chapter 4 –impedance matching - universiti malaysia...

Antenna & Propagation Impedance Matching

1

Chapter 4 – Impedance

Matching

1 – Quarter-wave transformer, series

section transformer

2 – Stub matching, lumped element

networks, feed point location

3 – Gamma match

4 – Delta- and T-match, Baluns


2

REVISION

1 – 2-port network

2 – Smith Chart


3

2-Port Network Representation

• Six ways to represent a two-port network in terms of V and I

at each port:

– Z-matrix open-circuit impedance

– Y-matrix short-circuit admittance

– ABCD-matrix chain or transmission parameters

– B-matrix inverse transmission parameters

– H-matrix hybrid parameters

– G-matrix inverse hybrid parameters

V2V1

I1 I2Two-Port

Network


4

2 -Port Network Representation

• 7th way to represent a two-port network in terms of waves

entering and leaving each port:

– S-matrix scattering parameters

• In general, there could be n ports in a network:

V2V1

I1 I2Two-Port

Network

1

2

3

n

n-port

network


5

[S] for Two-Port Network

Definition of normalized voltage waves:

1

1

11 i

o

i PZ

Va ==

1

1

11 r

o

r PZ

Vb ==

Incident wave:

Reflected wave:

~

a1

b1 Two-Port

NetworkZo1

Scattering parameters are then defined as:

ZLZo2

a2

b2

2

2

22 i

o

i PZ

Va ==

2

2

22 r

o

r PZ

Vb ==

2121111 aSaSb +=

2221212 aSaSb +=

=

2

1

2221

1211

2

1

a

a

SS

SS

b

bor

outputs inputs

[ ] [ ][ ]aSb =or

1 2


6

S11 for a One-Port Network

Measured S11 of an 800 MHz square microstrip

patch antenna on a 1.6 mm FR4 substrate

Antenna

Bandwidth

( RL=13 dB)


7

Measured S21 of a 542 MHz hair-pin bandpass

filter on a 1.6 mm FR4 substrate

S21 for a One-Port Network


8

Smith Chart Impedance or

Admittance


9

Smith Chart: Exercises

1. Use as a Z-chart, locate s/c, o/c, Zo , 1+ j1

2. Use as a Y-chart, locate s/c, o/c, Yo , 1+ j1

3. Move towards generator l/l

4. Move towards load l/l

5. Read VSWR, |G |, Vmax , Vmin

6. Read attenuation


10

IMPEDANCE MATCHING

1 – Introduction

2 – Importance of Impedance Matching


11

INTRODUCTION

• The operation of an antenna system over a frequency range is

not completely dependent upon the frequency response of

the antenna element itself but rather on the frequency

characteristics of the transmission line-antenna element

combination.

• In practice, the characteristic impedance of the transmission

line is usually real whereas that of the antenna element is

complex.

• Also the variation of each as a function of frequency is not the

same.

• The efficient coupling-matching networks must be designed

which attempt to couple-match the characteristics of the two

devices over the desired frequency range.


12

Feed-Point Impedance: Za

• Za = antenna impedance at its feed-point.

• Za is complex generally.

aaa jXRZ +=Za


13

Importance of Impedance

Matching

• Increased power throughput (e.g. maximum power transfer)

• Increased power handling capability in a transmission line

(due to reduced VSWR)

• Reduced effects on impedance matching sensitive circuits

(e.g. frequency pulling effect on the signal source)

• With "controlled mismatch", an RF amplifier can operate with

minimum noise generation (i.e. minimum noise figure)


14

Concept of maximum power

transferZo

ZLV

iI V

L

LoL

iLLL Z

ZZ

VZIIVP

2

2

2

1

2

1

2

1

+===

Power maximum whence ZL = Zo

Power deliver at ZL is

In lump circuitPL

ZL

Zo


15

continueIn transmission line

oL

oL

ZZ

ZZ

+−

=ρ

No reflection whence ZL = Zo , hence 0=ρ

The load ZL can be matched as long as ZL not equal to zero (short-

circuit) or infinity (open-circuit)

