Review of Smith Chart Chapter 5 Impedance Matching and ...thschang/notes/MWPA5.pdf · Chapter 5...

Chapter 5 Impedance Matching and TuningChapter 5 Impedance Matching and Tuning

• The matching network is ideally lossless and is usuallyThe matching network is ideally lossless, and is usually designed so that the impedance seen looking into the matching network is Z0.

• Only if Re[ZL] ≠ 0, a matching network can always be found.• The quarter-wave impedance transformer is for read load.q p• Important factors in selecting matching networks:

(1)complexity, (2)bandwidth, (3)implementation, (4)adjustability.( ) p y, ( ) , ( ) p , ( ) j y

1

Review of Smith Chart

zL = ZL / Z0zL ZL / Z0= rL + jxL

2

5.1 Lumped Element Matching (L-networks)p g ( )L-type or L-network matching is the simplest matching network.

(1) zL = ZL/Z0 is outside the r=1(1) zL = ZL/Z0 is inside the r=1 (1) zL ZL/Z0 is outside the r 1 circle (rL < 1).

(2) Z i i i h B h

(1) zL ZL/Z0 is inside the r 1 circle (rL > 1).

(2) Z i h i h B h (2) ZL is series with jB, then shunt with jX.

(2) ZL is shunt with jB, then series with jX.

3

Analytical Solutions (rL>1)

• When rL > 1, let ZL = RL + jXL and zL = ZL / Z0 = rL + jxL

01

1inZ jX ZjB

R jX

0 0( )L L

L L L

R jXB XR X Z R Z

0

2 2

(1 )L L L

L

X BX BZ R X

RX R X Z R

00

2 2

L L L L

L L

X R X Z RZ

BR X

inZ

0 01 L

L L

X Z ZXB R BR

4• Note that when rL > 1, the square root in B has real results.

Analytical Solutions (rL<1)

• When rL < 1, let ZL = RL + jXL and zL = ZL / Z0 = rL + jxL

0

1 1 1( )in L L

jBZ R j X X Z

0 0

0

( )( )

L L

L L

BZ X X Z RX X BZ R

0Z

0

0

( ) /L LZ R RB

Z

inZ0

0( )L L L

Z

X X R Z R

• Two analytical solutions are physically realizable.

• One of the two solutions may result in smaller reactive elements and

5

One of the two solutions may result in smaller reactive elements, and may have better matching, bandwidth, or better SWR on the line.

Example 5 1 L-Section Impedance MatchingExample 5.1 L Section Impedance Matching

ZL = 200-j100 Ω, Z0 = 100 Ω, j , ,f = 500 MHz, zL = ZL/Z0 = 2-j1, rL = 2 >1 inside the r = 1 circle.

Exact solutions:

2 2R 2 20

02 2

LL L L L

L L

RX R X R ZZ

BR X

3 12 2

2 2 3 1

2.899 10100 2 200 100 100 200 200 100 6 899 10

L LR X

3 1

0 0

200 100 6.899 10

122.471 LX Z ZX

6

122.47L LB R BR

Example 5 1 L-Section Impedance MatchingExample 5.1 L Section Impedance Matching

Th l iThe two solutions:

(1)3

6

2.899 10 0.9228 pF2 500 10

BC

( ) 6

6

p2 500 10

122.47 38.98 nH2 500 10

XL

62 500 10

1 1(2)6

1 1122.47 2 500 10

2 599 pF

CX

3 6

2.599 pF1 1

6 899 10 2 500 10L

B

7

6.899 10 2 500 1046.14 nH

B

Example 5 1 L-Section Impedance MatchingExample 5.1 L Section Impedance Matching

The two solutions:The two solutions:

8

Lumped Elements for Microwave Integrated Circuits (MICs)

9

5.2 Single-Stub TuningShunt‐stub tuning Series‐stub tuning

• No lumped element is required. Convenient for MIC fabrication.• Characteristic impedances of the lines and the stub can be different.• Keep the matching stub as close as possible to the load. • Solution procedure for shunt-stub and series-stub tunings.(1) Locate zL (yL) on the Smith Chart.(1) Locate zL (yL) on the Smith Chart.(2) Move along the const-Γ circle with 2d/λ wavelengths to

reach the g = 1 (r = 1) circle.(3) Sh t ith t i ith t t(3) Shunt with a susceptance or series with a reactance to

cancel the imaginary part. 10

Shunt-Stub Tuning Analytical Solution

0 00 tan

( )1 1L L Lin t dno stub

Z jZ t R j X Z tZ ZZ jZ t Y Z X t jR t G jB

0 0 0

220 0

022 2 2

( )( )(1 ) ,

L L L

L L LLin in

in inZ jZ t Y Z X t jR t

R t Z X t Z t XR tG B Y

G jB

022 20 0

2 2 20 0 0

2,( ) ( )

