Lecture 02

ECE 344

Microwave Fundamentals

Spring 2017

Transmission Line Theory

1

2

– Transmission Line Theory

– The lumped-element circuit model for a transmission line.

– Wave Equations for TL model and their solutions.

– Propagation in lossy & lossless TL.

– Characteristic Impedance.

– Reflection & Transmission Coefficients.

– Standing Wave.

– Input Impedance.

Agenda

Lumped circuits: resistors, capacitors, inductors

Neglects time delays (phase change)

Account for propagation and time delays

(phase change)

Transmission-Line Theory

Distributed circuit elements: transmission lines

We need transmission-line theory whenever the length of a line

is significant compared to a wavelength. 3

The key difference between circuit theory and transmission

line theory is electrical size.

4

Transmission Lines

Transmission Lines

2 conductors

4 per-unit-length parameters:

Dz

5

R = series resistance per unit length, for both conductors, in Ω/m.

L = series inductance per unit length, for both conductors, in H/m.

G = shunt conductance per unit length, in S/m or Ʊ/m.

C = shunt capacitance per unit length, in F/m.

Transmission Line (cont.)

zD

,i z t

+ + + + + + + - - - - - - - - - -

,v z tx x x B

6

RDz LDz

GDz CDz

z

v(z+Dz,t)

+

-

v(z,t)

+

-

i(z,t) i(z+Dz,t)

Note: There are equal and opposite currents on the two conductors. (We only need to work with the current on the top conductor, since we have chosen to put all of the series elements there.)

( , )( , ) ( , ) ( , )

( , )( , ) ( , ) ( , )

i z tv z t v z z t i z t R z L z

t

v z z ti z t i z z t v z z t G z C z

t

D D D

D D D D D

Transmission Line (cont.)

7

RDz LDz

GDz CDz

z

v(z+Dz,t)

+

-

v(z,t)

+

-

i(z,t) i(z+Dz,t)

Applying KVL

Applying KCL

(2.1)

(2.2)

Hence

( , ) ( , ) ( , )( , )

( , ) ( , ) ( , )( , )

v z z t v z t i z tRi z t L

z t

i z z t i z t v z z tGv z z t C

z t

D

D

D D D

D

taking the limit as Dz 0:

v iRi L

z t

i vGv C

z t

“Telegrapher's

Equations”

TEM Transmission Line (cont.)

8 These are the time domain form of the transmission line equations,

(2.3)

(2.4)

To combine these, take the derivative of the first one with respect to z:

TEM Transmission Line (cont.)

9

From instantaneous to phasor form

)()()(

)()()(

zVCjGz

zI

zILjRz

zV

)())(()(

)()(

)(

2

2

2

2

zVCjGLjRz

zV

z

zILjR

z

zV

(2.5)

(2.6)

The same differential equation also holds for i.

jt

Solution: ( ) z zV z Ae Be

TEM Transmission Line (cont.)

10

0)()(

0)()(

2

2

2

2

2

2

zIz

zI

zVz

zV

(2.7)

(2.8)

This yields to give wave equations for V(z) and I (z):

))(( CjGLjRj Where

is the complex propagation constant, which is a function of frequency.

Equations (2.7) and (2.8) are 2nd order differential equations

Denote:

TEM Transmission Line (cont.)

11

( )( )R j L G j C

[np/m] attenuationcontant

j

rad/m phaseconstant 1/m propagation constant

Traveling wave solutions to (2.7) and (2.8) can be found as

zz

zz

eIeIzI

eVeVzV

00

00

)(

)( (2.9)

(2.10)

TEM Transmission Line (cont.)

0 0( ) z z j zV z V e V e e

Forward travelling wave (a wave traveling in the positive z direction):

0

0

0

( , ) Re

Re

cos

z j z j t

j z j z j t

z

v z t V e e e

V e e e e

V e t z

g 0t

z

0

zV e

2

g

2g

The wave “repeats” when:

Hence:

12

Phase Velocity

Let’s track the velocity of a fixed point on the wave (a point of constant

phase), e.g., the crest of the wave.

0( , ) cos( )zv z t V e t z

z

vp (phase velocity)

13

Phase Velocity (cont.)

