Lecture 11 - Impedance Matching

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010

Impedance Matching

A. Nassiri -ANL

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 2

Impedance Matching

• Impedance matching is important for the following reasons:– Maximum power is delivered when the load is matched to

the line (assuming the generator is matched), and power loss in is minimized.

– Impedance matching sensitive receiver components (antenna, low-noise amplifier, etc.) improves the signal-to-noise ratio of the system.

– Impedance matching in a power distribution network (such as an antenna array feed network will reduce amplitude and phase errors.


Impedance Matching

• Factors that may be important in the selection of aparticular matching network include the following:– Complexity:

• A simpler matching network is usually cheaper, more reliable, and less lossy than a more complex design.

– Bandwidth:• Narrow or broadband.

– Implementation:• According to the technology used the matching network can be

decided on.

– Adjustability:• Required if dealing with variable load.


Matching with lumped elements (L-Networks)

Network for zL outside the 1 + jx circle

Network for zL inside the 1 + jx circle

jx+1

8 possibilities for matching

jX and jB Can be capacitors or

inductors



Analytic Solutions

LLL jXRZ +=o

LL Z

Zz =

jXjB

ZZ Lo +

=

1//

oL ZR >

)/(11

LLo jXRjB

jXZ++

+=

( ) LL

LL

LL

LLo jBRBX

jXRjXjXRjBjXRjXZ

+−+

+=++

++=

11)(( )( )

( ) LL

LLLLo jBRBX

jXRjBRBXjXZ+−

+++−=

11

( ) ( )( )( ) LL

LLLLo jBRBX

BXXXjXBRRZ+−

−++−=

11



( ) oLoLL ZRZXXRB −=−

( ) LLoL XRBZBXX −=−1

( ) ( )LLLo XBRRBXZ −=−1

RHLH ee ℜ=ℜ

RHLH mm ℑ=ℑ

( )LLoL BXXXZBR −+= 1

22

22

LL

LoLLoLL

XRRZXRZRX

B+

−+±=

Solving the 2nd order equation

oL ZR >But

( ) oLoLL ZRZXXRB −=−L

o

L

oL

BRZ

RZX

BX −+=

1



oL ZR <

Analytic Solutions

LLL jXRZ +=

)(11

LLo XXjRjB

Z +++=

( ))(

1)(1

LL

LL

o XXjRXXjRjB

Z +++++

=

( )( )LLoLL jBRXXBZXXjR ++−=++ )(1)(

RHLH ee ℜ=ℜ LoLo RZXXBZ −=+ )(

RHLH mm ℑ=ℑ LoL RBZXX =+ )(

( ) LLoL XRZRX −−±=

( )o

LLo

ZRRZ

B−

±=



Smith Chart Solutions

Example

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

Solution

Ω−= 100200 jZL

Ω−= 12 jzL

oL ZR >

© Dr. Hany Hammad, German University in Cairo

lz

1st Step Ω−= 12 jzL


lz

ly180o

P

O

Q

2st Step

OP==OQ

2.04.0 jyl +=


3rd Step 2.04.0 jyl +=

ly


4th Step

ly

5.04.01 jy +=

5.04.02 jy −=

2.04.0 jyl +=

3.01 jjb =7.02 jjb −=


4th Step

ly

5.04.01 jy +=

2.04.0 jyl +=

3.01 jjb =Sol. 1

2.111 jz −=

2.11 jjx +=


4th Step

ly

5.04.01 jy +=

2.04.0 jyl +=

7.02 jjb −=Sol. 2

2.111 jz +=2.12 jjx −=


Example

3.01 jjb = 2.11 jjx += 7.02 jjb −= 2.12 jjx −=

CjZ

jbjBo

ω=×=1

( ) pFZbC

o

955.0100105002

3.06 =

×==

πω

LjZjxjX o ω=×=

( ) nHxZL o 2.38105002

1002.16 =

××

==πω

Lj

ZjbjB

o ω−

=×=1

( )( ) nHbZL o 5.45

7.0105002100

6 =−×

−=

−=

πω

CjZjxjX o ω

−=×=

( ) ( )pF

zZC

o

65.21002.1105002

116

=×−××

−=

−=

πω


Example


The Quarter Wave Transformer

ljZZljZZZZ

Lo

oLoin β

βtantan

++

=

oL

L

L

L

L

in ZZZ

jZZ

jZZ

ZjZ

lZ

jZl

Z

ZZ ==+

+

=+

+=

21

1

1

11

1

1

2tan

2tan

tan

tan

π

π

β

β

LoZZZ =1

For only real loads

Need to drive the mismatch versus frequency?