(1 ) ( ) real part

in inL L L L

in L L L

GR X Z t R X Z t

G Y Z R t R X Z t

0 0 0in L L L

00

If , and tan2

LL

XR Z t dZ

0

11 tan , 02 2

LL

X XZd

0

1

2 2

1 tan , 0LL

ZdX X

0

tan , 02 2 LX

Z

11

Shunt-Stub Tuning Analytical Solution0

2 2 20 0 0 0 0

If Z ,

( ) ( )( )L

L L L L L L

R

X Z X Z Z R Z R Z R Xt

0 0 0 0 0

0 0

2 2 2

( ) ( )( )( )

( ) [( ) ]

L L L L L L

L

L

tZ R Z

RX R Z X

00

0

( ) [( ) ]L L L

L

X R Z XZ

R Z

here tant lLet the stub susceptance Bstub (or Bs) = -Bin,

(a) open-stub: and0stubZ jZ t sstubstub jBjBtjYY 0

here tant l

1 1

0 0

1 1tan ( ) tan2 2

o s inl B BY Y

(b) Short-stub: and tjZZstub 0 sstubstub jBjBtjYY 0

1 10 01 1tan ( ) tansl Y Y tan ( ) tan

2 2s inB B

If lo < 0 or ls < 0, λ/2 must be added to have a realistic result. 12

Example 5.2 Shunt Single-Stub Tuning

Match ZL = 15 + j10 Ω to a 50-Ω line. Use a shunt open-stub.Sol: RL ≠ Z0,Sol: RL ≠ Z0,

2 2 20( ) ( )L

L L LRX R Z X 0

0

0

( ) ( )10 397.5

35

L L L

L

Zt

R Z

d

11 1

10 397.5 1 ( tan ) 0.38735 2

1

dt t

l B

1

0

1 ... tan2

10 397 5 1

o inin

l BBY

d

12 2

1

10 397.5 1 tan 0.04435 2

1o in

dt t

l BB

13

1

0

... tan2

o ininB

Y

Example 5.2 Shunt Single-Stub Tuning (Cont’d)Use a shunt open-stub to match ZL = 15 + j10 Ω line.

Solution 1: Solution 2:

14

5.3 Double-Stub Tuning – Analytical Solution

• In single-stub tuning, the length d is not tunable.

• If this causes difficulty in circuit implementation, use double-stub tuning.

0 tanLY jY bY Y G jB

15

00

0 tanL

L L L

L

jY Y G jBY jY b

5.3 Double-Stub Tuning – Analytical Solution2 2 2 2 2

0 1 0

2

0 4 ( ) (1 )

1Lt Y B t B t Y t

Y

20

0 2 2

102 sin

has an upper limit

LYtG Y

t dG

( )Y G j B B 2 2 2

has an upper limit. Given , one can obtain:

(1 )

LGd

Y t G Y G t 1 1

1 02 0

0 1

( )( ) , tan

( )

L L

L L

L L

Y G j B BG j B B Y tY Y t d

Y jt G jB jB

0 01

2 2 2

(1 )

(1 )

L LL

Y t G Y G tB B

tG t G Y G t

0 1

2 0 2 222

2 0 1

( )Re[ ] , Im[ ]

( )1 0

L L

L

Y jt G jB jBY Y Y B

Y B t B ttG Y G

01 0

10

(1 )

1

L L LG t G Y G tB Y

tB

2 0 10 2 2

2 220 1

( ) 0

4 ( )1 1 1

LL L

L

G Y Gt t

t Y B t B ttG Y

101 2

0

1 0

1 tan , ,2

1s

B B B BY

Y

B B B16

0 10 2 2 2

0

( )1 1(1 )

LLG Y

t Y t

1 0tan2

s

B

1 2, ,B B B

Example 5.4 Double-Stub Tuning - PerformanceUse double-stub matching scheme to match ZL = 60 – j80Ω at 2.0 GHz to a 50-Ω line.

17

5.4 The Quarter-Wave Transformer• A single-section transformer may suffice for a narrow-band

impedance matching.• Single-section quarter-wave impedance matching λ= λ0 / 4 at the

desired frequency. (See Chap 2)

• Multisection quarter-wave transformer designs can be synthesized to yield optimum matching characteristics over a broad bandwith.

