(Chapter 5)

Asst Prof Dr. Zeeshan Zahid

 Introduction
 The purpose is to deliver max. power to the load (Γ = 0)
 This improves SNR and reduces amplitude or phase errors
 The process is a part of larger system design
 Matching is also called tuning
 We shall discuss two types of matching networks using
(1) Lumped elements (2) Stubs (Lossless networks)
 The basic idea of impedance matching is shown in fig.

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Asst Prof Zeeshan Zahid, EE 342, MCS
 The Idea of Matching
 The Zin must be equal to Zo of the trans. line

ZL

Zin

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Asst Prof Zeeshan Zahid, EE 342, MCS
 Matching with Lumped Elements
 Three categories are L,T & π networks shown in fig.
 Narrow band matching technique
 The networks contain reactive elements (C or L)
 L-network is most convenient and low cost

(L-network) (T-network) (π-network)

Asst Prof Zeeshan Zahid, EE 342, MCS 4

 Matching with L-Network
 Two possible combinations of the network are shown
 Unit resistance circle (1 + jx circle)
 Network (a) is selected if zL is inside unit circle
 Network (b) is selected if zL is outside unit circle
 Smith chart will be very useful for design of this network

Asst Prof Dr. Zeeshan Zahid, EE 342, MCS 5

Combinations for (a) Combinations for (b)

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Asst Prof Dr. Zeeshan Zahid, EE 342, MCS
 Analytical Solution
 Input impedance of network (a)
1
Z in = Z o = jX +
jB + 1/( RL + jX ) Zin

 Input impedance of network (b)

1 1 1
= = jB + Zin
Z o Z in RL + j ( X + X L )

Asst Prof Dr. Zeeshan Zahid, EE 342, MCS 7

 Possible Combinations
 Effective movements (shaded region is forbidden region)

(A) (B) (C) (D)

(E) (F) (G) (H)

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Asst Prof Dr. Zeeshan Zahid, EE 342, MCS
 Admittance & Impedance Conversion
 Impedance/admittance Smith charts are used very
frequently in impedance matching
 To understand the inter-conversion, try to locate
following loads on Impedance and admittance charts
 z1 = 0, z2 = ∞, z3 = 3, z4 = 2 - j
 y1 = ∞, y2 = 0, y3 = 0.33, y4 = 0.4 + 0.2j

Asst Prof Dr. Zeeshan Zahid, EE 342, MCS 9

Impedance Chart
z1 = 0, (Unit r circle)
z2 = ∞,
z3 = 3,
z4 = 2 - j Z1 Z3 Z2

Z4

Asst Prof Dr. Zeeshan Zahid, EE 342, MCS 10

 Admittance Chart
(Unit GL circle)

y1 = ∞,
y2 = 0,
y3 = 0.33, Y1 Y3 Y2
y4 = 0.4 + 0.2j
Y4

Asst Prof Dr. Zeeshan Zahid, EE 342, MCS 11

 Point to Remember
 zL and yL correspond to same point on combined chart

Z4/Y4

Asst Prof Dr. Zeeshan Zahid, EE 342, MCS 12

 Examples (Try yourself)
 Justify zin and yin for following cases using Smith chart

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Asst Prof Dr. Zeeshan Zahid, EE 342, MCS
 Key Points
Z-chart (Series reactance)
 Add series L (reduce C) to rotate clockwise (blue)
 Add series C (reduce L) to rotate anti-clockwise (black)
Y-chart (Shunt reactance)
 Add shunt C (reduce L) to rotate clockwise (green)
 Add shunt L (reduce C) to rotate anti-clockwise (red)

Shunt C Series C

Series L Shunt L

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Rule of thumb: Inductor always move to upper half plane and C to lower half of chart
Find Zin using Smith Chart

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Asst Prof Dr. Zeeshan Zahid, EE 342, MCS
1) Mark zL=1

2) Inductor 2j
(Blue line)

3) Capacitor -0.735j
(Black line)

4) Inductor 0.7j
(Red line)

Asst Prof Dr. Zeeshan Zahid, EE 342, MCS 16

 Rotations for Matched Point

1. Series L 1. Shunt L
2. Shunt C 2. Series C
(Network b) (Network a)

1. Series C 1. Shunt C
2. Shunt L 2. Series L
(Network b) (Network a)
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Asst Prof Dr. Zeeshan Zahid, EE 342, MCS
 Required Reactance
 The value of required reactance can be calculated as
 x = img{Z2} – img{Z1}
−1
 For example 0.6 – 1.5 = -0.9, (series capacitor,) C =
xZo xZo
 1.5 – 0.6 = 0.9, (series inductor) L =

b
Shunt C =
Z o
Z1
− Zo
Shunt L =
x Z 2

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Asst Prof Dr. Zeeshan Zahid, EE 342, MCS
 Example 5.1 (Using Smith chart )
 Design L-section matching network to match a series RC
load with an impedance ZL = 200 – j100 Ω to a 100 Ω
line at a frequency 500 MHz.
Solution
 Normalize impedance zL = 2 – j (inside r = 1 circle)
 Network (a) is selected
 jB is shunt susceptance, first
convert zL into yL (to add)
 jX is in series therefore convert the result back to
impedance to add again
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Asst Prof Dr. Zeeshan Zahid, EE 342, MCS
 (Unit GL circle)
1) Mark zL=2-j