The important parameter is reflection coefficient


16

Features of Impedance Matching

Networks• Complexity

– use the simplest --> cheap, low loss

• Bandwidth

– perfect match usually at a single frequency

– wider bandwidth increases complexity

• Implementation

– select matching components to suit an application:

– lumped-element (L, C, R)

– distributed-element (transmission line, waveguides)

• Adjustability

– for loads with variable impedance


17

MATCHING METHODS

1 – calculations

2 – smith chart


18

Matching with lumped elements

The simplest matching network is an L-section using two reactive elements

jX

jB ZL

Zo

jX

jB ZL

Zo

Configuration 1

Whence RL>Zo

Configuration 2

Whence RL<Zo

ZL=RL+jXL


19

continue

Configuration 1Configuration 2

If the load impedance

(normalized) lies in unity

circle, configuration 1 is

used.Otherwise

configuration 2 is used.

The reactive elements are

either inductors or

capacitors. So there are 8

possibilities for matching

circuit for various load

impedances.

Matching by lumped elements

are possible for frequency

below 1 GHz or for higher

frequency in integrated

circuit(MIC, MEM).


20

Impedances for serial lumped elements

Serial circuit

Reactance relationship values

+ve X=2πfL L=X/(2πf)-ve X=1/(2πfC) C=1/(2πfX)

L C

RR


21

Impedances for parallel lumped elements

Parallel circuit

Susceptance relationship values

+ve B=2πfC C=B/(2πf)-ve B=1/(2πfL) L=1/(2πfB)

LC

RR


22

Lumped elements for microwave integrated

circuit

Spiral inductorLoop inductor

Interdigital

gap capacitor

Planar resistor Chip resistor

Lossy film

Metal-insulator-

metal capacitor Chip capacitor

εr

Dielectric

εr


23

Matching by calculation for

configuration 1jX

jB ZL

Zo

( )LLo

jXRjBjXZ

+++=

/1

1

For matching, the total impedance of L-section plus ZL should

equal to Zo,thus

Rearranging and separating into real and imaginary parts gives us

( ) oLoLL ZRZXXRB −=−

( ) LLoL XRBZBXX −=−1

*

**


24

continue

22

22/

LL

LoLLoLL

XR

RZXRZRXB

+

−+±=

Since RL>Zo, then argument of the second root is always positive,

the series reactance can be found as

L

o

L

oL

BR

Z

R

ZX

BX −+=

1

Note that two solution for B are possible either positive or negative

+ve inductor

-ve capacitor

+ve capacitor

-ve inductor

Solving for X from simultaneous equations (*) and (**) and substitute X

in (**) for B, we obtain


25

Matching by calculation for

configuration 2

( )LLo XXjRjB

Z +++=

11

For matching, the total impedance of L-section plus ZL should equal to

1/Zo, thus

Rearranging and separating into real and imaginary parts gives us

( ) LoLo RZXXBZ −=+

( ) LoL RBZXX =+

*

**

jX

jB ZL

Zo


26

continueSolving for X and B from simultaneous equations (*) and (**) , we obtain

( )o

LLo

Z

RRZB

/−±=

Since RL<Zo,the argument of the square roots are always positive,

again two solution for X and B are possible either positive or

negative

( ) LLoL XRZRX −−±=+ve inductor

-ve capacitor

+ve capacitor

-ve inductor


27

parallel capacitance

parallel inductanceserial capacitance

serial inductance

(-ve)

(-ve)

(+ve)

(+ve)


28

C1

L1

C2

L2

50ΩC2

L!

10Ω

C1L

2

10Ω 50Ω

Serial C

Parallel

L

Serial

L

Parallel

C

Matching using lumped

components


29

ExampleDesign an L-section matching network to match a series RC load with an

impedance ZL=200-j100 W, to a 100 W line, at a frequency of 500 MHz.

