Q factor of an AM radio RLC circuit

In summary, the conversation discusses using a series RLC circuit to tune in a desired radio station while avoiding interference from a nearby station. The minimum Q for the circuit is calculated using various equations, with a final answer of approximately 600. The LC product is also discussed as a way to achieve resonance at the desired frequency.
  • #1
RyanP
17
0

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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  • #2
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.
 
  • #3
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...
 
  • #4
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.
 
  • #5
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
 

FAQ: Q factor of an AM radio RLC circuit

What is the Q factor of an AM radio RLC circuit?

The Q factor, also known as quality factor, of an AM radio RLC circuit is a measure of the selectivity or bandwidth of the circuit. It is defined as the ratio of the energy stored in the circuit to the energy dissipated per cycle.

How is the Q factor calculated?

The Q factor can be calculated using the formula Q = ω₀L/R, where ω₀ is the resonant frequency of the circuit, L is the inductance, and R is the resistance. Alternatively, it can also be calculated as Q = 1/R√(C/L), where C is the capacitance of the circuit.

What is the significance of the Q factor in an AM radio RLC circuit?

The Q factor is important in an AM radio RLC circuit as it determines the bandwidth of the circuit. A higher Q factor means a narrower bandwidth, resulting in better selectivity and less interference from other radio stations. This allows for clearer reception of a specific radio frequency.

Can the Q factor be adjusted in an AM radio RLC circuit?

Yes, the Q factor can be adjusted by varying the values of the inductance and capacitance in the circuit. Generally, a higher Q factor can be achieved by increasing the inductance or decreasing the capacitance, and vice versa.

What are the factors that can affect the Q factor of an AM radio RLC circuit?

The Q factor can be affected by various factors, including the quality of the components used, the temperature of the circuit, and external interference. Additionally, any changes in the circuit's inductance or capacitance can also affect the Q factor.

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