2) Draw SWR circle

3) Convert zL to yL
4) Move to unit yL yL
(along fixed ‘r’)

5) Convert back to Z

6) Read new point zL

7) Add reactance to
move to center

Asst Prof Dr. Zeeshan Zahid, EE 342, MCS 20

Effective Movement
Recall Slide # 17

Series C Shunt L

zL
Series L
Shunt C

Asst Prof Dr. Zeeshan Zahid, EE 342, MCS 21

 Example 5.1 (Solution)
 Adding a susceptance of 0.3j will move yL to unit
conductance circle
 Converting back to impedance gives z = 1 - 1.2j
 A series reactance of +1.2 j brings us to the center
 For f = 500 MHz, values of Cshunt and Lseries will be
C = b/(ω Zo) = 0.92 pF
L = xZo/ω = 38.8 nH

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Asst Prof Dr. Zeeshan Zahid, EE 342, MCS
Matching on Combined Chart
 Example 5.1: Convert zL to yL (same point)
 Use constant conductance circles for rotation
 Move to 1+jx circle and consider Z-chart.

zL

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Asst Prof Dr. Zeeshan Zahid, EE 342, MCS
Simulation in ADS (Example 5.1)

Note: The smith chart shows Γ as function of frequency not matching circles
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Asst Prof Dr. Zeeshan Zahid, EE 342, MCS
 Quality Factor
 LC networks are resonant circuits described by Q factor
 It is 2π × the ratio of energy stored to energy dissipated
 Q = fo/BW; narrow band means higher Q and vice versa
 Series RLC |x|
Qn =
r
r
 Parallel RLC Qn =
|x|

 For a circuit with z =0.2-0.4j,

Qn = 0.4/0.2 = 2; 1
fo =
LC
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Asst Prof Dr. Zeeshan Zahid, EE 342, MCS
 T and Pi Matching Networks
 L-matching network gives 2 degrees of freedom
 L-network at a time can match either ZL>Zo or ZL<Zo
 Q factor cannot be controlled in L networks
 T and π matching networks give 3 degrees of freedom
 T and π networks can be decomposed into 2 L networks
as shown

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Asst Prof Dr. Zeeshan Zahid, EE 342, MCS
 Constant Q Circles
 T and π networks are designed for specific Q factor

Rsourse
Q= −1
RLoad

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Asst Prof Dr. Zeeshan Zahid, EE 342, MCS
Lumped Elements for Microwave IC
 Available for frequencies up to 3 GHz.
 The resistors made using thin films of lossy (Nichrome)
 Inductors are spirals of micro-strip lines
 Capacitors are implemented in several ways as shown

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Asst Prof Dr. Zeeshan Zahid, EE 342, MCS
Surface Mount Devices (SMDs)

(Chip Resistors) (Chip Capacitors) (Chip Inductors)

(SMD Reel)