0

constant

t z

dz

dt

dz

dt

Set

Hence p

v

14

For any point on the wave the term ωt-βz should stay constant,

00

0 0

j z j j z

z z

z j zV e e

V z V e V

V e e e

e

e

0

0 cos

c

, R

os

e j t

z

z

V e t

v z t V z

z

V z

e

e t

In the time domain:

Wave in +z

direction Wave in -z

direction

General Case (Waves in Both Directions)

15

Applying (2.5) to the voltage of (2.9) gives the current on the line:

so

Characteristic Impedance Z0 (cont.)

16

)()()(

zILjRz

zV

zz eVeVzV 00)(

)()(

)(

)()(

00

00

zz

zz

eVeVLjR

zI

zILjReVeV

(2.9)

(2.5)

Comparison with (2.10) shows that a characteristic impedance, Z0, can be

defined as

)(

)()(0

CjG

LjRLjRZ

(2.11)

Characteristic Impedance Z0 (cont.)

17

0

00

0

0

I

VZ

I

V

zz eZ

Ve

Z

VzI

0

0

0

0)( (2.12)

Characteristic Impedance Z0

0

( )

( )

V zZ

I z

0

0

( )

( )

z

z

V z V e

I z I e

so 00

0

VZ

I

+

V+(z)

-

I+ (z)

z

Assumption: A wave is traveling in the positive z direction.

(Note: Z0 is a number, not a function of z.)

18

Backward-Traveling Wave

0

( )

( )

V zZ

I z

0

( )

( )

V zZ

I z

so

+

V -(z)

-

I - (z)

z

A wave is traveling in the negative z direction.

Note: The reference directions for voltage and current are

chosen the same as for the forward wave.

19

General Case

0 0

0 0

0

( )

1( )

z z

z z

V z V e V e

I z V e V eZ

A general superposition of forward and

backward traveling waves: Most general case:

Note: The reference

directions for voltage

and current are the

same for forward and

backward waves.

20

+

V (z)

-

I (z)

z

0 0

0 0

0 0

0

z z

z z

V z V e V e

V VI z e e

Z

j R j L G j C

R j LZ

G j

Z

C

I(z)

V(z)+

-z

2

mg

[m/s]pv

Guided wavelength

Phase velocity

Summary of Basic TL formulas

21

dB/m 8.686Attenuation in

1

10 dB/m 20log 8.686e

Attenuation in

10log 0.4343lnx x

Note :

Lossless Case

0, 0R G

( )( )j R j L G j C

j LC

so 0

LC

0

R j LZ

G j C

0

LZ

C

1pv

LC

pv

(independent of freq.) (real and independent of freq.)

22

In many practical cases, however, the loss of the line is very small and

so can be neglected, resulting in a simplification of the results.

23

zjzj eZ

Ve

Z

VzI

0

0

0

0)(

zjzj eVeVzV 00)(

Lossless Case

0 0

z zV z V e V e

Where do we assign z = 0 ?

The usual choice is at the load.

Amplitude of voltage wave

propagating in negative z

direction at z = 0.

Amplitude of voltage wave

propagating in positive z

direction at z = 0.

Terminated Transmission Line

25

V (z) +

- z

ZL

z = 0

Terminating impedance (load)

I (z)

0 0

z zV z V e V e

What is V(-d) ?

0 0

d dV d V e V e

0 0

0 0

d dV VI d e e

Z Z

Propagating

forwards

Propagating

backwards

Terminated Transmission Line (cont.)

d distance away from load

The current at z = -d is

28

V (-d) +

- z

ZL

z = 0

Terminating impedance (load)

I (-d)

z = -d

(This does not necessarily have to be the length of the entire line.)

20

0

1d d

L

VI d e e

Z

200 0 0

0

1d d d dVV d V e V e V e e

V

Similarly,

L Load reflection coefficient

Terminated Transmission Line (cont.)

29

V (-d) +

- z

ZL

z = 0

I (-d)

z = -d

2

0 1d d

LV d V e e

or

0

0

L

V

V

2

0

20

0

1

1

d d

L

d d

L

V d V e e

VI d e e

Z

Z(-d) = impedance seen “looking” towards load at z = -d.

Terminated Transmission Line (cont.)

30

V (-d) +

- z

ZL

z = 0

I (-d)

z = -d Z(-d)

2

0 2

1

1

d

L

d

L

V d eZ d Z

I d e

Note:

If we are at the

beginning of the

line, we will call this

“input impedance”.

