1James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

Smith�Chart�&�

Matching�Network

James Lu

2James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

Non�matched�Impedance�(��0)

• Reflections�lead�to�Zin variations�with�line�length�and�frequency

• Power�is�wasted�because�of�reactive�power,�which�can�also�damage�equipment�during�short�circuit�(for�example)

• Only�partial�power�is�delivered�to�the�load.• SWR�>�1:��there�will�be�voltage�maxima�on�the�line,�voltage�breakdown�at�high�power�levels

• Noises�(bounces�or�echoes)

3James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

Benefits�of�Matching�(�=0)

• Zin =�ZO, independent�of�line�length,�and�frequency�(over�the�bandwidth��of�the�matching�network)

• Maximum�power�transfer�to�the�load�is�achieved

• SWR�=�1:�no�voltage�peaks�on�the�line• No�bounces�(echoes)

UQ

4James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

Load�MatchingWhat if the load cannot be made equal to Zo for some other reasons? Then, we need to build a matching network so that the source effectively sees a match load.

0��

LZsP 0Z M

Typically we only want to use lossless devices such as capacitors, inductors, transmission lines, in our matching network so that we do not dissipate any power in the networkand deliver all the available power to the load.

5James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

Normalized�Impedance

jxrZZzo

���

It will be easier if we normalize the load impedance to the characteristic impedance of the transmission line attached to the load.

����

�11z

Since the impedance is a complex number, the reflection coefficient will be a complex number

jvu ���

� � 22

22

vu1vu1r��

��� � � 22 vu1

v2x��

�

6James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

Smith�ChartsThe impedance as a function of reflection coefficient can be re-written in the form:

� � 22

22

vu1vu1r��

���

� � 22 vu1v2x��

�

� �22

2

r11v

r1ru

���

�

� �

��

� � 2

22

x1

x1v1u ��

� � ���

These are equations for circles on the (u,v) plane

� � 222 )( ayyxx oo ����

7James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

Smith�Chart�– Real�Circles

1 0.5 0 0.5 1

1

0.5

0.5

1

� ��Re

� ��Im

r=0 r=1/3r=1 r=2.5

1�� Circle

8James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

Smith�Chart�– Imaginary�Circles

1 0.5 0 0.5 1

1

0.5

0.5

1

� ��Re

� ��Im

x=1/3 x=1 x=2.5

x=-1/3 x=-1 x=-2.5

9James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

WTG

WTL

Smith�ChartImpedances, voltages, currents, etc. all repeat every half wavelength

z=1+j

Capacitive

Purely imaginary impedancesalong the periphery

Purely real impedances along the horizontal centre line

Open(z=�)

Short(z=0)

z=1

Inductive

y=1/(1+j)=0.5-j0.5

SWR max)at(11

PrS o�����

�

LLL jxrz ������

�11

l

llz

����

�11

11��

����L

LL z

z

lj

ljl

e

e

��

�

4

2

�

�

��

���

�=1�=�1

�=0

rroo

PPmaxmaxPPminmin

10James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

WTG

WTL

Smith�Chart�Example�1Given:

�� 50Zo

���� 455.0L

What is ZL?

� �����

���5.675.69

35.139.150j

jZL

11James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

Smith�Chart�Example�2Given:

�� 50Zo

���� 25j15ZL

What is �L?

5.0j3.050

25j15zL

���

����

����� 1236.0L

WTG

WTL

z1 = 2 + jz2 = 1.5 -j2z3 = j4z4 = 3z5 = infinityz6 = 0z7 = 1z8 = 3.68 -j18

12James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

WTG

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Smith�Chart�Example�3Given:

�� 50Zo

���� 50j50ZL

What is Zin at 50 MHz?

0.1j0.150

50j50zL

���

����

���� 64445.0L

ns78.6��

����

�����

339.01078.61050 96

2/42

������

��������

���

fl

eee jL

ljL

ljLin

��180in����� 180445.0in

� � ����� 190.038.050 jZin

���� 2442

13James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

AdmittanceA matching network is going to be a combination of elements connected in series AND parallel.

Impedance is NOT well suited when working with parallel configurations.

21L ZZZ ��

2Z1Z

2Z1Z

21

21L ZZ

ZZZ�

�

ZIV �

For parallel loads it is better to work with admittance.

YVI �2Y1Y

21L YYY ��11 Z

1Y �

Impedance is well suited when working with series configurations. For example:

14James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

Normalized�AdmittancejbgYZ

YYy o

o����

����

�11y

� � 22

22

vu1vu1g��

���

� � 22 vu1v2b��

��

� �22

2

g11v

g1gu

���

���

��

�

� � 2

22

b1

b1v1u ��

� � ���

These are equations for circles on the (u,v) plane

15James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

1 0.5 0 0.5 1

1

0.5

0.5

1

1 0.5 0 0.5 1

1

0.5

0.5

1

Admittance�Smith�Chart

� ��Re

� ��Im

g=1/3

b=-1 b=-1/3

g=1g=2.5 g=0

b=2.5 b=1/3

b=1

b=-2.5

� ��Im

� ��Reg=�

b=�

Conductance Circles Susceptance Circles

16James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

Impedance�and�Admittance�Smith�Charts• For a matching network that contains elements

connected in series and parallel, we will need two types of Smith charts� impedance Smith chart� admittance Smith Chart

• The admittance Smith chart is the impedance Smith chart rotated 180 degrees.– We could use one Smith chart and flip the reflection

coefficient vector 180 degrees when switching between a series configuration to a parallel configuration.

