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ENE 428 Microwave Engineering Lecture 6 Transmission lines problems and microstrip lines RS 1 Review • Input impedance for finite length line – Quarter wavelength line – Half wavelength line • Smith chart – A graphical tool to solve transmission line problems – Use for measuring reflection coefficient, VSWR, input impedance, load impedance, the locations of Vmax and Vmin 2 Ex1 ZL = 25+j50 , given Z0 = 50 and the line length is 60 cm, the wavelength is 2 m, find Zin. Ex2 A 0.334 long TL with Z0 = 50 is terminated in a load ZL = 100-j100 . Use the Smith chart to find a) L b) VSWR c) Zin d) the distance from load to the first voltage minimum 3 Ex3 ZL = 80-j100 is located at z = 0 on a lossless 50 line, given the signal wavelength = 2 m, find a) If the line is 0.8 m in length, find Zin. b) VSWR c) What is the distance from load to the nearest voltage maximum d) what is the distance from the input to the nearest point at which the remainder of the line could be replaced by a pure resistance? 4 Ex4 A 0.269- long lossless line with Z0 = 50 is terminated in a load ZL = 60+j40 . Use the Smith chart to find a) L b) VSWR c) Zin d) the distance from the load to the first voltage maximum 5 Impedance matching • To minimize power reflection from load • Zin = Z0 • Matching techniques 1. Quarter - wave transformers Z S Z 0 RL for real load 2. single - stub tuners 3. lumped – element tuners • The capability of tuning is desired by having variable reactive elements or stub length. 6 Quarter-Wave Transformer Z S2 Z in RL Z S Z0 RL 7 Simple matching by adding reactive elements (1) EX5, a load 11+j25 is terminated in a 50 line. In order for 100% of power to reach a load, ZLoad must match with Z0, that means ZLoad = Z0 = 50 . This value corresponds to center of the chart. 8 Simple matching by adding reactive elements (1) 9 Distance d WTG = (0.188-0.076) = 0.112 to point 1+ j2 Therefore cut TL and insert a reactive element that has a normalized reactance of -j2. The normalized input impedance becomes 1+ j2 - j2 = 1 which corresponds to the center or the Smith chart. 10 Simple matching by adding reactive elements (1) EX5, a load 10-j25 is terminated in a 50 line. In order for 100% of power to reach a load, ZLoad must match with Z0, that means ZLoad = Z0 = 50 . Distance d WTG = (0.5-0.424) +0.189 = 0.265 to point 1+ j2.3. Therefore cut TL and insert a reactive element that has a normalized reactance of -j2.3. The normalized input impedance becomes 1+ j2.3 - j2.3 = 1 which corresponds to the center or the Smith chart. 11 Simple matching by adding reactive elements (2) The value of capacitance can be evaluated by known frequency, for example, 1 GHz is given. 1 j 2.3 50 j115 jC 1 C 1.38 pF j115 XC 12 Single stub tuners Working with admittance (Y) (Yo = 1/Zo) since it is more convenient to add shunt elements than series elements Stub tuning is the method to add purely reactive elements 13 Ex6 let zL = 2 + j1, what is the admittance? y L 1 / z L 1 /(2 j1) 0.40 j 0.20 Where is the location of y on Smith chart? We can easily find the admittance on the Smith chart by moving 180 from the location of z. Smith Chart is also a chart of normalized admittance. The normalized load admittance is YL 1 yL YO z L 14 15 Stub tuners on Y-chart (Admittance chart) (1) There are two types of stub tuners 1. Shorted end, y = (the rightmost of the Y chart) 2. opened end, y = 0 (the leftmost of the Y chart) l l Short-circuited shunt stub Open-circuited shunt stub 16 Short-circuited Shunt Stub Z in jZ O tan( d ) zin j tan( d ) Purely reactive tuning element follows the constant || circle along the periphery of Smith chart (z = 0 jx). Proper selection of the T-line length d allows us to choose any value of reactance that we want, whether it’s capacitive or inductive. The admittance of the short-circuited starts on the right side of the chart ( yshort = ∞ + j∞) 17 Stub tuners on Y-chart (Admittance chart) (2) Procedure 1. Locate zL and then yL. From yL, move clockwise to 1 jb circle, at which point the admittance yd = 1 jb. On the WTG scale, this represents length d. 2. For a short-circuited shunt stub, locate the short end at 0.250 then move to 0 jb, the length of stub is then l and then yl = jb. 3. For an open-circuit shunt stub, locate the open end at 0, then move to 0 jb. 4. Total normalized admittance ytot = yd+yl = 1. 18 Ex: Construct a short-circuited stub matching network for a 50- line terminated in a load ZL = 20 – j55 First, normalize load impedance, zL = ZL/ZO = 0.4 – j1.1 Next, locate yL Move along the constant-|| circuit in WTG (clockwise) direction ‘till it intersects with the 1 jb circle (in this case 1 + j2.0). We travel from 0.112g to 0.187g on the WTG scale, so our through-line length (d) is 0.075g Next, we insert the short-ckted shunt stub. Admittance is on the right side of the chart at WTG = 0.250g. We move clockwise until we reach the point 0 – j2.0, located at WTG = 0.324g. This give us stub length (l) = 0.324g- 0.250g = 0.074g. 19 20 Ex: We want to construct an open-ckted shunt-stub matching network for a 50- line terminated in a load ZL =150 + j100 . Solution: First, normalize load impedance, zL = ZL/ZO = 3.0 + j2.0 Next, locate yL Move along the constant-|| circuit in WTG (clockwise) direction ‘till it intersects with the 1 jb circle (in this case 1 + j1.6). We travel from 0.474g to 0.178g on the WTG scale. We add 0.500g to the end point, so our through-line length (d) is (0.500g) – 0.474g + 0.178g = 0.204g Next, we insert the open-ckted shunt stub. On the admittance chart, the location of the open-ckted is on the left side of the chart at WTG = 0.000g. We move clockwise until we reach the point 0 – j1.6, located at WTG = 0.339g. This give us stub length (l) = 0.339g. 21 First, normalize load impedance, zL = ZL/ZO = 0.4 – j1.1 Next, locate yL Move along the constant-|| circuit in WTG (clockwise) direction ‘till it intersects with the 1 jb circle (in this case 1 + j2.0). We travel from 0.112g to 0.187g on the WTG scale, so our through-line length (d) is 0.075g Next, we insert the short-ckted shunt stub. Admittance is on the right side of the chart at WTG = 0.250g. We move clockwise until we reach the point 0 – j2.0, located at WTG = 0.324g. This give us stub length (l) = 0.324g- 0.250g = 0.074g. 22 23 Homework Prob. 2.41: A load impedance ZL = 25 + j90 is to be matched to a 50- line using an open-ended shunt-stub tuner. Find the solution that minimizes the length of the shorted stub. 24 Microstrip (1) • The most popular transmission line since it can be fabricated using printed circuit techniques and it is convenient to connect lumped elements and transistor devices. • By definition, it is a transmission line that consists of a strip conductor and a grounded plane separated by a dielectric medium 25 Microstrip (2) • The EM field is not contained entirely in dielectric so it is not pure TEM mode but a quasi-TEM mode that is valid at lower microwave frequency. • The effective relative dielectric constant of the microstrip is related to the relative dielectric constant r of the dielectric and also takes into account the effect of the external EM field. Typical electric field lines Field lines where the air and dielectric have been replaced by a medium of effective relative 26 permittivity, eff Microstrip (2.1) Some typical dielectric substrates are RT/Duroid® (a trademark of Rogers Corporation, Chandler, Arizona), which is available with several values of εr (e.g. ε = 2.23εo, ε = 6εo, ε = 10.5εo, etc.); quartz (ε = 3.7εo); alumina (ε = 9εo) and Epsilam-109® (ε = 10εo). Various substrate materials are available for the construction of microstrip lines, with practical values of εr ranging from 2 to 10. The substrate material comes plated on both sides with copper, and an additional layer of gold plating on top of the copper is usually added after the ckt pattern is etched in order to prevent oxidation. Typical plating thickness of copper is from ½ mils to 2 mils (1 inch = 1000 mils). The value of εr and the dielectric thickness (h) determine the width of the microstrip line for a given Zo. These parameters also determine the speed of propagation in the line, and consequently its length. Typical thickness are 25, 30, 40, 50 and 100 mils. 27 Microstrip (3) Therefore in this case up c eff 2 f up and Z O L C and u p 1 1 ZO u pC LC m/s rad / m up 0 g f eff m. The evaluation of up, Zo and λ in microstrip line requires the evaluation of εeff and C. There are different methods for determining εeff and C and, of course, closedform expressions are of great importance in microstrip-line design. The evaluation of εeff and C based on a quasi-TEM mode is accurate for design purposes at lower microwave freq. However, at higher microwave freq, the longitudinal components of the EM fields are significant and the quasi-TEM assumption is no longer valid. 28 Evaluation of the microstrip configuration (1) • Consider t/h < 0.005 and assume no dependence of frequency, the ratio of w/h and r are known, we can calculate Z0 as 29 Evaluation of the microstrip configuration (2) • Assume t is negligible, if Z0 and r are known, the ratio w/h can be calculated as for w / h 2, w 8e A 2A h e 2 for w / h 2, r 1 w 2 0.61 B 1 ln(2 B 1) ln( B 1) 0.39 h 2 r r where and Z0 r 1 r 1 0.11 A (0.23 ) 60 2 r 1 r B 377 2Z 0 r The value of r and the dielectric thickness (h) determines the width (w) of the microstrip for a given Z0. 30 Characteristic impedance of the microstrip line versus w/h 31 Normalized wavelength of the microstrip line versus w/h 32 Ex8 A microstrip material with r = 10 and h = 1.016 mm is used to build a TL. Determine the width for the microstrip TL to have a Z0 = 50 . Also determine the wavelength and the effective relative dielectric constant of the microstrip line. 33 34 35 Wavelength in the microstrip line Assume t/h 0.005, 1/ 2 r for w / h 0.6, 0 r 1 0.6( 1)( w )0.0297 r h 1/ 2 r for w / h 0.6, 0 r 1 0.63( 1)( w )0.1255 r h 36 37 Attenuation (losses in microstrip lines) conductor loss dielectric loss radiation loss tot c d where c = conductor attenuation (Np/m) d = dielectric attenuation (Np/m 38 Conductor attenuation Rskin c Zo w ( Np / m) Rskin c 8.686 Zo w Rskin 1 (dB / m) If the conductor is thin, then the more accurate skin resistance can be shown as Rskin 1 . t / (1 e ) 39 Dielectric attenuation 2 f r ( eff 1) d tan c 2 eff ( r 1) Np / m 40