Chapter 5 Impedance Matching and Tuning. Why need Impedance Matching Maximum power is delivered and...

Chapter 5

Impedance Matching and Tuning

Why need Impedance MatchingMaximum power is delivered and power loss is minimum.

Impedance matching sensitive receiver components improves the signal-to-noise ratio of the system.

Impedance matching in a power distribution network will reduce amplitude and phase errors.

Basic IdeaThe matching network is ideally lossless and is placed between a load and a transmission line, to avoid unnecessary loss of power, and is usually designed so that the impedance seen looking into the matching network is Z0. ( Multiple reflections will exist between the matching network and the load)

The matching procedure is also referred to as “tuning”.

Design Considerations of Matching Network

As long as the load impedance has non-zero real part (i.e. Lossy term), a matching network can always be found.

Factors for selecting a matching network:

1) Complexity: a simpler matching network is more preferable because it is cheaper, more reliable, and less lossy. 2) Bandwidth: any type of matching network can ideally give a perfect match at a single frequency. However, some complicated design can provide matching over a range of frequencies. 3) Implementation: one type of matching network may be preferable compared to other methods. 4) Adjustability: adjustment may be required to match a variable load impedance.

Lumped Elements Matching L-Shape (Two-Element) MatchingCase 1: ZL inside the 1+jx circle (RL>Z0)

LL

L

LL

LLLLL

LLL

LLL

LLL

LL

BR

Z

R

ZX

BX

XR

XZXRZRXB

X

XRBZBXX

ZRZXXRB

jXRZjXRjB

jXZ

00

220

220

0

00

0

1

gives (2) into ngsubstituti and for (1) Solving

(2) )1(

(1) )(

partsimaginary and real into separating and gRearrangin

;)(

11

Use impedance identity method

Example5.1: Design 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?

Solution ( Use Smith chart)

nH8.382

pF;92.02

0

0

f

XZL

fZ

BC

1. Because the normalized load impedance ZL= 2-j1 inside the 1+jx circle, so case 1 network is select.2. jB close to ZL, so ZL YL.3. Move YL to 1+jx admittance circle, jB =j 0.3, where YL 0.4+j 0.5.4. Then YL ZL, ZL 1+j 1.2. So jX =j 1.2.5. Impedance identity method derives jB =j 0.29 and jX =j 1.22.

6. Solution 2 uses jB =-j 0.7, where YL 0.4-j 0.5.7. Then YL ZL, ZL 1-j 1.2. So jX =-j 1.2.8. Impedance identity method derives jB =-j 0.69 and jX =-j 1.22. nH1.46

2 pF;61.2

2

1 0

0

fB

ZL

fXZC

011

1

Because

11

1)1(

(1) into substitute Q

(2) From

(2) 0)(

(1) 1)(


1 ;

)(

1

2

0

SLLS

SLSL

LL

LS

S

LLS

LSL

LLL

LSLL

RGGR

RGQQRG

QG

BBQ

R

X

XGBBR

BBXRG

jBGZ

YRjBGjB

jXZ

Use resonator method (Case 1: RS<1/GL)

Goal: Zin=Rs =S11=(Zin-Rs)/(Zin+Rs)=0

LYjB

jX

AC

SR

inZ

L L LY G jB

LG( )Lj B B

jX

AC

SR

Series

or LR RC

Parallel

or LR RC

SS

XQ

R L

LL

B BQ

G

LSL

LLL

SLSS

SLLS

LSL

LLL

SLSS

SLLS

BRG

GBQGB

CRGRQRX

RGQQQ

BRG

GBQGB

LRG

RQRX

RGQQQ

11

ce)Capacitan1

( 0; 11

11

Choose

11

e)Inductanc( 0; 11

11

Choose

e)Inductanc1

( 0;

ce)Capacitan( 0;

L

C

e)Inductanc1

( 0;

ce)Capacitan( 0;

L

C

Case 2: ZL outside the 1+jx circle (RL<Z0)

0

0

0

0

00

00

)(

)(

gives and for Solving

(2) )(

(1) )(


;)(

11

Z

RRZB

XRZRX

BX

RBZXX

RZXXBZ

jXRZXXjR

jBZ

Y

LL

LLL

LL

LL

LLLLL

Use admittance identity method

01 Because

1 R-

(1) into substitute Q

(2) From

(2) )(

(1) R-)(


1

)(

1

L2

L

L

SLS

L

SSL

SSLS

L

LSL

SLS

SLLin

R

RRR

R

RQRQR

QBRQR

XX

BRRXX

RXXBR

RjXRjXjBY

Use resonator method (Case 2: RS>RL)

