5-1
CHAPTER 5
Impedance Matching and Smith Chart
* Pozar MW (Ch 5), “Impedance Matching and Tuning”
* Pozar RF (Ch 2), “Itransmission Lines & Microwave Networks”
*Ludwig, (Ch 3, Ch 8), “Matching and Biasing Networks”
*Rogers, (Ch 4), “Radio Frequency Integrated Circuit Design”

Matching with Lumped Elements
- L Network
- T &  Networks
- Lumped Elements for MIC : Chip R, L, C.



Microstrp Single-Stub and Double-Stub Tuning
Quarter-Wave Transformer
* The Bode-Fano Criteria
Appendix Smith Chart
ZM
L
Transmitter
C
ZT
2011-12
ZA
H.-R. Chuang EE NCKU
5-2
Impedance matching (or tuning) is important for the following reasons :
incident
(or input)
Matching
Network
Z0
reflection
Load
ZL
Z in
Reflection coefficien t (or Return Loss) :
in (or S11 )  ( Z in  Z 0 ) /( Z in  Z 0 )




minimum power loss in the feed line & maximum power delivery
linearizing the frequency response of the circuit
improving the S/N ratio of the system for sensitive receiver components (lownoise amplifier, etc.)
reducing amplitude & phase errors in a power distribution network (such as
antenna array-feed network)
* Factors in the selection of matching networks
- complexity -bandwidth requirement (such as broadband design)
- adjustability - implementation (transmission line, chip R/L/C elements ..)

/4 microstrip RF Choke
ShortCirucited
(S.C.)
Z0
* At high freq.,
capacitance is like
Short-cirucited
2 X sc / Z 0
l
4
RF
signal X
0.5
0.25
-2
-4
RF
signal X
RF Choke
扼流圈
RF
signal X
2011-12
H.-R. Chuang EE NCKU
5-3
Matching Network Types: L-/T-/-section Networks
 L-section Networks (Two-component ): Lumped elements: L/C
C
ZS
L
ZL
ZS
(a )
L
ZL
L1
ZL
ZS
ZL
ZS
L2 Z L
(f)
C2
L
ZS
ZL
(d )
(c )
L1
(e)
C1
ZS
(b)
L2
ZS
C1
C2
C
L
C
(g)
ZL
C
ZS
ZL
(h)
In any particular region on the Smith chart,
several matching circuits will work and others
will not.
The figure shows what matching networks will
work in which regions.
How does one choose?
There are a number of popular reasons for
choosing one over another.
1. Sometimes matching components can be
used as dc blocks (capacitors) or to provide
bias currents (inductors).
2. Some circuits may result in more reasonable
component values.
3. Personal preference. Sometimes when all
paths look equal, you just have to shoot
from the hip and pick one.
4. Stability. Since transistor gain is higher at
lower frequencies, there may be a lowfrequency stability problem. In such a case,
sometimes a highpass network (series
capacitor, parallel inductor) at the input
may be more stable.
2011-12
5. Harmonic filtering can be done with a
lowpass matching network (series L,
parallel C ). This may be important, for
example, for powerH.-R.
amplifiers
(PA).
Chuang EE NCKU
5-4
L-section Network
(1) complex ZL to real Z0 matching
jX
jX
jB
Z0
ZL
jB
Z0
admittance
(b)
admittance
(a)
How to determine jX & jB ?
ZL
Let zL = ZL / Zo = (RL + jXL) / Zo = r + jx
or
1. Analytic Solutions
2. Smith Chart Solution
(1) if RL Z 0 ( z  1) [zL is inside the (1 + jx) circle ] => choose (a) why?
1
for impedance matching (to Z 0 ) => jX 
 Z0
jB  1 ( R L  jX L )

X  RL Z 0 RL2  X L2  Z 0 RL
 B( XR L  X L Z 0 )  R L  Z 0
B  L
=> 
 
RL2  X L2
X
BX
BZ
R
X
(
)
1



L
0 L
L


 X  (1 B )  ( X L Z 0 RL )  ( Z 0 B RL )
(2) if RL Z 0 ( z  1) [zL is outside the (1 + jx) circle ] => choose (b) why?
1
1
for impedance matching (to Z 0 ) => jB 

