EE223 Microwave Circuits Fall2014 Lecture5

Click here to load reader

  • date post

    02-Oct-2015
  • Category

    Documents

  • view

    234
  • download

    9

Embed Size (px)

description

,

Transcript of EE223 Microwave Circuits Fall2014 Lecture5

  • EE-223 Microwave Circuits (Fall 2014) Lecture 5

    Dr. Atif Shamim EE Program King Abdullah University of Science and Technology (KAUST) 1

  • Impedance Matching with Smith Chart

    The objective of a matching network is to move to the center of the Smith Chart

    The normalized input impedance will be located at the center of the chart, where =0

    Matching network must therefore move us from the load to the center of the Smith chart

    Moving away from the load along the Zo line generates the constant | | circle

    Of special interest are two points where the circle crosses the 1 jx circle

    At either of these points, inserting the appropriate reactive element (that is adding jx ) will then move us to the matched condition at the center of the chart

  • Consider a 50- T-line terminated in an 11+ j25 load.

    zL= 0.22 +j 0.5

    Move a distance d of 0.112 along the constant | | circle to point 1+j 2.0

    We insert a series capacitive element of normalized reactance j 2.0, corresponding to j 100

    Zin = 1+ j2.0 - j2.0= 1+j0

    The value of capacitance depends on frequency

    j100 = jC

    Example

  • (a) A 50- T-line terminated in an 11+ j25 load (b) T-line with tuning capacitor added at the appropriate distance from the load.

  • Problem

    Suppose that the line in Example 1 is a coaxial cable made with a Teflon dielectric, and it must operate at 800 MHz. determine (a) the length of the coaxial line between the load and the capacitor and (b) the value of the series capacitor added to provide an impedance match.

    Eps = 2.1 (for Teflon) Answer: (a) d=2.9 cm, (b) C=2.0 pF

  • The previous matching networks employed series elements. It is often desirable to add shunt elements instead for matching. Now we must use the Smith Chart as a normalized admittance chart, with characteristic admittance Yo = 1/Zo The convenience of admittances is that shunt values may be added.

    Admittance of Shunt Stubs

    (a) Admittance relationship to impedance. (b) Adding shunt elements using admittances.

  • Fig. 6.26

    Admittance of Shunt Stubs

  • Problem

    A 50 line is terminated in a pair of parallel load impedances of 50 + j 100 and 50 j 100 . Determine the total load admittance and impedance seen by the line.

    Answer: 8 mS, 125

  • Lossless T-line Stubs (either shorted or open ended)

    Consider a shorted line of length d (as shown below):

    Fig. 6.27 (a) A shorted T-line stub

  • Fig. 6.27 (b) Smith Chart view of zin and yin for d = /8.

  • Shunt-Stub Matching

    The goal is to move to the centre of the Smith Chart, used as the admittance chart. From the normalized load admittance, a section of through line is traversed to arrive at the 1 jb circle. At this point we add a shunt stub of normalized admittance 0 jb . The sum of these admittances take us to the center of the chart where Yin = Yo and Zin = Zo, matching is complete.

    Fig. 6.28 (a) shorted shunt stub matching network

  • Fig. 6.28 (b) Adding shunt admittances. (c) Using the Smith Chart to find through line and stub lengths. Values on the chart apply to Example 2.

  • Shunt-Stub Matching Procedure

  • Shorted Shunt-Stub Example

    Example 2

  • Open-Ended Shunt-Stub Example

    Example 3

  • Fig. 6.29 (a) The generic layout of the open-ended shunt-stub matching network

  • Fig. 6.29 (b) Smith Chart solution to Example 3

  • Problem

    In Example 2, we chose the first intersection with the 1 jb circle (at 1+ j 2.0) in designing the matching network. We could also have continued on to the second intersection, occurring at 1 j 2.0. Determine the through line length d, and the stub length l for the matching network using this second intersection.

    Answer: d= 0.200 g, l=0.426 g

  • Problem

    For Example 3, determine the through line length d, and the stub length l for the open-ended shunt matching network, if the other intersection with the 1 jb circle is chosen.

    Answer: d= 0.348 g, l=0.161 g

    EE-223 Microwave Circuits(Fall 2014)Lecture 5Slide Number 2Example(a) A 50- T-line terminated in an 11+ j25 load (b) T-line with tuning capacitor added at the appropriate distance from the load.ProblemSlide Number 6Slide Number 7ProblemSlide Number 9Fig. 6.27 (b) Smith Chart view of zin and yin for d = /8.Slide Number 11Slide Number 12Slide Number 13Slide Number 14Slide Number 15Slide Number 16Slide Number 18ProblemProblem