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ENE 490 Applied Communication Systems Lecture 2 circuit matching on Smith chart 16/11/50 1 Review (1) High frequency operation and its applications Transmission line analysis (distributed elements) – Use Kirchholff’s law to obtain general equations for transmission lines – Voltage and current equations are the combination of incident and reflected waves. V ( z ) V0 e jz V0 e jz V0 jz V0 jz I ( z) e e Z0 Z0 where Z0 is a characteristic impedance of a transmission line. 16/11/50Assume the line is lossless. 2 Review (2) Terminated lossless line – voltage reflection coefficient V0 j 2d ( d ) e V0 – impedance along a transmission line Z L jZ 0 tan d Z (d ) Z 0 Z 0 jZ L tan d or 1 ( d ) Z (d ) 1 ( d ) 16/11/50 3 Review (3) - voltage standing wave ratio V (d ) max V (d ) min I (d ) max I (d ) min 1 L VSWR 1 L source and loaded transmission line Z in Z 0 in (d l ) 0e 2 jl Z in Z 0 Z S Z0 S Z S Z0 16/11/50 4 Review (4) power transmission of a transmission line 1 Pav Re VI * 2 for lossless and a matched condition 2 1 VS Pin Pavs 8 ZS power in decibels P W P dBm 10 log 1 mW 16/11/50 5 Impedance matching network (1) The need for matching network arises because amplifiers, in order to deliver maximum power to a load or to perform in a certain desired way, must be properly terminated at both the input and the output ports. Z1=50 W + VS Input matching network Output matching network Transistor Z2=50 W - 16/11/50 ZS ZL 6 Impedance matching network (2) Effect of adding a series reactance element to an impedance or a parallel susceptance are demonstrated in the following examples. Adding a series reactance produces a motion along a constant-resistance circle in the ZY Smith chart. Adding a shunt susceptance produces a motion along a constant-conductance circle in the ZY Smith chart. 16/11/50 7 Ex1 Adding a series inductor L (zL = j0.8) to an impedance z = 0.3-j0.3. z L= j0.8 z = 0.3-j0.3 16/11/50 8 Ex2 Adding a series capacitor C (zC = j0.8) to an impedance z = 0.3-j0.3. zC =-j0.8 z = 0.3-j0.3 16/11/50 9 Ex3 Adding a shunt inductor L (yL = j2.4) to an admittance y = 1.6+j1.6. y L=-j2.4 16/11/50 y = 1.6+j1.6 10 Ex4 Adding a shunt capacitor C (yC = j3.4) to an admittance y = 1.6+j1.6. yC=j3.4 16/11/50 C y = 1.6+j1.6 11 Examples of matching network design Ex5 Design a matching network to transform the load Zload = 100+j100 W to an input impedance of Zin = 50+j20 W. 16/11/50 12 Ex6 Design the matching network that provides YL = (4-j4)x10-3 S to the transistor. Find the element values at 700 MHz. C L 50 W y L=(4-j4)x10-3S 16/11/50 13 L L matching networks L C C L C C Z load L C Z load Z load C L2 Z load C1 Z load L2 Z load C1 Z load L1 Z load L2 L Z load C2 L C 16/11/50 Z load C2 C Z load Z load L1 L L1 C Z load Z load 14 Forbidden regions Sometimes a specific matching network cannot be used to accomplish a given match. 16/11/50 15 Load quality factor The developed matching networks can also be viewed as resonance circuits with f0 being a resonance frequency. These networks may be described by a loaded quality factor, Q L. f0 QL BW The estimation of QL is simply accomplished through the use of a so-called nodal quality factor Qn. At each node of the L-matching networks, there is an equivalent series input impedance, denoted by RS +jXS. Hence a circuit node Qn can be defined at each node as 16/11/50 Qn XS RS BP GP 16 Circuit node Qn and loaded QL Qn QL 2 L-matching network is not a good choice for a design of high QL circuit since it is fixed by Qn. For more complicated configurations (T-network, Pi- network), the loaded quality factor of the match network is usually estimated as simply the maximum circuit node quality factor Qn. 16/11/50 17 The example of Q calculation L=3.18 nH L=3.18 nH 50W C=12.7 pF VS 10 W Z IN = 50 W At 500 MHz Qn = 2 then QL = 1. and 16/11/50 f 0 = 500 MHz BW QL 18 Ex7 The low pass L network shown below was designed to transform a 200 W load to an input resistance of 200 W. Determine the loaded Q of the circuit at f = 500 MHz. AC 16/11/50 R = 200 W 20 W C =4.775 pF L = 19.09 nH 19 Constant Qn contours 16/11/50 20 The upper and lower part of Q contours satisfy a circle equation. Since 1 1U 2 V 2 2 z r jx j 1 (1 U )2 V 2 (1 U )2 V 2 then x 2U Qn r 1U 2 V 2 which can be written as U 2 (V 16/11/50 1 2 1 ) 1 2 Qn Qn 21 Contour equations The equations for these contours can be derived from the general derivation of the Smith chart. By following the derivation, Qn contours follow this circle equation, 2 i2 2 1 1 r 1 2 Qn Qn where the plus sign is taken for positive reactance x and the minus sign for negative x. 16/11/50 22 Qn circle parameters For x > 0, the center in the plane is at (0, -1/Qn). For x < 0, the center in the plane is at (0, +1/Qn). the radius of the circle can be written as 1 r 1 2 . Qn 16/11/50 23 Ex8 Design two T networks to transform the load impedance ZL = 50 W to the input impedance Zin = 10-j15 W with a Qn of 5. 16/11/50 24 Ex9 Design a Pi network to transform the load impedance Zload = 50 W to the input impedance Zin = 150 W with a Qn of 5. 16/11/50 25