R L  j ( X  X L ) Z0
 X   RL ( Z 0  RL )  X L
=> 
 B   ( Z 0  RL ) RL Z 0
 BZ ( X  X L )  Z 0  R L
  0
( X  X L )  BZ 0 R L
r
(1+jx) circle
x
r<1
r>1
2011-12
z L  Z L / Z o  r  jx
L  ( Z L  Z o ) /( Z L  Z o ) | L | 
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5-5
EX (Pozar MW EX 5.1) (Pozar RF EX 2.5)
jX
jB
Z 0  100
ZL
Z L  R L  jX L
 200  j100
Since RL = 200 > Z0 = 100 (zL is inside the (1 + jx) circle)
We choose (a) The solutions are
RL
200
XL
B1
XL
100
RL 
Zo
2
1
B1
XL
B2
2
2
2
2
1
B2
Zo
XL
RL
Zo RL
B1  2.899  10
XL
Zo
B1 RL
RL
RL
X2
2
XL
Zo
XL
RL
RL 
Zo
100
RL
RL
X1
Zo
3
X1  122.474
2
XL
Zo RL
B2  6.899  10
2
XL
3
Zo
B2 RL
X2  122.474
At f = 500 MHz
38.8nH
Z 0  100
ZL
0.92pF
(high pass)
 200  j100
(low pass)
Solution 1 (low pass)
BW
2.61pF
Z 0  100
46.1nH
ZL
 200  j100
Solution 2 (high pass)
2
  0.33 ( &   0.1) SWR  2 RL  9.5 dB
Solution (1) bandwidth
BW  0.3 GHz  0.3 / 0.5  60%
2011-12
H.-R. Chuang EE NCKU
5-6
Smith Chart Representation of the Matching Process
38.8nH
Z 0  100
ZL
0.92pF
 200  j100
Solution 1 (low pass)
ZL
 200  j100
38.8nH
0.92pF
38.8nH
Z 0  100
0.92pF
2011-12
zL  Z L / Zo  2  j
ZL
 200  j100
L  ( Z L  Z o ) /( Z L  Z o )
 0.45  26.6 o
H.-R. Chuang EE NCKU
5-7
(2) Complex to complex conjugate matching (Ludwig, RF Circuit Design P401)
(Conjugate Matching for maximum power transfer )
jX
jB
ZL
Z0
admittance
complex ZL to real Z0 matching
ZT  150  j 75 

Z A  75  j15 
f  2 GHz
Transmitter
ZT
Z M  Z A* Z A
complex ZA to complex ZT
conjugate matching
RT ( 150)  R A ( 75)
 choose ""
 argument () of
&
2011-12
()  0
vice versa
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5-8
Transmitter
ZT
Z M  Z A* Z A
complex ZT to complex ZA
conjugate matching
Z T  150  j 75 

Z A  75  j15 
f  2GHz

 zT  Z T / Z 0  (150  j 75) / 75  2  j1
Let Z 0  75  
 z A  Z A / Z 0  (75  j15) / 75  1  j 0.2
2011-12
H.-R. Chuang EE NCKU
5-9
(3) General L-section matching network (complex to complex)
complex Zs to complex ZL : conjugate matching
A, B,C , D (four paths)
z s       z*L
zL *
zL
A
C
B
D
Z s  50  j 25 

Z L  25  j 50 
f  2 GHz
 z s  1  j 0.5

 z L  0.5  j1
 z *  0.5  j1
 L
Transmitter
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2011-12
*
Zs  ZL
Zs
ZL
5-10
Ex: L-section Lumped-Elements & Microstrip Matching Networks
Conjugately Matched Amplifier Design
(Pozar MW EX11-3 or RF EX6-3 )
Design an amplifier for maximum gain at 4.0 GHz using single-stub matching
sections. Calculate and plot the input return loss & the gain from 3 to 5 GHz. The
GaAs FET has the following S parameters (Z0=50 ):
f (GHz)
S11
S21
S12
S22
3.0
0.80  89
2.86 99
0.0356
0.76  41
4.0
0.72  116
2.60 76
0.0357
0.73  54
5.0
0.66  142
2.3954
0.03 62
0.72  68
 FET S-parameters Touchstone file: Poz_11-3.s2p
! poz_11-3.s2p : Pozar Ex. 11-3 transistor S parameters
! Typical s-parameters at minimum attenuation setting, Ta=25°C
# ghz s ma r 50
3.00 0.800 -89.0 2.860 99.0 0.030 56.0 0.760 -41.0
4.00 0.720 -116.0 2.600 76.0 0.030 57.0 0.730 -54.0
5.00 0.660 -142.0 2.390 54.0 0.030 62.0 0.720 -68.0
It cab be derived that (see chapter of RF Amplifier Design)
in  S*  0.87  123o
s  0.872123
& 