18

yield optimum matching characteristics over a broad bandwith.

2.5 The Quarter-Wave TransformerReview

A quarter-wave transformer is an impedance matching circuit

2tanR jZ l Z 1 11

12

tan tan

Lin

L Ll

R jZ l ZZ ZZ jR l R

Z Z

0

0

inin

in

Z ZZ Z

19

0 1 0If 0 is required, , then in in LZ Z Z Z R

Estimate the Bandwidth of a Single-Section Impedance Transformera Single Section Impedance Transformer

1 0 LZ Z Z

11

1

, tan tan Lin

L

Z jZ tZ Z tZ jZ t

Z jZ t

11 0 2

0 1 0 1 012

10 1 0 1 01 0

( ) ( )( ) ( )

L

in L LL

Lin L L

Z jZ tZ ZZ Z Z Z Z j Z Z Z tZ jZ t

Z jZ tZ Z Z Z Z j Z Z Z tZ Z

1 01

002 2

1 2 4( ) 4

L

LL

Z jZ tZ ZZ Z

Z Z j t Z Z Z ZZ Z t Z Z

2 20 0 00 02 4( ) 4 1L L LL L

L

Z Z j t Z Z Z ZZ Z t Z ZZ Z

22

0

sec

0LZ Z 0

0

0

When , cos2 2

L

L

L

Z ZZ Z

Z Z

0

0

Set a max , or cos2

Lm m m

L

Z ZZ Z

20

Bandwidth of the Matching Transformer

20

2( / 2 )41 1

m

LZ Z

202 2

0

: 1 sec( )

2

Lm m

m L

m m

Z Zff f f

00 0

0

, ,2 2

2

m mm m

L

ff f ff f

Z Z

0

20

cos1

Lmm

LmZ Z

F ti l b d idth10 or 0.1

0

Fractional bandwidth :2( ) 42m mf ff

f f

4 or 0.25

0 0

01 242 cos Lm

f f

Z Z

2 or 0.5

21

20

2 cos1 Lm

Z Z

Example 5.5Single Section Quarter Wave Transformer BandwidthSingle-Section Quarter-Wave Transformer Bandwidth

ZL = 10 Ω, Z0 = 100 Ω, SWR = 1.2S lSol:

1 0 31.6LZ Z Z

1 0.11m

SWRSWR

max 0 0

min 0 0

11

V V VSWRV V V

01

2

242 cos1

Lm Z Zff Z Z

20 0

1

1

4 0.1 2 31.6 62 cos 2 9%

Lmf Z Z

22

2 cos 2 9%900.99

5.5 Theory of Small Reflections• For applications requiring more bandwidth than that a single quarter-

wave section can provide, multi-section transformers can be used.

(1) Single-Section Transformer

212

121

ZZZZZZ

2

23

2ZZZZZ

L

L

1

12

2121

21

21

ZT

ZZZT

12

1212 1

ZZT

23

5.5 Theory of Small Reflections

(1) Multi-Reflections

2 2 41 12 21 3 12 21 3 2 ...j jT T e T T e

1 12 21 3 12 21 3 2

22 2 41 12 21 3 3 2 3 2(1 ...)j j jT T e e e

2 21 12 21 3 3 2

02 2

nj j n

nj j

T T e e

T T e e

12 21 3 1 31 2 2

3 2 1 31 1

j j

j j

T T e ee e

• If is small, • Γ the reflection from the initial discontinuity between Z1 and Z2 + the

312

1 3je

24

first reflection from the discontinuity between Z2 and ZL

Multisection Transformer

• N commensurate (equal-length) sections of transmission lines. Let ( q g )the total reflection be Γ.

• Assume that all Z increase or decrease monotonically ZL is real• Assume that all Zn increase or decrease monotonically, ZL is real, and “The theory of small reflections” holds.

• It can be validated that

NL ZZZZZZZZ

23

212

101

0

25

NLN ZZZZZZZZ

...,,,23

212

101

0

Multisection Transformer

2 2

2 4 20 1 2

[ ( ) ( ) (if s mmetr]

.

)

..j j j

j N j NjN jN jN

NNe e e

0 1

0 1 /2

[ ( ) ( ) ... (if symmetry]

cos cos 2 ... , 2

)j jjN jN jN

NjN

e e e e e

N N N evene

0 1 1 /2

2cos cos 2 ... cos ,

j

N

eN N N odd

• We can synthesize any desired response as a function of frequency• We can synthesize any desired response as a function of frequency or θ, by properly choosing the Γn’s and enough number of sections N.

