Consider the following series-tuned antenna circuit consisting of:
- A variable capacitor (40 pF → 360 pF)
- A 240 μH antenna coil having a DC resistance of 12 Ω
Calculate the following:
- A) The frequency range of radio signals for which the radio is tunable.
- B) The quality factor (Q) at each end of the frequency range.
A) Determining the tunable range of frequencies
From our discussion of Series Resonance, recall that resonant frequency is defined as: $$ \omega_o = \frac{1}{\sqrt{LC}} \quad,Resonant\;frequency\; in \frac{rad}{sec} \quad,(Eqn\;1) $$ When the variable capacitor is at its lowest capacitance we have : $$ C = 40\;pF $$ We now substitute this value (as well as the given inductance of the antenna) into equation #1: $$ \omega_o = \frac{1}{\sqrt{(240\times 10^{-6})(40\times 10^{-12})}} $$ $$ \omega_o = (10.21\times10^6) \frac{rad}{s} $$ Recall that to convert angular frequency to Herz we have: $$ f = \frac{\omega}{2\pi} $$ ...therefore: $$ f_o = \frac{10.21\times10^6}{2\pi} $$
$$ f_o = 1.62\;MHz \quad,(frequency\;at\;40\;pF) $$
When the variable capacitor is at its highest capacitance we have : $$ C = 360\;pF $$ We again substitute this value (as well as the given inductance of the antenna) into equation #1: $$ \omega_o = \frac{1}{\sqrt{(240\times 10^{-6})(360\times 10^{-12})}} $$ $$ \omega_o = (3.4\times10^6) \frac{rad}{s} $$ converting this angular frequency to Herz we have: $$ f_o = \frac{3.4\times10^6}{2\pi} $$
$$ f_o = 541\;kHz \quad,(frequency\;at\;360\;pF) $$
We now know the range of tunable frequencies for this receiver to be: $$ 541\;kHz \lt f_o \lt 1.62 \; MHz$$
B) Determining the quality factor (Q) at each end of the frequency range
Additionally, we know that the quality factor for a series-resonant circuit is defined as:
$$ Q = \frac{2\pi f_o L}{R} $$
...Note that the formulas used to determine "Q" for a series resonant circuit are the same as for a passive Band-pass filter
So, for 541 kHz we have:
$$ Q_{541\;kHz} = \frac{2\pi (541\times 10^3)(240\times 10^{-6})}{12} $$
$$ Q_{541\;kHz} = 68 $$
For 1.62 MHz we have: $$ Q_{1.62\;MHz} = \frac{2\pi (1.62\times 10^6)(240\times 10^{-6})}{12} $$
$$ Q_{1.62\;MHz} = 203 $$