At the load (d = 0):

0

10

1

LL

L

Z Z Z

Thus,

20

0

0

20

0

1

1

dL

L

dL

L

Z Ze

Z ZZ d Z

Z Ze

Z Z

Terminated Transmission Line (cont.)

0

0

LL

L

Z Z

Z Z

2

0 2

1

1

d

L

d

L

eZ d Z

e

Recall

31

Simplifying, we have:

0

0

0

tanh

tanh

L

L

Z Z dZ d Z

Z Z d

Terminated Transmission Line (cont.)

20

2

0 0 0

0 0 2

2 0 00

0

0 0

0

0 0

0

0

0

1

1

cosh sinh

cosh sinh

dL

d

L L L

d

d L LL

L

d d

L L

d d

L L

L

L

Z Ze

Z Z Z Z Z Z eZ d Z Z

Z Z Z Z eZ Ze

Z Z

Z Z e Z Z eZ

Z Z e Z Z e

Z d Z dZ

Z d Z d

Hence, we have

32

2

0

20

0

1

1

j d j d

L

j d j d

L

V d V e e

VI d e e

Z

Impedance is periodic

with period g/2:

2 1

2 1

2 1

2

/ 2

g

g

d d

d d

d d

Terminated Lossless Transmission Line

j j

Note: tanh tanh tand j d j d

The tan function repeats when

0

0

0

tan

tan

L

L

Z jZ dZ d Z

Z jZ d

33

Lossless:

2

0 2

1

1

j d

L

j d

L

eZ d Z

e

Note: For the remainder of our transmission line discussion we will

assume that the transmission line is lossless.

2

0 2

0

0

0

1

1

tan

tan

j d

L

j d

L

L

L

eZ d Z

e

Z jZ dZ

Z jZ d

0

0

2

LL

L

g

p

Z Z

Z Z

v

Summary for Lossless Transmission Line

34

V (-d) +

- z

ZL

z = 0

I (-d)

z = -d Z(-d)

35

Reflection and transmission at the junction of two transmission lines with

different characteristic impedances.

the voltage for z < 0 is

reflection coefficient is given by

0z ),()( 0 zjzj eeVzV

01

01

ZZ

ZZ

the voltage for z > 0 is 0z ,)( 0 zjTeVzV

Equating these voltages at z = 0 gives the transmission coefficient, T , as

01

121

ZZ

ZT

The transmission coefficient between two points in a circuit is often expressed

in dB as the insertion loss, IL, dB ||log20 TIL

RL = −20 log |Γ| dB,

0

0

0LL

L

Z Z

Z Z

No reflection from the load

Matched Load (ZL=Z0)

0 Z d Z z for any

36

V (-d) +

- z

ZL

z = 0

I (-d)

z = -d Z(-d)

2

0 2

1

1

j d

L

j d

L

eZ d Z

e

1L

Always imaginary!

Short-Circuit Load (ZL=0)

37

V (-d) +

- z

z = 0

I (-d)

z = -d Z(-d)

0

0

0

tan

tan

L

L

Z jZ dZ d Z

Z jZ d

0 tanZ d jZ d

0

0

LL

L

Z Z

Z Z

Note: 2g

dd

scZ d jX

S.C. can become an O.C. with a

g/4 transmission line.

Short-Circuit Load (ZL=0)

0 tanscX Z d

38

0 1 / 4 1 / 2 3 / 4

Xsc

Inductive

Capacitive

d/g

0 tanZ d jZ d

0

0

L

Z

Z

Always imaginary!

Open-Circuit Load (ZL=)

39

V (-d) +

- z

z = 0

I (-d)

z = -d Z(-d)

1L

0

0

LL

L

Z Z

Z Z

0

0

0

tan

tan

L

L

Z jZ dZ d Z

Z jZ d

0 cotZ d jZ d

0

0

0

1 / tan

/ tan

L

L

j Z Z dZ d Z

Z Z j d

or

2g

dd

O.C. can become a S.C. with a

g/4 transmission line.

