Frequency selective networks

1. Frequency selective networks There are man frequency selective networks, the most common two networks are the series and parallel resonant circuits 1

2. Series resonant circuit • In this kind of circuits, at the resonant frequency the inductor and capacitors has equal impedances with opposite signs • At the resonant frequency vO will be in phase with vi • The quality factor for this circuit is given by 2

3. Series resonant circuit • The quality factor gives an indication of the bandwidth of the circuit • Narrower bandwidth means that the circuit has a larger quality factor as shown in the Figure 3-dB f2 3 f1

4. Series resonant circuit • The quality factor is defined according to the following equation • The bandwidth is defined as the difference between f1 and f2 • At these two particular frequencies the signal amplitude is less by 3 dB as indicated by the figure in slide 3 4

5. Series resonant circuit example 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 Solution From the quality factor we find that 5

6. Series resonant circuit example Solution The capacitance value can be found from 6

7. Series resonant circuit • The circuit gain at any frequency can be determined from • The attenuation at any harmonic can be found from 7

8. Series resonant circuit example 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 Solution 40 dB corresponds to a voltage ratio of 100:1, n=5 This means that Q=20.83 8

9. Parallel resonant circuit • In parallel resonant circuit two susceptances are added in parallel, so that the admittance, instead of impedance is a minimum at fO 9

10. Parallel resonant circuit • In practical life the inductor usually has a finite resistance, therefore a more accurate model for the parallel resonant circuit is shown below 10

11. Impedance matching and harmonic filtering using reactive networks • Reactive networks can be used to match impedances over a narrow frequency range • These networks can also be used to filter out some harmonics 11

12. Impedance matching • The input impedance of the two networks shown in the next slide are equal provided • If we take the real and imaginary parts and equate them we get the following 12

13. Impedance matching • The previous equations can be rewritten as 13

14. Impedance matching 14

15. Impedance matching and harmonic filtering using reactive networks 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 Solution: 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 15

17. Impedance matching and harmonic filtering using reactive networks Solution: Or 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. 17

18. Impedance matching and harmonic filtering using reactive networks Solution: 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 18

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20. Impedance matching and harmonic filtering using reactive networks Example: Design a lossless matching network to couple the impedance shown in the fig. below to a 50 Ω source impedance at 20 MHz 20

21. Impedance matching and harmonic filtering using reactive networks 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 21

22. Impedance matching and harmonic filtering using reactive networks Solution: 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 22

23. Impedance matching and harmonic filtering using reactive networks 23