Appendix C.Double-tuned matching circuit example

Radio Frequency Circuit Design. W. Alan Davis, Krishna Agarwal

Copyright 2001 John Wiley & Sons, Inc.

Print ISBN 0-471-35052-4 Electronic ISBN 0-471-20068-9

APPENDIX C

Double-Tuned Matching Circuit

Example

Assume that an impedance transformation is required between a 50 source and a 15 load. The matching is to be done using the double-tuned matching circuit described in Chapter 3 for the program DBLTUNE. The center frequency is at 4 MHz, the bandwidth is 100 kHz, and the pass band ripple is 0.5 dB. The capacitances and transformer parameters are to be determined. In the following computer output, the bold characters are the responses the program expects from the user. Furthermore, in this example, the verbose mode is chosen by choosing to display the intermediate results. An analysis of this circuit using SPICE is shown in Fig. C.1.

Display intermediate results? < Y/N > Y Center Freq, Bandwidth (Hz) = ? 4.E6, 100.E3 Fm1 = 0.396480E+07 Fm2 = 0.403551E+07

GTMIN = 0.99992E+00

Passband ripple in dB = ? 0.5 Resistance Ratio r = 0.19841E+01

Q2 m1 = 0.97432E+00 Q2 m2 = 0.10097E+01

Generator and Load resistances values = 50., 15. L2’ = 0.56259E+02 H C2’ = 0.28140E+02pF

RL’ = 0.79332E+05 Bm1 = 0.19480E-01 Bm2 = -0.20193E-01 Given terminal resistances: RG = 0.500E+02 RL

= 0.150E+02

Input Circuit: C1 = 0.446554E+05pF L11 = 0.354637E-01 H Output Circuit: C2 = 0.148828E+06pF L22

= 0.106441E-01 H

Transformer coupling coefficient k = 0.250991E-01

290

DOUBLE-TUNED MATCHING CIRCUIT EXAMPLE

291

Insertion Loss, dB

0.00
–5.00
–10.00
–15.00			Double-Tuned Parallel Matching
–20.00						C1 = 44.66 nF
	Design Bandwidth = 100 kHz					L11 = 35.46 nH
	Actual Bandwidth = 139.2 kHz					C2 = 148.8 nF
–25.00	Rg = 50 Ω					L22 = 148.8 nF
	RL = 15 Ω					k = 0.0251
–30.00
3.8	3.8	3.9	4.0	4.0	4.1	4.1	4.2	4.2

Frequency, MHz

FIGURE C.1 Double-tuned matching circuit example.