Frequency selective networks. There are man frequency selective networks, the most common two networks are the series and parallel resonant circuits. 1. Series resonant circuit.
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There are man frequency selective networks, the most common two networks are the series and parallel resonant circuits
Example: Design a filter to couple a voltage source, with negligible source impedance, to a 50 ohm load resistance. The specifications are that the filter center frequency be 5 MHz and the bandwidth be 100 kHz
From the quality factor we find that
The capacitance value can be found from
Example: A series tuned circuit is to be used to filter out harmonics of a waveform. What must be the minimum Q for the amplitude of the fifth harmonic to be 40 dB below the amplitude of the fundamental frequency
40 dB corresponds to a voltage ratio of 100:1, n=5
This means that Q=20.83
Example: The input impedance of a transistor amplifier is equal to 10 Ω in series with 0.2 μH. Design a matching network so that the input impedance is 50 Ω at 20 MHz
At 20 MHz the inductive reactance of a 0.2 μH inductor is 25.1 Ω. The combined impedance of the circuit became 10+j25.1 Ω.
To do the impedance transformation we can use the equation
in order to convert the real part of the impedance to 50 Ω, the imaginary part became Xs=20 Ω.
In order to convert the 25.1 Ω to 20 Ω a series reactance of -j5.1Ω must be added to the circuit.
Now Xp can be found from
if a reactance of –j25 is added in parallel with the circuit as shown in the figure of the next slide then input impedance became exactly 50 Ω
The –j25 reactance results from the impedance of a capacitor 318 pF
Example: Design a lossless matching network to couple the impedance shown in the fig. below to a 50 Ω source impedance at 20 MHz
solution: Since the series to parallel transformation always results in a parallel resistance larger than 50 Ω, there is no series reactance that will directly transform the 100 Ω resistor to a parallel equivalent of 50 Ω
The matching circuit can be solved based on the following equation
The input impedance of the circuit is
This can be brought to 50 Ω only by adding a –j50Ω reactance in series with the above mentioned circuit.
The completed solution will similar to the one shown in the next circuit