Chapter 3Matching and Tuning

By

Professor Syed Idris Syed Hassan

Sch of Elect. & Electron Eng

Engineering Campus USM

Nibong Tebal 14300

SPS Penang

©Prof Syed Idris Syed Hassan

Why matching or tuning is important?

•To maximize power delivery and minimize power loss.

•To improve signal to noise ratio as in sensitive receiver components such as LNA, antenna, etc.

•To reduce amplitude and phase error as in distributed network such as antenna array.

Basic idea of impedance matching

Load Z LMatchingNetwork

Z o

Concept of maximum power transfer

Z o

Z LV iI V L

LoL

iLLL Z

ZZ

VZIIVP

22

2

1

2

1

2

1

Power maximum whence ZL = Zo

Power deliver at ZL is

In lump circuitP L

Z LZ o

continueIn transmission line

oL

oL

ZZ

ZZ

No reflection whence ZL = Zo , hence 0

The load ZL can be matched as long as ZL not equal to zero (short-circuit) or infinity (open-circuit)

The important parameter is reflection coefficient

Factors in selecting matching network

• Complexity: simpler, cheaper, more reliable and low loss circuit is preferred.

• Bandwidth: match over a desirable bandwidth.• Implementation: depend on types of transmission

line either cable, stripline, microstripline, waveguide, lump circuit etc.

• Adjustability:some network may need adjustment to match a variable load.

Matching with lumped elementsThe simplest matching network is an L-section using two reactive elements

jX

jB Z LZ o

jX

jB Z LZ o

Configuration 1Whence RL>Zo

Configuration 2Whence RL<Zo

ZL=RL+jXL

continue

Configuration 1Configuration 2

If the load impedance (normalized) lies in unity circle, configuration 1 is used.Otherwise configuration 2 is used.

The reactive elements are either inductors or capacitors. So there are 8 possibilities for matching circuit for various load impedances.

Matching by lumped elements are possible for frequency below 1 GHz or for higher frequency in integrated circuit(MIC, MEM).

Impedances for serial lumped elements

Serial circuit

Reactance relationship values

+ve X=2fL L=X/(2f)-ve X=1/(2fC) C=1/(2fX)

L C

RR

Impedances for parallel lumped elements

Parallel circuit

Susceptance relationship values

+ve B=2fC C=B/(2f)-ve B=1/(2fL) L=1/(2fB)

LC

RR

Lumped elements for microwave integrated circuit

Spiral inductorLoop inductor

Interdigital gap capacitor

Planar resistor Chip resistor

Lossy film

Metal-insulator-metal capacitor

Chip capacitor

rDielectric

r

Matching by calculation for configuration 1

jX

jB Z LZ o

LLo jXRjB

jXZ

/1

1

For matching, the total impedance of L-section plus ZL should equal to Zo,thus

Rearranging and separating into real and imaginary parts gives us

oLoLL ZRZXXRB

LLoL XRBZBXX 1

*

**

continue

22

22/

LL

LoLLoLL

XR

RZXRZRXB

Since RL>Zo, then argument of the second root is always positive, the series reactance can be found as

L

o

L

oL

BR

Z

R

ZX

BX

1

Note that two solution for B are possible either positive or negative

+ve inductor-ve capacitor

+ve capacitor-ve inductor

Solving for X from simultaneous equations (*) and (**) and substitute X in (**) for B, we obtain

Matching by calculation for configuration 2

LLo XXjRjB

Z

11

For matching, the total impedance of L-section plus ZL should equal to 1/Zo,thus

Rearranging and separating into real and imaginary parts gives us

LoLo RZXXBZ

LoL RBZXX

*

**

jX

jB Z LZ o

continueSolving for X and B from simultaneous equations (*) and (**) , we obtain

o

LLo

Z

RRZB

/

Since RL<Zo,the argument of the square roots are always positive, again two solution for X and B are possible either positive or negative

LLoL XRZRX +ve inductor-ve capacitor

+ve capacitor-ve inductor

parallel capacitance

parallel inductanceserial capacitance

serial inductance

Matching using Smith chart

(-ve)

(-ve)

(+ve)

(+ve)

C 1 L 1

C 2

L 2

50C 2

L ! 10

C 1L 2

10 50

Serial C

ParallelL

SerialL

ParallelC

Matching using lumped components

ExampleDesign an L-section matching network to match a series RC load with an impedance ZL=200-j100 , to a 100 line, at a frequency of 500 MHz.