0
.
876
61
 L
out  L*  0.87  61o
2011-12
H.-R. Chuang EE NCKU
5-11
 Microstrip Matching Networks
0.206
50

0.206
0.206
s in
50

0120
. 
50
50
out L
(f = 4 GHz)
in  S*  0.87  123o
s  0.872123
& 





0
.
876
61
out  L*  0.87  61o
 L
Z in  4.43  j 26.97

( Z 0  50)


Z
12
.
68
j
83
.
5
 L
Lumped Elements Matching Networks
50 
2 4.19nH
3
4
1.63nH 1
50 
1.32pF
2.54pF
0
2011-12
s in
0
out L
0
H.-R. Chuang EE NCKU
5-12
* By Smith-Chart tool
DP-Nr. 1(4.4 - j27.0)Ohm Q = 6.1 4.000 GHz
DP-Nr. 2(4.4 + j14.1)Ohm Q = 3.2 4.000 GHz
DP-Nr. 3(49.4 - j0.2)Ohm Q = 0.0 4.000 GHz
rtransmission-line
matching network
(open-circuited stub)
DP-Nr. 1(4.4 - j27.0)Ohm Q = 6.1 4.000 GHz
DP-Nr. 2(3.6 + j13.0)Ohm Q = 3.6 4.000 GHz
DP-Nr. 3(50.4 + j1.4)Ohm Q = 0.0 4.000 GHz
2011-12
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5-13
Transmission-line matching network
(shorted-circuited stub)
DP-Nr. 1(4.4 - j27.0)Ohm Q = 6.1 4.000 GHz
DP-Nr. 2(3.6 - j12.7)Ohm Q = 3.5 4.000 GHz
DP-Nr. 3(48.0 - j0.0)Ohm Q = 0.0 4.000 GHz
2011-12
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5-14
 Forbidden Regions for L-type Matching Networks with Z s  Z 0  50
=> The shaded areas denote values of load impedance that cannot be matched to 50 Ω
2011-12
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5-15
Design Example: Forbidden Regions for L-type Matching Networks
RL  80 
X L  60 
z L  1.6  j1.2
(for Z 0  50 )
f 0  1 GHz
Since z L = 1.6 > 1
=> choose (c) or (d) from
forbidden regions of
L - network
(with Z S = 50)

2011-12
H.-R. Chuang EE NCKU
5-16
 Quality factor & Bandwidth (BW) (there are much more to be discussed!)
Z s  Rs  jX s or YP  GP  jBP

Qn 
| Xs |
| BP |
or
Rs
GP
f
f
 Q 
QL   n   o  BW  o
QL
 2  BW
* TMatching Network (discussed next)
2011-12
H.-R. Chuang EE NCKU
5-17
* T & Matching Network: The.addition of 3rd element into the two-element (L) matching
network introduces an additional degree of freedom in the çi!çuit, and allows us to control
the value of QL (to be discussed)by choosing an appropriate intermediate impedance. => wider
(matching) bandwidth
T Matching Network
 Matching Network
Z in  10  j 20 

Z L  60  j 30 
f  1 GHz

Z in  10  j 20 

Z L  60  j 30 
f  1 GHz
2011-12
H.-R. Chuang EE NCKU
5-18
Comparison between L-, T - &  - network
Design a match circuit at the center frequency of 100 MHz
* Prof. C.-F. Chang course note (NCCU)
51
0.1 H
10 pF
2011-12
L

T
4-element ladder
510 
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5-19
Microstrip Line Matching Networks (Ludwig P431)

 In the mid-GHz and higher frequency range, the wavelength becomes
sufficiently small and the distributed components are widely used. Also, the
discrete R/L/C lumped elements will have more noticeable parasitic effects (see
chapter 2) and let to complicating the circuit design process
 Distributed componenets (such as transmission line segments) can be used
to mix with lumped elements
From Discrete Components to Microstrip Lines

 Avoid using inductors (if possible) due to higher resistive loss (& higher price)
 In general, one shunt capacitor & two series transmission lines is
sufficiently to transform any load to any input impedance.