θβ tantan == ltLet2πθ == offat tjZZ

tjZZZZL

Lin +

+=

1

11

( ) ( )( ) ( )tjZZZtjZZZ

tjZZZtjZZZ

ZtjZZtjZZZ

ZtjZZtjZZZ

ZZZZ

LoL

LoL

oL

L

oL

L

oin

oin

++++−+

=+

++

−

++

=+−

=Γ111

111

1

11

1

11



( ) ( )( ) ( )LooL

LooL

ZZZjtZZZZZZjtZZZ

+++−+−

=Γ 211

211

LoZZZ =21&

LooL

oL

ZZtjZZZZ2++

−=Γ Γ

( )( ) ( ) ( ) ( )( )( ) 2/122222/122 4

1

4 oLLooLoLLooL

oL

ZZZZtZZZZZZtZZ

ZZ

−+−+=

++

−=Γ

( )( ) ( )22

2 41oL

Lo

oL

oL

ZZZZ

ZZZZ

−+=

−+

θθ 222 sectan11 =+=+ t

( ) ( )( )( ) ( )( )( ) 2/1222/1222 sec41

1

441

1

θoLLooLLooLLo ZZZZZZZZtZZZZ −+=

−+−+=Γ

( )( ) 2

11

1

2tZjZZZZZZ

oL

oL

++−

=Γ

( )( ) 1sec4 22 >− θoLLo ZZZZ



If we are operating close to fo 4ol λ

≈2πθ ≈ 1sec2 >>θ

θcos2 Lo

oL

ZZZZ −

≈Γ

−=∆ mθπθ2

2

Γm= Maximum reflection coefficient magnitude that can be tolerated

2

2 sec2

11

−+=

Γ moL

Lo

m ZZZZ

θ

oL

Lo

m

mm ZZ

ZZ−Γ−

Γ=

2

1cos

2θ

oo

p

p ff

fv

vfl

242 ππβθ ===

πθ om

mff 2

=Quarter

Wavelength at fo



( )

−Γ−

Γ−=−=−=

−=

∆ −

oL

Lo

m

mm

o

m

o

mo

o ZZZZ

ff

fff

ff 2

1cos4242222

2

1

ππθ


Example

Design a single-section quarter-wave matchingtransformer to match a 10 Ω load to a 50 Ω line, at fo =3 GHz. Determine the percent bandwidth for which theSWR ≤ 1.5.Solution

Ω=== 36.22)10)(50(1 LoZZZ

Length 4/λ GHz 3@ 2.015.115.1

11

=+−

=+−

=ΓSWRSWR

m

−Γ−

Γ−=∆ −

oL

Lo

m

m

o ZZZZ

ff 2

1cos42

2

1

π

( )( )( )

%29,29.0501010502

2.01

2.0cos422

1 orff

o

=

−−−=

∆ −

π


The Theory of Small Reflections

• Single-Section Transformer

12

121 ZZ

ZZ+−

=Γ

12 Γ−=Γ

?Γ

2

23 ZZ

ZZ

L

L

+−

=Γ

21

2121

21ZZ

ZT+

=Γ+=

21

1212

21ZZ

ZT+

=Γ+=



+ΓΓ+Γ+Γ=Γ −− θθ jj eTTeTT 42

232112

2321121 ∑

∞

=

−− ΓΓΓ+Γ=0

232

2321121

n

jnnnj eeTT θθ

∑∞

= −=

0 11

n

n

xx 1<xfor

θ

θ

j

j

eeTT

232

232112

1 1 −

−

ΓΓ−Γ

+Γ=Γ 12 Γ−=Γ

21

2121

21ZZ

ZT+

=Γ+=

21

1212

21ZZ

ZT+

=Γ+=

θ

θθ

θ

θ

θ

θ

j

jj

j

j

j

j

eee

ee

ee

231

23

21

23

211

231

23

21

1231

2311

1 1)1(

1)1(

1)1)(1(

−

−−

−

−

−

−

ΓΓ+ΓΓ−+ΓΓ+Γ

=ΓΓ+ΓΓ−

+Γ=ΓΓ+

ΓΓ−Γ++Γ=Γ



θ

θ

j

j

ee

231

231

1 −

−

ΓΓ+Γ+Γ

=Γ

If the discontinuities between the impedances Z1, Z2, and Z2, ZL aresmall, then |Γ1 Γ3| <<1.

θje 231

−Γ+Γ≅Γ


Multi-section Transformer

θje 231

−Γ+Γ≅Γo

o

ZZZZ

+−

=Γ1

10

on

nnn ZZ

ZZ+−

=Γ+

+

1

1

NL

NLN ZZ

ZZ+−

=Γ

Assume that the transformer is symmetrical

NΓ=Γ0 11 −Γ=Γ N 22 −Γ=Γ N etc.

( ) θθθθθθ jNN

NjN

NjN

jj eeeee 2)1(21

)2(22

42

210

−−−−

−−−

−− Γ+Γ+Γ++Γ+Γ+Γ≅Γ=Γ

( )( ) ( ) ( ) ( )

++Γ++Γ++Γ=Γ

Γ+Γ+Γ++Γ+Γ+Γ=Γ−−−−−−−

−−−−−−−

θθθθθ

θθθθθ

θ

θ)2(24

2)1(22

12

0

20

)1(21

)2(22

42

210

1 NjjNjjjN

jNNjNjjj

eeeeeeeeee


Multi-section Transformer

( ) ( ) ( ) ( )[ ]++Γ++Γ++Γ=Γ −−−−−−−− θθθθθθθθ )4()4(2

)2()2(10

NjNjNjNjjNjNjN eeeeeee

( ) ( ) ( )

Γ++−Γ++−Γ+Γ=Γ −

2/10 212cos2coscos2 Nn

jN nNNNe θθθθ θ

For N even

( ) ( ) ( )

Γ++−Γ++−Γ+Γ=Γ −

− θθθθθ θ cos212cos2coscos2 2/)1(10 Nn

jN nNNNe

For N oddFinite Fourier Cosine Series

By choosing the Γns and enough sections (N) we can achieve the required response.

Binomial MultisectionMatching Transformers

Chebyshev MultisectionMatching Transformers

Lecture 11 - Impedance Matching

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Transcript of Lecture 11 - Impedance Matching