SolutionNormalized ZL we have : ZL= 2-j1

ZL= 2-j1Parallel C(+j0.3)Serial L

(j1.2)

Serial C(-j1.2)

Parallel L(-j0.7)

Solution 1

Solution 2

continue

ZL=200-j100

2.61pF

46.1nH

200-j1000.92pF

38.8nHpFZf

bC

o92.0

2

nHf

ZxL o 8.38

2

pFZfx

Co

61.22

1

nHbf

ZL o 1.46

2

Reflection coefficient

0

0.2

0.4

0.6

0.8

1

1.2

0 0.5 1 1.5

freq (GHz)

refl

ec

tio

n c

oe

ffic

ien

t

solution 1

solution 2Solution 2 seems to be better matched at higher frequency

Single stub-matching

Z L

open-stub

Z o

d

x

Z L

short-stub

Z o

d

x

Short-stub matching

Open-stub matching

Parallel configuration

Example

Smith Chart

Design two single–stub shunt tuning networks to match a load ZL=15+j10 to 50 at 2GHz. The load consists of a resistor and inductor in series. Plot the reflection magnitude from 1 GHz to 3 GHz for each solution.

Solution

•Normalized the load zL=0.3+j0.2•Construct SWR circle and convert load to admittance•Then move from load to generator to meet a circle (1+jb) to obtain two points i.e y1=1-j1.33 and y2=1+j1.33. •The distance d from the load to stub is obtained either of these two points i.e. d1= 0.044 and d2=0.387.• To improve bandwidth, the stub is chosen as close as possible to the load.•The tuning requires a stub of j1.33 for y1 and –j1.33 for y2.•For open circuit-stub 1 =0.147 and 2=0.353.

Continue

15

0.796nH

50

0.044

0.147

50

Solution 1

15

0.796nH

50

0.387

0.353

50

Solution 2

Reflection coefficient

0

0.2

0.4

0.6

0.8

1

1 1.5 2 2.5 3

f (GHz)

refl

. co

eff

.Solution 1

Solution 2

nHL 796.01022

502.09

Convert j0.2 to inductance value, thus

Use this value to calculate reflection coefficient

oL

oL

ZZ

ZZ

Formulas for calculation

22

2

tan

tan1

dZXR

dRG

oLL

L

Let ZL=RL+jXL, then the impedance at distance d from the load is

djXRjZ

djZjXRZ

djZZ

djZZZZ

LLo

oLLo

Lo

oLod

tan)(

tan)(

tan

tan

Admittance at this point is

dd Z

jBGY1

Thus, by substituting Zd and separating real and imaginary, we have

22

2

tan

tantantan

dZXRZ

dZXdXZdRB

oLLo

oLLoL

continueEquating G = Yo = 1/Zo,and , thus we have

22

211

tZXR

tR

Z oLL

L

o

01 222 tRZtZXR LooLL

0)(2)( 222 LLoLoLoLo XRZRtZXtZRZSolving

oL

oL

oLLoLL ZRforZR

ZXRZRXt

/22

oLo

L ZRforZ

Xt

2

dt tan

continueThe two principle solution are

0tan2

1

0tan2

1

1

1

tfort

tfortd

To find the stub length,

stubopenforY

B

o

1tan

2

1

stubshortforB

Yo

1tan

2

1

Single stub-matching

Short-stub matching

Open-stub matching

Serial configurationZ L

short-stub

Z o

d

Z o

Z L

open-stub

Z o

d

Z o

Example

Smith Chart

Match a load impedance of ZL=100+j80 to a 50 W line using a single series open-stub.Assuming the load consists of resistor and inductor in series at 2GHz. Plot the reflection coefficient from 1 GHz to 3 GHz.

Solution

•Normalized the load zL=2+j1.6•Construct SWR circle•Then move from load to generator to obtain two points on unity circle(1+jx) z1=1-j1.33 and z2=1+j1.33. •The distance d from the load to stub is obtained either of these two points i.e. d1= 0.120 and d2=0.463.• To improve bandwidth, the stub is chosen as close as possible to the load.•The tuning requires a stub of j1.33 for z1 and –j1.33 for z2.•For open circuit-stub 1 =0.397 and 2=0.103.