Open-Circuit Load (ZL=)

40

ocZ d jX

0 cotocX Z d

0 cotZ d jZ d

Note:

0 1/4 1/2

Xoc Inductive

Capacitive

d/g

2

0 1 Lj j z

LV z V e e

2

0

2

0

1

1 L

j z j z

L

jj z j z

L

V z V e e

V e e e

max 0

min 0

1

1

L

L

V V

V V

Voltage Standing Wave

z

1+ L

1

1- L

0

( )V z

V

/ 2g

z D 0z

41

2 2 / 2z z g D D

V (z) +

- z

ZL

z = 0

I (z)

z

𝜙L + 2βz = 0,− 2π,…….,− 2nπ

Maximum Occurs When

Minimum Occurs When

𝜙L + 2βz = −π, − 3π, ….,− (2n+1)π

max

min

V

VVoltage Standing Wave Ratio VSWR

Voltage Standing Wave Ratio

1

1

L

L

VSWR

z

1+ L

1

1- L

0

( )V z

V

/ 2gz D 0z

42

max 0

min 0

1

1

L

L

V V

V V

V (z) +

- z

ZL

z = 0

I (z)

z

Voltage Standing Wave Ratio

Voltage Standing Wave Ratio

45

VSWR=(1+0)/|1-0|=1

VSWR=(1+0.2)/|1-0.2|=1.5

VSWR=(1+0.5)/|1-0.5|=3 VSWR=(1+1)/|1-1|=∞

Voltage Standing Wave Ratio

Using Transmission Lines to Synthesize Loads

A microwave filter constructed from microstrip line.

This is very useful in microwave engineering.

46

We can obtain any reactance that we want from a short or open transmission line.

47

Find the voltage at any point on the line.

ZTh

VTh

+

Zin

+

- -

V (-d)

0

0

0

tan

tan

L

in

L

Z jZ dZ Z d Z

Z jZ d

inTh

in Th

ZV d V

Z Z

At the input:

Voltage on a Transmission Line

I (z)

+ Z L

-

Z Th

V Th

d

Zin

+

-

0 ,Z V (z)

z=0

0

21j z zj

Lz V eV e 0

0

LL

L

Z Z

Z Z

2

0 1j d j d inL Th

in Th

ZV d V e e V

Z Z

2

2

1

1

j zj d zin L

Th j d

in Th L

Z eV z V e

Z Z e

At z = -d :

Hence

0 2

1

1

j dinTh j d

in Th L

ZV V e

Z Z e

48

Voltage on a Transmission Line (cont.)

Let’s derive an alternative form of the result.

2

0 2

1

1

j d

Lin j d

L

eZ Z d Z

e

2

20 20

2 22

00 2

2

0

2

0 0

2

0

20 0

0

1

11

1 11

1

1

1

1

j d

Lj dj d

LL

j d j dj d

L Th LLThj d

L

j d

L

j d

Th L Th

j d

L

j dTh ThL

Th

in

in Th

eZ

Z ee

Z e Z eeZ Z

e

Z e

Z Z e Z Z

eZ

Z

Z

Z Z

Z Z Ze

Z Z

Z

2

0

20 0

0

1

1

j d

L

j dTh ThL

Th

e

Z Z Z Ze

Z Z

49

Start with:

Voltage on a Transmission Line (cont.)

2

0

2

0

1

1

j zj d z L

Th j d

Th s L

Z eV z V e

Z Z e

2

0

2

0

1

1

j d

in L

j d

in Th Th s L

Z Z e

Z Z Z Z e

where 0

0

Ths

Th

Z Z

Z Z

Therefore, we have the following alternative form for the result:

Hence, we have

50

(source reflection coefficient)

Voltage on a Transmission Line (cont.)

2

2

1

1

j zj d zin L

Th j d

in Th L

Z eV z V e

Z Z e

Recall:

Substitute

2

0

2

0

1

1

j zj z d L

Th j d

Th s L

Z eV z V e

Z Z e

Voltage wave that would exist if there were no reflections from

the load (a semi-infinite transmission line or a matched load). 51

Voltage on a Transmission Line (cont.)

This term accounts for the multiple (infinite) bounces.

I (z)

+ZL

-

ZTh

VTh

d

Zin

+

-

0 ,Z V (z)

z=0

2 2

2 2 2 20

0

1 j d j d

L L s

j d j d j d j d

Th L s L L s L s

Th

e e

ZV d V e e e e

Z Z

Wave-bounce method (illustrated for z = -d ):

52

ZL

Z Th

V Th

d

+

-

0 ,Z

Voltage on a Transmission Line (cont.)