Goal: Yin=1/Rs =S11= =0in

S

inS

YR

YR

1

1

LZjB

jX

AC

SR

inY

L L LZ R jX

LR

( )Lj X X

jBAC

SR

Series

or LR RC

Parallel

or LR RC

S SQ BR LL

L

X XQ

R

LL

SLLL

L

S

SS

L

SLS

LL

SLLL

L

S

SS

L

SLS

XR

RRXQRX

LR

R

RR

QB

R

RQQQ

XR

RRXQRX

CR

R

RR

QB

R

RQQQ

1

e)Inductanc1

( 0; 11

1 Choose

1

e)Inductanc( 0; 11

1 Choose

e)Inductanc1

( 0;

ce)Capacitan( 0;

C

L

e)Inductanc1

( 0;

ce)Capacitan( 0;

C

L

Matching Bandwidth

Series-to-Parallel Transformation

PS

PS

ZZ

QQQ

SSS

S

SS

jXRZ

R

XQ

PP

PPP

P

PP

jXR

jXRZ

X

RQ

2

2

2

)1(

)1(

Q

QXX

QRR

SP

SP

sjXsR

ss

s

s s s

XQ

R

Z R jX

pjX

pR

pp

p

p pp

p p

RQ

X

R jXZ

R jX

Case 1: ZL inside the 1+jx circle (RL>Z0)

11

)1(

1

1

, Let 1 01


1)(

at matching Impedance

2

0

00

0

0

LS

SL

Sg

ggg

gg

in

GRQ

RQG

RR

LC

CBLXXBjB

jX

RjB

jXZ

gR

gjBjX

AC

SR

LG( )Lj B B

jX

AC

SR

11

21222BW% in bandwidth 3dB

222

22 whenoccurs 2for bandwidth 3dB

22

2

21

1

,0,let , As

Let

for bandwidth 3dB Find

)1( :Proof

0

0

0

00

2

0

LS

L

S

SS

S

Ssg

g

Sin

Sin

g

LSg

in

in

GR

QQX

RX

R

L

R

RL

RjL

jL

RjBjX

jBjX

RZ

RZ

CBLX

QQGQRR

XQ

QQ

Define QL and Qin for RLC resonator

g

s

R

R

gjBjX

AC

SR

inQ

2

L

in

Q

Q

1

212BW

L

SL

RRQQ

Similarly, for case 2: ZL inside the 1+jx circle (RL>Z0).

Summary

1 :2 caseFor

11

:1 caseFor

1222BW

0

0

L

S

LS

L

S

R

RQ

GRQ

QQX

R

1

1

2

0

2

Example5.2: Design an L-section matching network to match load impedance RL = 2000, to a RS = 50, at a frequency of 100 MHz?Solution

Because RS <1/ GL, so case 1 network is select.

nH7.509 0 2000

245.6

102

1

1

pF097.5 )245.6(50102

1

1

245.6 when2 Solution

%32245.6

22 BW BandwidthdB 3

pF9696.4 0 2000

245.6102

nH96.496 245.650102

245.6 when1 Solution

245.6150

20001

1

80

80

80

80

LL

BQGL

B

CC

QRC

X

Q

Q

CCBQGCB

LLQRLX

Q

GRQ

LL

S

LL

S

LS

2000gR

Matching

Network

AC

50sR

( 100MHz) 50

( 100MHz) 0inZ f

f

2000AC

50 496.96nH

4.9696pF

( 100MHz) 50

( 100MHz) 0inZ f

f

2000AC

50 5.097pF

509.7nH

=S11

=S22

TermTerm1

Z=50 OhmNum=1

CC21C=4.9696 pF

LL2

R=L=496.96 nH

RR1R=2000 Ohm

LL3

R=L=509.7 nH

CC22C=5.097 pF

TermTerm2

Z=50 OhmNum=2

RR2R=2000 Ohm

m1freq=dB(S(2,2))=-86.879

100.0MHz

m2freq=dB(S(1,1))=-96.908

100.0MHz

m3freq=dB(S(2,2))=-3.053

87.00MHzm4freq=dB(S(2,2))=-3.111

121.0MHz

60 70 80 90 100 110 120 130 14050 150

-80

-60

-40

-20

-100

0

freq, MHz

dB(S

(1,1

))

Readout

m2

dB(S

(2,2

))

100.0M-86.88

m1

87.00M-3.053

m3

121.0M-3.111

m4

m1freq=dB(S(2,2))=-86.879

100.0MHz

m2freq=dB(S(1,1))=-96.908

100.0MHz

m3freq=dB(S(2,2))=-3.053

87.00MHzm4freq=dB(S(2,2))=-3.111

121.0MHz

BW=34%

Use resonator method for complex load impedance.Splitting into two “L-shape” matching networks.