EX: transform load Z L to an input impedance Z in







 Z L  30  j10  z L  0.6  j 0.2 

 Z in  60  j80  zin  1.2  j1.6 
f  1.5 GHz
 Identify input & load SWR circles
 Choose A (yA= 1-j0.6) & transform zL to
A by a series TL (l1)
=>Transform A to B (on the input SWR circle)
by a parallel C1
=> Transform B to zin by a series TL (l2)
zL
+ series-TL (l1)
=> A + shunt C1
=> B + series-TL (l2)
=> zin
2011-12
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5-20
Single-Stub Matching Networks

4 adjustable parameters:
(l s , Z 0 s ,l L , Z 0 L, )
Z L  60  j 45 

Z in  75  j 90 
Z 0  75 
 z L  Z L / Z 0  0.8  j 0.6

 y L  1 / z L  0.8  j 0.6
 z  Z / Z  1  j1.2 
in
0
 in
g = 0.8
conductance circle
Input SWR circle associated with zin
has two intersected points (A & B) with
g = 0.8 conductance circle
 yA= 0.8 + j1.05 yB= 0.8 - j1.05
g=0
(O.C.)
ibSA
= 0.45
zL to A (yA= 0.8 + j1.5) by adding a shunt open-circuited (O.C.)TL lSA
 The corresponding susceptance for the stub : jbSA= yA- yL = (0.8 + j1.05)-( 0.8 + j0.6)=0.45
2011-12
O.C. point (g=0) to the point of ibSA = 0.45 is lSA = 0.067 
H.-R. Chuang EE NCKU
A to zin is lLA = 0.266 
5-21
Balanced stub design (l sB )
 l s  l sB // l sB



2011-12
open - circuit stub :

2l s 


1 
l sB  2 tan  2 tan  



short - circuit stub :

2l s 


1  1
tan
tan
l



sB

2
2





H.-R. Chuang EE NCKU
5-22
Double-Stub Matching Networks
Z in  Z 0  50 

Z L  50  j 50 
l1   / 8 l 2  l3  3 / 8

l s1  0.074 l s1  0.051
2011-12
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5-23
Quarter-Wave Transformer 四分之一波長(傳輸線)阻抗轉換匹配
( only useful for pure-resistance matching )
transmission
line 1
quarter-wavelength
transmission line 2
l  /4
Z in ( Z 0 )  Z 0
l   &
2
tan 
   Z 0
V1 ( z )
V1 ( z )  0
ZL
X
(Z0 )
Z in
Z L  jZ 0 tan l
Z 0  jZ L tan l
l
4
Z L  jZ 0  ()
Z
 Z 0 0
Z 0  jZ L  ()
ZL
 Z 0 Z L  Z 0 2
( Z 0 )
Ex: A microstrip quarter-wave trasformer that matches a 50  miscrostrip line to
a 20  load at f = 4 GHz (substrate: r=2.5, thickness h = 0.75 mm)
12.73[mm]
20[]
2.13[mm]
50[]
4.03[mm]
31.62[]
* Double Quarter-Wave Transformer for wider (matching bandwidth)
2011-12
H.-R. Chuang EE NCKU
5-24
* (Matching) Bandwidth (f ) of a Quarter-Wave Transformer
Pozar, Mcrowave & RF Design of Wireless Systems
Approximate behavior of the reflection coefficient magnitude of a quarter-wave
transformer near the design frequency
Increased BW for
Smaller load mismatch (ZL/Z0)
It can be proved that
1
m2
 2 Z0Z L

 1 
sec 

 Z L  Z 0
2
 
2 Z0Z L 
2( f 0  f m )
2f
4
4
f
( BW ) 
 2  m  2  m  2  cos 1  m

f0
f0
f0


 1 m2 Z L  Z 0 
2011-12
 H.-R. Chuang EE NCKU
5-25
2011-12
 H.-R. Chuang EE NCKU