Lect6_ImpedanceMatching

8/7/2019 Lect6_ImpedanceMatching

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431-618 ( 2011)

Passive ComponentDesign & Simulation

Dr. Bo YangLecture 6: Impedance Matching


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At a glance

Last Class: Transmission Lines & SmithChartThis Class: Impedance Matching

B. Yang 2431-618, the University of Melbourne


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At Z Smith chart, curve from point Ato pint C indicates impedance

transformation from resistance 25 Ohm to inductive impedance (25

+j25) Ohm

At Y Smith chart, curve from point C to point D indicates admittance transformation from inductive admittance (20 - j20) mS to

conductance 20 mS (50 Ohm) 0 . 5

1 .

0

2 . 0

3 . 0

- 3

. 0

- 2. 0 - 0 .5

1 -10.51.02.05.0

B / Y = O

- 0. 2

B / Y O = 0 .2

x

C

D

G /

Y O = 0 . 2

-1 .0

0. 5 2 .0

3 . 0

-

3 . 0

- 2 . 0

- 1

. 0

- 0 . 5

0.5 1.0 2.0 5.0 r

x

X / Z O = 0 .2

X / Z O = - 0

. 2

-1

C

A1

R /

Z O

= 0

. 2

1 .0

431-618, the University of Melbourne 3B. Yang


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C

D A Y Smith chart

provides transformation from point C to point D

At combined Z-Y

Smith chart:

Z Smith chart provides

transformation from

point A to point C



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5

Load MatchingWhat if the load cannot be made equal to Z o for some otherreasons? Then, we need to build a matching network so that thesource effectively sees a match load.

0

LZ

sP

0Z

M

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



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6

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 of thepower would be dissipated in the matching network. We wantto use only lossless elements such as inductors andcapacitors so we dont dissipate any power in the matchingnetwork



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7

Matching Example

Add positiveimaginary admittanceto get to z=1+j0

ImpedanceChart

nH80L

LMHz1002 j500.1 j0.1 j jx

pF16 100

nH80



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8

Matching ExampleThis solution wouldhave also worked

ImpedanceChart

pF32

100nH160



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Matching Bandwidth

50 60 70 80 90 100 110 120 130 140 15040

35

30

25

20

15

10

5

0

Frequency (MHz)

R e

f l e c t

i o n

C o e

f f i c i e n

t ( d B )

pF16100

nH80

50 MHz

150 MHz

ImpedanceChart

Because the inductor and capacitorimpedances change with frequency, thematch works over a narrow frequency

range 431-618, the University of Melbourne 9B. Yang


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10

Single Stub Tuner

Match 100 load to a 50 system at 100MHzusing two transmission lines connected in parallel

0

100t 1

t 2



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11

Single Stub TunerAdding length toCable 1 rotates thereflection coefficientclockwise.

Enough cable isadded so that thereflection coefficientreaches the mirrorimage circle

nS49.3

MHz1003602251

1

1t

t

ImpedanceChart

2511



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12

Single Stub Tuner

The stub is going tobe added in parallelso flip to theadmittance chart.

The stub has to adda normalizedadmittance of 0.7 tobring the trajectory tothe center of the

Smith Chart

AdmittanceChart431-618, the University of Melbourne 12B. Yang


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13

Single Stub TunerAn open stub of zerolength has anadmittance=j0.0

By adding enough

cable to the openstub, the admittanceof the stub willincrease.

70 degrees will give

the open stub anadmittance of j0.7

AdmittanceChart

702

nS97.0

MHz100360270

2

2t

t



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14

Single Stub Tuner

AdmittanceChart

01003.5nS

0.97nS



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15

Single Stub Tuner

AdmittanceChart

01001.5nS

1.4nS

This solution wouldhave worked as well.



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50 60 70 80 90 100 110 120 130 140 1540

35

30

25

20

15

10

5

0

Frequency (MHz)

R e

f l e c t

i o n

C o e

f f i c i e n

t ( d B )

01003.5nS

0.97nS

Single Stub Tuner Matching Bandwidth

Because the cable phase changeslinearly with frequency, the matchworks over a narrow frequency range

50 MHz

150 MHz

Impedance

Chart431-618, the University of Melbourne 16B. Yang


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X 1

X 2

L-transformer

R1 Z 1 X 1 R2 Z 2

X 2

Equivalence when Z 1 = Z 2 : 21

21

12

12

12

1

211

22 X R X R

j X R

X R jX R

221 1 Q R R 221 1 Q X X

where Q = R 1 / X 1 = X 2 / R 2 - quality factor equal for series and parallel circuits

Impedance parallel and series circuits



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For conjugate matching with reactance compensation :

221 1 Q X X 2

21 1 Q R R

.1 /

/

21

22

11

R RQ

Q R X

Q R X

X 1

X 2

R2 Z in = R 1

Input impedance Z in will be resistive and equal to R 1 when :

where Q = R 1 / X 1 = X 2 / R 2 - quality factor equal for series and parallel circuits



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Matching with lumped elements

Bandwidth properties

Two L-type matching circuits

2211

2

1

22

11

/ 1

/ 1

/

RQC

Q R L

R

RQ

RQ L

RQC

L2

1 C 1 R2 R1

C 2

R2 L1

a ). b).