Solution

Normalized ZL we

have : ZL= 2-j1

ZL= 2-j1 Parallel C

(+j0.3)Serial L

(j1.2)

Serial C

(-j1.2)

Parallel L

(-j0.7)

Solution 1

Solution 2


30

continue

ZL=200-j100

2.61pF

46.1nH

200-j1000.92pF

38.8nHpF

Zf

bC

o

92.02

==π

nHf

ZxL o 8.38

2==

π

pFZfx

Co

61.22

1=

−=

π

nHbf

ZL o 1.46

2=

−=

πReflection coefficient

0

0.2

0.4

0.6

0.8

1

1.2

0 0.5 1 1.5

freq (GHz)

reflection c

oeffic

ient

solution 1

solution 2Solution 2 seems to be

better matched at higher

frequency


31

MATCHING TECHNIQUES

1 – Discrete lumped-element

2 – Transmission Line


32

Impedances for serial lumped elements

Serial circuit

Reactance relationship values

+ve X=2πfL L=X/(2πf)-ve X=1/(2πfC) C=1/(2πfX)

L C

RR


33

Impedances for parallel lumped elements

Parallel circuit

Susceptance relationship values

+ve B=2πfC C=B/(2πf)-ve B=1/(2πfL) L=1/(2πfB)

LC

RR


34

Lumped elements for

microwave integrated circuit

Spiral inductorLoop inductor

Interdigital

gap capacitor

Planar resistor Chip resistor

Lossy film

Metal-insulator-metal

capacitor Chip capacitor

εr

Dielectric

εr


35

Resistive L-Section (Using

Resistors)• Advantage: Broadband

• Disadvantage: Very lossy

35

Matched conditions:

(a)

(b)

Solving (a) and (b) to give R1 and R2 :

R1

R2 Ro2Ro1

21 oo RR >

(a) (b)

)//( 2211 oo RRRR +=

)//( 1122 oo RRRR +=

)( 2111 ooo RRRR −=

21

122

oo

oo

RR

RRR

−=


36

Resistive L-Section

(Using Resistors)• Advantage: Broadband

• Disadvantage: Very lossy

Attenuation:R1

R2 Ro2

Ro1

21 oo RR >

(a) (b)

)//(

//

2211

22

oo

o

in

out

RRRR

RR

V

V

++=

Vout~Vin


37

Resistive L-Section: Example

37

Design a broadband resistive L-section matching network to match a 75 ΩTV antenna output to a 50 Ω transmission line. Calculate the attenuation of

the matching network.

Solution:

Ro1 = 75 Ω, Ro2 = 50 Ω

Substituting these values into previous equations, we get:

R1 = 43.3 Ω, R2 = 86.6 Ω

= 0.2113 = –13.5 dB

Balun75 Ω-50 Ω matching

network

75 Ωtxn line

50 Ωtxn line

7.313.4375

7.31

++=

in

out

V

V


38

Reactive L-Section (Using L & C)

• Advantages: Low loss, simplicity in design

• Disadvantages: Narrow-band, fixed Q

Matched conditions:

(a)

(b)

Solving (a) and (b) to give Xs and Xp :

jXs

jXp RpRs

sp RR >

(a) (b)

)//( ppss RjXjXR +=

)//( sspp jXRjXR +=

1−=s

p

R

RQ

1. −==s

p

sssR

RRRQX

Xs & Xp: opposite sign

1−

==

s

p

pp

p

R

R

R

Q

RX Xs = +ve for L, -ve for C

Xp = -ve for C, +ve for L

38


39

Reactive LC L-Section: Example

Design an L-section matching network with L and C to match a 50 Ωsource to a 600 Ω load for maximum power transfer at 400 MHz.

Give two solutions.