Goal: Zin(=0)=RS , (=0)=0

Three Elements Matching (High-Q Matching)

Case A: - shape matching

Goal: Zin(=0)=RS ,

(=0)=0

Case B: T- shape matching

LZ2jB

3jX

AC

SR L L LZ R jX

1jX

LY3jB

2jX

AC

SR L L LY G jB

1jB

0

0

( ) 50

( ) 0inZ

0

0

( ) 50

( ) 0inZ

1

2

11

22

1 2

1

11

2BW

2BW

BW min(BW , BW )

S

V

V L

RQ

R

QR G

Q

Q


1jB

2ajX

AC

sR

LY3jB

2bjX

AC

sRL L LY G jB

1jB

0

0

( )

( ) 0in sZ R

2ajX

vRvR

2 2 2a bX X X

LY3jB

AC

vR

2bjX

vR

0

0

( )

( ) 0in vZ R

P.S. RV : Virtual resistance

Conditions: RV<1/GL , RV<RS


1

2

11

22

1 2

1

1

2BW

2BW

BW min(BW , BW )

V

S

V

L

RQ

R

RQ

R

Q

Q

Conditions: RV>RL , RV>RS

2ajB

1jX

AC

sR

LZ2bjB

3jX

AC

SR L L LZ R jX

2ajB

0

0

( )

( ) 0in sZ R

1jX

vRvR

2 2 2a bB B B

LZ2bjB

AC

vR

3jX

vR

0

0

( )

( ) 0in vZ R


Example5.3: Design a three elements matching network to match load impedance RL = 2000, to a RS = 50, at a frequency of 100 MHz, and to have BW<5%?Solution


2492.1 Choose

2492.1

%5

12000

2BW)BW,BWmin(BW

BWBW

12000

2

11

22BW

150

2

1

22BW

221

212

2

11

V

V

V

VLV

VV

S

R

R

R

RGR

Q

RRRQ

2000gR

Matching

Network

AC

50sR

( 100MHz) 50

( 100MHz) 0inZ f

f

Splitting into two “L-shape” matching networks

pF95.203 1

nH74.12 1

6.247 When

nH42.12

pF85.198

6.247 When

012

0

11

1

012

01

1

1

CC

QRX

LLR

QB

Q

LLQRX

CCR

QB

Q

Va

S

Va

S

6.24712492.1

5011

V

S

R

RQ

1jB

2ajX

AC

50

( 100MHz) 50

( 100MHz) 0inZ f

f

3jBAC

2bjX

1.2492

( 100MHz) 1.2492

( 100MHz) 0inZ f

f

1.2492

11

2000

1jB

2ajX

AC

50

( 100MHz) 50

( 100MHz) 0inZ f

f

1.2492

pF85.31 1

nH58.79 1

04 When

nH53.79

pF83.31

04 When

022

023

2

022

023

2

CC

QRX

LL

BQGB

Q

LLQRX

CCBQGB

Q

Vb

LL

Vb

LL

0412492.1

20001

12

VLRGQ

Four solutions for -shape matching networks

3jBAC

2bjX

( 100MHz) 1.2492

( 100MHz) 0inZ f

f

1.2492

11

2000

AC

50

2000

79.53nH

31.83pF

12.42nH

198.85pFAC

50

200079.58nH

31.85pF12.42nH

198.85pF

AC

50

2000

79.53nH

31.83pF

12.74nH

203.95pF

AC

50

200079.58nH

31.85pF

12.74nH

203.95pF

m1freq=dB(S(2,2))=-84.294

100.0MHzm2freq=dB(S(1,1))=-79.844

100.0MHz

m3freq=dB(S(2,2))=-3.327

98.00MHzm4freq=dB(S(2,2))=-3.424

102.0MHz

m5freq=dB(S(3,3))=-68.063

100.0MHzm6freq=dB(S(4,4))=-65.485

100.0MHz

10090 110

-80

-70

-60

-50

-40

-30

-20

-10

-90

0

freq, MHz

dB(S

(1,1

))

100.0M-96.91

m2

dB(S

(2,2

))

100.0M-86.88

m1

98.00M-3.327

m3

102.0M-3.424

m4

dB(S

(3,3

))

Readout

m5

dB(S

(4,4

))