Resistance R 1connected to

parallel reactive element must be

greater than resistance R 2 connected to

series reactive element

Z in

R1

0.707 R1

1 f / f 0 0

2 f 0 0 / 2Q f f



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- transformer

for each L-transformer, resistances R 1 and R 2 are transformed to some intermediate resistance R 0 with value of R 0 < (R 1, R 2 )

X 1

X 3

X 2

X 3

X 2 X 1

X 1

X 3

R0 R1 X 2 R2

X 3

X 1

X 3 R0 R1

X 2

R2 X 3

Connection of two L-transformers

T- transformer

for same resistances R 1 and R 2 , T- and -transformers have better filtering properties, but narrower bandwidth compared with single L-transformer



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-type matching circuits

111 / RQC 222 / RQC

212113 1 / QQQ R L

1 1 211

22 Q

R

RQ 1

2

121

R

RQ

L3

R1 C 1 R2 C 2

widely used as matching circuit

useful for interstage matching when active device input and

output capacitances can be easily

incorporated inside matching circuit

provides significant level of harmonic suppression



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111 / Q R L 222 / RQC

221223 1 / QQQ R L ,

1 1 222

11 Q

R

RQ 1

1

222

R

RQ


-type matching circuits

111 / Q R L

222 / RQC

212223 / 1 QQ RQC

1 1 211

22 Q

R

RQ 1

2

121

R

RQ

L3

R1 L1 R2 C 2

C 3

R1 L1 R2 C 2



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111 / 1 Q RC 222 RQ L

222123 1 / Q RQQC

1 1 221

21 Q

R RQ 1

2

122

R R

Q

L2

R1 R2 C 3

C 1

T-type matching circuits

widely used as input,interstage and output matching

circuits in high power amplifiers

can incorporate active device lead and bondwire

inductances within matching circuit

provides significant level of harmonic suppression



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111 / 1 Q RC 222 RQ L

212223 / 1 QQQ R L ,

1 1 212

12 Q R

RQ 1

1

221

R

RQ

L2

R1 L3 R2

C 1


T-type matching circuits

L2

R1 R2 C 3

L1 111 RQ L 222 RQ L

222213 1 / Q RQQC

1 1 221

21 Q R

RQ 1

2

122 R

RQ



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Matching with lumped elements Matching design example

L3 + L in

Rsource = R1

Rin C 3

L2 C in

Z in

C 1

L1 C 2 R2 R3

132-174 MHz 150 W MOSFET power amplifier:

three-section input matching

MHz152174132c f

2.1 9.0in j Z

Q = 152/(174 - 132) = 3.6

in

3

3

2

2

1 R R

R R

R R

For R in = 0.9 Ohm and R 1 = 50 Ohm: R 3 = 3.5 Ohm, R 2 = 13 Ohm

Q = 1.7

Two low-pass and one high-pass L-sections



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Matching with transmission lines

For conjugate matching with reactance compensation when Z S = Z in * :

Transmission-line transformer

For quarter-wave transmission line with = 90 :

tan tan

L0

0L0in jZ Z

jZ Z Z Z

Z 0,

Z L Z in Z S

L20in / Z Z Z

L

j j

Z Z

2exp 12exp 1

L

L

0

in

Impedance at input of loaded transmission line:

L

L

0

L

1 1

Z Z

Input impedance for loaded transmission line with electrical length of , normalized to its characteristic impedance Z

0 , can be found by rotating this impedance point clockwise by

2 around Smith chart center point with radius L

SL

2S

2SL

2L

2LS

0 R R

X R R X R R Z

LSLS

LS0

1

tan

R X X R R R

Z



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Matching with transmission lines For pure resistive source

impedance Z S = R S :

Z L Z S

Z 01, /8 Z 02, /4 Z 03, /8

0 tan tan 1 2L2 L2 020L R X Z Z X

2L

2LL0 X R Z Z

L0

0LS X Z

Z R R

For electrical length = 45

Any load impedance can be transformed into real source impedance using /8-transformer whose impedance is equal to magnitude of load impedance

To match any source impedance Z S and load impedance Z L, matching circuit can be designed with two /8-transformers

and one /4-transformer

L Z 0,

< 90 C

Z 0, < 90

tan 0 Z

L

Lumped and transmission line single-frequency equivalence

0

tan

Z C



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Conjugate matching: L-type transformer

Second implicit equation : numerical or graphical solution


R1 Z 1C R2 Z in

Z 0,

21

21

1212

12

1

211inin X R

X R j X R

X R jX R

2in1 1 Q R R 2in1 1 Q X X Real and imaginary parts of

tan tan

20

020in jR Z

jZ R Z Z

2220

2

220in tan

tan 1

R Z R Z R

2220

22

20

0intan

tan

R Z

R Z Z X

1 / RQC

2202

2

22

2

0

2

2

0

2

1

sin / cos

cos sin 1

Z R

Z R

R Z

R R

where X 1 = -1/ C

Matching for any ratio of

R 1 /R 2



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Matching design example 470-860 MHz 150 W LDMOSFET power amplifier:

three-section input matching

Q = 635/(860 - 470) = 1.63

in

3

3

2

2

1 R R

R R

R R

For R in = 1.7 Ohm and R 1 = 50 Ohm: R 3 = 5.25 Ohm, R 2 = 16.2 Ohm

Q = 1.2

For Z 01= Z 02 = Z 03 = 50 Ohm 1 = 30

, 2 = 7.5 , 3 = 2.4


MHz635860470c f

3.1 7.1in j Z

Lin Rsource = R1

R in C 3 Z inC 2 R2 R3

Z 03, 3

in

C 1

Z 02, 2 Z 01, 1

For 1 =

2 =

3 = 30

Z01 = 50 Ohm Z 02 = 15 7 Ohm Z 03 = 5 1 Ohm

431-618 the University of Melbourne 29B Yang

Lect6_ImpedanceMatching

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