17James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

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• Procedure:

•Plot 1+j1 on chart

•vector =

•Flip vector 180 degrees

Admittance�Smith�Chart�Example�1Given:

��64445.0

Find ����z

1j1y ��

����� 116445.0

Plot y

Flip 180 degrees

Read� & z5.05.0 jz ��

18James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

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• Procedure:

• Plot �

• Flip vector by 180 degrees

• Read coordinate

Admittance�Smith�Chart�Example�2Given:

Find Y

����� 455.0 �� 50Zo

Plot �

Flip 180 degrees

Read y

36.0j38.0y ��

� �

� � Sj

jY

3102.76.7

36.038.050

1

����

��

�

19James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

Matching�Example

0��

�100sP �� 50Z0 M

Match 100� load to a 50� system at 100MHz

A 100� resistor in parallel would do the trick, but ½ of the power would be dissipated in the matching network. We want to use only lossless elements such as inductors and capacitors so we don’t dissipate any power in the matching network

20James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

Matching�Example� We need to go from

z=2+j0 to z=1+j0 on the Smith chart

� We won’t get any closer by adding series impedance so we will need to add something in parallel.

� We need to flip over to the admittance chart

Impedance Chart

WTG

WTL

21James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

WTG

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Matching�Example� y=0.5+j0

� Before we add the admittance, add a mirror of the r=1 circle as a guide.

Admittance Chart

22James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

Matching�Example� y=0.5+j0

� Before we add the admittance, add a mirror of the r=1 circle as a guide

� Now add positive imaginary admittance.

Admittance Chart

WTG

WTL

23James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

Matching�Example� y=0.5+j0

� Before we add the admittance, add a mirror of the r=1 circle as a guide

� Now add positive imaginary admittance jb = j0.5

Admittance Chart

� �

pF16C

CMHz1002j50

5.0j5.0jjb

�

���

�

pF16 �100

WTG

WTL

24James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

Matching�Example� We will now add

series impedance

� Flip to the impedance Smith Chart

� We land at on the r=1 circle at x=-1, i.e. z = 1 – j1

Impedance Chart

WTG

WTL

25James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

Matching�Example� Add positive

imaginary admittance to get to z=1+j0

Impedance Chart

pF16�100

� � � �nH80L

LMHz1002j500.1j0.1jjx

����

�

nH80W

TGW

TL

26James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

Matching�Example� This solution would

have also worked

Impedance Chart

pF32

�100nH160

WTG

WTL

27James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

WTG

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50 60 70 80 90 100 110 120 130 140 15040

35

30

25

20

15

10

5

0

Frequency (MHz)

Ref

lect

ion

Coe

ffici

ent (

dB)

Matching�Bandwidth

50 MHz

150 MHz

Because the inductor and capacitor impedances change with frequency, the match works over a narrow frequency range

pF16�100

nH80

Impedance Chart

28James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

Single�Stub�Tuner

0��

l1

l2

Stub length l = � up=f��Phase shift: ��=2�l=2(2�/�)l=4�(l/�)

zin=1 (yin=1)

ZLZ0

Goal:

Z0

Z0

Openor

Short

yl1

yl2

yin = yl1 + yl2 = 1= (gl1 + jbl1) + jbl2

gl1 = 1 (real-part condition) bl1 = -bl2 (imaginary-part condition)(2 Degrees of freedom)

ll

l

l

ll

e

zy

�2

111

����

����

��

Open: Zin = -jZocot�l, or zin = -jcot�lShort: Zin = jZotan�l, or zin = jtan�l

29James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

WTG

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Single�Stub�Tuner

� Flip to Admittance chart

� y=0.5+j0

� Adding length to Cable 1 rotates the reflection coefficient clockwise to g=1.

Admittance Chart

l1 = 0.152�

Match 100� load to a 50� system at 100MHz using two transmission lines connected in parallel

� y1l=1+j0.72

30James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

Single�Stub�Tuner

� The stub has to add a normalized admittance of -0.72 to bring the trajectory to the center of the Smith Chart

WTG

WTL

Admittance Chart

31James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

WTG

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Single�Stub�Tuner

Admittance Chart

� An short stub of zero length has an admittance=j�

� By adding enough cable to the short stub, the admittance of the stub will reach to -0.72

l2 = (0.401-0.25)� = 0.151�

32James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

Single�Stub�Tuner

0��100�

l2 = 0.151�

l1 = 0.152�

WTG

WTL

Admittance Chart

33James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

Single�Stub�Tuner

Admittance Chart

0��100�

WTG

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l2 = 0.097�

l1 = 0.347�

This solution would have worked as well.

An open stub of zero length has an admittance=j0

34James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

WTG

WTL

50 60 70 80 90 100 110 120 130 140 1540

35

30

25

20

15

10

5

0

Frequency (MHz)

Ref

lect

ion

Coe

ffici

ent (

dB)

0��100�

Single�Stub�Tuner�Matching�Bandwidth

50 MHz

150 MHz

Because the cable phase changes linearly with frequency, the match works over a narrow frequency range

Impedance Chart

l2 = 0.097�

l1 = 0.347�

36James�J.�Q.�Lu ECSE�2100�Fields�&�Waves�I

Summary• Impedance matching is necessary to:

– reduce VSWR– obtain maximum power transfer

• Lump reactive elements and a single stub can be used.

• A quarter-wave line can also be used to transform resistance values, and act as an impedance inverter.

• These matching network types are narrow-band: they are designed to operate at a single frequency only.