Readout

m6

m1freq=dB(S(2,2))=-84.294

100.0MHzm2freq=dB(S(1,1))=-79.844

100.0MHz

m3freq=dB(S(2,2))=-3.327

98.00MHzm4freq=dB(S(2,2))=-3.424

102.0MHz

m5freq=dB(S(3,3))=-68.063

100.0MHzm6freq=dB(S(4,4))=-65.485

100.0MHz

LL11

R=L=79.58 nH

CC30C=31.85 pF

LL10

R=L=12.74 nH

CC29C=203.95 pF

LL8

R=L=12.74 nH

CC27C=203.95 pF

LL5

R=L=79.58 nH

CC24C=31.85 pF

RR4R=2000 Ohm

TermTerm4

Z=50 OhmNum=4

LL7

R=L=79.53 nH

CC26C=31.83 pFRR3R=2000 Ohm

TermTerm3

Z=50 OhmNum=3

CC25C=198.85 pF

LL6

R=L=12.42 nH

RR2R=2000 Ohm

TermTerm2

Z=50 OhmNum=2

TermTerm1

Z=50 OhmNum=1 R

R1R=2000 Ohm

CC21C=31.83 pF

LL4

R=L=79.53 nH

LL2

R=L=12.42 nHC

C23C=198.85 pF

BW=4%


80050 Choose

80050

%5

150

2BW)BW,BWmin(BW

BWBW

12000

2

1

22BW

150

2

1

22BW

121

212

2

11

V

V

V

V

L

V

V

S

V

R

R

R

RRRQ

RRRQ


H 185.3 1

pF796.0 1

04 When

pF795.0

H183.3

04 When

0

12

011

1

01

2

011

1

LLR

QB

CC

QRX

Q

CCR

QB

LLQRX

Q

Va

S

Va

S

04150

8005011

S

V

R

RQ

2ajB

1jX

AC

50

( 100MHz) 50

( 100MHz) 0inZ f

f

2bjBAC

3jX

80050

80050

( 100MHz) 80050

( 100MHz) 0inZ f

f

2000

2ajB

1jX

AC

50

( 100MHz) 50

( 100MHz) 0inZ f

f

80050

H394.20 1

pF127.0 1

247.6 When

pF124.0

H884.19

247.6 When

0

22

023

2

02

2

023

2

LLR

QB

CC

XQRX

Q

CCR

QB

LLXQRX

Q

Vb

LL

Vb

LL

247.612000

8005012

L

V

R

RQ

Four solutions for T-shape matching networks

2bjBAC

3jX

( 100MHz) 80050

( 100MHz) 0inZ f

f

2000

80050

AC

50

2000

19.884 H

0.124pF

3.183 H

0.795pFAC

50

2000

20.394 H

0.127pF3.183 H

0.795pF

AC

50

2000

19.884 H

0.124pF3.185 H

0.796pF

AC

50

2000

3.185 H

0.796pF

20.394 H

0.127pF

m1freq=dB(S(2,2))=-35.421

100.0MHzm2freq=dB(S(1,1))=-37.846

100.0MHz

m3freq=dB(S(2,2))=-3.237

98.00MHzm4freq=dB(S(2,2))=-3.497

102.0MHz

m5freq=dB(S(3,3))=-30.693

100.0MHzm6freq=dB(S(4,4))=-31.955

100.0MHz

10090 110

-30

-20

-10

-40

0

freq, MHz

dB(S

(1,1

))

100.0M-96.91

m2

dB(S

(2,2

))

100.0M-86.88

m1

98.00M-3.327

m3

102.0M-3.424

m4

dB(S

(3,3

))

Readout

m5

dB(S

(4,4

))

Readout

m6

m1freq=dB(S(2,2))=-35.421

100.0MHzm2freq=dB(S(1,1))=-37.846

100.0MHz

m3freq=dB(S(2,2))=-3.237

98.00MHzm4freq=dB(S(2,2))=-3.497

102.0MHz

m5freq=dB(S(3,3))=-30.693

100.0MHzm6freq=dB(S(4,4))=-31.955

100.0MHz

LL11

R=L=3.185 uH

LL9

R=L=3.185 uH

CC30C=0.795 pF

LL10

R=L=19.884 uH

CC28C=0.124 pF

CC29C=0.795 pF

RR4R=2000 Ohm

TermTerm4

Z=50 OhmNum=4

TermTerm3

Z=50 OhmNum=3 R

R3R=2000 Ohm

LL7

R=L=20.394 uH

CC26C=0.127 pF

CC24C=0.127 pF

LL6

R=L=20.394 uH

RR2R=2000 Ohm

TermTerm2

Z=50 OhmNum=2

CC25C=0.795 pF

LL5

R=L=3.183 uH

LL2

R=L=3.183 uH

LL4

R=L=19.884 uH

CC21C=0.124 pFC

C23C=0.795 pF

TermTerm1

Z=50 OhmNum=1 R

R1R=2000 Ohm

BW=4%

Use resonator method for complex load impedance.Splitting into two “L-shape” matching networks.Low Q value but large bandwidth.

Conditions: RL< RV<RS

Goal: Zin(=0)=RS , (=0)=0

Cascaded L-Shape Matching (Low-Q Matching)

Case A:


Conditions: 1/GL> RV>RS

Goal: Zin(=0)=RS ,

(=0)=0

Case B:

LY2jB

2jX

AC

sRL L LY G jB

1jB

0

0

( )

( ) 0in sZ R

1jX

vRvRLZ

2jB

2jX

AC

sR L L LZ R jX

1jB

0

0

( )

( ) 0in sZ R

1jX

vRvR

Splitting into two “L-shape” matching networks for case A

1 2

1 2

1 1

Want to have maximum bandwidth,

so let

2 2BW

21

S V

V L

V S L

S

V

R RQ Q

R R

Q Q Q R R R

Q RR

LZ2jB

2jX

AC

sR L L LZ R jX

1jB

0

0

( )

( ) 0in sZ R

1jX

vRvR

1jB

1jX

AC

sR

0

0

( )