Continue

nHL 37.61022

506.19

Convert j1.6 to inductance value, thus

Use this value to calculate reflection coefficient

oL

oL

ZZ

ZZ

Reflection coefficient

0

0.2

0.4

0.6

0.8

1

1 1.5 2 2.5 3

f (GHz)

refl

. co

eff

.Solution 1

Solution 2

100

6.37nH

50

0.120 0.397

50

Solution 1

50

100

6.37nH

50

0.463 0.103

50

Solution 1

50

Formulas for calculation

22

21

tYBG

tGR

oLL

L

Let YL=GL+jBL, then the load admittance at distance d from the load is

tjBGjG

tjYjBGY

djYY

djYYYY

LLo

oLLo

Lo

oLod )(

)(

tan

tan

Impedance at this point is

dd Y

jXRZ1

Thus, by substituting Yd and separating real and imaginary, we have

22

2

tYBGZ

tYBtBYtGX

oLLo

oLLoL

Where t = tan d

continueEquating R = Zo = 1/Yo , thus we have

22

211

tYBG

tG

Y oLL

L

o

01 222 tGYtYBG LooLL

0)(2)( 222 LLoLoLoLo BGYGtYBtYGYSolving

oL

oL

oLLoLL YGforYG

YBGYGBt

/22

oLo

L YGforY

Bt

2

continueThe two principle solution are

0tan2

1

0tan2

1

1

1

tfort

tfortd

To find the stub length,

stubshortforZ

X

o

1tan

2

1

stubopenforX

Zo

1tan

2

1

Double-stub matching

The advantage of this technique is the position of stubs ( d and x) are fixed. The matching are done by changingthe length of stubs. The disadvantage of this technique is not all impedances can be matched.

Z LZ o

d x

S 1S 2

Open or short stubs

d

x

admittance cannot bematched

x

S2

S1

A= load admittanceA’=admittance of A at stub 2B= Adjust of stub S2 tobring to the S2 circleB’=Admittance of B at S1Then by adjusting stub 1 will bring the admittance to the center of the chart

AA’

B

B’

ExampleDesign a double-stub shunt tuner to match a load ZL=60-j80 to a 50 line. The stubs are to be short circuited stubs, and are spaced/8 apart. The load consists of a series resistor and capacitor that match at 2 GHz. Plot the reflection coefficient magnitude versus frequency from 1 GHz to 3GHz.

Solution

•Plot normalized load zL=1.2 –j 1.6 and convert to admittance we have yL= 0.3 +j0.4.•Construct a rotated 1+ j b circle which is a distance d=/8 from a 1+jb circle. Get two possible points on the rotated 1+jb circle, y1 and y1’ by adding susceptance of the first stubs. In this case we take x=0(match section immediate after the load . I.e b1=1.314 and b1’=-0.114.The length of stub will be 1=0.396 and i1’=0.232.•Now transform both point onto 1+jb circle along SWR circles.This bring two solution y2 =1-j3.38 and y2’=1+j1.38.•Then the second stub should be b2= 3.38 and b2’=-1.38. The length of stub will be 2=0.454 and i2’=0.100.

continue

/8

yL

y1

y1’

b1

b1’

y2’

y2

b2’

b2

Rotated 1+jb circle

continueReflection Coefficient vs frequency

0

0.2

0.4

0.6

0.8

1

1 1.5 2 2.5 3

Frequency(GHz)

Re

fle

cti

on

co

eff

icie

nt

Solution 1

Solution 2

The shorter the stubs the wider will be the bandwidth

Formulas for calculationLet YL=GL+jBL, then the load admittance at distance x from the load is

'''

tan

tanLL

Lo

oLoL jBG

xjYY

xjYYYY

Admittance at first stub 1''

1 BBjGY LL

x is distance between stub and load

tjBjBGjY

tYBBjGYY

LLo

oLLo

1''1

''

2)(

Where t = tan d

After transforming to stub 2 , we have

continue

Solving

At this point the ral part of Y2 =Yo ,thus

0

'1'

2

21

2

2'2

t

tBtBY

t

tYGG LooLL

22'

21

2

2

2

1

'411

2

1'

tY

tBtBYt

t

tYG

o

LooL

Since GL’ is real, the quantity in square root must be nonnegative, thus

11

'40

22'