Solution:

Rs = 50 Ω, Rp = 600 Ω

Substituting these values into previous equations, we get:

Xs = Q.Rs = 166 Ω

Xp = Rp/Q = 181 Ω

317.311150

6001 ==−=−=

s

p

R

RQ


40

Reactive LC L-Section: Example (cont'd)

Solution 1:

Xs = +166 Ω (inductive), Xp = –181 Ω (capacitive)

Solution 2:

Xs = –166 Ω (capacitive), Xp = +181 Ω (inductive)

nHf

XL ss 66

10*400*2

166

2 6===

ππ

pFXf

Cp

p 2.2181*10*400*2

1

2

16

===ππ

nHf

XL

p

p 7210*400*2

181

2 6===

ππ

pFXf

Cs

s 4.2166*10*400*2

1

2

16

===ππ


41

Reactive LC L-Section: Example (cont'd)

Solution 1:

Xs = +166 Ω (inductive), Xp = –181 Ω (capacitive)

Solution 2:

Xs = –166 Ω (capacitive), Xp = +181 Ω (inductive)


42

L-Section for Complex Impedances

• If Rs has a reactance jX', simply introduce a series with an

equal but opposite-sign reactance (–jX').

jXs

jXp RpRs

jX'–jX'

extra reactances

• The –jX' reactance is then combined with jXs to give its

final value.

• Similar treatment can be applied to Rp by introducing a

parallel jX' and a parallel –jX'.

42


43

T-Section

• Advantages: R1 can be < or > R2Variable Q (higher than L-section)

• Disadvantages: More L C elements

Method:(a) Introduce a hypothetical resistance level Rh at jX2, such that

Rh > R1 and Rh > R2(Rh determines the Q of the matching network)

(b) Split jX2 into jX2' and jX2" (in parallel)

(c) Treat the T-section as two L-sections (L-1 and L-2)

NOTE: Q of L-1 section > Q of 2-element L-section (as Rh > R2)

jX1

jX2' R2R1

Rh

jX3

jX2 jX2"L-1 L-2

43


44

T-Section: Q Values

44

Using the resistance values of the previous 2-element L-section:

(a) Q of original L-section:

(b) Q of T-section if Rh = 5050 Ω (Note: Q1 > QL)

L-1 section:

L-2 section:

317.3150

6001

1

2 =−=−=R

RQL

10150

50501

1

1 =−=−=R

RQ h

723.21600

50501

2

2 =−=−=R

RQ h

50 Ω 600 Ω


45

T-Section: Design Example

45

Design a T-section LC matching network to match R1 = 50 Ωand R2 = 600 Ω at 400 MHz, and Q associated with R1 is 10.

(a) For L-1 network, select Q1 = 10.

(b)

(c)

(d)

Ω=+=+= 5050)110(50)1( 22

11 QRRh

50 Ω 600 Ω

Ω=== 50510

5050

1

'

2Q

RX h

Ω=== 50010*50111 QRX

–ve for C (arbitrarily selected)

+ve for L (opposite of X2' )


46


46

(a) For L-2 network,

(b)

(c)

(d)

(e)

(f)

(g)

50 Ω 600 Ω

Ω=== 1854723.2

5050"

2

2Q

RX h

Ω=== 1634723.2*600223 QRX

723.21600

50501

2

2 =−=−=R

RQ h

Ω−=+−

−=

+== 694

1854505

1854*505

"'

"'."//'

22

22222

XX

XXXXX

+ve for L (arbitrarily selected)

–ve for C (opposite of X2'' )

–ve means C

nHf

XLX 199

10*400*2

500

2 6

111 ===⇒

ππpF

XfCX 57.0

694*10*400*2

1

2

16

2

22 ===⇒ππ

pFXf

CX 24.01634*10*400*2

1

2

16

3

33 ===⇒ππ


47


(cont'd)

47

50 Ω 600 Ω

nHL 1991 = pFC 57.02 = pFC 24.03 =

"AUTO-TRANSFORMER"

another possible solution


48

ππππ-Section

• Advantages: R1 can be < or > R2Variable Q (higher than L-section)

• Disadvantage: More L C elements

Method:(a) Introduce a hypothetical resistance level Rh at jX2, such that

Rh < R1 and Rh < R2(Rh determines the Q of the matching network)

(b) Split jX2 into jX2' and jX2" (in series)

(c) Treat the π-section as two L-sections (L-1 and L-2)NOTE: Q of L-1 section > Q of 2-element L-section (as Rh < R2)

jX1

jX2'

R2R1

Rh

jX3

jX2

jX2"

L-1 L-2

48


49

ππππ-Section: Design Example

Design a π-section LC matching network to match R1 = 50 Ωand R2 = 600 Ω at 400 MHz, and Q associated with R1 is 10.