( ) 0in sZ R

LZ2jB

AC

vR

2jX

vR

0

0

( )

( ) 0in vZ R

Splitting into two “L-shape” matching networks for case B

1 2

1 2

11 1

Want to have maximum bandwidth,

so let

2 2BW

2 11

V

S V L

SV

L

S L

RQ Q

R R G

RQ Q Q R

G

Q

R G

LY2jB

2jX

AC

sRL L LY G jB

1jB

0

0

( )

( ) 0in sZ R

1jX

vRvR

1jB

1jX

AC

sR

0

0

( )

( ) 0in sZ R

LY2jB

AC

vR

2jXvR

0

0

( )

( ) 0in vZ R

Example5.4: Design a cascaded L-shape matching network to match load impedance RL = 2000, to a RS = 50, at a frequency of 100 MHz, and to have BW60%?Solution

Select 1/GL> RV>RS

%29.61

150

2000

2

1

2BW

316.23200050

V

S

L

SV

RR

G

RR


2000gR

Matching

Network

AC

50sR

( 100MHz) 50

( 100MHz) 0inZ f

f

1jB

1jX

AC

50

0

0

( ) 50

( ) 0inZ

2jBAC

2jX

0

0

( ) 316.23

( ) 0inZ

316.23

316.23 11

2000

3075.2150

23.31612

L

V

R

RQ nH 11.218

1

pF79.13 1

3075.2 When

pF61.11

nH63.183

3075.2 When

0

11

011

1

01

1

011

1

LLR

QB

CC

QRX

Q

CCR

QB

LLQRX

Q

V

S

V

S

pF181.2 1

H379.1 1

3075.2 When

H161.1

pF836.1

3075.2 When

022

022

2

022

022

2

CC

QRX

LL

BQGB

Q

LLQRX

CCBQGB

Q

V

LL

V

LL

3075.2123.316

20001

12

VLRGQ

1jB

1jX

AC

50

0

0

( ) 50

( ) 0inZ

316.23

2jBAC

2jX

0

0

( ) 316.23

( ) 0inZ

316.23

11

2000

AC

50

2000

1.161 H

1.836pF

183.63nH

11.61pFAC

50

2000

1.379 H

2.181pF183.63nH

11.61pF

AC

50

2000

1.161 H

1.836pF218.11nH

13.79pF

AC

50

2000

218.11nH

13.79pF 2.181pF

1.379 H

Four solutions for Cascaded L-shape matching networks

m1freq=dB(S(2,2))=-70.393

100.0MHzm2freq=dB(S(1,1))=-76.012

100.0MHz

m3freq=dB(S(2,2))=-3.028

74.00MHzm4freq=dB(S(2,2))=-3.068

135.0MHz

m6freq=dB(S(4,4))=-72.208

100.0MHz

60 70 80 90 100 110 120 130 14050 150

-70

-60

-50

-40

-30

-20

-10

-80

0

freq, MHz

dB(S

(1,1

))

100.0M-96.91

m2

dB(S

(2,2

))

100.0M-86.88

m1

74.00M-3.028

m3

135.0M-3.068

m4

dB(S

(3,3

))

Readout

m5

dB(S

(4,4

))

Readout

m6m1freq=dB(S(2,2))=-70.393

100.0MHzm2freq=dB(S(1,1))=-76.012

100.0MHz

m3freq=dB(S(2,2))=-3.028

74.00MHzm4freq=dB(S(2,2))=-3.068

135.0MHz

m5freq=dB(S(3,3))=-65.015

100.0MHzm6freq=dB(S(4,4))=-72.208

100.0MHzm5freq=dB(S(3,3))=-65.015

100.0MHz

CC30C=13.79 pF

LL11

R=L=218.11 nH

LL9

R=L=1.161 uHL

L10

R=L=218.11 nH

CC29C=13.79 pF

RR4R=2000 Ohm

TermTerm4

Z=50 OhmNum=4

CC28C=1.836 pF

TermTerm3

Z=50 OhmNum=3 R

R3R=2000 Ohm

CC26C=2.181 pF

LL7

R=L=1.379 uH

LL5

R=L=1.379 uH

CC24C=2.181 pF

RR2R=2000 Ohm

TermTerm2

Z=50 OhmNum=2

LL6

R=L=183.63 nH

CC25C=11.61 pF

CC21C=1.836 pF

LL4

R=L=1.161 uHC

C23C=11.61 pF

LL2

R=L=183.63 nH

TermTerm1

Z=50 OhmNum=1 R

R1R=2000 Ohm

BW=61%

Lower Q value but larger bandwidth follows the number of L-section increased.

Multiple L-Shape Matching

Conclusions:Lossless matching networks consist of inductances and capacitances but not resistances to avoid power loss.Four kinds of matching techniques including L-shape, -shape, T-shape, and cascaded L-shape networks can be adopted. Generally larger Q value will lead to lower bandwidth.A large range of frequencies (> 1GHz) and circuit size may not be realizable.

