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4.3 Matching two unequal resistive impedances 175 4.3.1 L-Section Matching Suppose that we have a generator with an internal resistance R SOURCE =100Ω and a much greater load resistance R LP =1000Ω. To achieve maximum power transfer we want the generator see a load of 100 Ω. Now if we put a capacitor in parallel with R LP (Figure 4.3-2), the source will see a greater load and will supply additional current which will flow through the capacitor to charge it. Since the overall current through the R-C parallel is higher than the current supplied by the source when only R LP was present, whilst the voltage remains the same, the overall impedance seen by source is lower than R LP . Figure 4.3-2 The shunt capacitor C P reduces the resistive part of the load impedance You may argue that, if in place of the capacitor C P , we put a resistor in parallel with R LP , we could easily make the load impedance equal to that of the source. This is true, however the value of this resistor would have to be 110Ω and we would end up having almost 90% of the current flowing through it and just over 10% being delivered to the load! The advantage of using a reactive element, like the capacitor, is that it stores but does not dissipate energy 31 . So we managed to lower the overall impedance seen by the source by introducing a shunt capacitor, but how much have we lowered it by? Well, to work this out, the first step is to determine the equivalent series circuit (Figure 4.3-3), for our parallel R-C. Figure 4.3-3 Series equivalent of the Parallel R-C of Figure 4.3-2 From Table 4.2-2 we may work out the resistive part of the equivalent series impedance, , as shown below. represents the reactance of C P (Figure 4.3-2). 31 Actual capacitors are lossy as explained in section 4.2.6. C P V SOURCE R SOURCE R LP C S V SOURCE R SOURCE R LS (4.3-1) Conquer Radio Frequency 175 www.cadence.com/go/awr

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