(a) For L-1 network, select Q1 = 10.

(b)

(c)

(d)

Ω=+

=+

= 495.0110

50

122

1

1

Q

RRh

50 Ω 600 Ω

Ω=== 0.510

50

1

11

Q

RX

Ω=== 95.410*495.01

'

2 QRX h +ve for L (arbitrarily selected)

–ve for C (opposite of X2' )

49


50

ππππ-Section: Design Example (cont'd)

(a) For L-2 network,

(b)

(c)

(d)

(e)

(f)

(g)

50 Ω 600 Ω

Ω=== 23.1780.34*495.0" 22 QRX h

Ω=== 24.1780.34

600

2

23

Q

RX

80.341495.0

60012

2 =−=−=hR

RQ

Ω=+=+= 18.2223.1795.4"' 222 XXX

+ve for L (arbitrarily selected)

–ve for C (opposite of X2'' )

+ve means L

nHf

XLX 83.8

10*400*2

18.22

2 6

222 ===⇒

ππ

pFXf

CX 58.795*10*400*2

1

2

16

1

11 ===⇒ππ

pFXf

CX 08.2324.17*10*400*2

1

2

16

3

33 ===⇒ππ

50


51

ππππ-Section: Design Example (cont'd)

50 Ω 600 Ω

pFC 58.791 = nHL 83.82 = pFC 08.233 =

This is just one of the four possible solutions.

51


52

Common Transmission-Line

Applications

• Inductor and capacitor

• Quarter-wave transformer

• Single-stub tuner

• Double-stub tuner

• Triple-stub tuner

• [A good reference for the Double- and Triple-stub tuners is:

R.E.Collin, Foundations for Microwave Engineering, McGraw

Hill]

52


53

Transmission Line Inductor

• The feed-point impedance of a terminated transmission line

is:

• If the load impedance is a short-circuit, ZL = 0, then

• Thus, the terminated transmission line behaves like an

inductor (inductance = L) if bl < p/2.

53

)ltan(jZZ

)ltan(jZZZZ

Lo

oLoin β

β++

=

Lj)ltan(jZZ oin ωβ ==

ωβ )ltan(Z

L o=


54

Example

What is the equivalent inductance at 1 GHz at the feed-

point of a 50 W, l/8 transmission line which is

terminated with a short-circuit?

nH.)(

.tan)ltan(Z

L o 067102

8

250

9=

==π

λλπ

ωβ ZL

l

Zo, β

Zin

LZin


55

Transmission Line Capacitor


is:

• If the load impedance is an open-circuit, ZL = , then

• Thus, the terminated transmission line behaves like a

capacitor (capacitance = C) if bl < p/2.

55

)ltan(jZZ

)ltan(jZZZZ

Lo

oLoin β

β++

=

Cj)ltan(j

ZZ oin ωβ

1==

oZ

)ltan(C

ωβ

=

∞


56

Example

What is the equivalent capacitance at 1 GHz at the feed-point of

a 50 W, l/8 transmission line which is terminated with an open-

circuit?

pFZ

lC

o

18.350)10(2

8.

2tan

)tan(9

=

==π

λλπ

ωβ

o/c

l

Zo, β

Zin

CZin


57

Quarter-Wave Transformer


is:

• When l = l/4, i.e. bl = p/2, then

• Thus, we can transform the load impedance ZL to another

value Zin by designing the l/4 transmission line with a

suitable characteristic impedance Zo:

57

)ltan(jZZ

)ltan(jZZZZ

Lo

oLoin β

β++

=

L

oin

Z

ZZ

2

=

Lino ZZZ =

Ω=== 2617550 .xZZZ Lino


58

Example

Match a 75 W load to a 50 W source at 300 MHz using a

l/4 transformer.

