Chapter 3
Impedance Matching
Outcomes of This Chapter
• What is impedance matching?
• Why do we need impedance matching?
• What kinds of impedance matching
techniques do we have?
• How to implement impedance matching?
Concept of Impedance Matching
Impedance matching:
to transform the impedance from one to the other
Z in
Zin  Z0 or other given values (LNA, PA)
Impedance matching between two transmission lines
Importance of Impedance Matching
• Maximum power is delivered when the load is
matched to the line.
• Impedance matching sensitive components
improves S/N ratio of the system when they are
properly matched. (LNA)
• Impedance matching in a power distribution
network will reduce amplitude and phase errors.
Categories of Impedance Matching
• Lumped element matching network
--capacitors and inductors
• Distributed element matching network
--open/short stub, and transmission line
• Hybrid matching network
--Both lumped and distributed elements
Categories of Impedance Matching
Matching
network
Z0
Load
ZL
Z in
Reactive but not resistive components are used in
matching network, Why?
Reactive components:
store energy but not consume energy, lossless
Resistive components:
consume energy but not store energy, lossy
Factors in Matching Network Selection
• Complexity
--as simple as possible when specifications are met
--simple matching network is more reliable, cheaper
and less lossy
• Bandwidth
--perfect matching only at a single frequency
--some systems operating over a band of frequency
Factors in Matching Network Selection
• Implementation
--some type of matching can not be realized, i.e.,
microstrip series open/short stub
--specific type of matching network maybe preferable
compared to others, depending on transmission line
• Adjustability
--matching network may require adjustment
--some networks are adjustable after fabrication,
open stub (adjustable) vs short stub (non-adjustable)
Matching Network Design Methods
jX
Z0
jB
jX
ZL
Z0
jB
Zin
Zin
(a)
(b)
Purpose:
Zin  Z0
Analytical solutions
Smith Chart solutions
ZL
Analytical solutions
Analytic solutions
jX
Z0
jB
ZL
Zin
Z L  RL  jX L
(known)
1
Zin  jX 
jB  1/( RL  jX L )
Z in  Z 0
(known)
(B,X: unknown)
Analytical solutions
1
Z 0  jX 
jB  1/( RL  jX L )
Rearrange above equation and separate into real and
imaginary parts
B( XRL  X L Z 0 )  RL  Z 0
X (1  BX L )  BZ 0 RL  X L
X L  RL / Z 0 R  X  Z 0 RL
B
RL2  X L2
1 X L Z0 Z0
X 

B
RL
BRL
2
L
Solutions
2
L
Drawback: complex, time-consuming, non-straightforward
Basics of Smith Chart Matching
Connection of reactive components with load
Series connection:
jX
Z L  RL  jX L
Zin
Z in  Z L  jX  RL  j ( X  X L )
Move on constant impedance circle
Inductor: add a positive reactance, move clockwise
Capacitor: add a negative reactance, move anti-clockwise
Basics of Smith Chart
Parallel connection:
jB
YL  GL  jBL
Yin
Yin  YL  jB  GL  j ( B  BL )
Move on constant impedance circle
Inductor: add a negative susceptance, move anti-clockwise
Capacitor: add a positive susceptance, move clockwise
Basics of Smith Chart
Shunt L
Series L
Series C
Shunt C
Basics of Smith Chart
Two important circles
Constant resistance circle:
z=1+jx
Constant conductance circle: y=1+jb
The real part has been transformed to 50Ω. Only the
imaginary part need to be cancelled out to realize 50-Ω
matching.
Basics of Smith Chart
Impedance: 1+jx
Admittance:1+jb
Two important circles
Basics of Smith Chart