22 2

22 2 20

0

1

1

j d j d

L S L S

j d j d j d

Th L L S L S

Th

e e

ZV d V e e e

Z Z

Geometric series:

2

0

11 , 1

1

n

n

z z z zz

2 2

2 2 2 20

0

1 j d j d

L L s

j d j d j d j d

Th L s L L s L s

Th

e e

ZV d V e e e e

Z Z

2j d

L sz e

53

Group together alternating terms:

Voltage on a Transmission Line (cont.)

or

2

0

20

2

1

1

1

1

j d

L s

Th

j dTh

L j d

L s

eZV d V

Z Ze

e

2

0

2

0

1

1

j d

LTh j d

Th L s

Z eV d V

Z Z e

This agrees with the previous result (setting z = -d ).

Note: The wave-bounce method is a very tedious method – not recommended.

Hence

54

Voltage on a Transmission Line (cont.)

At a distance d from the load:

*

*

2

0 2 2 * 2

*

0

1Re

2

1Re 1 1

2

z z z

L L

P z V z I z

Ve e e

Z

2

20 2 4

0

11

2

z z

L

VP z e e

Z

If Z0 real (low-loss transmission line)

Time-Average Power Flow

2

0

20

0

1

1

z z

L

z z

L

V z V e e

VI z e e

Z

j

*2 * 2

*2 2

z z

L L

z z

L L

e e

e e

pure imaginary

Note:

55

V (z) +

- z

ZL

z = 0

I (z)

z

Low-loss line

2

20 2 4

0

2 2

20 02 2

0 0

11

2

1 1

2 2

z z

L

z z

L

VP z e e

Z

V Ve e

Z Z

Power in forward-traveling wave Power in backward-traveling wave

2

20

0

11

2L

VP z

Z

Lossless line ( = 0)

Time-Average Power Flow (cont.)

56

V (-d) +

- z

ZL

z = 0

I (-d)

z = -d

Note:

In the general lossy case, we cannot say

that the total power is the difference of

the powers in the two waves.

00

0

tan

tan

L Tin T

T L

Z jZ dZ Z

Z jZ d

2

4 4 2

g g

g

d

00

Tin T

L

jZZ Z

jZ

0

2

00

0in in

T

L

Z Z

ZZ

Z

Quarter-Wave Transformer

2

0Tin

L

ZZ

Z

so

0 0T LZ Z Z

Hence

(This requires ZL to be real.)

57

Matching condition

ZL Z0 Z0T

Zin d

Conjugate matching

Home work

Prove that the condition for maximum power delivered to the load is

58

I (z)

+ZL

-

ZTh

VTh

d

Zin

+

-

0 ,Z V (z)

z=0

*

thin ZZ

Field Analysis of Transmission Lines

In this section we will derive the transmission line parameters (R, L, G, C) in terms of the electric and magnetic fields of the transmission line

59

Coaxial Cable

Here we present a “case study” of one particular transmission line, the coaxial cable.

a

b ,r

Find C, L, G, R

We will assume no variation in the z direction, and take a length of one meter in

the z direction in order to calculate the per-unit-length parameters.

60

For a TEMz mode, the shape of the fields is independent of frequency, and

hence we can perform the calculation using electrostatics and magnetostatics.

Coaxial Cable (cont.)

-l0

l0

a

b

r0 0

0

ˆ ˆ2 2 r

E

Find C (capacitance / length)

Coaxial cable

h = 1 [m]

r

From Gauss’s law:

0

0

ln2

B

AB

A

b

ra

V V E dr

bE d

a

61

-l0

l0

a

b

r

Coaxial cable

h = 1 [m]

r

0

0

0

1

ln2 r

QC

V b

a

Hence

We then have:

0 F/m2

[ ]

ln

rCb

a

Coaxial Cable (cont.)

62

ˆ2

IH

Find L (inductance / length)

From Ampere’s law:

Coaxial cable

h = 1 [m]

r

I

0ˆ

2r

IB

(1)

b

a

B d S

h

I

I z Center conductor Magnetic flux:

Coaxial Cable (cont.)

63

Note: We ignore “internal inductance” here, and

only look at the magnetic field between the two

conductors (accurate for high frequency.

Coaxial cable

h = 1 [m]

r

I

0

0

0

1

2

ln2

b

r

a

b

r

a

r

H d

Id

I b

a

0

1ln

2r

bL

I a

0 H/mln [ ]2

r bL

a

Hence

Coaxial Cable (cont.)

64

0 H/mln [ ]2

r bL

a

Observation:

0 F/m2

[ ]

ln

rCb

a

0 0 r rLC

This result actually holds for any transmission line (proof omitted).