The physical length of the transmission line is

determined by the wavelength, which depends on the

velocity factor of the transmission line.

ZLZo, β

Zin

l=λ/4


59

STUB MATCHING


60

Single-Stub Tuner

• In a single-stub tuner, a short transmission line (called

"stub") is added in parallel with the main transmission line.

• The stub usually has the same Zo as the main transmission

line.

• By choosing the appropriate stub length d and its distance l

from the load, it is possible to achieve zero reflection from

the source to the point where the stub is added.

• Therefore, the stub is always placed close to the load.

• The stub is usually a short-circuit stub to minimize radiation

loss (which is more likely to occur with an open-circuit stub).

60


61

Single-Stub Tuner

ZLZo

l

d

Short-

circuit stub

Zo

Source Load


62

Single-Stub Tuner (cont'd)

• At point A where the stub is added, it is desired to have the

impedance equal to ZA=Zin=Zo. It will be more convenient to

use the normalized impedance zA=zin=1.

• Also, as the stub is added in parallel, it is more convenient to

work in admittances:

• The matching process involves:

– using the distance l to transform yL to 1+jb, and

– using the stub to add –jb to cancel the +jb susceptance.

• Although the matching process can be carried out using

equations, a Smith Chart is most conveniently used for this

application.

62

11

==in

inz

y11

==o

oz

yL

Lz

y1

=


63

Single-Stub Tuner Using the Smith Chart• Locate zL and yL. (yL is diagonally opposite zL.)

• Using a compass, draw a locus towards the

Generator along a constant |Γ| circle until r = 1

circle is reached (Point A). The distance moved is l

(in units of λ).

• Read the jb value at point A. (Two possible

solutions.)

• The stub is required to have –jb.

• Choose either a short-circuit or open-circuit stub.

• Determine the stub length d to produce the

required susceptance.

EXAMPLE: (zL = 2)

• Locate zL = 2, and yL = 0.5.

• Move towards Generator until point PA or PB.

Read yA = 1 + j0.71 and yB = 1 – j0.71

zLyL

PA

PB


64

zLyL

PA

PB

ZLZo

l

dShort-

circuit

stub

Zo

Source Load

A


65

Single-Stub Tuner Using the Smith Chart (cont’d)

EXAMPLE: (zL = 2)

• Locate zL = 2, and yL =0.5.

• Move towards Generator until point PA or

PB.

• Read on the circumference: lA= 0.152λ and

lB= 0.348λ.

• Read at point PA and PB:

yA=1 + j0.71 yB=1 – j0.71

• At point PA, the stub must have a

susceptance of –j0.7. If it is a short-circuit

stub, its length dAS = 0.402λ – 0.25λ =

0.152λ. If it is an open-circuit stub, its

length dAO = 0.402λ. (see next slide.)

• At point PB, the stub must have a

susceptance of +j0.7. If it is a short-circuit

stub, its length dBS = 0.098λ + 0.25λ =

0.348λ. If it is an open-circuit stub, its

length dBO = 0.098λ.

ZLZo

l

d

Short-

circuit

stub

Zo

Source LoadA

zLyL

PA

PB

lA

lB

∞=stuby0=stuby

65


66

For a stub with a

susceptance of –j0.7, if it

is a short-circuit stub, its

length dAS = 0.402λ – 0.25λ= 0.152λ. If it is an open-circuit stub, its length dAO= 0.402λ.

∞=stuby0=stuby

λ250.

λ4020.

λ00.

open-circuit short-circuit

Single-Stub Tuner Using the Smith Chart (cont’d)

66


67

Single-Stub Tuner: Exercise

The impedance of a WiFi

monopole antenna at 2.48

GHz was 20 – j25 Ω. The

antenna was connected to

the transmitter via a 50 Ωmicrostrip line with an

effective dielectric constant

of 2. Design a single-stub

matching network to match

the antenna to the

transmission line. Use the

shortest short-circuit stub.