• The most commonly used passband responses are:

(1) Binomial or maximally flat response, and(2) Chebyshev or equal-ripple response. 26

H Fi d h C di R fl i ?How to Find the Corresponding Reflection?

2 4 20 1 2

Theory of Samll Reflection: ( ) ...j j j N

Ne e e 0 1 2( )

Binomial Multi-Section Transformer:

N

2

Binomial Multi-Section Transformer:(1 )j NA e

Chebyshev Multi-Section Transformer:

(sec cos )n mAT

27

5.6 Binomial Multisection Matching Transformer• Maximally flat response: the response is as flat as possible near the design

frequency. Also called Binomial matching transformer.• Let 2(1 ) A d tifi i llj NA • Let

2(1 ) Assumed artificially

2 cos

j N

NN

A e

A

0

0

00 2

2N L

NL

Z ZA AZ Z

0

2 2

0

!(1 ) ,( )! !

LN

j N N j n Nn n

n

NA e A C e CN n n

2 4 20 1 2

( ) ...

(0)

j j j NN

N N

e e e

AC C

N! ("enn factorial“)

Properly

( )2

configure the i

N Nn n nNAC C

mpedances to synthesize the needed

N! ( enn factorial )“Four factorial" is written as "4!”

reflections .n28

Binomial Multisection Matching Transformer

nnn xxxZZZ 11 1112lnln1

LNN

LNN

Nn

nnnn

ZCZZCACZ

xx

xZZZ

01

1

1

ln1122ln

11,1

2ln,ln2

NNnC

Ln

nNL

nNnnn

ZZ

ZC

ZZCAC

Z2/

1

001 ln

2222ln

n ZZ 0

• Exact results for Z 1 / Z0 for N = 2 thru 6 are given in Table 5 1• Exact results for Zn-1 / Z0 for N = 2 thru 6 are given in Table 5.1.

29

Binomial Transformer Design Table 5.1

30

Example Binomial Transformer Design

ZL/Z0

N=5

Z1 /Z0 Z2 /Z0 Z3 /Z0 Z4 /Z0 Z5 /Z0

1.0 1.0000 1.0000 1.0000 1.0000 1.0000

1.5 1.0128 1.0790 1.2247 1.3902 1.4810

2.0 1.0220 1.1391 1.4142 1.7558 1.9569

3 0 1 0354 1 2300 1 7321 2 4390 2 89743.0 1.0354 1.2300 1.7321 2.4390 2.8974

4.0 1.0452 1.2995 2.0000 3.0781 3.8270

6.0 1.0596 1.4055 2.4495 4.2689 5.6625

8.0 1.0703 1.4870 2.8284 5.3800 7.4745

10.0 1.0789 1.5541 3.1623 6.4346 9.2687

A 5th d bi i l f f Z 6Z• A 5th order binomial transformer for ZL = 6Z0

31• Also note that 4055.1

2689.46,0596.1

6625.56

E ample Binomial Transformer Design Z /Z <1Example Binomial Transformer Design ZL/Z0 <1

ZL/Z0

N=5

Z1 /Z0 Z2 /Z0 Z3 /Z0 Z4 /Z0 Z5 /Z0

6.0000 1.0596 1.4055 2.4495 4.2689 5.6625

0.1667 0.9438 0.7115 0.4082 0.2343 0.1766

• A 5th order binomial transformer for ZL = Z0 / 6

32

Bandwidth of the Binomial Transformer,cos2 m

NNm A i.e, tolerable max over the passband.m

N/1

m

m A1

21cos

N

mmm

Afff

ff

/1

1

0

0

0 21cos4242)(2

Example 5.6 N = 3, ZL = 2Z0 = 100 Ω, find the BW for Γm = 0.05.Sol: From Table 5.1, the required impedance Zn can be found to be

1 2 354.5 , 70.7 , 91.71 100 502 (0) 0 4167N

Z Z Z

A

11 3

2 (0) 0.41678 100 50

4 1 0.052 cos ( ) 70%

A

f

33

3

02 cos ( ) 70%

2 0.4167f

Binomial Transformer’s Frequency Response

Reflection coefficient magnitude versus frequency for multisectionbinomial matching transformer of Ex. 5.6 with ZL = 2Z0 = 100 Ω and

34

binomial matching transformer of Ex. 5.6 with ZL 2Z0 100 Ω and Γm = 0.05.