Coaxial Cable (cont.)

65

0 H/mln [ ]2

r bL

a

For a lossless cable:

0 F/m2

[ ]

ln

rCb

a

0

LZ

C

0 0

1ln [ ]

2

r

r

bZ

a

00

0

376.7303 [ ]

Coaxial Cable (cont.)

66

-l0

l0

a

b

0 0

0

ˆ ˆ2 2 r

E

Find G (conductance / length)

Coaxial cable

h = 1 [m]

From Gauss’s law:

0

0

ln2

B

AB

A

b

ra

V V E dr

bE d

a

Coaxial Cable (cont.)

67

-l0

l0

a

b

J E

We then have leakIG

V

0

0

(1) 2

2

22

leak a

a

r

I J a

a E

aa

0

0

0

0

22

ln2

r

r

aa

Gb

a

2[S/m]

ln

Gb

a

or

Coaxial Cable (cont.)

68

Observation:

F/m2

[ ]

ln

Cb

a

G C

This result actually holds for any transmission line (proof omitted).

2[S/m]

ln

Gb

a

0 r

Coaxial Cable (cont.)

69

G C

Hence:

tanG

C

tanG

C

Coaxial Cable (cont.)

As just derived,

70

This is the loss tangent that would

arise from conductivity.

Find R (resistance / length)

Coaxial cable

h = 1 [m]

Coaxial Cable (cont.)

,b rb

a

b

,a ra

a bR R R

1

2a saR R

a

1

2b sbR R

b

1sa

a a

R

1

sb

b b

R

0

2a

ra a

0

2b

rb b

Rs = surface resistance of metal

71

72

At high frequency, discontinuity effects can become important.

Limitations of Transmission-Line Theory

Bend

Incident

Reflected

Transmitted

The simple TL model does not account for the bend.

ZTh

ZL Z0

+ -

73

At high frequency, radiation effects can become important.

When will radiation occur?

We want energy to travel from the generator to the load, without radiating.

Limitations of Transmission-Line Theory (cont.)

74

ZTh

ZL Z0

+ -

This is explored next.

r a

b

z

The coaxial cable is a perfectly shielded system – there is never any radiation at

any frequency, as long as the metal thickness is large compared with a skin depth.

The fields are confined

to the region between

the two conductors.

Limitations of Transmission-Line Theory (cont.)

75

Coaxial Cable

The twin lead is an open type of transmission

line – the fields extend out to infinity.

The extended fields may cause interference

with nearby objects.

(This may be improved by using “twisted pair”.)

+ -

Limitations of Transmission-Line Theory (cont.)

Having fields that extend to infinity is not the same thing as having radiation, however!

76

The infinite twin lead will not radiate by itself, regardless of how far

apart the lines are (this is true for any transmission line).

h

Incident

Reflected

The incident and reflected waves represent an exact solution to

Maxwell’s equations on the infinite line, at any frequency.

*1ˆRe E H 0

2t

S

P dS

S

+ -

Limitations of Transmission-Line Theory (cont.)

No attenuation on an infinite lossless line

77

A discontinuity on the twin lead will cause radiation to occur.

Note: Radiation effects usually

increase as the frequency increases.

Limitations of Transmission-Line Theory (cont.)

h

Incident wave Pipe

Obstacle

Reflected wave

Bend h

Incident wave

Bend

Reflected wave 78

To reduce radiation effects of the twin lead at discontinuities:

h

1) Reduce the separation distance h (keep h << ).

2) Twist the lines (twisted pair).

Limitations of Transmission-Line Theory (cont.)

CAT 5 cable

(twisted pair)

79

Baluns

80

Baluns are used to connect coaxial cables to twin leads.

4:1 impedance-transforming baluns, connecting 75 TV coax to 300 TV twin lead

Coaxial cable: an “unbalanced” transmission line

Twin lead: a “balanced” transmission line

Balun: “Balanced to “unbalanced”

https://en.wikipedia.org/wiki/Balun

They suppress the common mode currents on the transmission lines.

Baluns (cont.)

81

https://en.wikipedia.org/wiki/Balun

“Baluns are present in radars, transmitters, satellites, in every

telephone network, and probably in most wireless network

modem/routers used in homes.”

Baluns are also used to connect coax (unbalanced line)

to dipole antennas (balanced).

82

Ground (zero volts)

+ -

Ground (zero volts)

+

0

The voltages are

unbalanced with

respect to ground.