L-Shape

Matching

NetworkAC

sR

Multi-section L-Shape networks

L-Shape

Matching

Network

L-Shape

Matching

Network

LZ

Transmission-Line Elements Matching

Single-Stub MatchingEasy fabrication in microstrip or stripline form, where open-circuit stub is preferable. While short-circuit stub is preferable for coax or waveguide.

Lumped elements are not required.

Two adjustable parameters are the distance d and the value of susceptance or reactance provided by the shunt or series stub.

Shunt Stub

Series Stub

Example5.5: Design two single-stub (short circuit) shunt tuning networks to match this load ZL = 60-j 80 to a 50 line, at a frequency of 2 GHz?

1. The normalized load impedance ZL= 1.2-j1.6. 2. SWR circle intersects the 1+jb circle at both points y1 = 1.0+j1.47 y2 = 1.0-j1.47. Reading WTG can obtain: d1= 0.176-0.065=0.11 d2= 0.325-0.065=0.26.3. The stub length for tuning y1 to 1 requires l1 = 0.095,and for tuning y1 to 1 needs l2 = 0.405.

Solution

1. ZL = 60-j 80 at 2 GHz can find R= 60,C=0.995pF.2. Solution 1 is better than solution 2; this is because both d1 and l1 are shorter for solution, which reduces the frequency variation of the match.

Analytic Solution for Shunt Stub

00

00

022

0

00

20

20

002

20

2

2

0

00

;2

;/])[(

/1satisfy tochosen is

])([

))((

)(

)1(

1Let

tan where;)(

)(

is load thefrom lengtha down impedance The

1 impedance Load

ZRZ

X

ZRZR

ZXRZRX

t

ZYGd

tZXRZ

tZXtXZtRB

tZXR

tRG

jBGZ

Y

dttjXRjZ

tjZjXRZZ

dZ

jXRY

Z

LL

LL

LLLL

LL

LLL

LL

L

LL

LL

LLL

L

0for ])(tan[2

1

0for )(tan2

1

stub circuited-shorta for while

0for ])(tan[2

1

0for )(tan2

1

stub circuited-open anfor Then, .satisfy tochosen is

0for )tan(2

1

0for tan2

1

therefore,2

tantan

0

1

0

1

0

1

0

1

1

1

BY

B

BY

Bl

BY

B

BY

Bl

BBl

tt

ttd

ddt

O

O

S

Problem 1Problem 1: Repeat example 5.5 using analytic solution.

Example5.6: Design two single-stub (open circuit) series tuning networks to match this load ZL = 100+j 80 to a 50 line, at a frequency of 2 GHz?

1. The normalized load impedance ZL= 2-j1.6. 2. SWR circle intersects the 1+jx circle at both points z1 = 1.0-j1.33 z2 = 1.0+j1.33. Reading WTG can obtain: d1= 0.328-0.208=0.12 d2= 0.672-0.208=0.463.3. The stub length for tuning z1 to 1 requires l1 = 0.397,and for tuning z1 to 1 needs l2 = 0.103.

Solution

1. ZL = 100+j 80 at 2 GHz can find R= 100,L=6.37nH.

Analytic Solution for Series Stub

00

00

022

0

00

20

20

002

20

2

2

0

00

;2

;/])[(

/1satisfy tochosen is

])([

))((

)(

)1(

1Let

tan where;)(

)(

is load thefrom lengtha down admittance The

1 admittance Load

YGY

B

YGYG

YBGYGB

t

YZRd

tYBGY

tYBtBYtGX

tYBG

tGR

jXRY

Z

dttjBGjY

tjYjBGYY

dY

jBGZ

Y

LL

LL

LLLL

LL

LLL

LL

L

LL

LL

LLL

L

0for ])(tan[2

1

0for )(tan2

1

stub circuited-opena for while

0for ])(tan[2

1

0for )(tan2

1

stub circuited-short anfor Then, .satisfy tochosen is

0for )tan(2

1

0for tan2

1

therefore,2

tantan

0

1

0

1

0

1

0

1

1

1

BY

B

BY

Bl

XZ

X

XZ

Xl

BBl

tt

ttd

ddt

O

S

S


Double-Stub Matching

Variable length of length d between load and stub to have adjustable tuning between load and the first stub.

Shunt stubs are easier to implement in practice than series stubs.

In practice, stub spacing is chosen as /8 or 3/8 and far away 0 or /2 to reduce frequency sensitive.

Original circuit

Equivalent circuit

adjustable tuning

Disadvantage is the double-stub tuner cannot match all load impedances. The shaded region forms a forbidden range of load admittances.

Two possible solutions

b1,b2 and b1’,b2’ with the same distance d.

Example5.7: Design a double-stub (open circuit) shunt tuning networks to match this load ZL = 60-j 80 to a 50 line, at a frequency of 2 GHz?

