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.

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 3

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.

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 4

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

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 5

Matching with lumped elements (L-Networks)

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

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 6

Matching with lumped elements (L-Networks)

( ) 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

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 7

Matching with lumped elements (L-Networks)

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−

±=

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 8

Matching with lumped elements (L-Networks)

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

© Dr. Hany Hammad, German University in Cairo

lz

ly180o

P

O

Q

2st Step

OP==OQ

2.04.0 jyl +=

© Dr. Hany Hammad, German University in Cairo

3rd Step 2.04.0 jyl +=

ly

© Dr. Hany Hammad, German University in Cairo

4th Step

ly

5.04.01 jy +=

5.04.02 jy −=

2.04.0 jyl +=

3.01 jjb =7.02 jjb −=

© Dr. Hany Hammad, German University in Cairo

4th Step

ly

5.04.01 jy +=

2.04.0 jyl +=

3.01 jjb =Sol. 1

2.111 jz −=

2.11 jjx +=

© Dr. Hany Hammad, German University in Cairo

4th Step

ly

5.04.01 jy +=

2.04.0 jyl +=

7.02 jjb −=Sol. 2

2.111 jz +=2.12 jjx −=

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 15

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

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 16

Example

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 17

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

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 18

The Quarter Wave Transformer

( ) ( )( ) ( )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

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 19

The Quarter Wave Transformer

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

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 20

The Quarter Wave Transformer

( )

−Γ−

Γ−=−=−=

−=

∆ −

oL

Lo

m

mm

o

m

o

mo

o ZZZZ

ff

fff

ff 2

1cos4242222

2

1

ππθ

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 21

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

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 22

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+=

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 23

The Theory of Small Reflections

+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

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 24

The Theory of Small Reflections

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

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 25

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

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 26

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