The voltages are

balanced with respect

to ground.

Coax Twin Lead

Baluns (cont.)

Baluns (cont.)

83

Baluns are necessary because, in practice, the two transmission lines

are always both running over a ground plane.

If there were no ground plane, and you only had the two lines connected to each

other, then a balun would not be necessary.

(But you would still want to have a matching network between the two lines if they have

different characteristic impedances.)

Coax Twin Lead

Baluns (cont.)

84

When a ground plane is present, we really have three

conductors, forming a multiconductor transmission line, and

this system supports two different modes.

+ -

Differential mode

(desired)

+ +

Common mode

(undesired)

The differential and common modes on a twin lead over ground

Baluns (cont.)

85

Differential mode

(desired)

Common mode

(undesired)

The differential and common modes on a coax over ground

+

- +

-

Baluns (cont.)

86

A balun prevents common modes from being excited at the junction

between a coax and a twin lead.

Coax Twin Lead

Ground

Common mode Common mode

No Balun

Differential mode

Differential mode

Baluns (cont.)

87

From another point of view, a balun prevents currents from flowing on the

outside of the coax.

Coax Twin Lead

Ground

No Balun

I1

I1

I2 I1

I2- I1

(direct current path to ground)

2 1I I

Baluns (cont.)

88

A simple model for the mode excitation at the junction

The coax is replaced by a solid tube with a voltage source at the end.

Solid tube Twin Lead

Ground

+ - V0

Baluns (cont.)

89

Next, we use superposition.

+

Solid tube Twin Lead

Ground

+ -

+ -

V0 / 2

V0 / 2

Differential mode

Solid tube Twin Lead

Ground

+ -

+ -

V0 / 2

V0 / 2

Common mode

Baluns (cont.)

90

A balun prevents common modes from being excited at the junction

between a coax and a twin lead.

With Balun

Coax Twin Lead

Ground

Differential mode Differential mode

Balun

Baluns (cont.)

91

One type of balun uses an isolation transformer.

https://en.wikipedia.org/wiki/Balun

The input and output are isolated

from each other, so one side can be

grounded while the other is not.

Circuit diagram of a 4:1 autotransformer

balun, using three taps on a single

winding, on a ferrite rod

Baluns (cont.)

92

Here is a more exotic type of 4:1 impedance transforming balun.

0

V0 V0

0

0

V0

-V0

0 0 -V0

I0/2

I0 I0/2

I0 V0

This balun uses two 1:1 transformers to achieve symmetric output voltages.

“Guanella balun”

Baluns (cont.)

93

Inside of 4:1 impedance transforming VHF/UHF balun for TV

Ferrite core

Baluns (cont.)

94

Another type of balun uses a choke to “choke off” the common mode.

https://en.wikipedia.org/wiki/Balun

A coax is wound around a ferrite core. This creates a large inductance for the common mode,

while it does not affect the differential mode (whose fields are confined inside the coax).

Baluns (cont.)

95

Baluns are very useful for feeding differential circuits with coax on PCBs.

No balun

Without a balun, there would be a

common mode current on the

coplanar strips (CPS) line.

Baluns (cont.)

96

+ - + - + -

+ - - +

= +

V0 / 2 V0 / 2

V0 / 2 V0 / 2

V0

Excites desired differential mode

Excites undesired common mode

Superposition allows us to see what the problem is.

Baluns (cont.)

97

Baluns can be implemented in microstrip form.

Tapered balun

Balun feeding a microstrip yagi antenna

Baluns (cont.)

98

Baluns can be implemented in microstrip form.

Marchand balun

180o hybrid rat-race coupler used as a balun

Input (unbalanced)

Output #2

Output #1

Baluns (cont.)

99

If you try to feed a dipole antenna directly with coax, there will be a

common mode current on the coax.

I1= I2

I3 0

No balun

Baluns (cont.)

100

Baluns are commonly used to feed dipole antennas.

A balun first converts the coax to a twin lead,

and then the twin lead feeds the dipole. A “sleeve balun” directly

connects a coax to a dipole.

Coaxial “sleeve” region

Coaxial “sleeve balun”

Baluns (cont.)

101

Feeding a dipole antenna over a ground plane

/ 4

This acts like voltage source for the dipole, which forces the differential mode.

The dipole is loaded by a short-circuited section of twin lead – an open circuit because of the /4 height.