1. The normalized load impedance YL= 0.3+j0.4 (ZL= 1.2-j1.6). 2. Rotating /8 toward the load (WTL) to construct 1+jb circle can find two values of first stub b1 = 1.314 b’1 = -0.114. 3. Rotating /8 toward the generator (WTG) can obtain y2= 1-j3.38 y’2= 1+j1.38.

Solution

4. The susceptance of the second stubs should be b2 = 3.38 b’2 = -1.38.5. The lengyh of the open-circuited stubs are found as l1 = 0.146, l2 = 0.482,or l1 = 0.204, l2 = 0.350.6.ZL = 60-j 80 at 2 GHz can find R= 60, C=0.995pF.

Analytic Solution for Double Stub

d

Y

t

tYG

dGG

tY

tBtBYt

t

tYG

YY

ZYdt

tjBjBGjY

tYBBjGYY

BjBGY

BBjGY

L

LL

LL

LL

LL

LLL

LL

20

2

2

0

2220

210

2

2

2

0

02

00

10

0102

1

11

sin

10

. givena matchingfor of range thegivesit real, is Since

)1(

)(411[

2

1

derive can }Re{}Let Re{

1,tan where;

)(

)(

admittance withstub second The

stub.first theof esusceptanc theis load, theis where

)(

admittance withstubfirst The

21

01

0

1

022

02

02

220

20

1

or where

)(tan2

1

stub circuited-shorta for while

)(tan2

1

stub circuited-open anFor

)1(

)1(

as found be can stubs both fixed, been hasAfter

BBBB

Yl

Y

Bl

tG

YGtGYGtYB

t

tGYGtYBB

S

O

L

LLL

LLL


Quarter-Wave transformer

It can only match a real load impedance.

The length l= /4 at design frequency f0.

The important characteristics

]2

1[cos

42 BandwidthFraction

d tolertatemaximunfor )2

(2

)sec2

(11

0

0

2

1

0

2

0

02

01

ZZ

ZZ

f

f

ZZ

ZZ

ZZZ

L

L

m

m

mm

mL

L

m

L

Example5.8: Design a quarter-wave matching transformer to match a 10 load to a 50 line? Determine the percent bandwidth for SWR1.5?Solution

%2929.0

]5010

50102

)2.0(1

2.0[cos

42

]2

1[cos

42

2.015.1

15.1

1

1

36.221050

2

1

0

0

2

1

0

01

ZZ

ZZ

f

f

SWR

SWR

ZZZ

L

L

m

m

m

L

Binomial Multi-section MatchingThe passband response of a binomial matching transformer is

optimum to have as flat as possible near the design frequency, and is known as maximally flat.

The important characteristics

])(2

1[cos

42

42

)(2 bandwidth Fraction

2 ;cos2t coefficien reflection Maximun

)4

( 2

at , !)!(

! , 2 where

;)1()( response Binomial

11

0

0

0

00

0

0

22

Nmm

mmNN

m

Nn

L

LN

Nnn

N

n

jnNn

Nj

Af

ff

f

f

A

lfnnN

NC

ZZ

ZZA

CAeCAeA

Binomial Transformer Design

If ZL<Z0, the results should be reversed with Z1 starting at the end.

00

01 ln2)2(22lnZ

ZCC

ZZ

ZZ

Z

Z LNn

NNn

L

LNn

n

n

Example5.9: Design a three-section binomial transformer to match a 50 load to a 100 line? And calculate the bandwidth for m=0.05?Solution

5.5400.4ln2lnln:2

7.7026.4ln2lnln:1

7.91518.4ln2lnln:0

3!2!1

!3 3

!1!2

!3 1

!0!3

!3

%707.0])0433.0

05.0(

2

1[cos

42

])(2

1[cos

42

0433.0ln2

1 2

3For

10

3223

20

3112

10

3001

32

31

30

311

11

0

01N

0

0

ZZ

ZCZZn

ZZ

ZCZZn

ZZ

ZCZZn

CCC

Af

f

Z

Z

ZZ

ZZA

N

LN

LN

LN

Nm

L

L

LN

Using table design for N=3 and ZL/Z0=2(reverse) can find coefficient as 1.8337, 1.4142, and 1.0907.

Chebyshev Multi-section MatchingThe Chebyshev transformer is optimum bandwidth to allow

ripple within the passband response, and is known as equally ripple.

Larger bandwidth than that of binomial matching.

The Chebyshev characteristics

(5.60)or (5.59) as rewritten

be can spolynomial Chebyshev

matching,for cosseclet

);()(2)(

spolynomial Chebyshev

121

m

nnn

x

xTxTxxTxT

n

nn

m

L

L

Lm

m

mNL

LmN

mNjn

Z

Z

f

f

ZZ

ZZ

ZZ

A

TZZ

ZZATA

NN

TAe

1

0

m

01-

0

0

m

1-

0

0

2

1-N

2

N

2

ln2

1formula accurate More

42 bandwidth Fraction

)]2

)/ln((cosh

N

1cosh[

)]1

(coshN

1cosh[sec

evaluate toLet

)(sec

1)(sec)0(

. oddfor , evenfor 2

1 is last term thewhere

)cos(sec)( response Chebyshev

Chebyshev Transformer Design

Example5.10: Design a three-section Chebyshev transformer to match a 100 load to a 50 line, with m=0.05?

