Matching Circuit Design

Document Sample
Matching Circuit Design Powered By Docstoc
					                                          Matching Circuit Design

      Design a matching circuit which matches a whip antenna to a pre-amplifier with the following
specifications

      a) Frequency: matching is to be achieved for f 0 = 8.1 MHz

      b) Bandwidth: ∆f = 0.7 MHz

      c) Antenna impadace: Z A = R A ∠ jX A = 36 ∠ j15 Ω at f 0 ;

      d) Input inpedance: pre-amplifier has input impedance of Z L = R L = 200 Ω .

                                                         Design

       Since the antenna has capacitive impedance and its resistance R A = 36 Ω is less than the the input
resistance of the pre-ampplifier, the Series C (type 2) circuit can be used (see Fig. 6.16 in the textbook [1]
or Fig on the page 4-5 in the course notes [2]). Notatios from [1] are used.

      In order to provide required bandwidth the circuit has to have the following quality factor Q

                                             f0    8.1MHz
                                        Q = ---- = ------------------- ≈ 11.6
                                               -                     -                                                          (1)
                                            ∆f     0.7MHz

       Using equation (6.-25) in [1] one can calculate intermediate parameters A and B of the matching
circuit

                                                                                                4.88 kΩ
    A = R A ( Q 2 + 1 ) = 36Ω × ( 11.6 2 + 1 ) ≈ 4.88 kΩ , B =               ( A ⁄ RL ) ∠ 1 =   ------------------- ∠ 1 ≈ 4.3
                                                                                                                  -             (2)
                                                                                                0.25 kΩ

      This allows to calculate necessary reactances of the matching circuit according to equations (6.26):

                            X L = QR A + X A = 11.6 × 36Ω + 15Ω ≈ 433 Ω                                                         (3)

                                             A             4.88 kΩ
                                 X C = -------------- = ------------------------ ≈ 669 Ω
                                                    -                          -                                                (4)
                                    1  Q∠B              11.6 ∠ 4.3

                                 X C = BR L = 4.3 × 250 Ω ≈ 1.08 kΩ                                                             (5)
                                    2



Finally, reactance allows to calculate values of the elements since the frequency is also known

                                                       433 Ω
                                 L = X L ⁄ ω = ------------------------------ ≈ 8.5 µH
                                                                            -                                                   (6)
                                               2π ⋅ 8.1MHz



                                                           -1-