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Published byJose Fullam Modified over 9 years ago
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Impedance Matching (2)
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Outline Three Element Matching – Motivation – Pi Network – T Network Low Q or Wideband Matching Network Impedance Matching on Smith Chart – Two-Element – Three-Element Matching Multi-Element Matching Genesis
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Component Q=4.73RV=4.424 Ohms
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Circuit Q Q of Vin/VS=102.2/(125.4-83.1)=2.416 Q of Vout/VS=98/(117.6-73.4)=2.21Circuit Q is different from component Q!
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Four Combinations of L-Match R L >R S R S >R L
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Split a Pi Network into Two L Networks
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Virtual Resistor R L >R S R S >R L Virtual Resistance must be smaller than source resistance! (Blocks DC)
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Design a Pi-Match RS=100 Ohms RL=1000 Ohms Resonant Frequency: 100 MHz Q2=15 R2/R=(Q 2 2 +1)/(Q 1 2 +1) (See attached) Q2Q2 Q1Q1
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Calculation Design Sequence: 1.Q1, Q2 2.RV 3.L2, C2 4.L1, C1
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Pi-Match Schematic
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Use Pi-Match to Produce Matching at 100 MHz
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T-Match
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R L >R S R S >R L
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Calculation in Matlab
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Schematic Q2=4.472Q1=10 RV=1050 Ohms
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Vin/VS
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Review of Smith Chart Adding an inductor in series Adding a capacitor in series Adding a capacitor in parallel Adding an inductor in parallel
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Adding an Inductor in Series Insertion of a series inductor to an impedance moves the impedance upward, causing a rotation clockwise along a constant circle of resistance
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Adding a Capacitor in Series Insertion of a series capacitor to an impedance move impedance downward, causes a rotation counter clockwise along a constant circle of resistance
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Adding a Shunt Capacitance Insertion of a shunt capacitor causes a rotation clockwise along a constant circle of admittance
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Adding a Shunt Inductance Insertion of a shunt inductor causes a rotation counter clockwise along a constant circle of admittance
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Example Design a matching network with a source impedance of 25+15j Ohm and output impedance of 100+25j Ohms. (We need to have match the source and load to their complex conjugates)
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Starting Smith Chart (load) (source)
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Four Combinations of L-Match R L >R S R S >R L (Series L causes clockwise Movement on constant R on smith chart… …) (The only one)
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C=39.46 pF; L=156.35 nH
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Constant Q Q of series impedance=ratio of reactance to resistance
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Constant Q Circle
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Example 4-4 Constant Q of 15 Q=15 The end of large terminating resistor will determine the Q.
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Example 4-4 (Adding a Series L) Get the admittance circuit with a series L
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Example 4.4 Get back with the center of chart with a shunt cap.
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Example 4-4 (Q=15)
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Compare the Smith Chart Design with Calculation in Matlab C2 (matlab) C2(Smith Chart) L1+L2 (matlab) L1+L2 (Smith chart) C1 (matlab)C1 (Smith chart) 23.87 pF22.36 pF138.63 nH147.9 nH68.55 pF75.49 pF
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Example 4.8 Design a T network to match Z=15+15j Ohm source to a 225 Ohm load at 30 MHz with a loaded Q of 5.
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Example 4.8 Get on Constant Q=5 curve
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