05a ImpedanceMatching-p-1++ - IMPEDANCE MATCHING CIRCUITS...

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Unformatted text preview: IMPEDANCE MATCHING CIRCUITS Concepts: L-section impedance matching (Pozar Sec. 5.1) Part II: Passive Circuit Design Circuit Elements Impedance Matching Circuits Power Dividers Directional Couplers Resonators Filters Impedance Matching Circuits IM 1 Lumped Element Matching Network IM 1.1 L Network IM 2.1 Single Stub Matching IM 2.2 Double Stub Matching IM 2 Transmission Line Elements based Matching Network IM 2.3 IM 2.4 IM 2.5 Multisection Quarter-wave Tapered Lines Transformers transformer ------------------------------------------ bu konulardan bahsedilmeyecek. IM Impedance Matching Circuits Γ=0 Z in Matching Condition Z in = Z o Primary Function Maximum Power is delivered to the load (assuming the generator is matched) Use Receiver components (antenna, low-noise amplifier etc.) Power distribution network (antenna array) Reflection Coefficient Γ= Z in − Z o =0 Z in + Z o Reflections between the feed line and the matching network are eliminated Selection Factors Complexity - Low loss, low cost, and more reliable design preferred Bandwidth - Most frequently, broad band matching desired Implementation - Ease of implementation in the chosen transmission line Adjustability - Require adjustment to match a variable load impedance 1 IM 1.1 L Section Matching Network: Configuration A Input Impedance Configuration A: RL > Z o Z in = jX + For matching 1 jB + 1 /( RL + jX L ) Z L = RL + jX L Z in = Z o 1 jB + 1 /( R L + jX L ) Z 0 = jX + Real Part Z in Imaginary Part B( XRL − X L Z 0 ) = RL − Z 0 X (1 − BX L ) = BZ 0 RL − X L 1 + jx circle on impedance chart RL > Z o Z 1 XZ X= + L 0− 0 B RL BRL B= 2 X L ± R L / Z 0 RL2 + X L − Z 0 RL 2 RL2 + X L ∞> RL >1 Zo Configuration A Two possible solutions for B and X X Positive: Inductor X Negative: Capacitor B Positive: Capacitor B Negative: Inductor z L = rL + jx L Why? (see next slide) IM 1.1(a) Inductor and Capacitor: Reactance and Susceptance Inductor Impedance Admittance Impedance Capacitor Admittance Z = j ωL Compare with Y= −j 1 1 = = Z j ωL ωL Compare with −j 1 = Z= jωC ωC Compare with Y= 1 = jωC Z Compare with Z = jX Y = jB Z = jX −1 X= ωC Reactance X is Negative Y = jB X = ωL Reactance X is Positive −1 B= ωL Susceptance B is Negative B = ωC Susceptance B is Positive 2 Configuration A: Example 5.1 (Pozar p. 254) Smith Chart Solution Z o = 100 Ω Z in Z L = 200 − j100 Ω Normalized Load Impedance z L = 2 − j1 1 + jx circle on impedance chart Add Shunt Susceptance: Convert Impedance to admittance Add jb = j0.3 to reach the 1+jx circle (on the admittance chart) Add Series Reactance: Convert admittance back to Impedance Add jx = j1.2 to reach Z in = Z o Solution 1: b=0.3 and x=1.2 If we move in the opposite direction, we can obtain Solution 2: b=-0.7 and x=-1.2 Z in =1 Zo f = 500 MHz Solution 1: b=0.3 and x=1.2 B Positive: Capacitor, X: Positive: Inductor C= b = 0.92 pF 2πfZ 0 L= xZ 0 = 38.8 nH 2πf Solution 2: b=-0.7 and x=-1.2 B negative: Inductor, X Negative: Capacitor L= − Z0 = 46.1 nH 2πfb C= −1 = 2.61 pH 2πfxZ 0 3 IM 1.1 L Section Matching Network: Configuration B Input Admittance Configuration B: RL < Z o Yin = jB + For matching 1 RL + j(X + X L ) Yo = 1 Zo Z L = RL + jX Yo = Yin 1 1 = jB + Z0 RL + j ( X + X L ) Real Part Yin Imaginary Part 1 + jx circle on impedance chart BZ 0 ( X + X L ) = Z 0 − R L Solving for X and B ( X + X L ) = BZ 0 RL RL < Z o 0< RL <1 Zo => X = ± RL ( Z 0 − R L ) − X L Two possible solutions for B and X X Positive: Inductor X Negative: Capacitor B Positive: Capacitor B Negative: Inductor Configuration B => B=± ( Z 0 − RL ) / R L Z0 z L = rL + jx L her ikisi içinde ayni isaret seçilmeli 4 ...
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