Impedance Matching

Chapter Impedance Matching

Radio Frequency Integrated Circuit Design Second Edition Book
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Radio Frequency Integrated Circuit Design Second Edition Book

  • 101C H A P T E R 5Impedance Matching5.1  IntroductionIn RF circuits, we very seldom start with the impedance that we would like. There-fore, we need to develop techniques for transforming an arbitrary impedance into the impedance of choice. For example, consider the RF system shown in Figure 5.1. Here the source and load are 50W (a very popular impedance), as are the transmis-sion lines leading up to the IC. For optimum power transfer, prevention of ringing and radiation, and good noise behavior, for example, we need the circuit input and output impedances matched to the system. In general, some matching circuit must usually be added to the circuit, as shown in Figure 5.2. Typically, reactive matching circuits are used because they do not add noise to the circuit. However, using reactive matching components means that the circuit will only be matched over a range of frequencies and not at others. If a broad-band match is required, then other techniques may need to be used. An example of matching a transistor amplifier with a capacitive input is shown in Figure 5.3. The series inductance adds an impedance of jwL to cancel the input capacitive imped-ance. Note that in general, when an impedance is complex (R + jX), to match it, the impedance must be driven from the complex conjugate (R - jX).A more general matching circuit is required if the real part is not 50W. For ex-ample, if the real part of Zin is less than 50W, then the circuit can be matched using the circuit in Figure 5.4 and described in Example 5.1.Example 5.1: Matching Using Algebra Techniques A possible impedance matching network is shown in Figure 5.4. Use the matching network to match the transistor input impedance Zin = 40 - j30W to Zo = 50W. Perform the matching at 2 GHz. Solution: We can solve for Z2 and Y3, where for convenience, we have chosen impedance for series components and admittance for parallel components. An expression for Z2 is: 2inZZj Lω=+ where Zin = Rin - jXin. Solving for Y3 and equating it to the reference admittance Yo: 321ooYYj CYZω=+==