(SMT board)
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Asst Prof Dr. Zeeshan Zahid, EE 342, MCS
 SMD Package Sizes
Empirical Metric Dimension in Power
Code Code (inches) mm Rating W
1005 0402 (0.1 ×0.05) 0.4 × 0.2 0.031
0201 0603 (0.02 ×0.01) 0.6 × 0.3 0.05
0402 1005 (0.04 ×0.02) 1.0 × 0.5 0.062
0603 1608 (0.06 ×0.03) 1.6 × 0.8 0.1
0805 2012 (0.08 ×0.05) 2.0 ×1.25 0.125
1008 2520 (0.01 ×0.08) 2.5 × 2.0 0.1
1206 3216 (0.12 ×0.06) 3.2 × 1.6 0.25
(Empirical codes are more market famous)
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Asst Prof Dr. Zeeshan Zahid, EE 342, MCS
Standard Capacitor Values
0.2 pF 10 pF 270 pF 5.6 nF 120 nF
0.3 pF 12 pF 330 pF 6.8 nF 150 nF
0.4 pF 15 pF 390 pF 8.2 nF 180 nF
0.5 pF 18 pF 470 pF 10 nF 220 nF
0.6 pF 22 pF 560 pF 12 nF 270 nF
0.7 pF 27 pF 680 pF 15 nF 330 nF
0.8 pF 33 pF 820 pF 18 nF 390 nF
1.0 pF 39 pF 1.0 nF 22 nF 470 nF
1.1 pF 47 pF 1.2 nF 27 nF 560 nF
1.2 pF 56 pF 1.5 nF 33 nF 680 nF
1.5 pF 68 pF 1.8 nF 39 nF 820 nF
1.7 pF 82 pF 2.2 nF 47 nF 1.0 μF
1.8 pF 100 pF 2.7 nF 56 nF 1.2 μF
2.0 pF 120 pF 3.3 nF 68 nF 1.5 μF
2.2 pF 150 pF 3.9 nF 82 nF 1.8 μF
2.7 pF 180 pF 4.7 nF 100 nF 2.2 μF
3.0 pF 220 pF
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Asst Prof Dr. Zeeshan Zahid, EE 342, MCS
Standard Inductor Values
1.0 nH 10 nH 100 nH 1.0 μH 10 μH 100 μH
1.1 nH 11 nH 110 nH 1.1 μH 11 μH 110 μH
1.2 nH 12 nH 120 nH 1.2 μH 12 μH 120 μH
1.3 nH 13 nH 130 nH 1.3 μH 13 μH 130 μH
1.5 nH 15 nH 150 nH 1.5 μH 15 μH 150 μH
1.6 nH 16 nH 160 nH 1.6 μH 16 μH 160 μH
1.8 nH 18 nH 180 nH 1.8 μH 18 μH 180 μH
2.0 nH 20 nH 200 nH 2.0 μH 20 μH 200 μH
2.2 nH 22 nH 220 nH 2.2 μH 22 μH 220 μH
2.4 nH 24 nH 240 nH 2.4 μH 24 μH 240 μH
2.7 nH 27 nH 270 nH 2.7 μH 27 μH 270 μH
3.0 nH 30 nH 300 nH 3.0 μH 30 μH 300 μH
3.3 nH 33 nH 330 nH 3.3 μH 33 μH 330 μH
3.6 nH 36 nH 360 nH 3.6 μH 36 μH 360 μH
3.9 nH 39 nH 390 nH 3.9 μH 39 μH 390 μH
4.3 nH 43 nH 430 nH 4.3 μH 43 μH 430 μH
4.7 nH 47 nH 470 nH 4.7 μH 47 μH 470 μH
5.1 nH 51 nH 510 nH 5.1 μH 51 μH 510 μH
5.6 nH 56 nH 560 nH 5.6 μH 56 μH 560 μH
6.2 nH 62 nH 620 nH 6.2 μH 62 μH 620 μH
6.8 nH 68 nH 680 nH 6.8 μH 68 μH 680 μH
7.5 nH 75 nH 750 nH 7.5 μH 75 μH 750 μH
8.2 nH 82 nH 820 nH 8.2 μH 82 μH 820 μH
8.7 nH 87 nH 870 nH 8.7 μH 87 μH 870 μH
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9.1 nH 91 nH 910 nH 9.1 μH 91 μH 910 μH
 Single Stub Matching
 Stub* is a section of open or shorted trans. Line
 The stub can either be in series or in shunt topology
 Moreover the stub can be open or shorted to ground
 Open stubs are easy to fabricate on micro-strip lines
 Types of stubs are (i) balanced (ii) radial (iii) butterfly

*Stub means small branch

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Asst Prof Dr. Zeeshan Zahid, EE 342, MCS
 Single Stub Matching
 Find distance ‘d’ from load where Zin = Zo + jB
 Susceptance of stub must be –jB for matching
 Reflections from load and stub add destructively to give
matched condition
 Narrow band matching solution

Asst Prof Dr. Zeeshan Zahid, EE 342, MCS 34

 Single Stub Matching
 Charact. impedance of stub need not be equal to line Zo
 Inverse relationship between stub Zo and its length
 Wider the stub (lower Zo) longer its length (poor band)
 Thinner the stub (higher Zo) shorter its length (good band)
 To achieve better bandwidth performance
 Inductive reactance; High Zo shorted stub
 Capacitive reactance; Low Zo open stub
 Range of stub Zo is 10 Ω to 150 Ω

Asst Prof Dr. Zeeshan Zahid, EE 342, MCS 35

 Example 5.2 (Shunt Stub Matching)
 A load consists of R and C, ZL = 60 – j80 Ω. Design
shorted single stub shunt tuning network to match this
load to a 50 Ω line at 2 GHz.
Solution:
 Normalized impedance zL = 1.2 – 1.6j
 Draw SWR circle and convert zL to yL
 SWR circle intersects with 1 + jb circle at points y1, y2
 On smith chart y1 = 1 + 1.47j, y2 = 1 – 1.47j
 Read stub location from load, d1 = 0.11λ and d2 = 0.26λ
 For stub length move clockwise from SC point to ±1.47j
Asst Prof Dr. Zeeshan Zahid, EE 342, MCS 36
 Along WTG Scale
1) Mark zL=1.2 – 1.6j
2) Consider Y chart
d1
3) Draw SWR circle yL y1

4) Find y1 and y2

6) Required reactance
1.47j, –1.47j d2

7) Admit. of SC→OC
zL
y2
8) Find length along
WTG scale l1, l2
l2 = 0.405λ
l1 = 0.095λ

Asst Prof Dr. Zeeshan Zahid, EE 342, MCS 37

 Example 5.2 (Solution)
 If y1 is selected, a stub of reactance –j1.47 is required
 l1 = 0.095λ, l2 = 0.405λ

Asst Prof Dr. Zeeshan Zahid, EE 342, MCS 38

The End