9 1 Microwave Circuit Design I Lecture 9 Topics: 1. Admittance Smith Chart 2. Impedance Matching 3. Single-Stub Tuning Reading: Pozar pp. 228 235 The Admittance Smith Chart Since the following is also true Z L = Z 1 + Γ 1 Γ, (1) Y L = Y 1 Γ 1 + Γ from which we can define the normalized admittance Y = Y L = 1 Γ Y 1 + Γ We ll follow a treatment similar to what we saw for impedances to create the admittance Smith chart. (2) (3)
9 2 The normalized admittance is a complex number in general. So, we can write Y = G + jb = 1 Γ 1 + Γ = 1 u jv 1 + u + jv In a manner similar to before, lines of constant conductance in the admittance plane map to circles defined by ( u + G ) 2 + v 2 = G + 1 in the Γ plane. These are circles of radius r = 1 G + 1 (4) 1 (G + 1) 2 (5)
and center at ( G ) G + 1, 0 as shown below 9 3 Lines of constant susceptance in the admittance plane map to circles defined by ( (1 + u) 2 + v + 1 ) 2 = 1 (6) B B 2 in the Γ plane. The circles have radius r = 1 B and center ( 1, 1 ) B Note the sign of the center. This negative sign implies that negative susceptance maps to the top half (positive real) of the Γ plane and positive susceptance maps to the bottom half (negative real) of the Γ plane as shown below
9 4 The admittance Smith chart is shown on the next page.
Notes: 9 5 All components of the chart operate as before. We can use a combination impedance/admittance chart to convert between Γ, Z, and Y simultaneously. Example) Given Z = 1.6 + j1.2, determine Y and Γ from the admittance chart. Solution: Y = 0.4 j0.3 (7) Impedance Matching Γ = 39 (8) Γ = 34/73 = 0.466 (9) Γ = 0.466 39 (10) Consider an arbitrary, complex load impedance Z L. We wish to match that load impedance to a transmission line using some 2 port matching network: Define Z = Z L /Z = Normalized Load Impedance which can be plotted on the Smith chart. Also, let Z in be the normalized input impedance to the matching network. With these definitions, Z L is matched when Z in = 1.0 + j0.0 Y L is matched when Y in = 1.0 + j0.0
9 6 If Z L is connected directly to the transmission line, the input impedance traverses a circle centered at (0, 0) on the Smith chart (Y L does the same). Let s consider the example of Z = 1.6 + j1.2, or Y = 0.4 j0.3. Option 1 (Series T Line): Attach the load directly to the feeding transmission line and see if we can choose a specific length of line to affect an impedance match. For this analysis, we will plot Y on the admittance Smith chart as shown on the next page. To obtain perfect matching, we want the input impedance to be located at the origin, but a single connecting transmission line only allows us to traverse the circle centered at the origin. There is no way the single connecting transmission line can improve the level of match of this load.
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9 8 Option 2 (Shunt Reactance): Use a shunt reactance at the load. Adding a shunt (parallel) reactance moves us along the circle of constant G as shown on a subsequent page. In general we cannot obtain a perfect match, but clearly this scenario differs from the previous one.
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9 10 Option 3 (Single Stub Tuner): A series transmission line followed by a shunt reactance. If we place a length of series transmission line until the effective input impedance intersects with the line of constant G = 1, then we can add a shunt admittance of G = 1 to locate the input admittance at the origin of the Smith chart. This scenario looks like the following: and it is viewed on the Smith chart on the following page.
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9 12 Option 4 (Double Stub Tuner): Two shunt reactances with a fixed series transmission line between them. If we place a variable shunt reactance in parallel with the load, a fixed series transmission line followed by a variable shunt reactance at the input to the matching network, the input admittance can be tuned to the origin of the Smith Chart in most cases. This scenario looks like the following: This option is typically used when variable load impedances are encountered. When it is unsure what the load impedance will be it is easier to create variable length stubs than it is to create variable length series elements. For example, if the double stub tuner is fabricated using air core coaxial cable, the stubs are simply short circuit elements with a plunger to independently change the lengths of the two shunt elements. Single Stub Tuning Any arbitrary (complex) load impedance can be matched to a specific transmission line characteristic impedance at a single frequency by performing the following steps: 1. Compute Y = Z /Z L = normalized load admittance 2. Plot Y on the admittance Smith chart
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3. Draw a circle, centered at the origin, that crosses Y 9 14
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4. Draw lines from the origin through the following points: 9 16 (a) Y (b) The negative intersection of the constant SWR circle of step 3 and the circle of constant G = 1.0 (c) The positive intersection of the constant SWR circle of step 3 and the circle of constant G = 1.0
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9 18 5. d 1 is the shortest clockwise distance (in λ) from Y to the line of step 4b or 4c
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9 20 6. d 2 is the clockwise distance (in λ) from Y to the remaining line of step 4b or 4c
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9 22 7. The required shunt admittance a distance d 1 from the load is the negative of the susceptance associated with the intersection of the constant SWR circle and the constant G = 1.0 circle assocated with step 5 8. The required shunt admittance a distance d 2 from the load is the negative of the susceptance associated with the intersection of the constant SWR circle and the constant G = 1.0 circle of step 6 9. To determine the appropriate length of stub to realize a particular shunt admittance, recall, { jz tan(βl), short Z in = (11) jz cot(βl), open { jy cot(βl), short Y in = (12) jy tan(βl), open Y { in j cot(βl), short = (13) Y j tan(βl), open (a) for open circuit stubs: [ 1 l = 2π tan 1 (b) for short circuit stubs: [ 1 l = 2π tan 1 ( Yin )] λ (14) jy ( Y jy in )] λ (15) Note: l cannot be negative. If the equations above give a negative value for l, simply add π to βl. This adds 0.5 to l/λ.
9 23 Example: Design an open circuit single stub tuner for a normalized load impedance of Z L = 0.5 j1.0 Solution: From the Smith chart that follows, Y L = 1/Z L = 0.4 + j0.8 (16) d 1 = 0.429λ 0.365λ = 0.064λ (17) d 2 = 0.5λ 0.365λ + 0.072λ = 0.207λ (18) The input admittances to these lengths of lines are Y in,d1 = 1 + j1.6 Y d1 = j1.6 (19) Y in,d2 = 1 j1.6 Y d2 = +j1.6 (20) (21)
9 24 Since we have chosen to use an open circuit stub, the length of the stubs are Thus, l d1 = and for the other length l d2 = [ 1 2π tan 1 [ 1 2π tan 1 and the circuits look like the following: ( )] j1.6 λ = 0.161λ (22) j1.0 l d1 = 0.3389λ (23) ( )] j1.6 λ = 0.161λ (24) j1.0 l d2 = 0.161λ (25)
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