Solution

1203

1m3

1

03

0

3

1-

m

01-

3322

; symmetry From

1037.0 )sec(sec32 :cos

0698.0 sec2 :3cos

for (5.60c) Using

44.7 1.408)]05.02

)50/100ln((cosh

3

1cosh[

)]2

)/ln((cosh

N

1cosh[sec

05.0

)cos(sec)cos(sec)(

3For

m

m

m

Lm

m

mj

mNjn

A

A

T

ZZ

A

TAeTAe

N

01.1)180

7.44(2

42 bandwidth Fraction

0.87

466.4)1037.0(27.70ln2lnln :3

7.70

259.4)1037.0(25.57ln2lnln :1

5.57

051.4)0698.0(250ln2lnln :0

0

3

023

2

112

1

001

m

f

f

Z

ZZn

Z

ZZn

Z

ZZn

Using table design for N=3 and ZL/Z0=2 can find coefficient as 1.1475, 1.4142, and 1.7429. So Z1=57.37, Z2=70.71, and Z3=87.15.

Tapered Lines MatchingThe line can be continuously tapered instead of discrete multiple sections to achieve broadband matching.

Changing the type of line can obtain different passband characteristics.

Relation between characteristic impedance

and reflection coefficient

l

Z

Z

dz

de

L

z

zj

2;

)ln(2

1)(

00

2

Three type of tapered line

will be considered here

1) Exponential

2)Triangular

3) Klopfenstein

Exponential TaperThe length (L)of line should be greater than /2(l>) to minimize the mismatch at low frequency.

L

Le

ZZ

dzedz

de

Z

Z

La

eZzZ

LjL

azL

z

zj

L

az

sin

2

)ln(

)(ln2

1)(

)ln(1

)(

0

0

2

0

0

Triangular TaperThe peaks of the triangular taper are lower than the corresponding peaks of the exponential case.

First zero occurs at l=2

2

0

2/for /ln)1)/(2/4(0

2/0for /ln)/(20

]2/

2/sin[)ln(

2

1)(

)(0

2

02

L

Le

Z

Z

eZ

eZzZ

LjL

LzLZZLzLz

LzZZLz

L

L

Klopfenstein TaperFor a maximum reflection coefficient specification in the passband, the Klopfenstein taper yields the shortest matching section (optimum sense).

The impedance taper has steps at z=0 and L, and so does not smoothly join the source and load impedances.

AZ

Z

ZZ

ZZA

ALe

A

AAx

xAxA

xI

xdyyA

yAIAxAx

LzAL

zA

AZZzZ

mL

L

L

Lj

x

L

cosh );ln(

2

1cosh

)(cos)(

1cosh),( ;

2),( ;0),0(

valuesspecial the withfunction Besselmodified theis )(

1;1

)1(),(),(

0);,12

(cosh

ln2

1)(ln

0

00

00

22

0

2

1

0 2

21

200

Example5.11: Design a triangular, exponential, and Klopfenstein tapers to match a 50 load to a 100 line?

Solution

Triangular taper

2

2/for 2/1ln)1)/(2/4(

2/0for 2/1ln)/(2

]2/

2/sin)[

2

1ln(

2

1)(

100

100)( 2

2

L

L

e

ezZ

LzLLzLz

LzLz

Exponential taper

L

LL

a

eZzZ az

sin

2

)21ln()(

)2

1ln(

1

)( 0

Klopfenstein taper

cosh

)(cos)(

)346.0

(cosh)(cosh

346.0)ln(2

1

22

0

101

00

00

A

AL

A

Z

Z

ZZ

ZZ

mm

L

L

L

Bode-Fano CriterionThe criterion gives a theoretical limit on the minimum

reflection magnitude (or optimum result) for an arbitrary matching network

The criterion provide the upper limit of performance to tradeoff among reflection coefficient, bandwidth, and network complexity.

For example, if the response ( as the left hand side of next page) is needed to be synthesized, its function is given by applied the criterion of parallel RC

1ln

1ln

1ln

0

RCdωdω

mm

For a given load, broader bandwidth , higher m.

m 0 unless =o. Thus a perfect match can be achieved only at a finite number of frequencies.

As R and/or C increases, the quality of the match ( and/or m) must decrease. Thus higher-Q circuits are intrinsically harder to match than are lower-Q circuits.

Chapter 5 Impedance Matching and Tuning. Why need Impedance Matching Maximum power is delivered and...

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Transcript of Chapter 5 Impedance Matching and Tuning. Why need Impedance Matching Maximum power is delivered and...