Q factor of an AM radio RLC circuit

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Homework Help Overview

The discussion revolves around determining the Q factor of a series RLC circuit designed to tune into a specific AM radio frequency of 1.20 MHz while minimizing interference from a nearby station at 1.10 MHz. Participants explore the relationship between circuit parameters and the required Q factor to achieve a desired current response at these frequencies.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants present various formulas related to Q factor and resonance, questioning their applicability and accuracy. Some attempt to derive Q using different methods, while others express confusion over conflicting results.

Discussion Status

The discussion is ongoing, with participants sharing different approaches and calculations for the Q factor. Some have provided numerical results, but there is no consensus on the correct value or method, indicating a productive exploration of the topic.

Contextual Notes

Participants note constraints such as a fixed resistance of R = 0.1 ohm and the need to minimize inductance L. There is also mention of the complexity of calculations involved in resonance and the potential for mathematical errors in the derivations presented.

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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