TC 412 Microwave Communications 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 V max and V min 2
Ex1 Z L = 25+j50, given Z 0 = 50 and the line length is 60 cm, the wavelength is 2 m, find Z in. Ex2 A 0.334 long TL with Z 0 = 50 is terminated in a load Z L = 100-j100. Use the Smith chart to find a) L b) VSWR c) Z in d) the distance from load to the first voltage minimum 3
Ex3 Z L = 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 Z in. 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 Z 0 = 50 is terminated in a load Z L = 60+j40. Use the Smith chart to find a) L b) VSWR c) Z in d) the distance from the load to the first voltage maximum 5
Impedance matching To minimize power reflection from load Z in = Z 0 Matching techniques 1. Quarter - wave transformers 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. Z S Z R 0 L 6
Quarter-Wave Transformer Z Z S in Z R Z R 0 2 S L L 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, Z Load must match with Z 0, that means Z Load = Z 0 = 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, Z Load must match with Z 0, that means Z Load = Z 0 = 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 XC j2.350 j115 jc 1 C 1.38 j115 pf 12
Single stub tuners Working with admittance (Y) (Y o = 1/Z o ) 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 z L = 2 + j1, what is the admittance? y L 1/ z 1/(2 j1) 0.40 j0.20 L 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 y L Y Y L O 1 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 z in jz O tan( d) 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 ( y short = + j ) 17
Stub tuners on Y-chart (Admittance chart) (2) Procedure 1. Locate z L and then y L. From y L, move clockwise to 1 jb circle, at which point the admittance y d = 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 y l = jb. 3. For an open-circuit shunt stub, locate the open end at 0, then move to 0 jb. 4. Total normalized admittance y tot = y d +y l = 1. 18
Ex: Construct a short-circuited stub matching network for a 50- line terminated in a load Z L = 20 j55 First, normalize load impedance, z L = Z L /Z O = 0.4 j1.1 Next, locate y L 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 0.112 g to 0.187 g on the WTG scale, so our through-line length (d) is 0.075 g Next, we insert the short-ckted shunt stub. Admittance is on the right side of the chart at WTG = 0.250 g. We move clockwise until we reach the point 0 j2.0, located at WTG = 0.324 g. This give us stub length (l) = 0.324 g - 0.250 g = 0.074 g. 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, z L = Z L /Z O = 3.0 + j2.0 Next, locate y L 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 0.474 g to 0.178 g on the WTG scale. We add 0.500 g to the end point, so our through-line length (d) is (0.500 g ) 0.474 g + 0.178 g = 0.204 g 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 = 0.000 g. We move clockwise until we reach the point 0 j1.6, located at WTG = 0.339 g. This give us stub length (l) = 0.339 g. 21
22
Homework Prob. 2.41: A load impedance Z L = 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. 23
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 permittivity, 26 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 ε r (e.g. ε = 2.23ε o, ε = 6ε o, ε = 10.5ε o, etc.); quartz (ε = 3.7ε o ); alumina (ε = 9ε o ) and Epsilam-109 (ε = 10ε o ). Various substrate materials are available for the construction of microstrip lines, with practical values of ε r 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 ε r 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) Z O Therefore in this case and Z O L C 1 u C p and u p 1 LC u p g 2 f u u f p p c eff m / s rad / m 0 eff m. The evaluation of u p, Zo and λ in microstrip line requires the evaluation of ε eff and C. There are different methods for determining ε eff and C and, of course, closedform expressions are of great importance in microstrip-line design. The evaluation of ε eff 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 Z 0 as 29
Evaluation of the microstrip configuration (2) Assume t is negligible, if Z 0 and r are known, the ratio w/h can be calculated as for w / h 2, w h A 8e 2 A e 2 w 2 r 1 0.61 for w / h 2, 1 ln(2 1) ln( 1) 0.39 B B 2 B h r r where and A B Z 1 1 0.11 (0.23 ) 0 r r 60 2 r 1 377 2Z0 r The value of r and the dielectric thickness (h) determines the width (w) of the microstrip for a given Z 0. r 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 Z 0 = 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, for w / h 0.6, for w / h 0.6, 0 r w 0.0297 r 10.6( r 1)( ) 0 r w 0.1255 r 10.63( r 1)( ) h h 1/ 2 1/ 2 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 R c c skin Rskin Zw o ( Np / m) Rskin 8.686 ( db / m) Zw o 1 If the conductor is thin, then the more accurate skin resistance can be shown as 1 R skin t / (1 e ). 39
Dielectric attenuation d 2 f c ( 1) r eff 2 ( 1) eff r tan Np / m 40