations include selecting appropriate values for inductance (L) and capacitance (C) to ensure resonance at the desired frequency. The consensus suggests that the Q value is likely just under 600.
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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