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.
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