Z  Z0
Z  Z0
VSWR 
Vmax 1  

Vmin 1  
Constant VSWR circles
Basics of Smith Chart
Q=f0/BW, low Q results in wide bandwidth.
Techniques of Impedance Matching
• Two lumped elements matching *
• Multiple lumped elements matching
• Single-stub matching *
• Quarter-wavelength matching *
• Tapered line matching
Two Lumped Elements Matching
Matching network: L-type
L
ZL
C
Low pass
L
High pass
C
ZL
Example
Design a lumped element matching network to match a 50 
source to the impedance of Zin = 10 - j 10  at 500 MHz.
Source
50 
LC Matching
Network
Zin=10 - j10 
Normalize Zin : (10 - j10)/50 = 0.2 - 0.2j
Always follow the constant resistance or conductance circles
Start from 50-Ω Source
2 solutions
Solution 1: paths 1 and 2
Solution 2: paths 3 and 4
ZL
Z0
Solution 1: path 1 and 2
Path 1: Shunt Inductor
Path 2: Series Capacitor
Path 1
Path 2
C
50 
L
Zin=10-j10
50 
L
Y0
Y1
Path1: Change in susceptance j(ΔB)= -2j-0
Change in susceptance comes from adding a shunt inductor
Z0
j B 
j L
50
2
2    500 106  L
L  7.96nH
C
50 
L
Z1
ZL=10-j10
Path 2: Change in reactance j(ΔX)= -0.2j-(0.4j)=-0.6j
Change in reactance comes from adding a series capacitor
1
j X 
jCZ 0
1
0.6 
2    500 106  C  50
C  10.61 pF
Solution 2: path 3 and 4
Path 3: Shunt Capacitor
Path 4: Series Inductor
Path 4
Path 3
L
50 
C
Zin=10-j10
50 
C
Y0
Path 3
Y1
Path 3: Change in susceptance j(ΔB)= 2j-0
Change in susceptance comes from adding a shunt capacitor
jB  jCZ0
2  2    500 106  C  50
C  12.73 pF
L
50 
Path 4
C
Z1
ZL=10-j10
Path 4: Change in reactance j(ΔX)= -0.2j-(-0.4j)=0.2j
Change in reactance comes from adding a series inductor
j L
j X 
Z0
2    500 106  L
0.2 
50
L  3.18nH
Start from Load
Z=
LC Matching
Network
10-j10
50 
L
Z=10-j10
C
50 
L
Z=10-j10
C
YS
Y1 , Z1
ZL
Z1
1
1
X 
0.2 
CZ 0
2    500 106  C  50
Y1
Y0
B 
Z0
L
2
50
2    500 106  L
Any Problems ???
Z=50
C  31.83 pF
L  7.96nH
Steps
Step 1: Normalize ZL* : (10+ j10)/50 = 0.2+ 0.2j, find the
point in Smith Chart
Step 2: Move on constant conductance circle until it intersect
constant resistance circle 1+jb circle, record the susceptance
change and determine the value of lumped C.
Step 3: Move on constant resistcance 1+jb circle to the center point,
record the reactance change and determine the value of L.
L
ZL=10+j10
C
Y1 , Z1
YS
X 
Z=50
YS *
L
ZL
Z1
Y1
Y0 B  CZ 0
Z0
2    500 106  L
0.2 
50
L  3.18nH
2  50  2    500 106 C
C  12.73 pF
Possible Three Elements Matching
Possible Four Elements Matching
What is the benefit of multiple elements matching? Bandwidth!
Basics of Smith Chart
Q=f0/BW, low Q results in wide bandwidth.
Single Stub Matching
Matching network consists of a transmission line and
stub (open or short stub, series or shunt)
Single Stub Matching-From Load
1+jb
-jb
θ
yL
Goal: match YL to Y0 using transmission line and stub
Transmission line:
choose proper distance d so that Y=Y0+jB
Stub:
choose proper length L so that the stub susceptance is -jB, then
the imaginary part is cancelled, resulting in a matched condition.
θ
L
L
L
The location of  i should be…
ZL
Zi  Z0
i 
Zi  Z0
…(1)
Z L  Z0
L 
Z L  Z0
…(2)
Transmission Line Equations
Z L  jZ 0 tan 
Zi  Z0
Z 0  jZ L tan 
…(3)
Sub. (3) into (2),
Z L  jZ 0 tan 
Z0
 Z0
Z 0  jZ L tan 
i 
Z L  jZ 0 tan 
Z0
 Z0
Z 0  jZ L tan 
Z L  jZ 0 tan   Z 0  jZ L tan 
i 
Z L  jZ 0 tan   Z 0  jZ L tan 
i 
Z L  Z 0   jZ L tan   jZ 0 tan  
Z L  Z 0  jZ L tan   jZ0 tan 
Z L  Z 0 1  j tan 
i 
Z L  Z 0 1  j tan 
cos   j sin 
i   L
cos   j sin 
L
i  e j 2  L
2
Observations:
1. The locus of Г is a circle with center of
the Smith Chart and radius |Г|.
i
2. Adding a transmission line with length θcorresponds to rotating by 2θ on
the circle.
1+jb
-jb
θ
yL
Step 1: transform yL to be 1+jb
Step 2: adding a shunt stub with input admittance -jb
Example
Design a distributed matching network to convert the impedance, ZL = 60j80 to 50 ohm.
θ1
Z0
θ2
Step 1: Normalize admittance :
50
yL 
 0.3  j 0.4
60  j80
ZL
Impedance: 1+jx
Admittance:1+jb
Two important circles
Step 2: Allocate yL on the Smith Chart, move on the constant
VSWR circle until it intersects the circle 1+jb
θ1
ZL
y1=1-j1.45
1  104
Step 2: Add a shunt open/short stub with susceptance -jb to cancel out
the imaginary part
y2= j1.45
Z0
104°
θ2
ZL
Open Circuit Stub: y2= tan θ2
θ2 =55.8°
Short Circuit Stub: y2= -cot θ2
θ2 =145.8°
Step 3: Determine the length of shunt stub with susceptance -jb
Open stub
θ2 =245.6 °/2= 122.8°
S.C.
O.C.
Short
stub
θ2 =65.6 ° /2 =32.8°
Single Stub Matching-From Source
Design a lumped element matching network to match a 50 
source to the impedance of Zin = 10 - j 10  at 500 MHz.
Source
50 
Single Stub
Network
Zin=10 - j10 
Normalize yL : 50/(10 - j10) = 2.5 + 2.5j
Single Stub Matching-From Source
yL
Step 1: allocate the point yLand draw constant VSWR circle.
Single Stub Matching-From Source
Z0
1
Path 1
2
θ1
Open stub θ1 =0.17λ=61.2o
Short stub θ1 =0.42λ=151.2o
Step 2: move on the constant conductance circle from 1 to 2 (path 1),
record the change in susceptance
Zin
Single Stub Matching-From Source
θ2
Z0
1
θ1
Z
L
yL
3
Path 1
Path 2
2
θ2 =0.034λ=12.2o
Step 3: move on the constant VSWR circle from 2 to 3 (path 2), find
the rotation angle.
Double Stub Matching
Single stub matching requires changing the location of the stub. In
practice, this is difficult for adjustable tuner. Double stub matching
removes the difficulty.
Quarter-Wavelength Matching
Matching between real impedances, no imaginary part.
Zl  Z0ZL
l
4
Quarter-Wavelength Matching
Multiple reflection analysis of quarter-wave transformer.
Reflected signals are cancelled out and the line is matched.
Quarter-Wavelength Matching
ZLZ, bandwidth increases.
broader bandwidth multisection transformers.
Tapered Line Matching
Benefit: Wide matching bandwidth is possible.
Summary
• Concept and importance of impedance matching
• Two lumped elements matching
• Single stub matching
• Quarter-wave impedance matching