21

2

tY

tBtBYt

o

Lo Simplified tod

Y

t

tYG ooL 22

2'

sin

10

continueSo susceptances of stubs are

t

tGYGtYBB LoLoL

22'2

1'1

'

tG

YGtGYGtYB

L

oLLoLo

'

''1 22'2

2

and

To find the stub length ,

stubopenforY

B

o

1tan

2

1

stubshortforB

Yo

1tan2

1

B either B1 or B2

continueThe two principle solution are

0tan2

1

0tan2

1

1

1

tfort

tfortd

To find the stub length,

stubshortforZ

X

o

1tan

2

1

stubopenforX

Zo

1tan

2

1

• Plot ZL normalised to 50 ohm

•Find bisector and point intersect the chart axis (I.e A)

•Draw circle at center A touching the center of the chart

•Obtain R1 and calculated new characteristic impedance of the line

•Renormalised the ZL to ZT, thus we have ZL2 . Then plot on the chart.

•Obtain x for the length of the matching section by moving towards generator.

Graphical method

Suitable for load impedance laying inside the unity circle.

oT ZRZ 1

Graphical method

Unity circle

R 1

Z L

Z L2

x

A R 2

Transmission line transformer

Quarter-wavetransformer

Z 2

21ZZZo Z1

Matching at one frequency .For other frequency the impedance at the input of matching section will be

tjZZ

tjZZZZ

o

ooin

2

2

Where t = tan d

At matched frequency tan d = tan (/2)

continueReflection coefficient

21

212

212

12

1

1

ZZZjtZZZ

ZZZjtZZZ

ZZ

ZZ

oo

oo

in

in

Since Zo2=Z1Z2, this reduces to

2112

12

2 ZZtjZZ

ZZ

2/cos

2

sec/41

1

21

12

21221221

nearforZZ

ZZ

ZZZZ

m

0 m m2

m 2

2

continue

12

21

2

1 2

1cos

42

2

ZZ

ZZ

f

ff

f

f

m

m

o

mo

o

Fractional bandwidth is given by

Reflection coeff. vs frequency

0

0.2

0.4

0.6

0.8

1

0 0.5 1 1.5 2

f/fo

Re

f. C

oe

ff.

10:01

4:01

2:01

Reflection coefficient with different ratio of Z1:Z2

ExampleDesign a single section quarter-wave matching transformer to match a 10 load to a 50 line at fo =3GHz. Determine the percentage bandwidth for which the SWR < 1.5

36.225010210 ZZZ

For a pure resistive load we can just calculate directly

and the length of transformer is /4

29.01050

50502

2.01

2.0cos

42

2

1cos

42

2

1

12

21

2

1

ZZ

ZZ

f

f

m

m

o

2.015.1

15.1

1

1

SWR

SWRm

Then determine reflection coefficient for SWR=1.5

Hence fractional bandwidth is

or 29%

Multisection transformer

Z LZ o

Z 1Z 2 Z 3

Z n No

•Binomial multisection•Chebyshev multisection

NjeA 21 NN A cos2

Two types of transformer

Reflection coefficient for multisection can be written as

Or its magnitude

02 At center frequency 2

(i.e =/4)

Binomial transformer

Solving for to obtain value of A0

oL

LN

ZZ

ZZA

020oL

LN

ZZ

ZZA

02or

For binomial expansion the reflection coefficient can be written as

N

n

njNn

Nj eCAeA0

22 )1( where !!

!

nnN

NCNn

Note CnN=CN-n

N, C0N=1 and C1

N=N=CN-1N

Let Nnn AC then

n

n

nn

nnn Z

Z

ZZ

ZZ 1

1

1 ln2

1

Henceo

LNn

NNn

oL

oLNNn

n

nn Z

ZCC

ZZ

ZZAC

Z

Zln22ln

2

1 )1()1(1

continue

For bandwidth

mNN

m A cos2

N

mm A

/11

2

1cos

Therefore

The fractional bandwidth, thus

N

mm

o

mo

o Af

ff

f

f/1

1

2

1cos

42

42

2

ExampleDesign a three section binomial transformer to match 50 load to a 100 line