ZLZo

l

dShort-

circuit

stub

Zo

Source Load

67


68

GAMMA MATCH


69

69

GAMMA MATCH

• Gamma Match

– Unbalanced transmission lines.

– Equivalent circuit• The gamma match is equivalent to half of the T-match

• Requires a capacitor in series with the gamma rod

• The input impedance is:

( )[ ]( ) ag

ag

cinZZ

ZZjXZ

2

2

12

1

α

α

++

++−=

Za is the center point free space input impedance of the antenna in the

absence of gamma-match connection.

[3.21]


70

70

GAMMA MATCH

• Design procedure

– Determine the current division factor α by using Eq. [3.13]

– Find the free space impedance (in the absence of the gamma match) of the driven element at the center point. Designate it as Za

– Divide Za by 2 and multiply by the step-up ratio (1+α)2. Designate the result as Z2

– Determine the characteristic impedance Z0 of the transmission line form by the driven element and the gamma rod using Eq. [ 3.15a]

( )2

12

222aZjXRZ α+=+= [3.22]


71

71

GAMMA MATCH

– Normalized Z2 of Eq. [3.22] by Z0 and designate it as z2

– Invert z2 of Eq. [3.23] and obtain its equivalent

admittance y2=g2+jb2

– The normalized value Eq. [3.24]

=

+=+

==

2

'tan

22

0

22

0

22

lkjz

jxrZ

jXR

Z

Zz

g

[3.23]

[3.24]


72

72

GAMMA MATCH

– Equivalent admittance yg=gg+jbg

– Add two parallel admittances to obtain the total input

admittance at the gamma feed.

– Invert the normalize input admittance yin to obtain the

equivalent normalized input impedance

( ) ( )ininin

gggin

jxrz

bbjggyyy

+=

+++=+= 222[3.25]

[3.26]


73

73

GAMMA MATCH

– Obtain the unnormalized input impedance by

multiplying zin by Z0

– Select the capacitor C so that its reactance is equal in

magnitude to Xin

in

inininin

XfCC

zZjXRZ

==

=+=

πω 2

11

0 [3.27]

[3.28]


74

BALUNS

How they work

How they are made


75

What is a balun?

• A Balun is special type of transformer that performs

two functions:

– Impedance transformation

– Balanced to unbalanced transformation

• The word balun is a contraction of “balanced to

unbalanced transformer”


76

Why do we need a balun?

• Baluns are important because many types of antennas

(dipoles, yagis, loops) are balanced loads, which are fed with

an unbalanced transmission line (coax).

• Baluns are required for proper connection of parallel line to a

transceiver with a 50 ohm unbalanced output

• The antenna’s radiation pattern changes if the currents in the

driven element of a balanced antenna are not equal and

opposite.

• Baluns prevent unwanted RF currents from flowing in the

“third” conductor of a coaxial cable.


77

Balanced vs Unbalanced Transmission Lines

• A balanced transmission line is one whose currents

are currents are symmetric with respect to ground

so that all current flows through the transmission

line and the load and none through ground.

• Note that line balance depends on the current

through the line, not the voltage across the line.


78

An example of a Balanced Line

• Here is an example of a balanced line. DC rather than AC is used to simplify the analysis:

• Notice that the currents are equal and opposite and the that the total current flowing through ground = 25mA-25mA = 0

I = 25 mA

V = +6 VDC

6 V

6 V

V = -6 VDC

I = -25 mA

240 Ω

240 Ω


79

FAQ’s

• Do I really need a balun?

– Not necessarily. If you feed a balanced antenna with

unbalanced line and you don’t want feed line

radiation, use a balun!

• What kind of balun is best?

– There is no best balun for all applications. The

choice of balun depends on the type of antenna and

the frequency range.

• Will you make a Balun for me?

– No. However, I will be happy to show how to make

your own.

chapter 4 –impedance matching - universiti malaysia...

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