Example (2nd Midterm)Design a Butterworth transformer of N = 2 for ZL = 4Z0. Let Γ0, Γ1 and ΓL be

DIYg L 0 0, 1 L

respectively the reflection coefficients at the Z0 – Z1, Z1 – Z2 and Z2 – ZLjunctions, and Γ0 = ΓL. (a) Based on the “theory of small reflection,” find the Γ terms of Γ Γ and θ (b) If Γ = 0 and are required when θ0/ fΓin terms of Γ0, Γ1 and θ. (b) If Γin = 0 and are required when θ= λ / 4, find a and b.

0/ fin

2cos2 104

22

10 ee jjin

202

2cos2 0110

10210

in

4444444

11

02sin4 0

abbababababb

aa

in

359/141141

0434

31)(2

112

412322

aaaabaababaabaa

abab

ba

2751.0,5707.1

4321.08135

99/14,9273.1

271

27112

94

61

3 233 23

rqrrqr

qr

The “exact” solution in Table 5.1 is a = 1.4142 and b =2.8285. The error is from

358435.2/4

4067.12751.05707.191

ab

a the approximation used in the “theory of small reflection.”

5.7 Chebyshev Multisection Matching Transformers

Chebyshev polynomials: Recurrence formula:)()(

1)(

1

0

xxTxT

34)(24

33 xxxT

2)()(2)( 21 nxTxxTxT nnn

12)( 22 xxT 188)( 24

4 xxxT

Tn(x)345

n=1

2

n=1

xpassband

36

p

5.7 Chebyshev Multisection Matching Transformers(1) mapped to passband

Let x = cosθ, it can be shown that (2) t id th b d

1)(,1 xTx n

nTn cos)(cos 1)(1 T(2) outside the passband

(3) In general, 1)coscos()( 1 xxnxTn

1)coshcosh( 1 xxn

1)(,1 xTx n

1)coshcosh( xxn

Tn(x)345

n( )

1

2

n=1

xpassband

37

passband

Magnitude of Chebyshev Polynomials

|Tn(x)|345

22

n=1

xx

passband

38

passband

Chebyshev Responses

39

Mapping of Passband and Stopband

Define is mapped to passbandthe upper passband edgethe lower passband edge

1cos/cos xx m,1 xm1 x the lower passband edge ,1 xm

1

1

( ) cos [cos (cos / cos )] (sec cos )(sec cos ) sec cos

n m n mT x n TT

1

22

3

(sec cos ) sec cos

(sec cos ) sec (1 cos 2 ) 1

(sec cos ) sec (cos3 3cos ) 3sec cos

m m

m m

TTT

40

3

4 24

(sec cos ) sec (cos3 3cos ) 3sec cos

(sec cos ) sec (cos 4 4cos 2 3) 4sec (1 cos 2 ) 1m m m

m m m

TT

Design of Chebyshev Transformer

jN NNN ]22[2)(

LmN

L

njN

ZZZZ

TAAT

ZZZZ

nNNNe

)(sec1)(sec)0(

...]2cos...2coscos[2)(

00

10

mmNm

LmNL

ATAZZTZZ

)sec(cos)(sec 00

L

L

mm

L

L

mL

LmN

f

ZZZZ

NZZZZ

ZZZZ

AT 1cosh1coshsec,11)(sec

0

01

0

0

0

0

41m

ff 420

Table 5.2 Chebyshev Transformer Design

42

Example 5.6Design a Chebyshev transformer of N=3 for ZL = 2Z0 = 100 Ωwith Γm = 0.05.

3j Sol: 30 1

33

( ) 2 [ cos3 cos ]

(sec cos )

j

jm

eAe T

3sec (cos3 3cos ) 3 sec cos

0.05m m

m

A AA

1 01 1sec cosh cosh

m

Lm

Z ZN Z Z

0

11 20cosh cosh 1.4083 3

m LN Z Z

30 0 3

3 3

2 sec 0.0698mA

4312 3A 3

1 2(sec sec ) 0.1037m m

Example 5.6Design a Chebyshev transformer of N=3

50.5711 0

01

ZZ

81.70111

112

0

ZZ

20.87111

223

1

ZZ

%10142

1

7.44

2

om

m

ff

0mf

44

Example: “Chebyshev轉換器特性之深入探討電腦模與實驗量p測”清大物理 碩士論文 張靜宜

0chebyshev4-section

measured

-20

ion(

dB)

simulation

-40Ref

lect

i

10 11 12-60Load/

Transformer

ModeFrequency(GHz)

Ch b h 四階轉換器電腦模擬與實驗結果之比較

terminationMode

converter

45

Chebyshev四階轉換器電腦模擬與實驗結果之比較

gx XÇw Éy VtÑA H Ñ

46