SolutionN=3 , ZL=50 , Z0=100then 0433.0ln

2

12

10

o

LN

oL

LN

Z

Z

ZZ

ZZA

1!0!3

!330 C 3

!1!2

!331 C 3

!2!1

!332 C

Nn

n

n ACZ

Z1ln

2

1But

1)0433.0(22ln 30

0

1 ACZ

Z Z1=91.7

3)0433.0(22ln 31

1

2 ACZ

Z

3)0433.0(22ln 32

2

3 ACZ

Z

Z2=70.7

Z3=54.5

Chebyshev transformer

Useful forms of Chebyshev polinomial are

cossec)cos(sec1 mmT

12cos1sec)cos(sec 22 mmT

cossec3cos33cossec)cos(sec 33 mmmT

112cossec4

32cos44cossec)cos(sec2

44

m

mmT

Chebyshev polinomial in general

1)coscos()( 1 xforxnxTn

1)coshcosh()( 1 xforxnxTn

continue

cossec

)2cos(....)2cos(cos2 1

mNjN

nojN

TAe

nNNNe

Solving for to obtain value of A0

oL

LmN ZZ

ZZAT

0sec0 mNoL

L

TZZ

ZZA

sec

10

or

For maximum reflection coefficient magnitude mA

Therefore o

L

mmoL

LmN Z

Z

ZZ

ZZT ln

2

11sec 0

For N section,Chebyshev expansion the reflection coefficient can be written as

or

o

L

mm Z

Z

Nln

2

1cosh

1coshsec 1

ExampleDesign a three section Chebyshev transformer to match a 100 load to a 50 line.Taking m=0.05.

Solution

cos3cos2cossec 13 o

jNm

jN eTAe

cossec3cos33cossec)cos(sec 33 mmmT

*

**Equating * and ** and A=m=0.05

cos

408.1

50

100ln

05.02

1cosh

3

1coshln

2

1cosh

1coshsec 11

o

L

mm Z

Z

N

408.1sec05.0sec2 33 mo A

0698.0o

continuecos )408.1sec(408.1sec05.03secsec32 33

1 mmA

1037.01

n

nn Z

Z 1ln2

1But Then nnn ZZ ln2ln 1

Start n=0 051.450ln0698.02ln2ln 1 oo ZZ

5.571Z

n=1 259.45.57ln1037.02ln2ln 112 ZZ

7.702Z

For symmetrical 1037.012 0698.003 and

n=2 466.47.70ln1037.02ln2ln 223 ZZ

0.873Z

Simple form of multisection transformers

Z 1Z 2

Z AZ B

2-quarter-wave transformer

4/1

12

1

ZZZZA

4/3

12

1

ZZZZB

Z 1 Z 2

Z A Z B Z C

3-quarter-wave transformer

121

eZZA

1

21 ln

2

1

8

1

Z

Z 122

eZZC1

21 Z

ZZZB

Tapered Transmission line transformer

Z 1Z 2

Exponential Tapertransformer

Triangular Tapertransformer

Klofenstein Tapertransformer

dzZ

Z

dz

de

o

L

oz

zj

ln2

1 2

o

L

az

Z

Z

Lawhere

LzforeZzZ

ln1

0)( 0L

Le

ZZ zjoL

sin

2

ln 2

2/0

2/0)(

/ln1/2/40

/ln/20

22

2

LzforeZ

LzforeZzZ

oL

oL

ZZLzLz

ZZLz 22

2/

2/sinln

2

1

L

L

Z

Ze

o

Lzj

Lzfor

ALzAA

ZZzZ oL

0

,1/2cosh

ln2

1)(ln 2

0

ALforA

ALe zj

o

cosh

cos 22

11

1,

2

21

xfordyyA

yAIAx

x

owhere

Impedance distribution Reflection

Transmission line transformer

Z 1Z 2 Z 2

Z 1

T

T

1arctan1

1

1

2 Z

Z

Z

ZT

Nonsynchronous transformer

E.g matching Z1=75 to Z2=50

T = 29.3o

Transformer Frequency Response for 100 /10

-6

-5

-4

-3

-2

-1

0

0.5 1 1.5

Normalised frequency f/fo

Am

pli

tud

e (

dB

)

Quarter-wave

Nonsynchronous

Complex to Complex Impedance Matching

jX a

Transmission lineZ o

jX b

R a R b

ba

abbbaao RR

RXRRXRZ

2222

abba

abo

RXRX

RRZ1tan